Fluid Mechanics solution

Chapter 7 Dimensional Analysis and Modeling Solutions Manual for Fluid Mechanics: Fundamentals and Applications Second Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill, 2010 CHAPTER 7 DIMENSIONAL ANALYSIS AND MODELING PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. 7-1 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Dimensions and Units, Primary Dimensions 7-1C Solution We are to explain the difference between a dimension and a unit, and give examples. Analysis A dimension is a measure of a physical quantity (without numerical values), while a unit is a way to assign a number to that dimension. Examples are numerous – length and meter, temperature and oC, weight and lbf, mass and kg, time and second, power and watt,… Discussion units. 7-2C Solution When performing dimensional analysis, it is important to recognize the difference between dimensions and We are to list the seven primary dimensions and explain their significance. Analysis The seven primary dimensions are mass, length, time, temperature, electrical current, amount of light, and amount of matter. Their significance is that all other dimensions can be formed by combinations of these seven primary dimensions. Discussion One of the first steps in a dimensional analysis is to write down the primary dimensions of every variable or parameter that is important in the problem. 7-3 Solution We are to append the given table with other parameters and their primary dimensions. Analysis Students’ tables will differ, but they should add entries such as angular velocity, kinematic viscosity, work, energy, power, specific heat, thermal conductivity, torque or moment, stress, etc. Discussion This problem should be assigned as an ongoing homework problem throughout the semester, and then collected towards the end. Individual instructors can determine how many entries to be required in the table. 7-2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-4 Solution We are to write the primary dimensions of several variables in the force-length-time system. Analysis We start with the dimensions of force in the mass-length-time (m-L-t) system: {F} = {m1L1t-2}, from which we solve for the dimensions of mass in the force-length-time (F-L-t) system, {F} = {m1L1 t −2 } Primary dimensions of mass in the F-L-t system: {ρ} = {m1L−3 } → {m} = {F1L−1 t 2 } {ρ} = {F1L−1 t 2 L−3 } = {F1L−4 t 2 } We plug in the above conversion to go from the m-L-t system to the F-L-t system for each of the variables: Density: Surface tension: Viscosity: {σ s } = {m1 t −2 } {μ} = {m1L−1 t −1} → → → {σ s } = {F1L−1 t 2 t −2 } = {F1L−1 } {μ} = {F1L−1t 2 L−1t −1 } = {F1L−2 t1 } Discussion Sometimes the F-L-t system is easier to use than the m-L-t system. Neither one is “right” or “wrong” – it is a matter of personal preference, although the m-L-t system is the more popular one, especially in fluid mechanics. 7-5 Solution We are to write the primary dimensions of the universal ideal gas constant. Analysis From the given equation, Primary dimensions of the universal ideal gas constant: ⎧ m ⎫ × L3 ⎪ ⎧ pressure × volume ⎫ ⎪⎪ t 2 L ⎪ {Ru } = ⎨ ⎬=⎨ ⎬= mol temperature N T × × ⎩ ⎭ ⎪ ⎪ ⎪⎩ ⎭⎪ ⎧ mL2 ⎫ ⎨ 2 ⎬ ⎩ t TN ⎭ Or, in exponent form, {Ru} = {m1 L2 t-2 T-1 N-1}. Discussion the result. The standard value of Ru is 8314.3 J/kmol⋅K. You can verify that these units agree with the dimensions of 7-6 Solution We are to write the primary dimensions of atomic weight. Analysis By definition, atomic weight is mass per mol, Primary dimensions of atomic weight: {M } = ⎧⎨ mass ⎫ ⎧ m ⎫ ⎬= ⎨ ⎬ ⎩ mol ⎭ ⎩ N ⎭ (1) Or, in exponent form, {M} = {m1 N-1}. Discussion In terms of primary dimensions, atomic mass is not dimensionless, although many authors treat it as such. Note that mass and amount of matter are defined as two separate primary dimensions. 7-3 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-7 Solution We are to write the primary dimensions of the universal ideal gas constant in the alternate system where force replaces mass as a primary dimension. Analysis From Newton’s second law, force equals mass times acceleration. Thus, mass is written in terms of force as Primary dimensions of mass in the alternate system: 2 force ⎫ ⎧ F ⎫ ⎧ Ft ⎫ {mass} = ⎧⎨ ⎬=⎨ 2⎬=⎨ ⎬ ⎩ acceleration ⎭ ⎩ L/t ⎭ ⎩ L ⎭ (1) We substitute Eq. 1 into the results of Problem 7-4, ⎧ Ft 2 2 ⎫ L ⎪ ⎧ mL ⎫ ⎪ FL {Ru } = ⎨ 2 ⎬ = ⎨⎪ L 2 ⎬⎪ = ⎧⎨ ⎫⎬ Nt T Nt T TN ⎭ ⎩ ⎭ ⎪ ⎪ ⎩ ⎪⎩ ⎭⎪ Primary dimensions of the universal ideal gas constant: 2 (2) Or, in exponent form, {Ru} = {F1 L1 T-1 N-1}. Discussion Eq. 2. The standard value of Ru is 8314.3 J/kmol⋅K. You can verify that these units agree with the dimensions of 7-8 Solution We are to write the primary dimensions of the specific ideal gas constant, and verify are result by comparing to the standard SI units of Rair. Analysis We can approach this problem two ways. If we have already worked through Problem 7-4, we can use our results. Namely, ⎧ mL2 ⎫ ⎧ R ⎫ ⎪⎪ 2 ⎪⎪ ⎧ L2 ⎫ = ⎨ u ⎬ = ⎨ Nt T ⎬ = ⎨ 2 ⎬ ⎩ M ⎭ ⎪ m ⎪ ⎩t T⎭ ⎪⎩ N ⎭⎪ Primary dimensions of specific ideal gas constant: {R } gas (1) Or, in exponent form, {Rgas} = {L2 t-2 T-1}. Alternatively, we can use either form of the ideal gas law, Primary dimensions of specific ideal gas constant: {R } gas ⎧ m ⎫ × L3 ⎪ ⎧ 2 ⎫ ⎧ pressure × volume ⎫ ⎪⎪ t 2 L L ⎪ =⎨ ⎬=⎨ ⎬= ⎨ 2 ⎬ t ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ T ⎭ ⎪⎩ ⎭⎪ (2) For air, Rair = 287.0 J/kg⋅K. We transform these units into primary dimensions, Primary dimensions of the specific ideal gas constant for air: ⎧ mL2 ⎫ ⎧ J ⎫ ⎪⎪ t 2 ⎪⎪ {Rair } = ⎨287.0 ⎬=⎨ ⎬= kg×K ⎭ ⎪ m × T ⎪ ⎩ ⎪⎩ ⎭⎪ ⎧ L2 ⎫ ⎨ 2 ⎬ ⎩t T⎭ (3) Equation 3 agrees with Eq. 1 and Eq. 2, increasing our confidence that we have performed the algebra correctly. Discussion Notice that numbers, like the value 287.0 in Eq. 3 have no influence on the dimensions. 7-4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-9 Solution We are to write the primary dimensions of torque and list its units. Analysis Torque is the product of a length and a force, Primary dimensions of torque: { } G length ⎫ ⎧ M = ⎨length × mass ⎬= time 2 ⎭ ⎩ ⎧ L2 ⎫ ⎨m 2 ⎬ ⎩ t ⎭ (1) G Or, in exponent form, { M } = {m1 L2 t-2}. The common units of torque are newton-meter (SI) and inch-pound (English). In primary units, however, we write the primary SI units according to Eq. 1, Units of torque = kg ⋅ m 2 /s 2 Primary SI units: and in primary English units, Units of torque = lbm ⋅ ft 2 /s 2 Primary English units: Discussion Since torque is the product of a force and a length, it has the same dimensions as energy. Primary units are not required for dimensional analysis, but are often useful for unit conversions and for verification of proper units when solving a problem. 7-10 Solution We are to determine the primary dimensions of each variable. Analysis (a) Energy is force times length (the same dimensions as work), Primary dimensions of energy: 2 mass × length ⎫ ⎧ mL ⎫ × length ⎬ = ⎨ 2 ⎬ 2 time ⎩ ⎭ ⎩ t ⎭ {E} = {force × length} = ⎧⎨ (1) Or, in exponent form, {E} = {m1 L2 t-2}. (b) Specific energy is energy per unit mass, Primary dimensions of specific energy: energy ⎫ ⎧ mass × length 2 1 ⎫ ⎧ L2 ⎫ × {e} = ⎧⎨ ⎬=⎨ ⎬=⎨ ⎬ mass ⎭ ⎩ t 2 ⎭ time 2 ⎩ mass ⎭ ⎩ (2) Or, in exponent form, {e} = { L2 t-2}. (c) Power is the rate of change of energy, i.e. energy per unit time, { } Primary dimensions of power: 2 1 ⎫ ⎧ mL2 ⎫ ⎧ energy ⎫ ⎧ mass × length W = ⎨ = × ⎬ ⎨ ⎬=⎨ ⎬ time 2 time ⎭ ⎩ t 3 ⎭ ⎩ time ⎭ ⎩ (3) Or, in exponent form, { W } = {m1 L2 t-3}. Discussion In dimensional analysis it is important to distinguish between energy, specific energy, and power. 7-5 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-11 Solution We are to determine the primary dimensions of electrical voltage. Analysis From the hint, ⎧ mass × length 2 power ⎫ ⎪⎪ time3 {voltage} = ⎧⎨ ⎬=⎨ current ⎩ current ⎭ ⎪ ⎩⎪ Primary dimensions of voltage: ⎫ 2 ⎪⎪ ⎧ mL ⎫ ⎬=⎨ 3 ⎬ ⎪ ⎩ t I ⎭ ⎭⎪ (1) Or, in exponent form, {E} = {m1 L2 t-3 I-1}. Discussion dimensions. 7-12 Solution Analysis current, We see that all dimensions, even those of electrical properties, can be expressed in terms of primary We are to write the primary dimensions of electrical resistance. From Ohm’s law, we see that resistance has the dimensions of voltage difference divided by electrical ⎧ mass × length 2 ⎫ ⎪ ⎪ ⎧ mL2 ⎫ 3 ΔE {R} = ⎨⎧ ⎬⎫ = ⎨⎪ time × current ⎬⎪ = ⎨ 3 2 ⎬ current ⎩ I ⎭ ⎪ ⎪ ⎩t I ⎭ ⎩⎪ ⎭⎪ Primary dimensions of resistance: Or, in exponent form, {R} = {m1 L2 t-3 I-2}, where we have also used the result of the previous problem. Discussion All dimensions can be written in terms of primary dimensions. 7-6 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-13 Solution We are to determine the primary dimensions of each variable. Analysis (a) Acceleration is the rate of change of velocity, {a} = ⎧⎨ Primary dimensions of acceleration: velocity ⎫ ⎧ length 1 ⎫ ⎧L ⎫ × ⎬=⎨ ⎬=⎨ 2⎬ time time time ⎩ ⎭ ⎩ ⎭ ⎩t ⎭ (1) Or, in exponent form, {a} = { L1 t-2}. (b) Angular velocity is the rate of change of angle, Primary dimensions of angular velocity: Or, in exponent form, {ω} = { t-1}. {ω} = ⎧⎨ angle ⎫ ⎧ 1 ⎫ ⎧ 1 ⎫ ⎬=⎨ ⎬=⎨ ⎬ ⎩ time ⎭ ⎩ time ⎭ ⎩ t ⎭ (2) (c) Angular acceleration is the rate of change of angular velocity, Primary dimensions of angular acceleration: angular velocity ⎫ ⎧ 1 1 ⎫ ⎧1⎫ × {α = ω } = ⎨⎧ ⎬=⎨ ⎬=⎨ 2⎬ time ⎩ ⎭ ⎩ time time ⎭ ⎩ t ⎭ (3) Or, in exponent form, {α} = { t-2}. Discussion In Part (b) we note that the unit of angle is radian, which is a dimensionless unit. Therefore the dimensions of angle are unity. 7-14 Solution We are to write the primary dimensions of angular momentum and list its units. Analysis Angular momentum is the product of length, mass, and velocity, { } Primary dimensions of angular momentum: G length ⎫ ⎧ mL2 ⎫ ⎧ (1) H = ⎨length × mass × ⎬=⎨ ⎬ time ⎭ ⎩ t ⎭ ⎩ G Or, in exponent form, { H } = {m1 L2 t-1}. We write the primary SI units according to Eq. 1, Primary SI units: Units of angular momentum = and in primary English units, Primary English units: Units of angular momentum = kg ⋅ m 2 s lbm ⋅ ft 2 s Discussion Primary units are not required for dimensional analysis, but are often useful for unit conversions and for verification of proper units when solving a problem. 7-7 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-15 Solution We are to determine the primary dimensions of each variable. Analysis (a) Specific heat is energy per unit mass per unit temperature, ⎧ mL2 ⎫ ⎧ ⎫ ⎪⎪ t 2 ⎪⎪ ⎧ L2 ⎫ energy =⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ t T ⎭ ⎩⎪ ⎭⎪ Primary dimensions of specific heat at constant pressure: {c } p (1) Or, in exponent form, {cp} = { L2 t-2 T-1}. (b) Specific weight is density times gravitational acceleration, ⎧ mass length ⎫ ⎧ m ⎫ =⎨ 2 2⎬ Primary dimensions of specific weight: { ρ g} = ⎨ 2 ⎬ ⎩ volume time ⎭ ⎩ L t ⎭ (2) Or, in exponent form, {ρg} = {m1 L-2 t-2}. (c) Specific enthalpy has dimensions of energy per unit mass, ⎧ mL2 ⎧ energy ⎫ ⎪⎪ t 2 Primary dimensions of specific enthalpy: {h} = ⎨ ⎬=⎨ ⎩ mass ⎭ ⎪ m ⎪⎩ ⎫ ⎪⎪ ⎧ L2 ⎫ ⎬=⎨ 2 ⎬ ⎪ ⎩t ⎭ ⎪⎭ (3) Or, in exponent form, {h} = { L2 t-2}. Discussion As a check, from our study of thermodynamics we know that dh = cpdT for an ideal gas. Thus, the dimensions of dh must equal the dimensions of cp times the dimensions of dT. Comparing Eqs. 1 and 3 above, we see that this is indeed the case. 7-16 Solution We are to determine the primary dimensions of thermal conductivity. Analysis The primary dimensions of Q conduction are energy/time, and the primary dimensions of dT/dx are temperature/length. From the given equation, Primary dimensions of thermal conductivity: ⎧ ⎫ ⎧ mL2 ⎫ energy ⎪⎪ ⎪⎪ ⎪⎪ t 3 ⎪⎪ ⎧ mL ⎫ time {k } = ⎨ ⎬=⎨ ⎬=⎨ 3 ⎬ ⎪ length 2 × temperature ⎪ ⎪ L × T ⎪ ⎩ t T ⎭ length ⎪⎭ ⎩⎪ ⎪⎩ ⎭⎪ (1) Or, in exponent form, {k} = {m1 L1 t-3 T-1}. We obtain a value of k from a reference book. E.g. kcopper = 401 W/m⋅K. These units have dimensions of power/length⋅temperature. Since power is energy/time, we see immediately that Eq. 1 is correct. Alternatively, we can transform the units of k into primary units, Primary SI units of thermal conductivity: kcopper = 401 W ⎛ N m ⎞⎛ kg m ⎞ kg ⋅ m = 401 3 ⎜ ⎟⎜ 2 ⎟ m K ⎝ s W ⎠⎝ N s ⎠ s ⋅K (2) Discussion We have used the principle of dimensional homogeneity to determine the primary dimensions of k. Namely, we utilized the fact that the dimensions of both terms of the given equation must be identical. 7-8 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-17 Solution We are to determine the primary dimensions of each variable. Analysis (a) Heat generation rate is energy per unit volume per unit time, Primary dimensions of heat generation rate: ⎧ mL2 energy ⎫ ⎪⎪ t 2 { g } = ⎧⎨ ⎬=⎨ 3 ⎩ volume × time ⎭ ⎪ L t ⎩⎪ ⎫ ⎪⎪ ⎧ m ⎫ ⎬=⎨ 3⎬ ⎪ ⎩ Lt ⎭ ⎭⎪ (1) Or, in exponent form, { g } = {m1 L-1 t-3}. (b) Heat flux is energy per unit area per unit time, Primary dimensions of heat flux: ⎧ mL2 energy ⎫ ⎪⎪ t 2 {q} = ⎧⎨ ⎬=⎨ 2 ⎩ area × time ⎭ ⎪ L t ⎪⎩ ⎫ ⎪⎪ ⎧ m ⎫ ⎬=⎨ 3⎬ ⎪ ⎩t ⎭ ⎪⎭ (2) Or, in exponent form, { q } = {m1 t-3}. (c) Heat flux is energy per unit area per unit time per unit temperature, Primary dimensions of heat transfer coefficient: ⎧ mL2 ⎫ ⎪⎪ ⎧ m ⎫ ⎧ ⎫ ⎪⎪ t 2 energy {h} = ⎨ ⎬=⎨ 2 ⎬=⎨ 3 ⎬ ⎩ area × time × temperature ⎭ ⎪ L × t × T ⎪ ⎩ t T ⎭ ⎩⎪ ⎭⎪ (3) Or, in exponent form, {h} = {m1 t-3 T-1}. Discussion variables. In the field of heat transfer it is critical that one be careful with the dimensions (and units) of heat transfer 7-9 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-18 Solution dimensions. We are to choose three properties or constants and write out their names, their SI units, and their primary Analysis There are many options. For example, Students may choose cv (specific heat at constant volume). The units are kJ/kg⋅K, which is energy per mass per temperature. Thus, ⎧ mL2 ⎫ ⎧ ⎫ ⎪⎪ t 2 ⎪⎪ ⎧ L2 ⎫ energy {cv } = ⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ t T ⎭ ⎩⎪ ⎭⎪ Primary dimensions of specific heat at constant volume: (1) Or, in exponent form, {cv} = { L2 t-2 T-1}. Students may choose v (specific volume). The units are m3/kg, which is volume per mass. Thus, volume ⎫ ⎧ L3 ⎫ ⎬=⎨ ⎬ ⎩ mass ⎭ ⎩ m ⎭ {v } = ⎧⎨ Primary dimensions of specific volume: (2) Or, in exponent form, {v} = {m-1 L3}. Students may choose hfg (latent heat of vaporization). The units are kJ/kg, which is energy per mass. Thus, ⎧ mL2 ⎧ energy ⎫ ⎪⎪ t 2 =⎨ ⎬=⎨ ⎩ mass ⎭ ⎪ m ⎪⎩ Primary dimensions of latent heat of vaporization: {h } fg ⎫ ⎪⎪ ⎧ L2 ⎫ ⎬=⎨ 2 ⎬ ⎪ ⎩t ⎭ ⎪⎭ (3) Or, in exponent form, {hfg} = {L2 t-2}. (The same dimensions hold for hf and hg.) Students may choose sf (specific entropy of a saturated liquid). The units are kJ/kg⋅K, which is energy per mass per temperature. Thus, Primary dimensions of specific entropy of a saturated liquid: ⎧ mL2 ⎫ ⎧ ⎫ ⎪⎪ t 2 ⎪⎪ ⎧ L2 ⎫ energy sf = ⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ t T ⎭ ⎪⎩ ⎪⎭ { } (4) Or, in exponent form, {sf} = { L2 t-2 T-1}. (The same dimensions hold for sfg and sg.) Discussion Students’ answers will vary. There are some other choices. 7-10 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-19E Solution dimensions. We are to choose three properties or constants and write out their names, their SI units, and their primary Analysis There are many options. For example, Students may choose cv (specific heat at constant volume). The units are Btu/lbm⋅R, which is energy per mass per temperature. Thus, ⎧ mL2 ⎫ ⎧ ⎫ ⎪⎪ t 2 ⎪⎪ ⎧ L2 ⎫ energy {cv } = ⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ t T ⎭ ⎩⎪ ⎭⎪ Primary dimensions of specific heat at constant volume: (1) Or, in exponent form, {cv} = { L2 t-2 T-1}. Students may choose v (specific volume). The units are ft3/lbm, which is volume per mass. Thus, volume ⎫ ⎧ L3 ⎫ ⎬=⎨ ⎬ ⎩ mass ⎭ ⎩ m ⎭ {v } = ⎧⎨ Primary dimensions of specific volume: (2) Or, in exponent form, {v} = {m-1 L3}. Students may choose hfg (latent heat of vaporization). The units are Btu/lbm, which is energy per mass. Thus, ⎧ mL2 ⎧ energy ⎫ ⎪⎪ t 2 =⎨ ⎬=⎨ ⎩ mass ⎭ ⎪ m ⎪⎩ Primary dimensions of latent heat of vaporization: {h } fg ⎫ ⎪⎪ ⎧ L2 ⎫ ⎬=⎨ 2 ⎬ ⎪ ⎩t ⎭ ⎪⎭ (3) Or, in exponent form, {hfg} = {L2 t-2}. (The same dimensions hold for hf and hg.) Students may choose sf (specific entropy of a saturated liquid). The units are Btu/lbm⋅R, which is energy per mass per temperature. Thus, Primary dimensions of specific entropy of a saturated liquid: ⎧ mL2 ⎫ ⎧ ⎫ ⎪⎪ t 2 ⎪⎪ ⎧ L2 ⎫ energy sf = ⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩ mass × temperature ⎭ ⎪ m × T ⎪ ⎩ t T ⎭ ⎪⎩ ⎪⎭ { } (4) Or, in exponent form, {sf} = { L2 t-2 T-1}. (The same dimensions hold for sfg and sg.) Discussion Students’ answers will vary. There are some other choices. 7-11 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Dimensional Homogeneity 7-20C Solution We are to explain the law of dimensional homogeneity. Analysis The law of dimensional homogeneity states that every additive term in an equation must have the same dimensions. As a simple counter example, an equation with one term of dimensions length and another term of dimensions temperature would clearly violate the law of dimensional homogeneity – you cannot add length and temperature. All terms in the equation must have the same dimensions. Discussion If in the solution of an equation you realize that the dimensions of two terms are not equivalent, this is a sure sign that you have made a mistake somewhere! 7-21 Solution We are to determine the primary dimensions of the gradient operator, and then verify that primary dimensions of each additive term in the equation are the same. Analysis (a) By definition, the gradient operator is a three-dimensional derivative operator. For example, in Cartesian coordinates, Gradient operator in Cartesian coordinates: G ⎛ ∂ ∂ ∂ ⎞ G ∂ G ∂ G ∂ ∇=⎜ , , ⎟=i +j +k ∂x ∂y ∂z ⎝ ∂x ∂y ∂z ⎠ Therefore its dimensions must be 1/length. Thus, Primary dimensions of the gradient operator: G Or, in exponent form, { ∇ } = { L-1}. {∇} = ⎧⎨⎩ L1 ⎭⎫⎬ G (b) Similarly, the primary dimensions of a time derivative (∂/∂t) are 1/time. Also, the primary dimensions of velocity are length/time, and the primary dimensions of acceleration are length/time2. Thus each term in the given equation can be written in terms of primary dimensions, {a} = ⎧⎨ G {a} = ⎧⎨ length ⎫ 2 ⎬ ⎩ time ⎭ G ⎧ length ⎫ G ⎧⎪ ∂V ⎫⎪ ⎪⎪ time ⎪⎪ ⎧ length ⎫ ⎨ ⎬=⎨ ⎬=⎨ 2 ⎬ ⎩⎪ ∂t ⎭⎪ ⎪ time ⎪ ⎩ time ⎭ ⎪⎩ ⎪⎭ L⎫ 2 ⎬ ⎩t ⎭ G ⎪⎧ ∂V ⎪⎫ ⎧ L ⎫ ⎨ ⎬=⎨ 2⎬ ⎩⎪ ∂t ⎭⎪ ⎩ t ⎭ 1 length ⎫ ⎧ length ⎫ × × ⎬=⎨ ⎬ {(V ⋅∇ )V } = ⎩⎨⎧ length time length time ⎭ ⎩ time ⎭ G G G 2 {(V ⋅ ∇ )V } = ⎧⎨⎩ tL ⎭⎫⎬ G G G 2 Indeed, all three additive terms have the same dimensions, namely {L1 t-2}. Discussion If the dimensions of any of the terms were different from the others, it would be a sure sign that an error was made somewhere in deriving or copying the equation. 7-12 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-22 Solution We are to determine the primary dimensions of each additive term in the equation, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of the time derivative (∂/∂t) are 1/time. The primary dimensions of the gradient vector are 1/length, and the primary dimensions of velocity are length/time. Thus each term in the equation can be written in terms of primary dimensions, ⎧ mass×length ⎫ G ⎧⎪ F ⎫⎪ ⎧ force ⎫ ⎪⎪ ⎪⎪ time 2 ⎨ ⎬=⎨ ⎬=⎨ ⎬ mass ⎪⎩ m ⎪⎭ ⎩ mass ⎭ ⎪ ⎪ ⎩⎪ ⎭⎪ G ⎧⎪ F ⎫⎪ ⎧ L ⎫ ⎨ ⎬=⎨ 2⎬ ⎩⎪ m ⎭⎪ ⎩ t ⎭ ⎧ length ⎫ G ⎧⎪ ∂V ⎫⎪ ⎪⎪ time ⎪⎪ ⎨ ⎬=⎨ ⎬ ⎩⎪ ∂t ⎭⎪ ⎪ time ⎪ ⎪⎩ ⎭⎪ G ⎪⎧ ∂V ⎪⎫ ⎧ L ⎫ ⎨ ⎬=⎨ 2⎬ ⎩⎪ ∂t ⎭⎪ ⎩ t ⎭ 1 length ⎫ × × ⎬ {(V ⋅∇ )V } = ⎧⎨⎩ length time length time ⎭ G G G {(V ⋅∇ )V } = ⎩⎧⎨ tL ⎭⎫⎬ G G G 2 Indeed, all three additive terms have the same dimensions, namely {L1 t-2}. Discussion The dimensions are, in fact, those of acceleration. 7-23 Solution We are to determine the primary dimensions of each additive term in Eq. 1, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of the material derivative (D/Dt) are 1/time. The primary dimensions of volume are length3, and the primary dimensions of velocity are length/time. Thus each term in the equation can be written in terms of primary dimensions, length 3 ⎫ ⎧ 1 ⎫ ⎧ 1 DV ⎫ ⎧ 1 = × ⎨ ⎬ ⎨ ⎬=⎨ ⎬ 3 time ⎭ ⎩ time ⎭ ⎩V Dt ⎭ ⎩ length ⎧ 1 DV ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩V Dt ⎭ ⎩ t ⎭ ⎧ length ⎫ ⎧ ∂u ⎫ ⎪⎪ time ⎪⎪ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬=⎨ ⎬ ⎩ ∂x ⎭ ⎪ length ⎪ ⎩ time ⎭ ⎩⎪ ⎭⎪ ⎧ ∂u ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩ ∂x ⎭ ⎩ t ⎭ ⎧ length ⎫ ⎧ ∂v ⎫ ⎪⎪ time ⎪⎪ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬=⎨ ⎬ ⎩ ∂y ⎭ ⎪ length ⎪ ⎩ time ⎭ ⎩⎪ ⎭⎪ ⎧ ∂v ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩ ∂y ⎭ ⎩ t ⎭ ⎧ length ⎫ ⎧ ∂w ⎫ ⎪⎪ time ⎪⎪ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬=⎨ ⎬ ⎩ ∂z ⎭ ⎪ length ⎪ ⎩ time ⎭ ⎪⎩ ⎪⎭ ⎧ ∂w ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩ ∂z ⎭ ⎩ t ⎭ Indeed, all four additive terms have the same dimensions, namely {t-1}. Discussion If the dimensions of any of the terms were different from the others, it would be a sure sign that an error was made somewhere in deriving or copying the equation. 7-13 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-24 Solution We are to determine the primary dimensions of each additive term, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of the velocity components are length/time. The primary dimensions of coordinates r and z are length, and the primary dimensions of coordinate θ are unity (it is a dimensionless angle). Thus each term in the equation can be written in terms of primary dimensions, length ⎫ ⎧ length ⎧⎪ 1 ∂ ( rur ) ⎫⎪ ⎪⎪ 1 time ⎪⎪ = ⎧ 1 ⎫ × ⎨ ⎬=⎨ ⎬ ⎨ ⎬ length ⎪ ⎩ time ⎭ ⎩⎪ r ∂r ⎭⎪ ⎪ length ⎩⎪ ⎭⎪ ⎪⎧ 1 ∂ ( rur ) ⎪⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩⎪ r ∂r ⎭⎪ ⎩ t ⎭ length ⎫ ⎧ ⎪⎪ 1 ⎪⎪ ⎧ 1 ⎫ ⎧ 1 ∂uθ ⎫ × time ⎬ = ⎨ ⎨ ⎬ == ⎨ ⎬ 1 ⎪ ⎩ time ⎭ ⎩ r ∂θ ⎭ ⎪ length ⎪⎩ ⎪⎭ ⎧ 1 ∂uθ ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩ r ∂θ ⎭ ⎩ t ⎭ ⎧ length ⎫ ⎧ ∂u z ⎫ ⎪⎪ time ⎪⎪ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬=⎨ ⎬ ⎩ ∂z ⎭ ⎪ length ⎪ ⎩ time ⎭ ⎩⎪ ⎭⎪ ⎧ ∂u z ⎫ ⎧ 1 ⎫ ⎨ ⎬=⎨ ⎬ ⎩ ∂z ⎭ ⎩ t ⎭ Indeed, all three additive terms have the same dimensions, namely {t-1}. Discussion If the dimensions of any of the terms were different from the others, it would be a sure sign that an error was made somewhere in deriving or copying the equation. 7-14 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-25 Solution We are to determine the primary dimensions of each additive term in the equation, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of heat transfer rate are energy/time. The primary dimensions of mass flow rate are mass/time, and those of specific heat are energy/mass⋅temperature, as found in Problem 7-14. Thus each term in the equation can be written in terms of primary dimensions, ⎧ mL2 ⎧ energy ⎫ ⎪⎪ t 2 Q = ⎨ ⎬=⎨ ⎩ time ⎭ ⎪ t ⎩⎪ {}  T } {mC p out  T } {mC p in ⎫ ⎪⎪ ⎧ mL2 ⎫ ⎬=⎨ 3 ⎬ ⎪ ⎩ t ⎭ ⎭⎪ {Q } = ⎨ mL ⎬ t ⎧ ⎩ 2 3 ⎫ ⎭ ⎧ ⎫ mL2 2 2 ⎧ mass ⎫ ⎪⎪ m energy ⎪⎪ ⎧ mL ⎫ =⎨ × × temperature ⎬ = ⎨ × t × T ⎬ = ⎨ 3 ⎬ ⎩ time mass × temperature ⎭ ⎪ t m×T ⎪ ⎩ t ⎭ ⎪⎩ ⎪⎭ mL  T }=⎨ {mC ⎬ t ⎧ ⎫ mL2 ⎪ ⎧ mL2 ⎫ 2 ⎧ mass ⎫ ⎪⎪ m energy ⎪ =⎨ × × temperature ⎬ = ⎨ × t × T ⎬ = ⎨ 3 ⎬ ⎩ time mass × temperature ⎭ ⎪ t m×T ⎪ ⎩ t ⎭ ⎩⎪ ⎭⎪ mL  T }=⎨ {mC ⎬ t ⎧ p out ⎩ 2 3 ⎧ p in ⎩ 2 3 ⎫ ⎭ ⎫ ⎭ Indeed, all three additive terms have the same dimensions, namely {m1 L2 t-3}. Discussion We could also have left the temperature difference in parentheses as a temperature difference (same dimensions as the individual temperatures), and treated the equation as having only two terms. 7-26 Solution We are to determine the primary dimensions of each additive term, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of the time derivative (d/dt) are 1/time. The primary dimensions of density are mass/length3, those of volume are length3, those of area are length2, and those of velocity are length/time. The primary G G dimensions of unit vector n are unity, i.e. {1} (in other words n has no dimensions). Finally, the primary dimensions of b, which is defined as B per unit mass, are {B/m}. Thus each term in the equation can be written in terms of primary dimensions, ⎧ dBsys ⎫ ⎧ B ⎫ ⎨ ⎬=⎨ ⎬ ⎩ dt ⎭ ⎩ time ⎭ {∫ ⎧d ⎨ ⎩ dt CS ∫ CV ⎧ dBsys ⎫ ⎧ B ⎫ ⎨ ⎬=⎨ ⎬ ⎩ dt ⎭ ⎩ t ⎭ ⎧ 1 ⎫ B mass × × × length 3 ⎬ 3 ⎩ time length mass ⎭ ⎫ ρ bdV ⎬ = ⎨ } ⎭ ⎧ mass ⎫ B length × × × 1× length 2 ⎬ 3 ⎩ length mass time ⎭ ρ bVr ⋅ ndA = ⎨ G G ⎧d ⎨ ⎩ dt {∫ CS ∫ CV ⎫ ⎧B⎫ ⎩t⎭ ρ bdV ⎬ = ⎨ ⎬ ⎭ } ⎧B⎫ ⎩t⎭ ρ bVr ⋅ ndA = ⎨ ⎬ G G Indeed, all three additive terms have the same dimensions, namely {B t-1}. Discussion The RTT for property B has dimensions of rate of change of B. 7-15 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-27 Solution We are to determine the primary dimensions of the first three additive term, and we are to verify that those terms are dimensionally homogeneous. Then we are to evaluate the dimensions of the adsorption coefficient. Analysis The primary dimensions of the time derivative (d/dt) are 1/time. Those of As are length2, those of V are length3, those of c are mass/length3, and those of V are length3/time. Thus the primary dimensions of the first three terms are ⎧ ⎨V ⎩ mass ⎫ ⎧ 3 ⎪ ⎪ dc ⎫ ⎪ ⎧ mass ⎫ 3 length ⎪ ⎬ = ⎨length ⎬=⎨ ⎬ dt ⎭ ⎪ time ⎪ ⎩ time ⎭ ⎪⎭ ⎩⎪ {S} = ⎧⎨ {V c} = ⎧⎩⎨ length time {S} = ⎧⎨ mass ⎫ ⎬ ⎩ time ⎭ 3 × ⎧ dc ⎫ ⎧ m ⎫ ⎨V ⎬ = ⎨ ⎬ ⎩ dt ⎭ ⎩ t ⎭ m⎫ ⎬ ⎩t ⎭ mass ⎫ ⎧ mass ⎫ ⎬=⎨ ⎬ length 3 ⎭ ⎩ time ⎭ {V c} = ⎧⎨⎩ mt ⎭⎫⎬ Indeed, the first three additive terms have the same dimensions, namely {m1 t-1}. Since the equation must be dimensionally homogeneous, the last term must have the same dimensions as well. We use this fact to find the dimensions of kw, mass ⎫ {cAs kw } = ⎧⎨ ⎬ ⎩ time ⎭ ⎫ mass ⎧ mass ⎫ ⎧ ⎪⎪ time ⎪⎪ ⎪⎪ ⎪⎪ time {k w } = ⎨ ⎬=⎨ ⎬ ⎪ cAs ⎪ ⎪ mass 3 × length 2 ⎪ ⎩⎪ ⎭⎪ ⎩⎪ length ⎭⎪ {kw } = ⎧⎨ L⎫ ⎬ ⎩t⎭ Or, in exponent form, {kw} = {L1 t-1}. The dimensions of wall adsorption coefficient are those of velocity. Discussion In fact, some authors call kw a “deposition velocity”. 7-16 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Nondimensionalization of Equations 7-28C Solution Analysis problem. We are to give the primary reason for nondimensionalizing an equation. The primary reason for nondimensionalizing an equation is to reduce the number of parameters in the Discussion As shown in the examples in the text, nondimensionalization of an equation reduces the number of independent parameters in the problem, simplifying the analysis. 7-29 Solution We are to nondimensionalize all the variables, and then re-write the equation in nondimensionalized form. Assumptions 1 The air in the room is well mixed so that c is only a function of time. Analysis (a) We nondimensionalize the variables by inspection according to their dimensions, Nondimensionalized variables: A c L2 S V V , t* = t 3 , As * = 2s , k w * = k w , and S * = V * = 3 , c* =  climit L L L V climitV d ( c*climit ) (b) We substitute these into the equation to generate the nondimensionalized equation, V *L3 ( )   − V c*c − ( c*c ) A *L2 ⎛⎜ k * V V S c * = limit limit limit s w ⎜ L2 ⎛ L3 ⎞ ⎝ d ⎜ t* ⎟  ⎝ V ⎠ ⎞ ⎟⎟ ⎠ (1) We notice that every term in Eq. 1 contains the quantity V climit . We divide every term by this quantity to get a nondimensionalized form of the equation, Nondimensionalized equation: V* dc* = S * − c* − c*As *kw * dt* No dimensionless groups have arisen in this nondimensionalization. Discussion Since all the characteristic scales disappear, no dimensionless groups have arisen. Since there are no dimensionless parameters, one solution in nondimensionalized variables is valid for all combinations of L, V , and climit. 7-17 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-30 Solution We are to nondimensionalize the equation, and identify the dimensionless parameters that appear in the nondimensionalized equation. Assumptions 1 The flow is steady. 2 The flow is incompressible. Analysis We plug the nondimensionalized variables into the equation. For example, u = u*U and x = x*L in the first term. The result is U ∂u * U ∂v * U ∂w * + + =0 L ∂x * L ∂y * L ∂z * or, after simplifying, Nondimensionalized incompressible flow relationship: ∂u * ∂v * ∂w * + + =0 ∂x * ∂y * ∂z * (1) There are no nondimensional parameters in the nondimensionalized equation. The original equation comes from pure kinematics – there are no fluid properties involved in the equation, and therefore it is not surprising that no nondimensional parameters appear in the nondimensionalized form of the equation, Eq. 1. Discussion We show in Chap. 9 that the equation given in this problem is the differential equation for conservation of mass for an incompressible flow field – the incompressible continuity equation. 7-31 Solution We are to nondimensionalize the equation of motion and identify the dimensionless parameters that appear in the nondimensionalized equation. Analysis First, we must expand the first material derivative term since the nondimensionalization is not identical for the individual terms. Then we plug in the nondimensionalized variables. For example, u = u*V and x = x*L in the first term on the right. The result is ( ) G G ⎞ ∂u ∂v ∂w 1 DV 1 ⎛ ∂V = ⎜ + V ⋅∇ V ⎟ = + + V Dt V ⎝ ∂t ⎠ ∂x ∂y ∂z ⎞ V ∂u * V ∂v * V ∂w * 1 ⎛ 3 ∂V * L3V JJJG JJJG + + + V * ⋅∇ * V * ⎟ = ⎜L f 3 ∂t * L LV * ⎝ ⎠ L ∂x * L ∂y * L ∂z * or, after simplifying (multiply each term by L/V), ( ) ( ) JJJG JJJG ⎞ ∂u * ∂v * ∂w * 1 ⎛ ⎛ fL ⎞ ∂V * + V * ⋅∇ * V * ⎟ = + + ⎜⎜ ⎟ V * ⎝ ⎝ V ⎠ ∂t * ⎠ ∂x * ∂y * ∂z * ( (1) ) We recognize the nondimensional parameter in parentheses in Eq. 1 as St, the Strouhal number, and we re-write Eq. 1 as Nondimensionalized oscillating compressible flow relationship: 1 ⎛ ∂V * JJJG JJJG ⎞ ∂u * ∂v * ∂w * + V *⋅ ∇ * V * ⎟ = + + St ⎜ V * ⎝ ∂t * ⎠ ∂x * ∂y * ∂z * Discussion We show in Chap. 9 that the given equation of motion is the differential equation for conservation of mass for an unsteady, compressible flow field – the general continuity equation. We may also use angular frequency ω (radians per second) in place of physical frequency f (cycles per second), with the same result. 7-18 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-32 Solution We are to determine the primary dimensions of the stream function, nondimensionalize the variables, and then re-write the definition of ψ in nondimensionalized form. Assumptions 1 The flow is incompressible. 2 The flow is two-dimensional in the x-y plane. ψ, (a) We use that fact that all equations must be dimensionally homogeneous. We solve for the dimensions of Analysis Primary dimensions of stream function: 2 L ⎫ ⎧L ⎫ × L⎬ = ⎨ ⎬ ⎩t ⎭ ⎩ t ⎭ {ψ } = {u} × { y} = ⎨⎧ Or, in exponent form, {ψ} = {L2 t-1}. (b) We nondimensionalize the variables by inspection according to their dimensions, Nondimensionalized variables: x y x* = y* = L L u* = u t L v* = v (c) We generate the nondimensionalized equations, ⎛ L2 ⎞ ∂ψ * ⎜ ⎟ ⎛ L⎞ ⎝ t ⎠ u *⎜ ⎟ = ∂ * t y ( L) ⎝ ⎠ t L ψ* =ψ t L2 ⎛ L2 ⎞ ∂ψ * ⎜ ⎟ ⎛ L⎞ ⎝ t ⎠ v *⎜ ⎟ = − ∂ * t x ( L) ⎝ ⎠ We notice that every term in both parts of the above equation contains the ratio L/t. We divide every term by L/t to get the final nondimensionalized form of the equations, Nondimensionalized stream function equations: u* = ∂ψ * ∂y * v* = − ∂ψ * ∂x * No dimensionless groups have arisen in this nondimensionalization. Discussion Since all the nondimensionalized variables scale with L and t, no dimensionless groups have arisen. 7-19 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-33 Solution We are to nondimensionalize the equation of motion and identify the dimensionless parameters that appear in the nondimensionalized equation. G G Analysis We plug the nondimensionalized variables into the equation. For example, t = t*/ω and V = V∞V * in the first term on the right hand side. The result is G G ∂V * U 2 G G G 2 V * ⋅∇ * V * + ω L F / m * = ωU L ∂t * ( ) or, after simplifying by multiplying each term by L/ V∞2, ( ) ( ) ( ) G 2 G G G ⎛ωL ⎞ G ⎛ ω L ⎞ ∂V * + V * ⋅∇ * V * ⎜ ⎟ F /m *=⎜ ⎟ ⎝ V∞ ⎠ ⎝ V∞ ⎠ ∂t * (1) We recognize the nondimensional parameter in parentheses in Eq. 1 as St, the Strouhal number. We re-write Eq. 1 as ( ) ( ) Nondimensionalized Newton’s second law for incompressible oscillatory G G G G G ∂V * 2 + V * ⋅∇ * V * flow: (St ) F / m * = (St ) ∂t * Discussion We used angular frequency ω in this problem. The same result would be obtained if we used physical frequency. Equation 1 is the basis for forming the differential equation for conservation of linear momentum for an unsteady, incompressible flow field. 7-34 Solution We are to nondimensionalize the Bernoulli equation and generate an expression for the pressure coefficient. Assumptions other terms. 1 The flow is incompressible. 2 Gravitational terms in the Bernoulli equation are negligible compared to the Analysis We nondimensionalize the equation by dividing each term by the dynamic pressure, Nondimensionalization: P 1 ρV∞ 2 2 + Rearranging, Pressure coefficient: Cp = 1 ρV∞ 2 , 2 P∞ V2 = +1 2 1 V∞ ρV∞ 2 2 P − P∞ V2 = 1− 2 1 V∞ ρV∞ 2 2 Discussion Pressure coefficient is a useful dimensionless parameter that is inversely related to local air speed – as local air speed V increases, Cp decreases. 7-20 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Dimensional Analysis and Similarity 7-35C Solution We are to list the three primary purposes of dimensional analysis. Analysis The three primary purposes of dimensional analysis are: 1. To generate nondimensional parameters that help in the design of experiments and in the reporting of experimental results. 2. To obtain scaling laws so that prototype performance can be predicted from model performance. 3. To (sometimes) predict trends in the relationship between parameters. Discussion Dimensional analysis is most useful for difficult problems that cannot be solved analytically. 7-36C Solution prototype. We are to list and describe the three necessary conditions for complete similarity between a model and a Analysis The three necessary conditions for complete similarity between a model and a prototype are: 1. Geometric similarity – the model must be the same shape as the prototype, but scaled by some constant scale factor. 2. Kinematic similarity – the velocity at any point in the model flow must be proportional (by a constant scale factor) to the velocity at the corresponding point in the prototype flow. 3. Dynamic similarity – all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow. Discussion Complete similarity is achievable only when all three of the above similarity conditions are met. 7-37 Solution We are to use CFD to calculate the drag coefficient around a rectangular block for three fluids – air, water, and kerosene, and we are to compare the results at the same Reynolds number. Assumptions 1 The flow is two-dimensional and incompressible. 2 The flow is symmetric about the x axis. 3 The flow is turbulent, but steady in the mean. Properties The density and viscosity of the default air are ρ = 1.225 kg/m3 and μ = 1.7894 × 10-5 kg/m⋅s. The density and viscosity of liquid water at T = 15oC are 998.2 kg/m3 and 1.003 × 10-3 kg/m⋅s. The density and viscosity of kerosene at T = 15oC are 780.0 kg/m3 and 2.40 × 10-3 kg/m⋅s. TABLE 1 Drag coefficient as a function of fluid type for the case of turbulent flow over a rectangular block. In all cases, the Reynolds number is the same. Fluid Air Water Kerosene Re 1.37 × 104 1.37 × 104 1.37 × 104 CD 1.34344 1.34343 1.34343 Analysis We run FlowLab using template Block_fluid. A comparison of the CFD calculations for all three fluids is given in Table 1. The drag coefficient is identical to about five digits of precision. We conclude that for incompressible flow without free surface effects, the Reynolds number is the critical parameter; the type of fluid is irrelevant provided that the Reynolds number is the same. This reinforces what we learned about dimensional analysis. Discussion Newer versions of FlowLab may give slightly different results. Some incompressible CFD codes work with normalized variables from the start, requiring input of a Reynolds number instead of dimensional quantities such as velocity, density, and viscosity. In such cases, the fluid would be irrelevant. 7-21 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-38 Solution discuss. We are to generate CFD solutions for drag coefficient as a function of Reynolds number and compare and Assumptions 1 The flow is two-dimensional and incompressible. 2 The flow is symmetric about the x axis. 3 The flow is turbulent, but steady in the mean. Properties kg/m⋅s. The fluid is air with ρ = 1.225 kg/m3 and μ = 1.7894 × 10-5 TABLE 1 Drag coefficient as a function of Reynolds number for turbulent flow over a rectangular block. Re 10000 50000 100000 500000 1.00E+06 3.00E+06 5.00E+06 CD 1.30788 1.40848 1.42215 1.43065 1.43194 1.43296 1.43431 Analysis We run FlowLab using template Block_Reynolds. We compare six cases in Table 1. We see that CD levels off to a value of 1.43 to three digits of precision for Re greater than about 5 × 105. Thus, we have achieved Reynolds number independence, although the required Reynolds number is somewhat larger than that required experimentally. The last two cases are peculiar in that the Mach numbers are well beyond the incompressible limit (around 0.3) since the speed of sound in air at room temperature is around 340 m/s. However, even though these flows are unphysical, the CFD code is run as incompressible, and is not “aware” of this problem since the speed of sound is treated as infinite in an incompressible flow solver. The comparison with Re is still valid since we can use any incompressible fluid for the calculations, as illustrated in the previous problem. Discussion Newer versions of FlowLab may give slightly different results. Reynolds number independence checks are not always as simple as that shown here, because as Re increases, boundary layer thicknesses tend to decrease, requiring a finer mesh near walls. In the present problem this is not really an issue because the flow separates at the sharp edges of the block, and boundary layer thickness is not an important parameter in calculation of the drag. 7-39 Solution For a scale model of a submarine being tested in air, we are to calculate the wind tunnel speed required to achieve similarity with the prototype submarine that moves through water at a given speed. Assumptions 1 Compressibility of the air is assumed to be negligible. 2 The wind tunnel walls are far enough away so as to not interfere with the aerodynamic drag on the model sub. 3 The model is geometrically similar to the prototype. Properties For water at T = 15oC and atmospheric pressure, ρ = 999.1 kg/m3 and μ = 1.138 × 10-3 kg/m⋅s. For air at T = o 25 C and atmospheric pressure, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/m⋅s. Similarity is achieved when the Reynolds number of the model is equal to that of the prototype, Analysis Similarity: Re m = ρ pVp Lp ρ mVm Lm = Re p = μm μp (1) We solve Eq. 1 for the unknown wind tunnel speed, ⎛μ Vm = Vp ⎜ m ⎜ ⎝ μp ⎞ ⎛ ρ p ⎞ ⎛ Lp ⎞ ⎛ 1.849 × 10−5 kg/m ⋅ s ⎞ ⎛ 999.1 kg/m3 ⎞ = 0.520 m/s 8 = 57.0 m/s ⎟⎜ ( ) ⎟⎜ ⎟ ⎜ ⎟⎜ 3 ⎟( ) −3 ⎟ ⎝ 1.138 × 10 kg/m ⋅ s ⎠ ⎝ 1.184 kg/m ⎠ ⎠ ⎝ ρ m ⎠ ⎝ Lm ⎠ Discussion At this air temperature, the speed of sound is around 346 m/s. Thus the Mach number in the wind tunnel is equal to 57.0/346 = 0.165. This is sufficiently low that the incompressible flow approximation is reasonable. 7-22 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-40 Solution For a scale model of a submarine being tested in air, we are to calculate the wind tunnel speed required to achieve similarity with the prototype submarine that moves through water at a given speed. Assumptions 1 Compressibility of the air is assumed to be negligible. 2 The wind tunnel walls are far enough away so as to not interfere with the aerodynamic drag on the model sub. 3 The model is geometrically similar to the prototype. Properties For water at T = 15oC and atmospheric pressure, ρ = 999.1 kg/m3 and μ = 1.138 × 10-3 kg/m⋅s. For air at T = o 25 C and atmospheric pressure, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/m⋅s. Analysis Similarity is achieved when the Reynolds number of the model is equal to that of the prototype, Re m = Similarity: ρ pVp Lp ρ mVm Lm = Re p = μm μp (1) We solve Eq. 1 for the unknown wind tunnel speed, ⎛μ Vm = Vp ⎜ m ⎜ ⎝ μp ⎞ ⎛ ρ p ⎞ ⎛ Lp ⎞ ⎛ 1.849 × 10 −5 kg/m ⋅ s ⎞⎛ 999.1 kg/m 3 ⎞ = 0.520 m/s ⎜ 24 ) = 171 m/s ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ −3 3 ⎟( ⎟ ⎝ 1.138 × 10 kg/m ⋅ s ⎠⎝ 1.184 kg/m ⎠ ⎠ ⎝ ρ m ⎠ ⎝ Lm ⎠ At this air temperature, the speed of sound is around 346 m/s. Thus the Mach number in the wind tunnel is equal to 171/346 = 0.495. The Mach number is sufficiently high that the incompressible flow approximation is not reasonable. The wind tunnel should be run at a flow speed at which the Mach number is less than one-third of the speed of sound. At this lower speed, the Reynolds number of the model will be too small, but the results may still be usable, either by extrapolation to higher Re, or if we are fortunate enough to have Reynolds number independence, as discussed in Section 7-5. Discussion It is also unlikely that a small instructional wind tunnel can achieve such a high speed. 7-41 Solution We are to estimate the drag on a prototype submarine in water, based on aerodynamic drag measurements performed in a wind tunnel. Assumptions 1 The model is geometrically similar. 2 The wind tunnel is run at conditions which ensure similarity between model and prototype. Properties For water at T = 15oC and atmospheric pressure, ρ = 999.1 kg/m3 and μ = 1.138 × 10-3 kg/m⋅s. For air at T = o 25 C and atmospheric pressure, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/m⋅s. Analysis Since the Reynolds numbers have been matched, the nondimensionalized drag coefficient of the model equals that of the prototype, FD ,m ρ mVm Lm 2 2 = FD ,p ρ pVp 2 Lp 2 (1) We solve Eq. 1 for the unknown aerodynamic drag force on the prototype, FD,p, 2 ⎛ ρ p ⎞ ⎛ Vp ⎞ ⎛ Lp ⎞ ⎛ 999.1 kg/m ⋅ s ⎞ ⎛ 0.520 m/s ⎞ 2 = FD ,p = FD ,m ⎜ 2.45 N ( ) ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ( 8 ) = 11.0 N ⋅ ρ V L 1.184 kg/m s 57.035 m/s ⎝ ⎠ ⎝ ⎠ ⎝ m ⎠⎝ m ⎠ ⎝ m ⎠ 2 2 where we have used the wind tunnel speed calculated in Problem 7-36. Discussion Although the prototype moves at a much slower speed than the model, the density of water is much higher than that of air, and the prototype is eight times larger than the model. When all of these factors are combined, the drag force on the prototype is much larger than that on the model. 7-23 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-42E Solution For a prototype parachute and its model we are to calculate drag coefficient, and determine the wind tunnel speed that ensures dynamic similarity. Then we are to estimate the aerodynamic drag on the model. Assumptions Properties 1 The model is geometrically similar to the prototype. For air at 60oF and standard atmospheric pressure, ρ = 0.07633 lbm/ft3 and μ = 1.213 × 10-5 lbm/ft⋅s. Analysis (a) The aerodynamic drag on the prototype parachute is equal to the total weight. We can then easily calculate the drag coefficient CD, Drag coefficient: CD = FD ,p 1 2 ρ pVp 2 Ap = 24 ft ( 0.07633 lbm/ft ) (18 ft/s ) π ( 4 ) 230 lbf 1 2 3 2 2 ⎛ 32.2 lbm ft ⎞ ⎜ lbf s 2 ⎟ = 1.32 ⎝ ⎠ (b) We must match model and prototype Reynolds numbers in order to achieve dynamic similarity, Re m = Similarity: ρ pVp Lp ρ mVm Lm = Re p = μm μp (1) We solve Eq. 1 for the unknown wind tunnel speed, Wind tunnel speed: ⎛μ Vm = Vp ⎜ m ⎜ μp ⎝ ⎞ ⎛ ρ p ⎞ ⎛ Lp ⎞ = 18 ft/s )(1)(1)(12 ) = 216 ft/s ⎟⎜ ⎟ ρ m ⎟ ⎜ Lm ⎟ ( ⎠⎝ ⎠ ⎠⎝ (2) (c) As discussed in the text, if the fluid is the same and dynamic similarity between the model and the prototype is achieved, the aerodynamic drag force on the model is the same as that on the prototype. Thus, Aerodynamic drag on model: FD ,m = FD ,p = 230 lbf (3) Discussion We should check that the wind tunnel speed of Eq. 2 is not too high that the incompressibility approximation becomes invalid. The Mach number at this speed is about 216/1120 = 0.193. Since this is less than 0.3, compressibility is not an issue in this model test. The drag force on the model is quite large, and a fairly hefty drag balance must be available to measure such a large force. 7-43 Solution We are to discuss why one would pressurize a wind tunnel. Analysis As we see in some of the example problems and homework problems in this chapter, it is often difficult to achieve a high-enough wind tunnel speed to match the Reynolds number between a small model and a large prototype. Even if we were able to match the speed, the Mach number would often be too high. A pressurized wind tunnel has higher density air. At the same Reynolds number, the larger density leads to a lower air speed requirement. In other words, a pressurized wind tunnel can achieve higher Reynolds numbers for the same scale model. If the pressure were to be increased by a factor of 1.8, the air density would also go up by a factor of 1.8 (ideal gas law), assuming that the air temperature remains constant. Then the Reynolds number, Re = ρVL/μ, would go up by approximately 1.8. Note that we are also assuming that the viscosity does not change significantly with pressure, which is a reasonable assumption. Discussion The speed of sound is not a strong function of pressure, so Mach number is not affected significantly by pressurizing the wind tunnel. However, the power requirement for the wind tunnel blower increases significantly as air density is increased, so this must be taken into account when designing the wind tunnel. 7-24 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-44E Solution The concept of similarity will be utilized to determine the speed of the wind tunnel. Assumptions 1 Compressibility of the air is ignored (the validity of this assumption will be discussed later). 2 The wind tunnel walls are far enough away so as to not interfere with the aerodynamic drag on the model car. 3 The model is geometrically similar to the prototype. 4 Both the air in the wind tunnel and the air flowing over the prototype car are at standard atmospheric pressure. Properties For air at T = 25oC and atmospheric pressure, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/m⋅s. Analysis Since there is only one independent Π in this problem, similarity is achieved if Π2,m = Π2,p, where Π2 is the Reynolds number. Thus, we can write Π 2,m = Re m = ρ pVp Lp ρ mVm Lm = Π 2,p = Re p = μm μp which can be solved for the unknown wind tunnel speed for the model tests, Vm, ⎛μ Vm = Vp ⎜ m ⎜ μp ⎝ ⎞ ⎛ ρ p ⎞ ⎛ Lp ⎞ = 60.0 mph )(1)(1)( 3) = 180 mph ⎟⎜ ⎟ ρ m ⎟ ⎜ Lm ⎟ ( ⎠⎝ ⎠ ⎠⎝ Thus, to ensure similarity, the wind tunnel should be run at 180 miles per hour (to three significant digits). Discussion This speed is quite high, and the wind tunnel may not be able to run at that speed. The Mach number is Ma = (180 mph)/(774 mph) = 0.23, so compressible effects can be reasonably ignored. We were never given the actual length of either car, but the ratio of Lp to Lm is known because the prototype is three times larger than the scale model. The problem statement contains a mixture of SI and English units, but it does not matter since we use ratios in the algebra. 7-45E Solution wind tunnel. We are to estimate the drag on a prototype car, based on aerodynamic drag measurements performed in a Assumptions 1 The model is geometrically similar. 2 The wind tunnel is run at conditions which ensure similarity between model and prototype. Properties For air at T = 25oC and atmospheric pressure, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/m⋅s. Analysis Following the example in the text, since the Reynolds numbers have been matched, the nondimensionalized drag coefficient of the model equals that of the prototype, FD ,m ρ mVm Lm 2 2 = FD ,p ρ pVp 2 Lp 2 (1) We solve Eq. 1 for the unknown aerodynamic drag force on the prototype, FD,p, ⎛ ρ p ⎞ ⎛ Vp ⎞ ⎛ Lp ⎞ ⎛ 60.0 mph ⎞ 2 FD ,p = FD ,m ⎜ ⎟⎜ ⎟ ⎜ ⎟ = ( 33.5 lbf )(1) ⎜ ⎟ ( 3) = 33.5 lbf V L 180 mph ρ ⎝ ⎠ ⎝ m ⎠⎝ m ⎠ ⎝ m ⎠ 2 2 2 where we have used the wind tunnel speed calculated in the previous problem. The predicted drag on the prototype car is 33.5 lbf (to three significant digits). Discussion Since the air properties of the wind tunnel are identical to those of the air flowing around the prototype car, it turns out that the aerodynamic drag force on the prototype is the same as that on the model. This would not be the case if the wind tunnel air were at a different temperature or pressure compared to that of the prototype. 7-25 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-46 Solution We are to discuss whether cold or hot air in a wind tunnel is better, and we are to support our answer by comparing air at two given temperatures. Properties For air at atmospheric pressure and at T = 10oC, ρ = 1.246 kg/m3 and μ = 1.778 × 10-5 kg/m⋅s. At T = 50oC, 3 ρ = 1.092 kg/m and μ = 1.963 × 10-5 kg/m⋅s. Analysis As we see in some of the example problems and homework problems in this chapter, it is often difficult to achieve a high-enough wind tunnel speed to match the Reynolds number between a small model and a large prototype. Even if we were able to match the speed, the Mach number would often be too high. Cold air has higher density than warm air. In addition, the viscosity of cold air is lower than that of hot air. Thus, at the same Reynolds number, the colder air leads to a lower air speed requirement. In other words, a cold wind tunnel can achieve higher Reynolds numbers than can a hot wind tunnel for the same scale model, all else being equal. We support our conclusion by comparing air at two temperatures, Comparison of Reynolds numbers: ρcoldVL μcold ρ μ Re cold 1.246 kg/m3 1.963 × 10−5 kg/m ⋅ s = = cold hot = = 1.26 ρ hotVL Re hot ρ hot μcold 1.092 kg/m3 1.778 × 10−5 kg/m ⋅ s μhot Thus we see that the colder wind tunnel can achieve approximately 26% higher Reynolds number, all else being equal. Discussion There are other issues however. First of all, the denser air of the cold wind tunnel is harder to pump – the cold wind tunnel may not be able to achieve the same wind speed as the hot wind tunnel. Furthermore, the speed of sound is proportional to the square root of temperature. Thus, at colder temperatures, the Mach number is higher than at warmer temperatures for the same value of V, and compressibility effects are therefore more significant at lower temperatures. 7-26 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-47E Solution We are to calculate the speed and angular velocity (rpm) of a spinning baseball in a water channel such that flow conditions are dynamically similar to that of the actual baseball moving and spinning in air. Properties For air at T = 20oC and atmospheric pressure, ρ = 1.204 kg/m3 and μ = 1.825 × 10-5 kg/m⋅s. For water at T = o 20 C and atmospheric pressure, ρ = 998.0 kg/m3 and μ = 1.002 × 10-3 kg/m⋅s. Analysis The model (in the water) and the prototype (in the air) are actually the same baseball, so their characteristic lengths are equal, Lm = Lp. We match Reynolds number, Re m = ρ pVp Lp ρ mVm Lm = Re p = μm μp (1) and solve for the required water tunnel speed for the model tests, Vm, ⎛μ Vm = Vp ⎜ m ⎜ μp ⎝ ⎞ ⎛ ρ p ⎞⎛ Lp ⎞ ⎛ 1.002 ×10 −3 kg/m ⋅ s ⎞ ⎛ 1.204 kg/m3 ⎞ 1 = 5.63 mph = ( 85.0 mph ) ⎜ ⎟⎜ ⎟⎜ ⎟ ⎟⎜ 3 ⎟( ) −5 ⎟ ρ m Lm ⎝ 1.825 × 10 kg/m ⋅ s ⎠ ⎝ 998.0 kg/m ⎠ ⎠⎝ ⎠ ⎠⎝ (2) We also match Strouhal numbers, recognizing that n is proportional to f, St m = f p Lp f m Lm = St p = Vm Vp → nm Lm np Lp = Vm Vp (3) from which we solve for the required spin rate in the water tunnel, ⎛ Lp nm = np ⎜ ⎝ Lm ⎞ ⎛ Vm ⎟ ⎜⎜ ⎠ ⎝ Vp ⎞ ⎛ 5.63 mph ⎞ ⎟ = ( 320 rpm )(1) ⎜ ⎟ = 21.2 rpm ⎟ ⎝ 85.0 mph ⎠ ⎠ (4) The water tunnel needs to be run at 5.63 mph, and the baseball needs to be spun at 21.2 rpm for dynamic similarity. Discussion Because of the difference in fluid properties between air and water, the required water tunnel speed is much lower than that in air. In addition, the spin rate is much lower, making flow visualization easier. 7-27 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Dimensionless Parameters and the Method of Repeating Variables 7-48 Solution We are to verify that the Archimedes number is dimensionless. Analysis Archimedes number is defined as Archimedes number: ρ s gL3 ( ρs − ρ ) μ2 Ar = (1) We know the primary dimensions of density, gravitational acceleration, length, and viscosity. Thus, Primary dimensions of Archimedes number: ⎧m L 3 ⎫ ⎪ 3 2 L m⎪ {Ar} = ⎪⎨ L t 2 3 ⎪⎬ = {1} L ⎪ ⎪ m ⎪⎭ ⎩⎪ L2 t 2 (2) Discussion If the primary dimensions were not unity, we would assume that we made an error in the dimensions of one or more of the parameters. 7-49 Solution We are to verify that the Grashof number is dimensionless. Analysis Grashof number is defined as Grashof number: Gr = g β ΔT L3 ρ 2 μ2 (1) We know the primary dimensions of density, gravitational acceleration, length, temperature, and viscosity. The dimensions of coefficient of thermal expansion β are 1/temperature. Thus, Primary dimensions of Grashof number: ⎧ L 1 3 m2 ⎫ ⎪ 2 TL 6 ⎪ {Gr} = ⎪⎨ t T 2 L ⎪⎬ = {1} m ⎪ ⎪ ⎪⎭ L2 t 2 ⎩⎪ (2) Discussion If the primary dimensions were not unity, we would assume that we made an error in the dimensions of one or more of the parameters. 7-28 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-50 Solution We are to verify that the Rayleigh number is dimensionless, and determine what other established nondimensional parameter is formed by the ratio of Ra and Gr. Analysis Rayleigh number is defined as Ra = Rayleigh number: g β ΔT L3 ρ 2 c p kμ (1) We know the primary dimensions of density, gravitational acceleration, length, temperature, and viscosity. The dimensions of coefficient of thermal expansion β are 1/temperature, those of specific heat cp are length2/time2⋅temperature (Problem 714), and those of thermal conductivity k are mass⋅length/time3⋅temperature. Thus, Primary dimensions of Rayleigh number: ⎧ L 1 3 m 2 L2 ⎫ ⎪ 2 TL 6 2 ⎪⎪ L t T = {1} {Ra} = ⎪⎨ t T mL m ⎬ ⎪ ⎪ t 3 T Lt ⎪⎭ ⎩⎪ We take the ratio of Ra and Gr: Ratio of Rayleigh number and Grashof number: g β ΔT L3 ρ 2 c p Ra kμ = g β ΔT L3 ρ 2 Gr μ2 We recognize Eq. 3 as the Prandtl number, Prandtl number: Pr = = cp μ k Ra c p μ ρ c p μ ν = = = ρk α k Gr (2) (3) (4) Discussion Many of the established nondimensional parameters are formed by the ratio or product of two (or more) other established nondimensional parameters. 7-29 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-51 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the h = f (ω , ρ , g , R ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 (1) Step 2 The primary dimensions of each parameter are listed, ω {L } ρ {t } h {m L } −1 1 {L t } g 1 −3 1 −2 {L } R 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we elect not to pick the viscosity. We choose ω, ρ, and R Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = hω a1 ρ b1 R c1 {m } = {m } mass: {t } = {t } {L } = { L L 1 −3b1 0 The dependent Π is thus Lc1 } Π 1: −1 ) (m L ) (L ) 1 −3 a1 b1 1 c1 } 0 = b1 b1 = 0 0 = − a1 a1 = 0 0 = 1 − 3b1 + c1 c1 = −1 − a1 0 length: 1 b1 0 time: {( L )( t {( L t Π1 = h R } The second Pi (the only independent Π in this problem) is generated: Π 2 = gω a2 ρ b2 R c2 mass: time: {m } = {m } b2 0 { t } = {t 0 −2 − a2 t } {Π 2 } = 1 −2 )( t ) ( m L ) ( L ) −1 a2 1 −3 b2 1 c2 0 = b2 b2 = 0 0 = −2 − a2 a2 = −2 7-30 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling length: {L } = {L L 0 1 −3b2 Lc2 } c2 = −1 0 = 1 − 3b2 + c2 0 = 1 + c2 which yields Π 2: Π2 = g ω2R If we take Π2 to the power –1/2 and recognize that ωR is the speed of the rim, we see that Π2 can be modified into a Froude number, Modified Π2: Step 6 We write the final functional relationship as Relationship between Πs: Π 2 = Fr = ωR gR h = f ( Fr ) R (2) Discussion In the generation of the first Π, h and R have the same dimensions. Thus, we could have immediately written down the result, Π1 = h/R. Notice that density ρ does not appear in the result. Thus, density is not a relevant parameter after all. 7-31 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-52 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the h = f ( ω , ρ , g , R, t , μ ) Step 1 There are seven parameters in this problem; n = 7, List of relevant parameters: n=7 (1) Step 2 The primary dimensions of each parameter are listed, ω {L } ρ {t } h {m L } −1 1 {L t } {L } g 1 −3 {t } t R 1 −2 1 1 μ {m L t } 1 −1 −1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 7−3 = 4 Step 4 We need to choose three repeating parameters since j = 3. For convenience we choose the same repeating parameters that we used in the previous problem, ω, ρ, and R Repeating parameters: Step 5 The first two Πs are identical to those of the previous problem: Π 1: Π1 = and Π 2: Π2 = h R ωR gR where Π2 is identified as a form of the Froude number. The third Π is formed with time t. Since repeating parameter ω has dimensions of 1/time, it is the only one that remains in the Π. Thus, without the formal algebra, Π 3: {( m L t Finally, Π4 is generated with liquid viscosity, Π 4 = μω a4 ρ b4 R c4 {m } = {m m } mass: 0 time: length: 1 {t } = { t 0 {L } = { L The final Π is thus 0 b4 −1 − a4 t −1 −3b1 L {Π 4 } = } Lc1 } 1 −1 −1 Π 3 = ωt )( t ) ( m L ) ( L ) −1 a4 1 −3 b4 1 c4 } 0 = 1 + b4 b4 = −1 0 = −1 − a4 a4 = −1 0 = −1 − 3b4 + c4 c4 = −2 c4 = 1 + 3b4 7-32 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Π 4: Π4 = μ ρΩR 2 (2) If we invert Π4 and recognize that ωR is the speed of the rim, it becomes clear that Π4 of Eq. 2 can be modified into a Reynolds number, Modified Π4: Step 6 We write the final functional relationship as Relationship between Πs: Π4 = ρω R 2 = Re μ h = f ( Fr, ωt , Re ) R (3) (4) Discussion Notice that this time density ρ does appear in the result. There are other acceptable answers, but this one has the most established dimensionless groups. 7-33 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-53 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the f k = f (V , ρ , μ , D ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed, {t } ρ V −1 μ {m L } {L t } fk 1 −1 {m L t } 1 −3 1 −1 −1 {L } D 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we elect not to pick the viscosity. We choose V, ρ, and D Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = f kV a1 ρ b1 D c1 {m } = { m } mass: {t } = { t 0 −1 − a1 t {L } = { L L length: −1 )( L t ) ( m L ) ( L ) 1 −1 a1 0 The dependent Π is thus −3b1 } Lc1 } 1 −3 a1 b1 1 c1 } 0 = b1 b1 = 0 0 = −1 − a1 a1 = −1 0 = a1 − 3b1 + c1 c1 = 1 b1 0 time: {( t Π 1: Π1 = {( m L t fk D = St V where we have identified this Pi as the Strouhal number. The second Pi (the only independent Π in this problem) is generated: Π 2 = μV a2 ρ b2 D c2 mass: time: {m } = {m m } 0 {t } = { t 0 1 b2 −1 − a2 t } {Π 2 } = 1 −1 − 1 )( L t ) ( m L ) ( L ) 1 −1 a2 1 −3 b2 1 c2 } 0 = 1 + b2 b2 = −1 0 = −1 − a2 a2 = −1 7-34 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling length: {L } = {L 0 L L−3b2 Lc2 −1 a2 } c2 = −1 0 = −1 + a2 − 3b2 + c2 0 = −1 − 1 + 3 + c2 which yields Π 2: Π2 = μ ρVD We recognize this Π as the inverse of the Reynolds number. So, after inverting, Modified Π2: Π2 = ρVD = Reynolds number = Re μ Step 6 We write the final functional relationship as Relationship between Πs: St = f ( Re ) Discussion We cannot tell from dimensional analysis the exact form of the functional relationship. However, experiments verify that the Strouhal number is indeed a function of Reynolds number. 7-35 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-54 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the f k = f (V , ρ , μ , D, c ) Step 1 There are six parameters in this problem; n = 6, List of relevant parameters: n=6 (1) Step 2 The primary dimensions of each parameter are listed, {t } ρ {L t } fk −1 μ {m L } V 1 −1 {m L t } 1 −3 {L } D 1 −1 −1 1 {L t } c 1 −1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 6−3 = 3 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we elect not to pick the viscosity. We choose V, ρ, and D Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = f kV a1 ρ b1 D c1 {m } = {m } mass: {t } = { t 0 −1 − a1 t {L } = {L L length: −1 )( L t ) ( m L ) ( L ) 1 −1 a1 0 The dependent Π is thus −3b1 } Lc1 } 1 −3 a1 b1 1 c1 } 0 = b1 b1 = 0 0 = −1 − a1 a1 = −1 0 = a1 − 3b1 + c1 c1 = 1 b1 0 time: {( t Π 1: Π1 = {( m L t fk D = St V where we have identified this Pi as the Strouhal number. The second Pi (the first independent Π in this problem) is generated: Π 2 = μV a2 ρ b2 D c2 mass: time: {m } = {m m } 0 {t } = { t 0 1 b2 −1 − a2 t } {Π 2 } = 1 −1 − 1 )( L t ) ( m L ) ( L ) 1 −1 a2 1 −3 b2 1 c2 } 0 = 1 + b2 b2 = −1 0 = −1 − a2 a2 = −1 7-36 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling length: {L } = {L L L−3b2 Lc2 −1 a2 0 } c2 = −1 0 = −1 + a2 − 3b2 + c2 0 = −1 − 1 + 3 + c2 which yields Π 2: Π2 = μ ρVD We recognize this Π as the inverse of the Reynolds number. So, after inverting, Modified Π2: Π2 = ρVD = Reynolds number = Re μ {( L t } The third Pi (the second independent Π in this problem) is generated: {Π 3 } = Π 3 = cV a3 ρ b3 D c3 {m } = {m } mass: {t } = { t time: length: b3 0 0 {L } = {L L 0 −1 − a3 1 a3 t } L−3b3 Lc3 } 1 −1 )( L t ) ( m L ) ( L ) 1 −1 a3 1 −3 b3 1 c3 0 = b3 b3 = 0 0 = −1 − a3 a3 = −1 0 = 1 + a3 − 3b3 + c3 c3 = 0 0 = 1 − 1 + c3 which yields Π 3: Π3 = c V We recognize this Π as the inverse of the Mach number. So, after inverting, Modified Π3: Π3 = Step 6 We write the final functional relationship as Relationship between Πs: V = Mach number = Ma c St = f ( Re, Ma ) Discussion We have shown all the details. After you become comfortable with the method of repeating variables, you can do some of the algebra in your head. 7-37 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-55 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Step 1 There are five parameters in this problem; n = 5, W = f (ω , ρ , μ , D ) List of relevant parameters: n=5 (1) Step 2 The primary dimensions of each parameter are listed, ω W {m L t } ρ {t } 1 2 −3 {m L } −1 1 −3 μ {m L t } 1 −1 −1 {L } D 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we elect not to pick the viscosity. We choose ω, ρ, and D Repeating parameters: Step 5 The dependent Π is generated: Π1 = W ω a1 ρ b1 D c1 {m } = {m m } mass: 0 1 {t } = { t time: 0 {L } = {L L length: 0 b1 −3 − a1 2 The dependent Π is thus t −3b1 {Π1} = } Lc1 } {( m L t 1 2 −3 )( t ) ( m L ) ( L ) −1 1 −3 a1 b1 1 c1 } 0 = 1 + b1 b1 = −1 0 = −3 − a1 a1 = −3 0 = 2 − 3b1 + c1 c1 = −5 Π 1: Π1 = {( m L t W = NP ρ D 5ω 3 } where we have defined this Pi as the power number (Table 7-5). The second Pi (the only independent Π in this problem) is generated: Π 2 = μω a2 ρ b2 D c2 mass: {m } = {m m } 0 1 b2 {Π 2 } = 1 −1 −1 )( t ) ( m L ) ( L ) −1 a2 1 −3 b2 1 c2 b2 = −1 0 = 1 + b2 7-38 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. time: length: {t } = { t 0 {L } = {L 0 −1 − a2 t −1 −3b2 L } Lc2 Chapter 7 Dimensional Analysis and Modeling a2 = −1 0 = −1 − a2 } c2 = −2 0 = −1 − 3b2 + c2 0 = −1 + 3 + c2 which yields Π 2: Π2 = μ ρ D 2ω Since Dω is the speed of the tip of the rotating stirrer blade, we recognize this Π as the inverse of a Reynolds number. So, after inverting, Modified Π2: Π2 = ρ D 2ω ρ ( Dω ) D = = Reynolds number = Re μ μ Step 6 We write the final functional relationship as Relationship between Πs: N P = f ( Re ) (2) Discussion After some practice you should be able to do some of the algebra with the exponents in your head. Also, we usually expect a type of Reynolds number when we combine viscosity with a density, a length, and some kind of speed, be it angular speed or linear speed. 7-39 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-56 Solution We are to determine the dimensionless relationship between the given parameters Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis The dimensional analysis is identical to the previous problem except that we add two additional independent parameters, both of which have dimensions of length. The two Πs of the previous problem remain. We get two additional Πs since n is now equal to 7 instead of 5. There is no need to go through all the algebra for the two additional Πs – since their dimensions match those of one of the repeating variables (D), we know that all the exponents in the Π will be zero except the exponent for D, which will be –1. The two additional Πs are Π3 and Π4: The final functional relationship is Relationship between Πs: Π3 = Dtank D Π4 = hsurface D D h ⎛ ⎞ N P = f ⎜ Re, tank , surface ⎟ D D ⎝ ⎠ (1) Discussion We could also manipulate our Πs so that we have other length ratios like hsurface/Dtank, etc. Any such combination is acceptable. 7-40 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-57 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the δ = f ( x, V , ρ , μ ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 (1) Step 2 The primary dimensions of each parameter are listed, δ {L } 1 {L } ρ μ {L t } { m L } {m L t } x V 1 −1 1 1 −3 1 −1 −1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We pick length scale x, density ρ, and freestream velocity V. x, ρ , and V Repeating parameters: Step 5 The Πs are generated. Note that for the first Π we can do the algebra in our heads since the relationship is very simple. Namely, the dimensions of δ are identical to those of one of the repeating variables (x). In such a case we know that all the exponents in the Π group are zero except the one for x, which is –1. The dependent Π is thus Π 1: Π1 = The second Π is formed with viscosity, {Π 2 } = Π 2 = μ x a ρ bV c {m } = { m m } mass: 0 time: length: {t } = { t 0 {L } = { L 0 1 b −1 − c } −1 a t L L−3b Lc } {( m L t 1 −1 −1 x )( L ) ( m L ) ( L t ) } 1 a 1 −3 b 1 −1 c 0 = 1+ b b = −1 0 = −1 − c c = −1 0 = −1 + a − 3b + c 0 = −1 + a + 3 − 1 a = −1 which yields Π 2: δ Π2 = We recognize this Π as the inverse of the Reynolds number, μ ρVx 7-41 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Modified Π2 = Reynolds number based on x: Step 6 We write the final functional relationship as Relationship between Πs: Π 2 = Re x = δ x ρVx μ = f ( Re x ) Discussion We cannot determine the form of the relationship by purely dimensional reasoning since there are two Πs. However, in Chap. 10 we shall see that for a laminar boundary layer, Π1 is proportional to the square root of Π2. 7-42 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-58 Solution We are to create a scale for volume flow rate and then define an appropriate Richardson number. Analysis By “back of the envelope” reasoning (or by inspection), we define a volume flow rate scale as L2V. Then the Richardson number can be defined as Richardson number: Ri = L5 g Δρ L5 g Δρ Lg Δρ = = 2 2  2 ρV 2 ρV ρ LV ( ) (1) Discussion It is perhaps more clear from the form of Eq. 1 that Richardson number is a ratio of buoyancy forces to inertial forces. 7-43 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-59 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the u = f ( μ , V , h, ρ , y ) Step 1 There are six parameters in this problem; n = 6, List of relevant parameters: n=6 (1) Step 2 The primary dimensions of each parameter are listed, μ {L t } {m L t } u 1 −1 {L t } 1 −1 −1 ρ {L } {m L } h V 1 −1 1 −3 1 {L } y 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 6−3 = 3 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we elect not to pick the viscosity. It is better to pick a fixed length (h) rather than a variable length (y); otherwise y would appear in each Pi, which would not be desirable. We choose V, ρ, and h Repeating parameters: Step 5 The dependent Π is generated: Π1 = uV a1 ρ b1 h c1 mass: length: {m } = {m } {t } = { t 0 −1 − a1 t {L } = {L L L 1 a1 0 {( L t 1 −1 The dependent Π is thus } −3b1 Lc1 } )( L t ) ( m L ) ( L ) 1 −1 1 −3 a1 b1 1 c1 } 0 = b1 b1 = 0 0 = −1 − a1 a1 = −1 0 = 1 + a1 − 3b1 + c1 c1 = 0 b1 0 time: {Π1} = Π 1: Π1 = {( m L t u V The second Pi (the first independent Π in this problem) is generated: Π 2 = μV a2 ρ b2 h c2 mass: {m } = {m m } 0 1 b2 {Π 2 } = 1 −1 − 1 )( L t ) ( m L ) ( L ) 1 −1 a2 1 −3 b2 1 c2 } b2 = −1 0 = 1 + b2 7-44 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. {t } = { t time: length: 0 {L } = {L −1 − a2 t } a2 = −1 0 = −1 − a2 L L−3b2 Lc2 −1 a2 0 Chapter 7 Dimensional Analysis and Modeling } c2 = −1 0 = −1 + a2 − 3b2 + c2 0 = −1 − 1 + 3 + c2 which yields Π 2: Π2 = μ ρVh We recognize this Π as the inverse of the Reynolds number. So, after inverting, Modified Π2: Π2 = ρVh = Reynolds number = Re μ {( L )( L t } The third Pi (the second independent Π in this problem) is generated: {Π 3 } = Π 3 = yV a3 ρ b3 hc3 {m } = { m } mass: {t } = {t } time: length: b3 0 − a3 0 {L } = {L L 0 1 a3 L−3b3 Lc3 } 1 1 −1 ) (m L ) (L ) 1 −3 a3 b3 1 c3 0 = b3 b3 = 0 0 = − a3 a3 = 0 0 = 1 + a3 − 3b3 + c3 c3 = −1 0 = 1 + c3 which yields Π 3: Step 6 We write the final functional relationship as Relationship between Πs: Π3 = y h u y⎞ ⎛ = f ⎜ Re, ⎟ V h ⎝ ⎠ (2) Discussion We notice in the first and third Πs that when the parameter on which we are working has the same dimensions as one of the repeating parameters, the Π is simply the ratio of those two parameters (here u/V and y/h). 7-45 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-60 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the u = f ( μ , V , h, ρ , y, t ) Step 1 There are seven parameters in this problem; n = 7, List of relevant parameters: n=7 (1) Step 2 The primary dimensions of each parameter are listed, μ {L t } {m L t } u 1 −1 1 −1 −1 {L t } ρ {L } {m L } h V 1 −1 {L } y 1 −3 1 1 {t } t 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 7−3 = 4 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines outlined in this chapter, we do not pick the viscosity. It is better to pick a fixed length (h) rather than a variable length (y); otherwise y would appear in each Pi, which would not be desirable. It would also not be wise to have time appear in each parameter. We choose V, ρ, and h Repeating parameters: Step 5 The Πs are generated. The first three Πs are identical to those of the previous problem, so we do not include the details here. The fourth Π is formed by joining the new parameter t to the repeating variables, {Π 4 } = Π 4 = tV a4 ρ b4 h c4 mass: time: length: This Π is thus {m } = {m } b4 0 {t } = {t t } 1 − a4 0 {L } = {L L 0 a1 −3b4 Lc4 } {( t )( L t 1 1 −1 ) (m L ) (L ) 1 −3 a4 1 c4 } 0 = b4 b4 = 0 0 = 1 − a4 a4 = 1 0 = a4 − 3b4 + c4 c4 = −1 Π4 = Π 4: b4 tV h Step 6 Combining this result with the first three Πs from the previous problem, Relationship between Πs: Discussion u y tV ⎞ ⎛ = f ⎜ Re, , ⎟ V h h ⎠ ⎝ (2) As t → ∞, Π4 becomes irrelevant and the result degenerates into that of the previous problem. 7-46 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-61 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the ( Step 1 There are four parameters in this problem; n = 4, c = f k , T , Rgas List of relevant parameters: ) n=4 (1) Step 2 The primary dimensions of each parameter are listed; the ratio of specific heats k is dimensionless. {L t } c {T } k {1} 1 −1 {L t T Rgas 2 −2 1 T −1 } Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (T, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Thus we expect only one Π. Step 4 We need to choose three repeating parameters since j = 3. We only have one choice in this problem, since there are only three independent parameters on the right-hand side of Eq. 1. However, one of these is already dimensionless, so it is a Π all by itself. In this situation we reduce j by one and continue, j = 3 −1 = 2 Reduction: If this revised value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−2 = 2 Repeating parameters: T and Rgas We now expect two Πs. We choose two repeating parameters since j = 2, Step 5 The dependent Π is generated: Π1 = cT a1 Rgas b1 time: temperature: length: { t } = {t 0 −1− 2 b1 } {T } = {T } 0 a1 − b1 {L } = {L L } 0 1 2 b1 {Π1} = {( L t 1 −1 )( T ) ( L t 1 a1 2 −2 T −1 ) b1 } 0 = −1 − 2b1 b1 = −1/ 2 a1 = b1 a1 = −1/ 2 0 = 1 + 2b1 b1 = −1/ 2 Fortunately the two results for exponent b1 agree. The dependent Π is thus Π 1: Π1 = c RgasT 7-47 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The independent Π is already known, Π 2: Chapter 7 Dimensional Analysis and Modeling Π2 = k Step 6 We write the final functional relationship as Relationship between Πs: c RgasT = f (k ) (2) Discussion We cannot tell from dimensional analysis the exact form of the functional relationship. However, in this case the result agrees with the known equation for speed of sound in an ideal gas, c = kRgasT . 7-48 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-62 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the c = f ( k , T , Ru , M ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 (1) Step 2 The primary dimensions of each parameter are listed; the ratio of specific heats k is dimensionless. {L t } c {T } k {1} 1 −1 {m L t T Ru 1 2 −2 1 T −1 N −1 } {m N } M 1 −1 Step 3 As a first guess, j is set equal to 5, the number of primary dimensions represented in the problem (m, T, L, N, and t). j =5 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−5 = 0 Obviously we cannot have zero Πs. We check that we have not missed a relevant parameter. Convinced that we have included all the relevant parameters we reduce j by 1: j = 5 −1 = 4 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−4 =1 Step 4 We need to choose four repeating parameters since j = 4. We only have one choice in this problem, since there are only four independent parameters on the right-hand side of Eq. 1. However, one of these is already dimensionless, so it is a Π all by itself. In this situation we reduce j by one (again) and continue, j = 4 −1 = 3 Reduction: If this revised value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Repeating parameters: T, M, and Ru We now expect two Πs. Since j = 3 we choose three repeating parameters, {( L t Step 5 The dependent Π is generated: {Π1} = Π1 = cT a1 M b1 Ru c1 time: mass: {t } = {t 0 −1 −2 c1 {m } = {m 0 t b1 } m c1 } 1 −1 )( T ) ( m n ) ( m L t 1 a1 1 −1 b1 0 = −1 − 2c1 0 = b1 + c1 b1 = −c1 1 2 −2 T −1 N −1 ) c1 } c1 = −1/ 2 b1 = 1/ 2 7-49 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. amount of matter: temperature: length: {N } = {N 0 − b1 {T } = {T 0 a1 N − c1 T − c1 } } {L } = {L L } 0 1 2 c1 0 = −b1 − c1 b1 = −c1 0 = a1 − c1 a1 = c1 Chapter 7 Dimensional Analysis and Modeling b1 = 1/ 2 a1 = −1/ 2 c1 = −1/ 2 0 = 1 + 2c1 Fortunately the two results for exponent b1 agree, and the two results for exponent c1 agree. (If they did not agree, we would search for algebra mistakes. Finding none we would suspect that j is not correct or that we are missing a relevant parameter in the problem.) The dependent Π is thus Π 1: Π1 = The independent Π is already known, Π 2: c M Π2 = k Step 6 We write the final functional relationship as Relationship between Πs: Discussion RuT Π1 = c M RuT = f (k ) (2) Since we know that Rgas = Ru/M, we see that the result here is the same as that of the previous problem. 7-50 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-63 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the ( Step 1 There are three parameters in this problem; n = 3, c = f T , Rgas List of relevant parameters: ) n=3 (1) Step 2 The primary dimensions of each parameter are listed, {L t } {T } c {L t T 1 −1 Rgas 2 −2 1 T −1 } Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (T, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3−3 = 0 Obviously this is not correct, so we re-examine our initial assumptions. We can add another variable, k (the ratio of specific heats) to our List of relevant parameters. This problem would then be identical to Problem 7-58. Instead, for instructional purposes we reduce j by one and continue, j = 3 −1 = 2 Reduction: If this revised value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 We now expect only one Π. Step 4 We need to choose two repeating parameters since j = 2. We only have one choice in this problem, since there are only two independent parameters on the right-hand side of Eq. 1, Repeating parameters: T and Rgas Step 5 The dependent Π is generated: Π1 = cT a1 Rgas b1 time: temperature: length: { t } = {t 0 −1− 2 b1 } {T } = {T } 0 a1 − b1 {L } = {L L } 0 1 2 b1 {Π1} = {( L t 1 −1 )( T ) ( L t 1 a1 2 −2 T −1 ) b1 } 0 = −1 − 2b1 b1 = −1/ 2 a1 = b1 a1 = −1/ 2 0 = 1 + 2b1 b1 = −1/ 2 Fortunately the two results for exponent b1 agree. The dependent Π is thus 7-51 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Π 1: Π1 = c RgasT Step 6 Since there is only one Π, it is a function of nothing. This is only possible if we set the Π equal to a constant. We write the final functional relationship as Relationship between Πs: Π1 = c RgasT = constant (2) Discussion Our result represents an interesting case of “luck”. Although we failed to include the ratio of specific heats k in our analysis, we nevertheless obtain the correct result. In fact, if we set the constant in Eq. 2 as the square root of k, our result agrees with the known equation for speed of sound in an ideal gas, c = kRgasT . 7-52 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-64 Solution We are to use dimensional analysis to find the functional relationship between the given parameters, and compare to the known equation for an ideal gas. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the c = f ( P, ρ ) Step 1 There are three parameters in this problem; n = 3, List of relevant parameters: n=3 (1) Step 2 The primary dimensions of each parameter are listed, {L t } ρ {m L t } c {m L } P 1 −1 1 −1 −2 1 −3 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3−3 = 0 Obviously this is not correct, so we re-examine our initial assumptions. If we are convinced that c is a function of only P and ρ, we reduce j by one and continue, j = 3 −1 = 2 Reduction: If this revised value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 We now expect only one Π. Step 4 We need to choose two repeating parameters since j = 2. We only have one choice in this problem, since there are only two independent parameters on the right-hand side of Eq. 1, P and ρ Repeating parameters: Step 5 The dependent Π is generated: Π1 = cP a1 ρ b1 time: mass: length: {t } = {t 0 −1 −2 a1 {m } = {m 0 {L } = {L L 0 t a1 } mb1 1 −2 a1 {Π1} = } L−3b1 } {( L t 1 −1 )( m L 1 −1 −2 t ) (m L ) a1 1 −3 b1 } 0 = −1 − 2a1 a1 = −1/ 2 0 = a1 + b1 b1 = 1/ 2 0 = 1 − a1 − 3b1 0=0 0 = 1+ − 1 2 3 2 Fortunately the exponents for length agree with those of mass and time. The dependent Π is thus 7-53 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Π 1: Π1 = c ρ Chapter 7 Dimensional Analysis and Modeling P Step 6 Since there is only one Π, it is a function of nothing. This is only possible if we set the Π equal to a constant. We write the final functional relationship as Relationship between Πs: Π1 = c ρ P = constant, or c = constant ρ P (2) The ideal gas equation is P = ρRgasT, or P/ρ = RgasT. Thus, Eq. 2 can be written as Alternative result using ideal gas law: c = constant RgasT (3) Equation 3 is indeed consistent with the equation c = kRgasT . Discussion There is no way to obtain the value of the constant in Eq. 2 or 3 solely by dimensional analysis, but it turns out that the constant is the square root of k. 7-54 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling We are to use dimensional analysis to find a functional relationship between FD and variables V, L, and μ. 7-65 Solution Assumptions 1 We assume Re << 1 so that the creeping flow approximation applies. 2 Gravitational effects are irrelevant. 3 No parameters other than those listed in the problem statement are relevant to the problem. We follow the step-by-step method of repeating variables. Analysis Step 1 There are four variables and constants in this problem; n = 4. They are listed in functional form, with the dependent variable listed as a function of the independent variables and constants: FD = f (V , L, μ ) List of relevant parameters: n=4 Step 2 The primary dimensions of each parameter are listed. {m L t } FD 1 1 −2 {L t } μ {m L t } {L } V L 1 −1 1 −1 −1 1 Step 3 As a first guess, we set j equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the number of Πs expected is Number of expected Πs: k = n− j = 4−3 =1 Step 4 Now we need to choose three repeating parameters since j = 3. Since we cannot choose the dependent variable, our only choices are V, L, and μ. Step 5 Now we combine these repeating parameters into a product with the dependent variable FD to create the dependent Π, Π1 = FDV a1 Lb1 μ c1 Dependent Π: (1) We apply the primary dimensions of Step 2 into Eq. 1 and force the Π to be dimensionless, Dimensions of Π1: {Π1} = {m0 L0 t 0 } = {FDV a Lb μ c } = 1 {m } = {m m } 1 1 {( m L t 1 1 −2 )( L t ) ( L ) ( m L 1 −1 a1 1 b1 1 −1 −1 t ) c1 } Now we equate the exponents of each primary dimension to solve for exponents a1 through c1. mass: time: length: 0 { t } = {t 0 c1 1 −2 − a1 − c1 t t 1 a1 b1 } {L } = {L L L L } 0 − c1 0 = 1 + c1 c1 = −1 0 = −2 − a1 − c1 a1 = −1 0 = 1 + a1 + b1 − c1 b1 = −1 Equation 1 thus becomes Π 1: Π1 = FD μVL (2) Step 6 We now write the functional relationship between the nondimensional parameters. In the case at hand, there is only one Π, which is a function of nothing. This is possible only if the Π is constant. Putting Eq. 2 into standard functional form, 7-55 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Relationship between Πs: or Result of dimensional analysis: Π1 = FD = f ( nothing ) = constant μVL FD = constant ⋅ μVL (3) (4) Thus we have shown that for creeping flow around an object, the aerodynamic drag force is simply a constant multiplied by μVL, regardless of the shape of the object. Discussion This result is very significant because all that is left to do is find the constant, which will be a function of the shape of the object (and its orientation with respect to the flow). 7-56 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-66 Solution We are to find the functional relationship between the given parameters and name any established dimensionless parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the ( ( Step 1 There are five parameters in this problem; n = 5, ) V = f d p , ρp − ρ , μ, g List of relevant parameters: ) n=5 (1) Step 2 The primary dimensions of each parameter are listed, {L t } V 1 −1 ( ρ p - ρ) μ {m L } {m L t } {L t } {L } dp 1 −3 1 g 1 −1 −1 1 −2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We pick length scale dp, density difference (ρp - ρ), and gravitational constant g. ( ) d p , ρ p -ρ , and g Repeating parameters: Step 5 The Πs are generated. Note that for the first Π we do the algebra in our heads since the relationship is very simple. The dependent Π is Π1 = a Froude number: Π1 = {( m L t V gd p )( L ) ( m L ) ( L t ) } This Π is a type of Froude number. Similarly, the Π formed with viscosity is generated, ( Π2 = μ d pa ρ p − ρ mass: b {Π 2 } = gc {m } = { m m } 0 1 {t } = { t time: length: ) 0 {L } = { L 0 t −1 a } L L−3b Lc 1 a 1 −3 0 = −1 − 2c } 1 −2 c c=− 0 = −1 + a − 3b + c 0 = −1 + a + 3 − b b = −1 0 = 1+ b b −1 −2 c 1 −1 −1 1 2 a=− 1 2 3 2 which yields 7-57 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Π 2: Π2 = (ρ Chapter 7 Dimensional Analysis and Modeling μ p ) − ρ dp2 g 3 We recognize this Π as the inverse of a kind of Reynolds number if we split the dp terms to separate them into a length scale and (when combined with g) a velocity scale. The final form is Modified Π2 = a Reynolds number: Π2 = Step 6 We write the final functional relationship as Relationship between Πs: (ρ p ) − ρ d p gd p ( μ ) ⎛ ρ p − ρ d p gd p ⎞ ⎟ = f⎜ ⎜ ⎟ μ gd p ⎝ ⎠ V (2) Discussion We cannot determine the form of the relationship by purely dimensional reasoning since there are two Πs. However, in Chap. 10 we shall see that Π1 is a constant times Π2. 7-58 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-67 Solution conditions. We are to develop an equation for the settling speed of an aerosol particle falling in air under creeping flow Assumptions 1 The particle falls at steady speed V. 2 The Reynolds number is small enough that the creeping flow approximation is valid. Analysis We start by recognizing that as the particle falls at steady settling speed, its net weight W must equal the aerodynamic drag FD on the particle. We also know that W is proportional to (ρp - ρ)gdp3. Thus, Equating forces: ( ) W = constant1 ρ p − ρ gd p 3 = FD = constant 2 μVd p (1) where we have converted the notation of the previous problem, and we have defined two different constants. The two constants in Eq. 1 can be combined into one new constant for simplicity. Solving for V, Settling speed: If we divide both sides of Eq. 2 by (ρ V = constant p ) − ρ gd p 2 μ (2) gd p we see that the functional relationship given by Eq. 2 of the previous problem is consistent. Discussion This result is valid only if the Reynolds number is much smaller than one, as will be discussed in Chap. 10. If the particle is less dense than the fluid (e.g. bubbles rising in water), our result is still valid, but the particle rises instead of falls. 7-68 Solution We are to determine how the settling speed of an aerosol particle falling in air under creeping flow conditions changes when certain parameters are doubled. Assumptions 1 The particle falls at steady speed V. 2 The Reynolds number is small enough that the creeping flow approximation is valid. Analysis From the results of the previous problem, we see that if particle size doubles, the settling speed increases by a factor of 22 = 4. Similarly, if density difference doubles, the settling speed increases by a factor of 21 = 2. Discussion This result is valid only if the Reynolds number remains much smaller than unity, as will be discussed in Chap. 10. As the particle’s settling speed increases by a factor of 2 or 4, the Reynolds number will also increase by that same factor. If the new Reynolds number is not small enough, the creeping flow approximation will be invalid and our results will not be correct, although the error will probably be small. 7-59 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-69 Solution We are to generate a nondimensional relationship between the given parameters. Assumptions problem. 1 The flow is fully developed. 2 The fluid is incompressible. 3 No other parameters are significant in the Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters. ΔP = f (V , ε , ρ , μ , D, L ) Step 1 All the relevant parameters in the problem are listed in functional form: List of relevant parameters: n=7 Step 2 The primary dimensions of each parameter are listed: ΔP {m L t } {L t } V 1 −1 −2 1 −1 ε ρ {L } {m L } 1 −3 1 μ {m L t } 1 −1 −1 {L } D 1 {L } L 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 7−3 = 4 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines listed in Table 7-3, we cannot pick the dependent variable, ΔP. We cannot choose any two of parameters ε, L, and D since their dimensions are identical. It is not desirable to have μ or ε appear in all the Πs. The best choice of repeating parameters is thus V, D, and ρ. V, D, and ρ Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = ΔPV a1 D b1 ρ c1 mass: {m } = {m m } 0 {t } = { t time: length: 1 0 {L } = {L 0 } L Lb1 L−3c1 −1 a1 The dependent Π is thus } 1 −1 −2 )( L t ) ( L ) ( m L ) 1 −1 a1 1 b1 1 −3 c1 } 0 = 1 + c1 c1 = −1 0 = −2 − a1 a1 = −2 0 = −1 + a1 + b1 − 3c1 b1 = 0 c1 −2 − a1 t {( m L t 0 = −1 − 2 + b1 + 3 Π 1: Π1 = ΔP ρV 2 From Table 7-5, the established nondimensional parameter most similar to our Π1 is the Euler number Eu. No manipulation is required. We form the second Π with μ. By now we know that we will generate a Reynolds number, Π 2 = μV a2 D b2 ρ c2 Π2 = ρVD = Reynolds number = Re μ 7-60 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling The final two Π groups are formed with ε and then with L. The algebra is trivial for these cases since their dimension (length) is identical to that of one of the repeating variables (D). The results are Π 3 = εV a3 D b3 ρ c3 Π 4 = LV a4 D b4 ρ c4 Π3 = Π4 = ε D = Roughness ratio L = Length-to-diameter ratio or aspect ratio D Step 6 We write the final functional relationship as Relationship between Πs: Eu = ΔP ε L⎞ ⎛ = f ⎜ Re, , ⎟ 2 D D⎠ ρV ⎝ (1) Discussion The result applies to both laminar and turbulent fully developed pipe flow; it turns out, however, that the second independent Π (roughness ratio) is not nearly as important in laminar pipe flow as in turbulent pipe flow. Since ΔP drops linearly with distance down the pipe, we know that ΔP is linearly proportional to L/D. It is not possible to determine the functional relationships between the other Πs by dimensional reasoning alone. 7-61 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-70 Solution We are to determine by what factor volume flow rate increases in the case of fully developed laminar pipe flow when pipe diameter is doubled. Assumptions 1 The flow is steady. 2 The flow is fully developed, meaning that dP/dx is constant and the velocity profile does not change downstream. The step-by-step method of repeating variables is employed to obtain the nondimensional parameters. Analysis Step 1 All the relevant parameters in the problem are listed in functional form: dP ⎞ ⎛ V = f ⎜ D, μ , dx ⎟⎠ ⎝ List of relevant parameters: n=4 Step 2 The primary dimensions of each parameter are listed: V {L t } μ {L } {m L t } {m L D 3 −1 dP/dx 1 −1 −1 1 1 −2 −2 t } Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Step 4 We need to choose three repeating parameters since j = 3. Here we must pick all three independent parameters, D, μ, and dP/dx Repeating parameters: Step 5 The Π is generated: a Π1 = V ( dP / dx ) 1 D b1 μ c1 mass: time: {m } = {m 0 {t } = {t 0 length: t 3 −2 a1 The dependent Π is thus Π 1: m c1 } −1 −2 a1 − c1 {L } = {L L 0 a1 t {Π1} = } Lb1 L− c1 {( L t 3 −1 )( m L 1 −2 −2 t ) (L ) (m L a1 1 b1 1 −1 −1 ) c1 } 0 = a1 + c1 c1 = −a1 0 = −1 − 2a1 − c1 a1 = −1 0 = 3 − 2a1 + b1 − c1 b1 = −4 0 = −1 − 2a1 + a1 } t c1 = 1 b1 = −3 + 2a1 + c1 Π1 = V μ dP D4 dx Step 6 Since there is only one Π, we set it equal to a constant. We write the final functional relationship as 7-62 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Relationship between Πs: Π1 = constant, D 4 dP V = constant μ dx (1) We see immediately that if the pipe diameter is doubled with all other parameters fixed, the volume flow rate will increase by a factor of 24 = 16. Discussion We will see in Chap. 9 that the constant is π/8. There is no way to obtain the value of the constant from dimensional analysis alone. 7-63 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-71 Solution We are to determine the units of a parameter, and then use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis (a) The units of G are easily obtained by knowing the units of the other variables. Since F = G m1m2 , r2 Fr 2 N ⋅ m 2 ⎛ kg ⋅ m/s 2 ⎞ m3 m3 = . Note: This can be easily . G has units of and the units of G are therefore ⎜ ⎟ 2 m1m2 kg ⋅ s 2 kg 2 ⎝ N ⎠ kg ⋅ s G= verified by looking up the universal gravitational constant in a textbook or on the Internet. (b) The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Πs). F = F ( G , m1 , m2 , r ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed, {m L t } {L m t } F {m } G 1 1 −2 -1 −2 3 {m } m1 m2 1 1 {L } r 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines, we cannot pick both masses since they have the same dimensions. We choose Repeating parameters: G, m1, and r Step 5 The dependent Π is generated: time: mass: length: {t } = { t 0 −2 −2 a1 t {m } = {m m 0 1 {L } = {L L 0 The dependent Π is thus Π 1: ⎧ mL ⎞ ⎛ L3 ⎞ a1 ⎫ b c ⎪ m) 1 ( L) 1 ⎬ 2 ⎟⎜ 2 ⎟ ( t ⎠ ⎝ mt ⎠ ⎩⎪⎝ ⎭⎪ {Π1} = ⎪⎨⎛⎜ Π1 = FG a1 m1b1 r c1 − a1 1 3 a1 } mb1 Lc1 } } 0 = −2 − 2a1 a1 = −1 0 = 1 − a1 + b1 b1 = −2 0 = 1 + 3a1 + c1 c1 = 2 Π1 = Fr 2 Gm12 7-64 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling The second Pi (the only independent Π in this problem) is generated using the remaining variable m2. This one is trivial and is done “by eye” since one of the repeating variables is also a mass, Π 2: Π2 = Step 6 We write the final functional relationship as Relationship between Πs: m2 m1 ⎛m ⎞ Fr 2 = f⎜ 2⎟ 2 Gm1 ⎝ m1 ⎠ Π1 = f ( Π 2 ) or (c) Finally, comparing to the known law of universal gravitation, we see that the relationship is in fact linear, namely, Relationship between Πs: Π1 = Π 2 or m Fr 2 = 2 2 Gm1 m1 which yields F = G m1m2 r2 Discussion Not only is the relationship linear, but the constant of proportionality is unity. There is no way to get this result from dimensional analysis alone. 7-65 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-72 Solution We are to use dimensional analysis to find the functional relationship between the given parameters, namely, damping coefficient ζ as a function of spring constant k, mass m, and damping coefficient c. Assumptions 1 The given parameters are the only relevant ones in the problem. 2 ζ is already dimensionless. Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Πs). In this situation, ζ is already dimensionless, and so it is automatically one of the Πs. We can therefore form the other Π from the remaining three variables. Or, we can go through the normal procedure. We elect to show the full procedure below, but it is simpler to form a nondimensional parameter from variables k, m, and c. ζ = ζ ( k , m, c ) Step 1 There are four parameters in this problem; n = 4, List of relevant parameters: n=4 Step 2 The primary dimensions of each parameter are listed, ζ {m L t } = {1} {m L t } {m } {m L t } 0 k 1 0 −2 0 0 c m 1 0 −1 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines, we have only one choice, Repeating parameters: k, m, and c Step 5 The dependent Π is generated: Π1 = ζ k a1 mb1 c c1 time: mass: length: {t } = {t 0 −2 a1 − c1 {m } = {m 0 t a1 mb1 m c1 {L } = { L } 0 } 0 { {Π1} = (1) ( m1t −2 ) } a1 (m ) (m t ) 1 b1 1 −1 c1 } 0 = −2a1 − c1 a1 = −c1 / 2 0 = a1 + b1 + c1 b1 = −c1 / 2 - - Here we have the curious situation that the length dimension drops out, and so we have only two equations, but three unknown exponents. Thus there is a whole family of possible answers. In other words, we are free to choose any value of c1 that we wish, and the other two coefficients can then be determined. For example, if we pick c1 = 1, then a1 = b1 = -1/2, and the Π is thus Π1 for c1 = 1: Π1 = ζc km If we choose a different value of c1 = 1, we get a different ∏. For example, if we pick c1 = -1, then a1 = b1 = 1/2, and the Π is thus 7-66 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Π1 for c1 = -1: Π1 = ζ km Chapter 7 Dimensional Analysis and Modeling c It turns out that the latter is more appropriate to match with standard dynamic system analysis. However, there is no way to know this from dimensional analysis alone. Jen must remember at least one other thing from her dynamic systems class – that the damping ratio increases with damping coefficient (which makes sense on physical grounds), and thus the latter equation for ∏1 makes the most physical sense. Step 6 Since there is only one ∏, it must be a constant and we write the final functional relationship as Relationship between Πs: Π1 = ζ km c = constant, or ζ = constant c km It turns out that the constant is ½, and the correct definition for ζ is ζ = c . Thus, our analysis yields 2 km the correct equation to within a constant. Note, however, that any other choice of exponent c1 is also valid from a purely dimensional point of view. For example, if we pick c1 = -2, then a1 = b1 = 1, and the result is simply the square of the c2 ζ km . While this result is correct dimensionally (ζ is still previous result, i.e., Π1 = 2 = constant, or ζ = constant km c dimensionless), it does not agree with the standard definition. Discussion 7-67 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-73 Solution We are to use dimensional analysis to find the functional relationship between the given parameters, namely, voltage drop ΔE as a function of electrical current I and electrical resistance R. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters. ΔE = function ( I , R ) Step 1 There are three parameters in this problem; n = 3, List of relevant parameters: n=3 Step 2 The primary dimensions of each parameter are listed [we get these dimensions from a previous problem], ΔE {m L t 1 2 −3 −1 I } {I } {m L t R I 1 2 −3 −2 1 I } Step 3 As a first guess, j is set equal to 4, the number of primary dimensions represented in the problem (m, L, t, and I). Reduction: j=4 If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3 − 4 = −1 This is clearly incorrect, so we reduce j by one, yielding k = n − j = 3 − 3 = 0 . This is again clearly incorrect (we cannot have zero Πs). So, Reduction: j=2 and the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 Step 4 We need to choose two repeating parameters since j = 2. Following the guidelines, we have only one choice, Repeating parameters: I and R Step 5 The dependent Π is generated: Π1 = ΔEI a1 R b1 time: mass: length: current: {t } = {t 0 −3 −3b1 t {Π1} = } {m } = { m m } 0 b1 1 {L } = { L L } 0 {I } = {I 0 2 2 b1 −1 a1 −2 b1 I I } {( m L t 1 2 −3 − 1 I )( I ) ( m L t a1 1 I ) b1 } 0 = −3 − 3b1 b1 = −1 0 = 1 + b1 b1 = −1 0 = 2 + 2b1 b1 = −1 0 = −1 + a1 − 2b1 a1 = −1 Thus, Π 1: 1 2 −3 − 2 Π1 = ΔE IR 7-68 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Step 6 Since there is only one ∏, it must be a constant and we write the final functional relationship as Relationship between Πs: Π1 = ΔE = constant, or ΔE = constant ⋅ IR IR Comparing to Ohm’s law, ΔE = IR , we see that the constant is unity. Discussion In this case, since the number of variables is small, we were able to generate the exact form of Ohm’s law to within an unknown constant. The constant cannot be obtained through dimensional analysis alone. 7-69 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-74 Solution We are to use dimensional analysis to find the functional relationship between the given parameters, namely, energy E as a function of mass m and speed of light c. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters. E = function ( m, c ) Step 1 There are three parameters in this problem; n = 3, List of relevant parameters: n=3 Step 2 The primary dimensions of each parameter are listed, {m L t } {m } E {L t } m 1 2 −2 c 1 −1 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). Reduction: j=3 If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3−3 = 0 This is clearly incorrect (we cannot have zero Πs). So we reduce j by one, Reduction: j=2 and the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 Step 4 We need to choose two repeating parameters since j = 2. Following the guidelines, we have only one choice, Repeating parameters: m and c Step 5 The dependent Π is generated: Π1 = Em a1 cb1 time: mass: length: {t } = { t 0 −2 − b1 t } {m } = {m m } 0 1 a1 {L } = {L L } 0 2 b1 {Π1} = {( m L t 1 2 −2 )( m ) ( L t ) 1 a1 1 −1 b1 } 0 = −2 − b1 b1 = −2 0 = 1 + a1 a1 = −1 0 = 2 + b1 b1 = −2 Thus, Π 1: Π1 = E mc 2 Step 6 Since there is only one ∏, it must be a constant and we write the final functional relationship as 7-70 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Relationship between Πs: Π1 = E = constant, or E = constant ⋅ mc 2 mc 2 Comparing to Einstein’s famous equation, E = mc 2 , we see that the constant is unity. Discussion In this case, since the number of variables is small, we were able to generate the exact form of Einstein’s equation to within an unknown constant. The constant cannot be obtained through dimensional analysis alone. 7-71 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-75 Solution We are to find the functional relationship between the given parameters and name any established dimensionless parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis The method of repeating variables is employed to obtain the nondimensional parameters (the Πs). ( Step 1 There are six parameters in this problem; n = 6, ΔP = f ρ , ω , D, μ ,V List of relevant parameters: ) n=6 Step 2 The primary dimensions of each parameter are listed, ρ ΔP ω {m L t } {m L } 1 -1 −2 {t } 1 −3 μ {L } −1 V {m L t } {L t } D 1 −1 −1 1 3 −1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 6−3 = 3 Repeating parameters: ρ , ω , and D Step 4 We need to choose three repeating parameters since j = 3. We pick fluid density ρ, length scale D, and angular velocity ω. {( m L t } Step 5 The Πs are generated. The dependent Π is generated using ΔP and the three repeating variables, Π1 = ΔP ρ a1 ω b1 D c1 time: mass: length: {t } = { t 0 −2 − b1 t } {m } = {m m } 0 1 {L } = { L a1 −1 −3 a1 0 {Π1} = Lc1 L } 1 −1 − 2 )( m L ) ( t ) ( L ) 1 −3 a1 −1 b1 c1 1 0 = −2 − b1 b1 = −2 0 = 1 + a1 a1 = −1 0 = −1 − 3a1 + c1 c1 = −2 Thus, Π1 = a kind of pressure coefficient: Π1 = {( m L t ΔP ρω 2 D 2 } The second Π is obtained using viscosity and the three repeating variables, Π 2 = μρ a2 ω b2 D c2 time: mass: length: {t } = { t 0 −1 − b2 t } {m } = {m m } 0 1 {L } = {L 0 −1 −3 a2 L {Π1} = a2 Lc2 } 1 −1 −1 )( m L ) ( t ) ( L ) 1 −3 a2 −1 b2 1 c2 0 = −1 − b2 b2 = −1 0 = 1 + a2 a2 = −1 0 = −1 − 3a2 + c2 c2 = −2 7-72 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Thus, Π 2: Π2 = μ ρω D 2 We recognize this Π as the inverse of a kind of Reynolds number. The modified form of the Π is thus Modified Π2 = a Reynolds number: Π2 = ρω D 2 μ In similar fashion we obtain the third nondimensional parameter, combining volume flow rate and the three repeating variables. This one is trivial and the algebra can be performed in our heads, yielding Π 3: Π3 = Step 6 We write the final functional relationship as Relationship between Πs: V ω D3 ⎛ ρω D 2 V ΔP = f ⎜ ⎜ μ , ω D3 ρω 2 D 2 ⎝ ⎞ ⎟⎟ ⎠ Discussion You may choose different repeating variables, and may generate different nondimensional groups. If you do the algebra correctly, your answer is not “wrong” – you just may not get the same dimensionless groups. 7-73 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-76 Solution We are to find the functional relationship between the given parameters and name any established dimensionless parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis The method of repeating variables is employed to obtain the nondimensional parameters (the Πs). T = f ( ρ , ω , D, μ ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed, ρ ω {m L t } { m L } T 1 2 −2 {t } 1 −3 μ {L } {m L t } D −1 1 −1 −1 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Repeating parameters: ρ , ω , and D Step 4 We need to choose three repeating parameters since j = 3. We pick fluid density ρ, length scale D, and angular velocity ω. {( m L t } Step 5 The Πs are generated. The dependent Π is generated by combining T with the three repeating variables, Π1 = T ρ a1 ω b1 D c1 {t } = { t {Π1} = } 1 2 −2 )( m L ) ( t ) ( L ) 1 −3 −1 a1 b1 1 c1 0 = −2 − b1 b1 = −2 } 0 = 1 + a1 a1 = −1 0 = 2 − 3a1 + c1 c1 = −5 Π1 = some kind of torque coefficient: Π1 = time: mass: length: 0 −2 − b1 t {m } = {m m } 0 1 {L } = {L L 0 2 −3 a1 a1 Lc1 Thus, {( m L t ρω 2 D 5 T } The second Π is obtained using viscosity and the three repeating variables, Π 2 = μρ a2 ω b2 D c2 time: mass: length: {t } = { t 0 −1 − b2 t } {m } = {m m } 0 1 {L } = {L 0 −1 −3 a2 L {Π1} = a2 Lc2 } 1 −1 −1 )( m L ) ( t ) ( L ) 1 −3 a2 −1 b2 1 c2 0 = −1 − b2 b2 = −1 0 = 1 + a2 a2 = −1 0 = −1 − 3a2 + c2 c2 = −2 Thus, 7-74 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Π 2: Π2 = μ ρω D 2 Chapter 7 Dimensional Analysis and Modeling We recognize this Π as the inverse of a kind of Reynolds number. The modified form of the Π is thus Modified Π2 = a Reynolds number: Step 6 We write the final functional relationship as Relationship between Πs: Π2 = ρω D 2 μ ⎛ ρω D 2 ⎞ = f⎜ ⎟ ρω D ⎝ μ ⎠ T 2 5 Discussion You may choose different repeating variables, and may generate different nondimensional groups. If you do the algebra correctly, your answer is not “wrong” – you just may not get the same dimensionless groups. 7-75 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-77 Solution We are to repeat the previous problem, but with the fluid being a compressible gas rather than a liquid. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. 2 The speed of sound in the gas is an additional relevant parameter. Analysis The method of repeating variables is employed to obtain the nondimensional parameters (the Πs). In this case, the key is to add speed of sound c to the list of independent variables since the fluid is compressible. T = f ( ρ , ω , D, μ , c ) Step 1 There are now six parameters in this problem; n = 6, List of relevant parameters: n=6 Step 2 The primary dimensions of each parameter are listed, ρ ω {m L t } { m L } T 1 2 −2 {t } 1 −3 μ {L } {m L t } D −1 1 −1 −1 1 {L t } c 1 −1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 6−3 = 3 Repeating parameters: ρ , ω , and D Step 4 We need to choose three repeating parameters since j = 3. We pick fluid density ρ, length scale D, and angular velocity ω. {( m L t } Step 5 The Πs are generated. The dependent Π is generated by combining T with the three repeating variables, Π1 = T ρ a1 ω b1 D c1 {t } = { t {Π1} = } 1 2 −2 )( m L ) ( t ) ( L ) 1 −3 −1 a1 b1 1 c1 0 = −2 − b1 b1 = −2 } 0 = 1 + a1 a1 = −1 0 = 2 − 3a1 + c1 c1 = −5 Π1 = some kind of torque coefficient: Π1 = time: mass: length: 0 −2 − b1 t {m } = {m m } 0 1 {L } = {L L 0 2 −3 a1 a1 Lc1 Thus, {( m L t ρω 2 D 5 T } The second Π is obtained using viscosity and the three repeating variables, Π 2 = μρ a2 ω b2 D c2 time: mass: length: {t } = { t 0 −1 − b2 t } {m } = {m m } 0 1 {L } = {L 0 −1 −3 a2 L {Π1} = a2 Lc2 } 1 −1 −1 )( m L ) ( t ) ( L ) 1 −3 a2 −1 b2 1 c2 0 = −1 − b2 b2 = −1 0 = 1 + a2 a2 = −1 0 = −1 − 3a2 + c2 c2 = −2 7-76 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Thus, Π 2: Π2 = μ ρω D 2 We recognize this Π as the inverse of a kind of Reynolds number. The modified form of the Π is thus Modified Π2 = a Reynolds number: Π2 = {( L t ρω D 2 μ } In similar fashion we obtain the third nondimensional parameter by combining speed of sound with the three repeating variables. Π 3 = c ρ a3 ω b3 D c3 time: mass: length: {t } = { t 0 −1 − b3 t } {m } = { m } a3 0 {L } = {L L 0 {Π 3 } = 1 −3 a3 Lc3 } 1 −1 )( m L ) ( t ) ( L ) 1 −3 a3 −1 b3 1 c3 0 = −1 − b3 b3 = −1 0 = a3 a3 = 0 0 = 1 − 3a3 + c3 c3 = −1 Thus, Π 3: Π3 = c ωD Since ωD is a speed – the tip speed of the propeller, we recognize this Π as the inverse of a kind of Mach number. The modified form of the Π is thus Modified Π3 = a Mach number: Π3 = ωD c Step 6 We write the final functional relationship as Relationship between Πs: ⎛ ρω D 2 ω D ⎞ = f⎜ , ⎟ c ⎠ ρω D ⎝ μ T 2 5 Discussion We notice that the first two Πs are identical to those of the previous problem, since we have the same variables and we chose the same repeating parameters. The addition of speed of sound leads to a third Π. 7-77 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-78 Solution We are to use dimensional analysis to find the functional relationship between the given parameters, namely, turbulent viscous dissipation rate ε (rate of energy loss per unit mass) as a function of length scale A and velocity scale u′ of the large-scale turbulent eddies. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the ε = function of ( A, u ′ ) Step 1 There are three parameters in this problem; n = 3, List of relevant parameters: n=3 Step 2 The primary dimensions of each parameter are listed, ε {m L t } {m L t } 0 u′ A 2 −3 {m L t } 0 1 −1 0 1 0 Step 3 As a first guess, j is set equal to 2, the number of primary dimensions represented in the problem (L and t). Reduction: j=2 If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 Step 4 We need to choose three repeating parameters since j = 2. Following the guidelines, we have only one choice, A and u′ Repeating parameters: Step 5 The dependent Π is generated: Π1 = ε A a ( u ′ ) time: length: b {t } = {t 0 −3 − b t } {Π1} = {L } = { L L L } 0 2 a b So, the dependent (and only) Π becomes {( L t 2 −3 )( L ) ( L t ) } 1 a 1 −1 b 0 = −3 − b b = −3 0 = 2+a+b a = −2 − b = 1 Π1 = ε Π 1: ( u′) A 3 Step 6 Since there is only one ∏, it must be a constant and we write the final functional relationship as Relationship between Πs: Π1 = ε (u′) A 3 = constant, or ε = constant ( u′) 3 A 7-78 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Discussion The constant cannot be obtained from dimensional analysis. Indeed, the constant is not really a constant at all, but depends on the specifics of the turbulent flow being examined. It is common in the study of turbulence to write the final result in order-of-magnitude form, namely, ε ∼ (u′) A 3 , where the tilde means “is of order of magnitude”. 7-79 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-79 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the ( Step 1 There are four parameters in this problem; n = 4, Q = f m , c p , (Tout − Tin ) List of relevant parameters: ) n=4 (1) Step 2 The primary dimensions of each parameter are listed, Q m {m L t } {m t } 1 2 −3 {L t 1 −1 cp 2 −2 T −1 Tout- Tin } {T } 1 Step 3 As a first guess, j is set equal to 4, the number of primary dimensions represented in the problem (m, T, L, and t). j=4 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−4 = 0 Obviously this is not correct, so we re-examine our initial assumptions. We are convinced that our list of parameters is sufficient, so we reduce j by one and continue, j = 4 −1 = 3 Reduction: If this revised value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 We now expect only one Π. Step 4 We need to choose three repeating parameters since j = 3. We only have one choice in this problem, since there are only three independent parameters on the right-hand side of Eq. 1, m , c p , and (Tout − Tin ) Repeating parameters: Step 5 The dependent Π is generated: {Π1} =   a1 c b1 (T − T )c1 Π1 = Qm p out in mass: length: temperature: time: {m } = {m } 1+ a1 0 {L } = {L L } 0 2 {T } = {T 0 {t } = {t 0 2 b1 − b1 + c1 } −3 − a1 − 2 b1 } {( m L t 1 2 −3 )( m t ) ( L t 1 −1 a1 2 −2 T −1 ) (T ) b1 1 c1 } 0 = 1 + a1 a1 = −1 0 = 2 + 2b1 b1 = −1 c1 = b1 c1 = −1 0 = −3 − a1 − 2b1 3 = 1+ 2 Fortunately the result for the time exponents is consistent with that of the other dimensions. The dependent Π is thus 7-80 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Π 1: Π1 = Q  p (Tout − Tin ) mc Step 6 Since there is only one Π, it is a function of nothing. This is only possible if we set the Π equal to a constant. We write the final functional relationship as Relationship between Πs: Π1 = Q = constant  p (Tout − Tin ) mc (2) Discussion When there is only one Π, we know the functional relationship to within some (unknown) constant. In this  p (Tout − Tin ) . There is no way particular case, comparing to Eq. 1 of Problem 7-24, we see that the constant is unity, Q = mc to obtain the constant in Eq. 2 from dimensional analysis; however, one experiment would be sufficient to determine the constant. Experimental Testing and Incomplete Similarity 7-80C Solution We are to define wind tunnel blockage and discuss its acceptable limit. We are also to discuss the source of measurement errors at high values of blockage. Analysis Wind tunnel blockage is defined as the ratio of model frontal area to cross-sectional area of the testsection. The rule of thumb is that the blockage should be no more than 7.5%. If the blockage were significantly higher than this value, the flow would have to accelerate around the model much more than if the model were in an unbounded situation. Hence, similarity would not be achieved. We might expect the aerodynamic drag on the model to be too high since the effective freestream speed is too large due to the blockage. Discussion increases. 7-81C Solution There are formulas to correct for wind tunnel blockage, but they become less and less reliable as blockage We are to discuss the rule of thumb concerning Mach number and incompressibility. Analysis The rule of thumb is that the Mach number must stay below about 0.3 in order for the flow field to be considered “incompressible”. What this really means is that compressibility effects, although present at all Mach numbers, are negligibly small compared to other effects driving the flow. If Ma is larger than about 0.3 in a wind tunnel test, the model flow field loses both kinematic and dynamic similarity, and the measured results are questionable. Of course, the error increases as Ma increases. Discussion Compressible flow is discussed in detail in Chap. 12. There you will see where the value 0.3 comes from. 7-81 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-82C Solution We are to discuss some situations in which a model should be larger than its prototype. Analysis There are many possible situations, and students’ examples should vary. Generally, any flow field that is very small and/or very fast benefits from simulation with a larger model. In most cases these are situations in which we want the model to be larger and slower so that experimental measurements and flow visualization are easier. Here are a few examples: − − − − − − Modeling a hard disk drive. Modeling insect flight. Modeling the settling of very small particles in air or water. Modeling the motion of water droplets in clouds. Modeling flow through very fine tubing. Modeling biological systems like blood flow through capillaries, flow in the bronchi of lungs, etc. Discussion You can think of several more examples. 7-83C Solution We are to discuss the purpose of a moving ground belt and suggest an alternative. Analysis From the frame of reference of a moving car, both the air and the ground approach the car at freestream speed. When we test a model car in a wind tunnel, the air approaches at freestream speed, but the ground (floor of the wind tunnel) is stationary. Therefore we are not modeling the same flow. A boundary layer builds up on the wind tunnel floor, and the flow under the car cannot be expected to be the same as that under a real car. A moving ground belt solves this problem. Another way to say the same thing is to say that without the moving ground belt, there would not be kinematic similarity between the underside of the model and the underside of the prototype. If a moving ground belt is unavailable, we could instead install a false wall – i.e., a thin flat plate just above the boundary layer on the floor of the wind tunnel. A sketch is shown in Fig. 1. At least then the boundary layer will be very thin and will not have as much influence on the flow under the model. Discussion V Wind tunnel test section False wall FIGURE 1 A false wall along the floor of a wind tunnel to reduce the size of the ground boundary layer. We discuss boundary layer growth in Chap. 10. 7-84C Solution We are to discuss whether Reynolds number independence has been achieved, and whether the researchers can be confident about it. Analysis We remove the last four data points from Table 7-7 and from Fig. 7-41. From the remaining data it appears that the drag coefficient is beginning to level off, but is still decreasing with Re. Thus, the researchers do not know if they have achieved Reynolds number independence or not. Discussion The wind tunnel speed is too low to achieve Reynolds number independence. 7-82 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-85 Solution We are to show that Froude number and Reynolds number are the dimensionless parameters that appear in a problem involving shallow water waves. 1 Wave speed c is a function only of depth h, gravitational acceleration g, fluid density ρ, and fluid Assumptions viscosity μ. We perform a dimensional analysis using the method of repeating variables. Analysis c = f ( h, ρ , μ , g ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed, {L t } c 1 −1 ρ {L } μ { m L } {m L t } {L t } h 1 −3 1 g 1 −1 −1 1 −2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We pick length scale h, density difference ρ, and gravitational constant g. h, ρ , and g Repeating parameters: Step 5 The Πs are generated. Note that for the first Π we do the algebra in our heads since the relationship is very simple. The dependent Π is Π1 = Froude number: Π1 = Fr = {( m L t c (1) gh )( L ) ( m L ) ( L t ) } This Π is the Froude number. Similarly, the Π formed with viscosity is generated, {Π 2 } = Π 2 = μ ha ρ b g c {m } = { m m } mass: 0 {t } = { t time: length: 1 0 {L } = { L 0 t −1 a } L L−3b Lc a 1 1 −3 0 = −1 − 2c } c b = −1 c=− 0 = −1 + a − 3b + c 0 = −1 + a + 3 − a=− 1 2 which yields Π 2: 1 −2 b 0 = 1+ b b −1 − 2 c 1 −1 −1 Π2 = ρh 1 2 3 2 μ 3 2 g 7-83 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling We can manipulate this Π into the Reynolds number if we invert it and then multiply by Fr (Eq. 1) The final form is Modified Π2 = Reynolds number: Π 2 = Re = Step 6 We write the final functional relationship as Relationship between Πs: Discussion Fr = c gh ρ ch μ = f ( Re ) where Re = ρ ch μ As discussed in this chapter, it is often difficult to match both Fr and Re between a model and a prototype. 7-84 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-86 Solution We are to nondimensionalize experimental pipe data, plot the data, and determine if Reynolds number independence has been 0.309 achieved. We are then to extrapolate to a higher speed. Assumptions 1 The flow is fully developed. 2 The flow is steady and incompressible. Properties For water at T = 20 C and atmospheric pressure, ρ = 998.0 kg/m3 and μ = 1.002 × 10-3 kg/m⋅s. 0.308 0.307 o Analysis (a) We convert each data point in the table from V and ΔP to Reynolds number and Euler number. The calculations at the last (highest speed) data point are shown here: ρVD ( 998.0 kg/m ) ( 50 m/s )( 0.104 m ) = = 5.18 × 106 μ 1.002 × 10−3 kg/m ⋅ s Reynolds number: Re = 3 0.305 0.304 0.303 0 (1) Re × 10-6 2 4 6 FIGURE 1 Nondimensionalized experimental data from a section of pipe. and Euler number: 758, 700 N/m 2 ΔP ⎛ kg m ⎞ Eu = = ⎟ = 0.304 2 2 ⎜ 2 3 ρV 998.0 kg/m ( 50 m/s ) ⎝ s N ⎠ ( Reynolds number independence Eu 0.306 ) (2) We plot Eu versus Re in Fig. 1. Although there is experimental scatter in the data, it appears that Reynolds number independence has been achieved beyond a Reynolds number of about 2 × 106. The average value of Eu based on the last 6 data points is 0.3042. (b) We extrapolate to higher speeds. At V = 80 m/s, we calculate ΔP, assuming that Eu remains constant to higher values of Re, Extrapolated value: ( ) 2 2⎛ s N ⎞ 2 ΔP = Eu × ρV 2 = 0.3042 998.0 kg/m3 ( 80 m/s ) ⎜ ⎟ = 1, 940, 000 N/m kg m ⎝ ⎠ (3) Discussion It is shown in Chap. 8 that Reynolds number independence is indeed achieved at high-enough values of Re. The threshold value above which Re independence is achieved is a function of relative roughness height, ε/D. 7-85 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-87 Solution We are to calculate the wind tunnel blockage of a model truck in a wind tunnel and determine if it is within acceptable limits. Assumptions 1 The frontal area is equal to truck width times height. (Note that the actual area of the truck may be somewhat smaller than this due to rounded corners and the air gap under the truck, but a truck looks nearly like a rectangle from the front, so this is not a bad approximation.) Analysis Blockage: ( 0.159 m )( 0.257 m ) = 0.034 = 3.4% (1.2 m )(1.0 m ) Wind tunnel blockage is defined as the ratio of model frontal area to cross-sectional area of the test-section, Blockage = Amodel Awind tunnel = (1) The rule of thumb is that the blockage should be no more than 7.5%. Since we are well below this value, we need not worry about blockage effects. Discussion model. The length of the model does not enter our analysis since we are only concerned with the frontal area of the 7-86 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-88E Solution We are to calculate the size and scale of the model truck to be constructed, and calculate its maximum Reynolds number. Then we are to determine whether this model in this wind tunnel will achieve Reynolds number independence. Assumptions 1 The model will be constructed carefully so as to achieve approximate geometric similarity. 2 The wind tunnel air is at the same temperature and pressure as that flowing over the prototype truck. Properties For air at T = 80oF and atmospheric pressure, ρ = 0.07350 lbm/ft3 and μ = 1.248 × 10-5 lbm/ft⋅s. Analysis (a) The rule of thumb about blockage is that we should keep the blockage below 7.5%. Thus, the frontal area of the model truck must be no more than 0.075 × Awind tunnel. The ratio of height to width of the full-scale truck is Hp/Wp = 12/8.33 = 1.44. Thus, for the geometrically similar model truck, Wm = Equation for model truck width: Am 7.5% Awind tunnel = 1.44Wm Hm ( ) = 4.56 in (2) Wm = 4.56 in, H m = 6.57 in, Lm = 28.5 in (3) We solve Eq. 1 for Wm, Model truck width: Wm = (1) 7.5% Awind tunnel = 1.44 0.075 400 in 2 1.44 Scaling the height and length geometrically, Model truck dimensions: These dimensions represent a model that is scaled at approximately 1:22. (b) At the maximum speed, with Re based on truck width, ρWmVmax ( 0.07350 lbm/ft ) ( 4.56 in )(145 ft/s ) ⎛ 1 ft ⎞ 5 = ⎜ 12 in ⎟ = 3.25×10 1.248 × 10−5 lbm/ft ⋅ s μ ⎝ ⎠ Maximum Re: Re = 3 (4) (c) Based on the data of Fig. 7-41, this Reynolds number is shy of the value needed to achieve Reynolds number independence. Discussion The students should run at the highest wind tunnel speed. Their measured values of CD will probably be higher than those of the prototype, but the relative difference in CD due to their modifications should still be valid. 7-87 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-89 Solution We are to calculate and plot CD as a function of Re for a given set of wind tunnel measurements, and determine if dynamic similarity and/or Reynolds number independence have been achieved. Finally, we are to estimate the aerodynamic drag force acting on the prototype car. Assumptions 1 The model car is geometrically similar to the prototype car. 2 The aerodynamic drag on the strut holding the model car is negligible. For air at atmospheric pressure and at T = 25oC, ρ = 1.184 kg/m3 and μ = 1.849 × 10-5 kg/(m s). Properties Analysis We calculate CD and Re for the last data point listed in the given table (at the fastest wind tunnel speed), Model drag coefficient at last data point: F CD ,m = 1 D ,m2 2 ρ mVm Am = (1.184 kg/m ) ( 55 m/s ) 4.91 N 1 2 3 = 0.319 2 0.6 0.55 ⎛ kg m ⎞ (1.69 m )(1.30 m ) ⎜⎝ s 2 N ⎟⎠ 162 0.5 CD 0.45 0.4 0.35 and Model Reynolds number at last data point: ρ VW Rem = m m m ( μm 0.3 0 ) ⎛ 1.69 ⎞ 1.184 kg/m3 ( 55 m/s ) ⎜ m⎟ ⎝ 16 ⎠ = 3.72 × 105 = 1.849 × 10−5 kg/m ⋅ s 1 2 Re × 10 3 4 -5 FIGURE 1 Aerodynamic drag coefficient as a function of Reynolds number – results nondimensionalized from wind tunnel test data on a model car. We repeat the above calculations for all the data points in the given table, and we plot CD verses Re in Fig. 1. Have we achieved dynamic similarity? Well, we have geometric similarity between model and prototype, but the Reynolds number of the prototype car is Reynolds number of prototype car: Rep = 3 ρ pVpWp (1.184 kg/m ) ( 31.3 m/s )(1.69 m ) = = 3.39 × 106 1.849 × 10−5 kg/m ⋅ s μp (2) where the width and speed of the prototype are used in the calculation of Rep. Comparison of Eqs. 1 and 2 reveals that the prototype Reynolds number is more than eight times larger than that of the model. Since we cannot match the independent Πs in the problem, dynamic similarity has not been achieved. Have we achieved Reynolds number independence? From Fig. 1 we see that Yes, Reynolds number independence has indeed been achieved – at Re greater than about 3 × 105, CD has leveled off to a value of about 0.32 (to two significant digits). Since we have achieved Reynolds number independence, we can extrapolate to the full scale prototype, assuming that CD remains constant as Re is increased to that of the full scale prototype. ( ) Aerodynamic drag on the prototype: ⎛ s2 N ⎞ 1 1 2 FD ,p = ρ pVp 2 Ap CD ,p = 1.184 kg/m 3 ( 31.3 m/s ) (1.69 m )(1.30 m ) 0.32 ⎜ ⎟ = 408 N 2 2 ⎝ kg m ⎠ Discussion We give our final result to two significant digits. 7-88 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Review Problems 7-90C Solution (a) False: Kinematic similarity is a necessary but not sufficient condition for dynamic similarity. (b) True: You cannot have dynamic similarity if the model and prototype are not geometrically similar. (c) True: You cannot have kinematic similarity if the model and prototype are not geometrically similar. (d) False: It is possible to have kinematic similarity (scaled velocities at corresponding points), yet not have dynamic similarity (forces do not scale at corresponding points). 7-91C Solution We are to think of and describe a prototype and model flow in which there is geometric but not kinematic similarity even though Rem = Rep. Students’ responses will vary. Here are some examples: Analysis − − − A model car is being tested in a wind tunnel such that there is geometric similarity and the wind tunnel speed is adjusted so that Rem = Rep. However, there is not a moving ground belt, so there is not kinematic similarity between the model and prototype. A model airplane is being tested in a wind tunnel such that there is geometric similarity and the wind tunnel speed is adjusted so that Rem = Rep. However, the Mach numbers are quite different, and therefore kinematic similarity is not achieved. A model of a river or waterfall or other open surface flow problem in which there is geometric similarity and the speed is adjusted so that Rem = Rep. However, the Froude numbers do not match and therefore the velocity fields are not similar and kinematic similarity is not achieved. Discussion There are many more acceptable cases that students may imagine. 7-92C Solution We are to find at least three established nondimensional parameters not listed in Table 7-5, and list these following the format of that table. Analysis Students’ responses will vary. Here are some examples: Name Definition Bingham number Bm = Elasticity number El = Galileo number Ga = τL μV Ratio of significance yield stress viscous stress tc μ ρ L2 gD 3 ρ 2 μ 2 elastic force inertial force gravitational force viscous force In the above, tc is a characteristic time. Discussion There are many more established dimensionless parameters in the literature. Some sneaky students may make up their own! 7-89 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-93 Solution We are to determine the primary dimensions of each variable, and then show that Hooke’s law is dimensionally homogeneous. Analysis (a) Moment of inertia has dimensions of length4, Primary dimensions of moment of inertia: {I } = {length 4 } = {L4 } (1) (b) Modulus of elasticity has the same dimensions as pressure, Primary dimensions of modulus of elasticity: force ⎫ ⎧ mass × length 1 ⎫ ⎧ m ⎫ × {E} = ⎧⎨ ⎬=⎨ ⎬=⎨ ⎬ 2 time length 2 ⎭ ⎩ Lt 2 ⎭ ⎩ area ⎭ ⎩ (2) Or, in exponent form, {E} = {m1 L-1 t-2}. (c) Strain is defined as change in length per unit length, so it is dimensionless. ⎧ length ⎫ ⎬ = {1} ⎩ length ⎭ {ε } = ⎨ Primary dimensions of strain: (3) (d) Stress is force per unit area, again just like pressure. Primary dimensions of stress: force ⎫ ⎧ mass × length ⎫ ⎧ m ⎫ =⎨ 2⎬ ⎬=⎨ 2 2 ⎬ ⎩ area ⎭ ⎩ time × length ⎭ ⎩ Lt ⎭ {σ } = ⎧⎨ Or, in exponent form, {σ} = {m1 L-1 t-2}. (4) (e) Hooke’s law is σ = Eε. We write the primary dimensions of both sides: ⎧ m ⎫ ⎧ m ⎫ ⎧ m ⎫ Primary dimensions of Hooke’s law: {σ } = ⎨ 2 ⎬ = { Eε } = ⎨ 2 × 1⎬ = ⎨ 2 ⎬ ⎩ Lt ⎭ ⎩ Lt ⎭ ⎩ Lt ⎭ (5) Or, in exponent form, the dimensions of both sides of the equation are {m1 L-1 t-2}. Thus we see that Hooke’s law is indeed dimensionally homogeneous. Discussion somewhere. If the dimensions of Eq. 5 were not homogeneous, we would surely expect that we made an error 7-90 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-94 Solution We are to find the functional relationship between the given parameters and name any established dimensionless parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the z d = f ( F , L, E , I ) Step 1 There are five parameters in this problem; n = 5, List of relevant parameters: n=5 (1) Step 2 The primary dimensions of each parameter are listed, {L } zd 1 {m L t } {m L t } {L } F L 1 1 −2 E 1 −1 −2 1 {L } I 4 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We cannot pick both length L and moment of inertia I since their dimensions differ only by a power. We also notice that we cannot choose F, L, and E since these three parameters can form a Π all by themselves. So, we set j = 3 – 1 = 2, and we choose two repeating parameters, expecting 5 – 2 = 3 Πs, Repeating parameters: L and E Step 5 The Πs are generated. Note that for the first Π we do the algebra in our heads since zd has the same dimensions as L. The dependent Π is Π 1: Π1 = zd L This Π is not an established dimensionless group, although it is a ratio of two lengths, similar to an aspect ratio. We form the second Π with force F: Π 2 = FLa E b mass: time: length: {m } = { m m } 0 {t } = {t 0 1 b −2 −2 b t } {L } = { L L L } 0 1 a −b {Π 2 } = {( m L t 1 1 −2 )( L ) ( m L 1 a 1 −1 −2 t )} b 0 = 1+ b b = −1 0 = −2 − 2b b = −1 0 = 1+ a − b a = −1 + b a = −2 which yields Π 2: Π2 = F L2 E 7-91 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling We do not recognize Π2 as a named dimensionless parameter. The final Π is formed with moment of inertia. Since {I} = {L4}, there is no need to go through the algebra – we write Π 3: Π3 = I L4 Again, we do not recognize Π2 as a named dimensionless parameter. Step 6 We write the final functional relationship as Relationship between Πs: Discussion zd ⎛ F I ⎞ = f⎜ 2 , 4⎟ L ⎝LE L ⎠ (2) We cannot determine the form of the relationship by purely dimensional reasoning since there are three Πs. 7-92 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-95 Solution We are to generate dimensionless relationships among given parameters, and then we are to discuss how ΔP decreases if the time is doubled. 1 The given parameters are the only relevant ones in the problem. Assumptions (a) We perform dimensional analyses using the method of repeating variables. First we analyze ΔP: Analysis ΔP = f ( t , c, E ) Step 1 There are four parameters in this problem; n = 4, List of relevant parameters: n=4 (1) Step 2 The primary dimensions of each parameter are listed, ΔP {m L t } 1 −1 −2 {t } {L t } {m L t } c t E 1 −1 1 1 2 −2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Step 4 We need to choose three repeating parameters since j = 3. We pick all the independent parameters – time t, speed of sound c, and energy E, Repeating parameters: t , c, and E Step 5 The dependent Π is generated: {Π1} = Π1 = ΔP × t a1 cb1 E c1 mass: length: time: {m } = { m m } 0 1 {L } = {L 0 {t } = {t 0 c1 −1 b1 L L2 c1 } −2 a1 − b1 −2 c1 t t t The dependent Π is thus } {( m L t 1 −1 −2 )( t ) ( L t ) ( m L t ) 1 1 −1 a1 b1 1 2 −2 c1 } 0 = 1 + c1 c1 = −1 0 = −1 + b1 + 2c1 b1 = 3 0 = −2 + a1 − b1 − 2c1 a1 = 3 b1 = 1 − 2c1 a1 = 2 + b1 + 2c1 Π1 for ΔP: Π1 = This Π is not an established one, so we leave it as is. t 3 c 3 ΔP E Step 6 We write the final functional relationship as Relationship between Πs: ΔP = constant E t 3c3 (2) 7-93 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling We perform a similar dimensional analysis using the same repeating variables, but this time for radius r. We do not show the algebra since the Π can be found by inspection. We get Π1 for r: Since this is the only Π, it must be equal to a constant, Relationship between Πs: Π1 = r ct r = constant ⋅ ct (3) (b) From Eq. 2 we see that if t is doubled, ΔP decreases by a factor of 23 = 8. Discussion The pressure rise across the blast wave decays rapidly with time (and with distance from the explosion). The speed of sound depends on temperature. If the explosion is of sufficient strength, T will increase significantly and c will not remain constant. 7-94 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-96 Solution We are to find an alternate definition of Archimedes number, and list it following the format of Table 7-5. Then we are to find an established Π group that is similar. Analysis Students’ responses will vary. There seems to be a plethora of definitions of Archimedes number. Here is the one most appropriate for buoyant fluids: Name Definition Archimedes number Ar = gLΔρ ρV 2 Ratio of significance buoyant force inertial force In the above, Δρ is a characteristic density difference in the fluid (due to buoyancy) and ρ is a characteristic or average density of the fluid. A glance through Table 7-5 shows that the Richardson number is very similar to this alternative definition of Ar. In fact, the alternate form of Ri (Problem 7-55) is identical to our new Ar. Discussion Some students may find other definitions that are also valid. For example, Δρ/ρ may be replaced by ΔT/T. 7-95 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-97 Solution We are to generate a dimensionless relationship between the given parameters. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. 3 The flow is fully developed. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Step 1 There are five parameters in this problem; n = 5, ⎛ dP ⎞ u = f ⎜ h, , μ, y ⎟ dx ⎝ ⎠ List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed, {L t } u 1 −1 {L } {m L h μ dP/dx 1 −2 −2 1 (1) t } {m L t } 1 −1 −1 {L } y 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We cannot pick both h and y since they have the same dimensions. We choose h, dP/dx, and μ Repeating parameters: Step 5 The dependent Π is generated: {Π1} = ⎛ dP ⎞ c1 Π1 = uh a1 ⎜ ⎟ μ ⎝ dx ⎠ b1 mass: time: length: {m } = {m 0 {t } = {t 0 The dependent Π is thus Π 1: m c1 } −1 −2 b1 − c1 {L } = {L L 0 b1 t t } 1 a1 −2b1 − c1 L L } {( L t 1 −1 )( L ) ( m L 1 a1 1 −2 −2 t ) (m L 1 −1 −1 b1 t ) c1 } 0 = b1 + c1 c1 = −b1 0 = −1 − 2b1 − c1 b1 = −1 0 = 1 + a1 − 2b1 − c1 a1 = −2 0 = −1 − b1 c1 = 1 0 = 1 + a1 + 1 μu Π1 = h 2 dP dx The independent Π is generated with variable y. Since {y} = {L}, and this is the same as one of the repeating variables (h), Π2 is simply y/h, 7-96 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Π 2: Π2 = Step 6 We write the final functional relationship as Relationship between Πs: Π1 = y h μu ⎛ y⎞ = f⎜ ⎟ dP ⎝h⎠ h2 dx (2) Discussion If we were to solve this problem exactly (using the methods of Chap. 9) we would see that the functional relationship of Eq. 2 is correct. 7-97 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-98 Solution We are to generate a dimensionless relationship between the given parameters and then analyze the behavior of umax when an independent variable is doubled. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. 3 The flow is fully developed. Analysis (a) A step-by-step dimensional analysis procedure could be performed. However, we notice that umax has the same dimensions as u. Therefore the algebra would be identical to that of the previous problem except that there is only one Π instead of two since y is no longer a parameter. The result is Relationship between Πs: Π1 = μ umax h2 dP dx = constant = C (1) or Final relationship for umax: umax = C h 2 dP μ dx (2) Alternatively, we can use the results of the previous problem directly. Namely, since we know that the maximum velocity occurs at the centerline, y/h = 1/2 there, and is a constant. Hence, Eq. 2 of the previous problem reduces to Eq. 1 of the present problem. (b) If h doubles, we see from Eq. 2 that umax will increase by a factor of 22 = 4. (c) If dP/dx doubles, we see from Eq. 2 that umax will increase by a factor of 21 = 2. (d) Since there is only one Π in this problem, we would need to conduct only one experiment to determine the constant C in Eq. 2. Discussion The constant turns out to be -1/8, but there is no way to determine this from dimensional analysis alone. To obtain the constant, we would need to either do an experiment, or solve the problem exactly using the methods discussed in Chap. 9. 7-98 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-99 [Also solved using EES on enclosed DVD] Solution We are to generate a relationship for Darcy friction factor f in terms of Euler number Eu. We are then to plot f as a function of Re and discuss whether Reynolds number independence has been achieved. Assumptions 1 The flow is fully developed. 2 The flow is steady and incompressible. 0.0495 Properties For water at T = 20 C and atmospheric pressure, ρ = 998.0 kg/m3 and μ = 1.002 × 10-3 kg/m⋅s. 0.0494 o 0.0493 0.0492 Analysis (a) Since the flow is fully developed, the control volume cuts through two cross sections in which the velocity profiles are identical. The flow is also steady, so the control volume momentum equation in the horizontal (x) direction reduces to ∑F = ∑F 0.0491 f 0.0489 0.0488 + ∑ Fx , shear stress = 0 Conservation of momentum: x x , pressure 0.0487 (1) 0.0486 0.0485 We multiply pressure by cross-sectional area to obtain the pressure force, and wall shear stress times inner pipe wall surface area to obtain the shear stress force, ∑ Fx, pressure = ΔP π D2 4 ∑F x , shear stress Reynolds number independence 0.049 = −τ wπ DL (2) Note the negative sign in the shear stress term since τw points to the left. We substitute Eq. 2 into Eq. 1. After some algebra, ΔP = Result: 4τ w L D 0 Re × 10-6 2 4 6 FIGURE 1 Nondimensionalized experimental data from a section of pipe. (3) Finally, we divide both sides of Eq. 3 by ρV2 to convert ΔP into an Euler number, Eu = Nondimensional relationship: 4τ w L 1 L ⎛ 8τ w ⎞ ΔP = = ⎜ ⎟ 2 ρV ρV 2 D 2 D ⎝ ρV 2 ⎠ (4) We recognize the term in parentheses on the right as the Darcy friction factor. Thus, Final nondimensional relationship: Eu = 1 L f 2D or f =2 D Eu L (5) (b) We use Eq. 5 to calculate f at each data point of Table P7-74. We plot f as a function of Re in Fig. 1. We see that the behavior of f mimics that of Eu (as it must because of Eq. 5 where we see that f is just a constant times Eu). Since Eu shows Reynolds number independence for Re greater than about 2 × 106, so does f. We see Reynolds number independence for Re greater than about 2 × 106. From the plot, the extrapolated value of f at large Re is about 0.04867, which agrees with Eq. 5 when we plug in the Re-independent value of Eu, Extrapolated value of f: f =2 D 0.104 m Eu = 2 ( 0.3042 ) = 0.0487 L 1.3 m (6) Discussion We show in Chap. 8 (the Moody chart) that f does indeed flatten out at high enough values of Re, depending on the relative roughness height, ε/D. 7-99 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-100 Solution We are to create characteristic scales so that we can define a desired established dimensionless parameter. Analysis (a) For Froude number we need a velocity scale, a length scale, and gravity. We already have a length scale and gravity. We create a velocity scale as V ′ / L . We then define a Froude number as Fr = Froude number: V gL = V ′ L gL = V ′ gL3 (b) For Reynolds number we need a velocity scale, a length scale, and kinematic viscosity. Of these we only have the kinematic viscosity, so we need to create a velocity scale and a length scale. After a “back of the envelope” analysis, we create a velocity scale as V ′ / L where L is some undefined characteristic length scale. Thus, Re = Reynolds number: ν LV = LV ′ V ′ = Lν ν Note that in this case, the length scales drop out, so it doesn’t matter that we could not define a length scale from the given parameters. (c) For Richardson number we need a length scale, the gravitational constant, a volume flow rate, a density, and a density difference. Of these we have all but the volume flow rate, so we create a volume flow rate scale as V ′ L . Thus, Richardson number: Discussion Ri = L5 g Δρ L5 g Δρ L3 g Δρ = = 2 2 2 ρV ρ V ′L ρ V ′ ( ) ( ) You can verify that each of the parameters above is dimensionless. 7-100 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-101 Solution We are to find the functional relationship between the given parameters and name any established dimensionless parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the V = f ( d , D, ρ , μ , h, g ) Step 1 There are seven parameters in this problem; n = 7, List of relevant parameters: n=7 (1) Step 2 The primary dimensions of each parameter are listed, {L t } V 1 −1 {L } d 1 ρ {L } μ {m L } {m L t } D 1 −3 1 {L } h 1 −1 −1 1 {L t } g 1 −2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 7−3 = 4 Step 4 We need to choose three repeating parameters since j = 3. We pick length scale h, fluid density ρ, and gravitational constant g. h, ρ , and g Repeating parameters: Step 5 The Πs are generated. Note that in this case we do the algebra in our heads since these relationships are very simple. The dependent Π is Π1 = a Froude number: Π1 = V gh This Π is a type of Froude number. Similarly, the two length-scale Πs are obtained easily, Π2 = Π 2: d h and Π 3: Π3 = {( m L t Finally, the Π formed with viscosity is generated, Π 4 = μ h a4 ρ b4 g c4 mass: time: {m } = {m m } 0 {t } = { t 0 1 b4 −1 −2 c4 t } {Π 4 } = 1 −1 −1 D h )( L ) ( m L ) ( L t ) 1 a4 1 −3 b4 1 −2 c4 } b4 = −1 0 = 1 + b4 0 = −1 − 2c4 c4 = − 1 2 7-101 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. length: {L } = {L 0 L L−3b4 Lc4 −1 a4 } Chapter 7 Dimensional Analysis and Modeling 0 = −1 + a4 − 3b4 + c4 0 = −1 + a4 + 3 − a4 = − 1 2 which yields Π 4: Π4 = 3 2 μ ρh2 g 3 We recognize this Π as the inverse of a kind of Reynolds number. We also split the h terms to separate them into a length scale and (when combined with g) a velocity scale. The final form is Modified Π4 = a Reynolds number: Π4 = ρ h gh μ Step 6 We write the final functional relationship as Relationship between Πs: ⎛ d D ρ h gh = f⎜ , , ⎜h h μ gh ⎝ V ⎞ ⎟ ⎟ ⎠ (2) Discussion You may choose different repeating variables, and may generate different nondimensional groups. If you do the algebra correctly, your answer is not “wrong” – you just may not get the same dimensionless groups. 7-102 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-102 Solution We are to find a dimensionless relationship among the given parameters. Assumptions 1 The given parameters are the only ones relevant to the flow at hand. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the tempty = f ( d , D, ρ , μ , h, g ) Step 1 There are seven parameters in this problem; n = 7, List of relevant parameters: n=7 (1) Step 2 The primary dimensions of each parameter are listed, {t } tempty 1 {L } d 1 ρ {L } μ {m L } {m L t } D 1 −3 1 1 −1 −1 {L } h 1 {L t } g 1 −2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: j =3 k = n− j = 7−3 = 4 Step 4 We need to choose three repeating parameters since j = 3. We pick length scale h, fluid density ρ, and gravitational constant g. (Note: these are the same repeating parameters as in the previous problem.) h, ρ , and g Repeating parameters: Step 5 The Πs are generated. We leave out the details since the algebra is trivial and can be done by inspection in most cases. The dependent Π is Π 1: Π1 = tempty The rest of the Πs are identical to those of the previous problem. Step 6 We write the final functional relationship as Relationship between Πs: tempty g h ⎛ d D ρ h gh ⎞ g = f⎜ , , ⎟ ⎜h h h μ ⎟⎠ ⎝ (2) Discussion You may choose different repeating variables, and may generate different nondimensional groups. If you do the algebra correctly, your answer is not “wrong” – you just may not get the same dimensionless groups. 7-103 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-103 Solution We are to calculate the temperature of water in a model test to ensure similarity with the prototype, and we are to predict the time required to empty the prototype tank. Assumptions 1 The parameters specified in the previous problem are the only parameters relevant to the problem. 2 The model and prototype are geometrically similar. Properties For ethylene glycol at 60oC, ν = μ/ρ = 4.75×10-6 m2/s (given). Analysis (a) We use the functional relationship obtained in the previous problem, Dimensionless relationship: tempty ⎛ d D ρ h gh ⎞ g = f⎜ , , ⎟ ⎜h h h μ ⎟⎠ ⎝ (1) Since the model and prototype are geometrically similar, (d/h)model = (d/h)prototype and (D/h)model = (D/h)prototype. Thus, we are left with only one Π to match to ensure similarity. Namely, the Reynolds number parameter in Eq. 1 must be matched between model and prototype. Since g remains the same in either case, and using “m” for model and “p” for prototype, ⎛ ρ h gh ⎞ ⎛ ρ h gh ⎜ ⎟ =⎜ ⎜ μ ⎟ ⎜ μ ⎝ ⎠m ⎝ Similarity: ⎞ ρ p ⎛ hp ⎞ 2 ρ ⎟ or m = ⎜ ⎟ ⎟ μ m μ p ⎝ hm ⎠ ⎠p 3 (2) We recognize that ν = μ/ρ, and we know that hp/hm = 4. Thus, Eq. 2 reduces to −3 Similarity: −3 ⎛ h ⎞2 ν m = ν p ⎜ p ⎟ = 4.75 × 10−6 m 2 /s ( 4 ) 2 = 5.94 × 10−7 m 2 /s ⎝ hm ⎠ (3) For similarity we need to find the temperature of water where the kinematic viscosity is 5.94×10-7 m2/s. By interpolation from the property tables, the designers should run the model tests at a water temperature of 45.8oC. (b) At dynamically similar conditions, Eq. 1 yields At dynamically similar conditions: ⎛ g⎞ ⎛ g⎞ ⎜⎜ tempty ⎟⎟ = ⎜⎜ tempty ⎟ h ⎠p ⎝ h ⎟⎠ m ⎝ → tempty,p = tempty,m hp hm = 4.53 min 4 = 9.06 min (5) Discussion We set up Eqs. 3 and 5 in terms of ratios of hp to hm so that the actual dimensions are not needed – just the ratio is needed, and it is given. 7-104 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-104 Solution doubled. For the simplified case in which V depends only on h and g, we are to determine how V increases when h is Assumptions 1 The given parameters are the only ones relevant to the problem. Analysis We employ the dimensional analysis results of Problem 7-90. Dropping d, D, ρ, and μ from the list of parameters, we are left with n = 3, V = f ( h, g ) List of relevant parameters: n=3 (1) We perform the analysis in our heads –only one Π remains, and it is therefore set to a constant. The final result of the dimensional analysis is Relationship between Πs: V gh = constant (2) Thus, when h is doubled, we can easily calculate the factor by which V increases, V2 Increase in V: gh2 = V1 gh1 or V2 = V1 h2 = V1 2 h1 Thus, when h increases by a factor of 2, V increases by a factor of (3) 2. Discussion We don’t need to know the constant in Eq. 2 to solve the problem. However, it turns out that the constant is 2 (see Chap. 5). 7-105 Solution We are to verify the dimensions of particle relaxation time τp, and then identify the established dimensionless parameter formed by nondimensionalization of τp. Analysis First we obtain the primary dimensions of τp, Primary dimensions of τp: {τ } p ⎧m 2 ⎫ ⎪⎪ L3 × L ⎪⎪ =⎨ ⎬ = {t} ⎪ m ⎪ ⎩⎪ Lt ⎭⎪ A characteristic time scale for the air flow is L/V. Thus, we nondimensionalize τp, Nondimensionalized particle relaxation time: τ p* = τ p = From Table 7-5 we recognize this as the Stokes number, Stk, Stokes number: Discussion Stk = ρ p d p2 V 18μ L ρ p d p 2V 18μ L Stokes number is useful when studying the flow of aerosol particles. 7-105 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-106 Solution We are to compare the primary dimensions of each given property in mass-based and force-based primary dimensions, and discuss. Analysis From previous problems and examples in this chapter, we can write down the primary dimensions of each property in the mass-based system. We use the fundamental definitions of these quantities to generate the primary dimensions in the force-based system: (a) For pressure P the primary dimensions are Mass-based primary dimensions Force-based primary dimensions {P} = ⎧⎨ {P} = ⎧⎨ m ⎫ 2 ⎬ ⎩t L⎭ force ⎫ ⎧ F ⎫ ⎬=⎨ 2⎬ ⎩ area ⎭ ⎩ L ⎭ G (b) For moment M the primary dimensions are { } {M } = ⎧⎨⎩m Lt ⎫⎬⎭ Mass-based primary dimensions G Force-based primary dimensions G M = {force × moment arm} = {FL} 2 2 (c) For energy E the primary dimensions are {E} = {force × distance} = {FL} Mass-based primary dimensions ⎧ {E} = ⎨m ⎩ Force-based primary dimensions L2 ⎫ ⎬ t2 ⎭ We see that (in these three examples anyway), the forced-base cases have only two primary dimensions represented (F and L), whereas the mass-based cases have three primary dimensions represented (m, L, and t). Some authors would prefer the force-based system because of its reduced complexity when dealing with forces, pressures, energies, etc. Discussion Not all variables have a simpler form in the force-based system. Mass itself for example has primary dimensions of {m} in the mass-based system, but has primary dimensions of {Ft2/L} in the force-based system. In problems involving mass, mass flow rates, and/or density, the force-based system may not have any advantage. 7-106 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-107 Solution The pressure difference between the inside of a soap bubble and the outside air is to be analyzed with dimensional analysis and the method of repeating variables using the force-based system of primary dimensions. Assumptions 1 The soap bubble is neutrally buoyant in the air, and gravity is not relevant. 2 No other variables or constants are important in this problem. Analysis The step-by-step method of repeating variables is employed. ΔP = f ( R, σ s ) Step 1 There are three variables and constants in this problem; n = 3, n=3 (1) Step 2 The primary dimensions of each parameter are listed. The dimensions of pressure are force per area and those of surface tension are force per length. ΔP σs {L } {F L } {F L } R 1 −2 1 −1 1 Step 3 As a first guess, j is set equal to 2, the number of primary dimensions represented in the problem (F and L). j=2 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n − j = 3− 2 =1 Step 4 We choose two repeating parameters since j = 2. Our only choice is R and σs since ΔP is the dependent variable. Step 5 The dependent Π is generated: {Π1} = {F0 L0 } = Π1 = ΔPR a1 σ s b1 force: length: {F } = {F F } {L } = {L 0 −2 La1 L− b1 } 1 −2 ) L (F L ) a1 1 −1 b1 } 0 = 1 + b1 b1 = −1 0 = −2 + a1 − b1 a1 = 1 1 b1 0 {( F L a1 = 2 + b1 Eq. 1 thus becomes Π 1: Π1 = ΔPR σs (2) From Table 7-5, the established nondimensional parameter most similar to Eq. 2 is the Weber number, defined as a pressure times a length divided by surface tension. There is no need to further manipulate this Π. Step 6 We now write the functional relationship between the nondimensional parameters. Since there is only one Π, it is a function of nothing, which means it must be a constant, Relationship between Πs: Π1 = ΔPR σs = f ( nothing ) = constant → ΔP = constant σs R (3) The result using force-based primary dimensions is indeed identical to the previous result using the mass-based system. Discussion Because only two primary dimensions are represented in the problem when using the force-based system, the algebra is in fact a lot easier. 7-107 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-108 Solution We are to a third established nondimensional parameter that is formed by the product or ratio of two given established nondimensional parameters. Analysis (a) The product of Reynolds number and Prandtl number yields Reynolds number times Prandtl number: Re × Pr = ρ LV cP μ ρ LVcP × = μ k k We recognize Eq. 1 as the Peclet number, Pe = Re × Pr = Peclet number: (2) k Sc ρ DAB = = μ c ρ cP DAB Pr P k (3) We recognize Eq. 3 as the Lewis number, Le = Lewis number: = α (b) The ratio of Schmidt number and Prandtl number yields Schmidt number divided by Prandtl number: ρ LVcP (1) k LV μ k Sc α = = Pr ρ cP DAB DAB (4) (c) The product of Reynolds number and Schmidt number yields Reynolds number times Schmidt number: Re × Sc = ρ LV μ LV × = μ ρ DAB DAB (5) We recognize Eq. 5 as the Sherwood number, Sherwood number: Discussion Sh = Re × Sc = LV DAB (6) Can you find any other such combinations from Table 7-5? 7-108 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-109 Solution We are to determine the relationship between four established nondimensional parameters, and then try to form the Stanton number by some combination of only two other established dimensionless parameters. Analysis We manipulate Re, Nu, and Pr, guided by the known result. After some trial and error, Stanton number: Lh h Nu k = = St = c pV ρ Re × Pr ρVL μ cP × k μ (1) We recognize from Table 7-5 (or from the previous problem) that Peclet number is equal to the product of Reynolds number and Prandtl number. Thus, Stanton number: Lh Nu h k St = = = ρ LVcP ρ c pV Pe k (2) Discussion Not all named, established dimensionless parameters are independent of other named, established dimensionless parameters. 7-110 Solution We are to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis First we do some thinking. If we imagine traveling at the same speed as the bottom plate, the flow would be identical to that of Problem 7-56 except that the top plate speed would be (Vtop – Vbottom) instead of just V. The step-bystep method of repeating variables is otherwise identical to that of Problem 7-56, and the details are not included here. The final functional relationship is Relationship between Πs: Vtop u y⎞ ⎛ = f ⎜ Re, ⎟ h⎠ − Vbottom ⎝ where Reynolds number: Re = ρ (Vtop − Vbottom ) h μ (1) (2) Discussion It is always wise to look for shortcuts like this to save us time. 7-111 Solution We are to determine the primary dimensions of electrical charge. Analysis The fundamental definition of electrical current is charge per unit time. Thus, Primary dimensions of charge: {q} = {current × time} = {I t} (1) Or, in exponent form, {q} = {t1 I1}. Discussion dimensions. We see that all dimensions, even those of electrical properties, can be expressed in terms of primary 7-109 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-112 Solution We are to determine the primary dimensions of electrical capacitance. Analysis Electrical capacitance C is measured in units of farads (F). By definition, a one-farad capacitor with an applied electric potential of one volt across it will store one coulomb of electrical charge. Thus, ⎧ ⎪ It {C} = {charge / voltage} = ⎪⎨ 2 ⎪ mL ⎩⎪ t 3 I Primary dimensions of capacitance: ⎫ ⎪⎪ ⎧ I 2 t 4 ⎫ ⎬=⎨ 2 ⎬ ⎪ ⎩ mL ⎭ ⎭⎪ (1) where the primary dimensions of voltage are obtained from Problem 7-10, and those of electric charge are obtained from the previous problem. Or, in exponent form, {C} = {m-1 L-2 t4 I2}. Discussion dimensions. 7-113 Solution our result. We see that all dimensions, even those of electrical properties, can be expressed in terms of primary We are to determine the primary dimensions of electrical time constant RC, and discuss the significance of Analysis The primary dimensions of electrical resistance are obtained from Problem 7-11. Those of electrical capacitance C are obtained from the previous problem. Thus, Primary dimensions of electrical time constant RC: ⎧ mL2 {RC} = {resistance × capacitance} = ⎨ ⎩t I 3 2 × I2 t 4 ⎫ ⎬ = {t} mL2 ⎭ (1) Thus we see that the primary dimensions of RC are those of time. This explains why a resistor and capacitor in series is often used in timing circuits. Discussion The cut-off frequency of the low-pass filter is proportional to 1/R⋅C. If the resistor and the capacitor were to swap places we would have a high-pass rather than a low-pass filter. 7-114 Solution We are to determine the primary dimensions of both sides of the equation, and we are to verify that the equation is dimensionally homogeneous. Analysis The primary dimensions of the time derivative (d/dt) are 1/time. The primary dimensions of capacitance are current2×time4 / (mass×length2), as obtained from Problem 7-101. Thus both sides of the equation can be written in terms of primary dimensions, {I } = {current} ⎧ mass × length 2 2 4 dE ⎪⎪ current × time current × time3 C =⎨ time dt ⎪ mass × length 2 ⎩⎪ {I } = {I} ⎫ ⎪⎪ ⎬ = {current} ⎪ ⎭⎪ ⎧ dE ⎫ ⎨C ⎬ = {I} ⎩ dt ⎭ Indeed, both sides of the equation have the same dimensions, namely {I}. Discussion Current is one of our seven primary dimensions. These results verify our algebra in Problem 7-101. 7-110 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-115 Solution We are to find the functional relationship between the given parameters, and then answer some questions about scaling. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis (a) The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Πs). ( Step 1 There are four parameters in this problem; n = 4, δ P = f ρ ,V , D List of relevant parameters: ) n=4 (1) Step 2 The primary dimensions of each parameter are listed, δP {m L t } 1 −1 −2 ρ V {L t } {m L } 1 −3 {L } D 3 −1 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Step 4 We need to choose three repeating parameters since j = 3. We only have one choice in this problem, since there are only three independent parameters on the right-hand side of Eq. 1, ρ, V , and D Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = δ P ρ a1V b1 D c1 {m } = {m } mass: time: length: 1+ a1 0 {t } = {t } −2 − b1 0 {L } = {L 0 The dependent Π is thus −1 −3a1 3b1 c1 L L L } {( m L t 1 −1 −2 )( m L ) ( L t ) ( L ) 1 −3 3 −1 a1 b1 1 c1 } 0 = 1 + a1 a1 = −1 0 = −2 − b1 b1 = −2 0 = −1 − 3a1 + 3b1 + c1 c1 = 4 Π 1: Π1 = D 4δ P ρV 2 Step 6 Since there is only one Π, it is a function of nothing. This is only possible if we set the Π equal to a constant. We write the final functional relationship as Relationship between Πs: Π1 = D 4δ P = constant ρV 2 (2) 7-111 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling (b) We re-write Eq. 2 as Equation for δP: δ P = constant ρV 2 D4 (3) Thus, if we double the size of the cyclone, the pressure drop will decrease by a factor of 24 = 16 (Answer: The pressure drop will change by a factor of 1/16.). (c) Also from Eq. 3 we see that if we double the volume flow rate, the pressure drop will increase by a factor of 22 = 4 (Answer: The pressure drop will change by a factor of 4.). Discussion The pressure drop would be smallest for the largest cyclone operating at the smallest volume flow rate. (This agrees with our intuition.) 7-112 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-116 Solution We are to find the functional relationship between the given parameters, and then answer some questions about scaling. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis (a) The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the Πs). ( Step 1 There are five parameters in this problem; n = 5, w = f q p , E f , μ , Dp List of relevant parameters: ) n=5 Step 2 The primary dimensions of each parameter are listed, {L t } {m L t {I t } w qp 1 −1 μ Ef 1 1 −3 − 1 1 1 (1) I } {m L t } {L } Dp 1 −1 −1 1 where the primary dimensions of voltage are obtained from Problem 7-10, and those of electric charge are obtained from Problem 7-100. Step 3 As a first guess, j is set equal to 4, the number of primary dimensions represented in the problem (m, L, t, and I). j=4 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−4 =1 Step 4 We need to choose four repeating parameters since j = 4. We only have one choice in this problem, since there are only four independent parameters on the right-hand side of Eq. 1, qp, E, Dp, and μ Repeating parameters: {( L t Step 5 The dependent Π is generated: Π1 = wq p a1 E f b1 μ c1 D p d1 {I } = {I current: mass: time: length: 0 a1 − b1 {m } = {m 0 {t } = {t 0 I b1 0 } m c1 } −1 a1 −3b1 − c1 t t {L } = {L L L The dependent Π is thus Π 1: {Π1} = t 1 b1 − c1 d1 L 1 −1 } } )( I t ) ( m L t 1 1 a1 1 1 −3 −1 I ) (m L 1 −1 −1 b1 t ) (L ) c1 1 d1 } 0 = a1 − b1 a1 = b1 0 = b1 + c1 c1 = −b1 = − a1 0 = −1 + a1 − 3b1 − c1 c1 = 1 d1 = 1 0 = 1 + b1 − c1 + d1 Π1 = a1 = b1 = −1 wμ D p qp E f 7-113 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Step 6 Since there is only one Π, it is a function of nothing. This is only possible if we set the Π equal to a constant. We write the final functional relationship as Relationship between Πs: Π1 = wμ D p qp E = constant (2) (b) We re-write Eq. 2 as Equation for w: w = constant qp E f μ Dp (3) Thus, if we double the electric field strength, the drift velocity will increase by a factor of 2. (c) Also from Eq. 3 we see that if we double the particle size, the drift velocity will decrease by a factor of 2. Discussion These results agree with our intuition. Certainly we would expect the drift velocity to increase if we increase the field strength. Also, larger particles have more aerodynamic drag, so for the same charge, we would expect a larger dust particle to drift more slowly than a smaller dust particle. 7-114 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-117 Solution We are to generate a nondimensional relationship between the given parameters. Assumptions 1 The fluid is incompressible. 2 No other parameters are significant in the problem. Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters. F = f (V1 , ΔP, ρ , μ , A1 , A2 , L ) Step 1 All the relevant parameters in the problem are listed in functional form: List of relevant parameters: n=8 Step 2 The primary dimensions of each parameter are listed: ρ ΔP μ { m L t } {L t } {m L t } { m L } F V1 1 1 −2 1 −1 1 −1 −2 {m L t } 1 −3 1 −1 −1 {L } A1 2 {L } A1 2 {L } L 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 8−3 = 5 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines listed in Table 7-3, we cannot pick the dependent variable, F. We cannot choose any two of parameters A1, A2, and L since length1 and length2 are related by an exponent. It is not desirable to have μ or ΔP appear in all the Πs. The best choice of repeating parameters is thus V1, ρ, and one of the length scales. We choose A1. V1, A1, and ρ Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = FV1a1 A1b1 ρ c1 mass: {m } = {m m } 0 {t } = { t time: length: 1 0 t {L } = {L L L 0 1 a1 The dependent Π is thus 2 b1 } L−3c1 } 1 1 −2 )( L t ) ( L ) ( m L ) 1 −1 a1 2 b1 1 −3 c1 } 0 = 1 + c1 c1 = −1 0 = −2 − a1 a1 = −2 0 = 1 + a1 + 2b1 − 3c1 b1 = −1 c1 −2 − a1 {( m L t 0 = 1 − 2 + 2b1 + 3 Π 1: Π1 = ρV12 A1 F From Table 7-5, the established nondimensional parameter most similar to our Π1 is a kind of force coefficient (similar to a lift or drag coefficient) which we shall call CF. No manipulation is required, although a constant of ½ is often placed in the denominator in parameters like this. We form the second Π with ΔP. 7-115 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. {Π 2 } = Π 2 = ΔPV1a2 A1b2 ρ c2 {m } = {m m } mass: 0 { t } = {t time: length: 1 0 {L } = {L 0 1 −1 −2 } L L2b2 L−3c2 −1 a2 } )( L t ) ( L ) ( m L ) 1 −1 a2 2 1 −3 b2 c2 } 0 = 1 + c2 c2 = −1 0 = −2 − a2 a2 = −2 0 = −1 + a2 + 2b2 − 3c2 b2 = 0 c2 −2 − a2 t {( m L t Chapter 7 Dimensional Analysis and Modeling 0 = −1 − 2 + 2b2 + 3 The second Π (first independent Π) is thus Π 2: Π2 = ΔP ρV12 From Table 7-5, the established nondimensional parameter most similar to our Π2 is the Euler number Eu. No manipulation is required. We form the third Π with μ. By now we know that we will generate a type of Reynolds number. Here, since we chose area A1 instead of a length, the length scale in our Reynolds number is the square root of A1. Π 3 = μV1a3 A1b3 ρ c3 Π3 = ρV1 A1 = Reynolds number = Re μ Note: In hindsight, it would probably have been better to use diameter d1 as a parameter instead of A1. The final two Π groups are formed with A2 and then with L. The algebra is trivial for these cases since their dimensions contain nothing but length). The results are Π4 = A2 = Area ratio A1 Π5 = L A1 = A type of length-to-diameter ratio Step 6 We write the final functional relationship as Relationship between Πs: Or, if the constant ½ is applied, Alternate relationship between Πs: ⎛ ΔP ρV A A F L ⎞ ⎟ where Re = 1 1 = f⎜ ,Re, 2 , 2 2 ⎜ ρV1 A1 ρV1 A1 μ A1 ⎟⎠ ⎝ CF = CF = ⎛ ΔP ρV A A2 L ⎞ F ⎜ ⎟ where Re = 1 1 f = ,Re, , 2 2 1 ⎜ ⎟ V A V A ρ ρ μ A1 ⎠ 1 1 1 2 ⎝ 1 Or, if we had chosen the inlet diameter d1 as one of the variables instead of area A1, we would have gotten Alternate relationship between Πs: CF = ρV12 d12 F 1 2 ⎛ ΔP A L⎞ ρV d = f⎜ ,Re, 22 , ⎟ where Re = 1 1 2 μ d ρ V d 1 ⎠ 1 ⎝ 1 Discussion The result applies to both laminar and turbulent flow. Any of the above is acceptable as there is never really only one “right” answer in dimensional analysis. 7-116 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-118 Solution We are to generate CFD solutions for force and force coefficient on a fireman’s nozzle as functions of inlet velocity and Reynolds number, respectively, and compare and discuss. Assumptions 1 The fluid is incompressible. 2 No other parameters are significant in the problem. Analysis (a) We use the standard step-by-step analysis. Step 1 All the relevant parameters in the problem are listed in functional form: F = f (V1 , ρ , μ , A1 ) List of relevant parameters: n=5 Step 2 The primary dimensions of each parameter are listed: ρ { m L t } {L t } F 1 1 −2 μ {m L t } {m L } V1 1 −1 1 −3 1 −1 −1 {L } A1 2 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j =3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines, we cannot pick the dependent variable, F. It is not desirable to have μ appear in all the Πs. The best choice of repeating parameters is thus V1, ρ, and A1. V1, A1, and ρ Repeating parameters: Step 5 The dependent Π is generated: {Π1} = Π1 = FV1a1 A1b1 ρ c1 mass: {m } = {m m } 0 {t } = { t time: length: 1 0 t {L } = {L L L 0 1 a1 The dependent Π is thus 2 b1 } L−3c1 } 1 1 −2 )( L t ) ( L ) ( m L ) 1 −1 a1 2 b1 1 −3 c1 } 0 = 1 + c1 c1 = −1 0 = −2 − a1 a1 = −2 0 = 1 + a1 + 2b1 − 3c1 b1 = −1 c1 −2 − a1 {( m L t 0 = 1 − 2 + 2b1 + 3 Π 1: Π1 = ρV12 A1 F The established nondimensional parameter most similar to our Π1 is a kind of force coefficient (similar to a lift or drag coefficient) which we shall call CF. A constant of ½ is often placed in the denominator in parameters like this. We form the third Π with μ. By now we know that we will generate a type of Reynolds number. Here, since we chose area A1 instead of a length, the length scale in our Reynolds number is the square root of A1. Π 2 = μV1a2 A1b2 ρ c2 Π2 = ρV1 A1 = Reynolds number = Re μ 7-117 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Step 6 We write the final functional relationship as Relationship between Πs: CF = ρV A F = f ( Re ) where Re = 1 1 2 μ ρV1 A1 Or, if the constant ½ is applied, Alternate relationship between Πs: CF = 1 2 ρV A F = f ( Re ) where Re = 1 1 2 μ ρV1 A1 (b) We run FlowLab using template Fireman_nozzle_velocity. The CFD results are tabulated and plotted here in both physical variables (F as a function of V1) and nondimensional variables (CF as a function of Re). The force coefficient rises, reaches a maximum, and then falls slightly. However, the change in CF with Re is very small, with the difference between minimum and maximum being less than 1%. So, within the accuracy of the calculations, and for the range of parameters studies, we can say that CF is at most a weak function of Re, and in fact is nearly independent of Re, with an average value around 4.7. Discussion Dimensional analysis has greatly reduced both the cost and time requirements of the experiment. It is possible that these small differences may be due to convergence issues in the CFD program. 7-118 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-119 Solution We are to generate a dimensionless functional relationship between the given parameters and then compare our results with a known exact analytical solution. Assumptions problem. Analysis 1 There is no flow (hydrostatics). 2 The parameters listed here are the only relevant parameters in the (a) We perform a dimensional analysis using the method of repeating variables. h = f ( ρ , g , σ s , D, φ ) Step 1 There are five parameters in this problem; n = 6, List of relevant parameters: n=6 Step 2 The primary dimensions of each parameter are listed, ρ {L } {m t } g 1 −3 1 σs {L t } {m L } h 1 −2 (1) {L } D 1 −2 1 φ {1} Note that the dimensions of the contact angle are unity (angles are dimensionless). Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 6−3 = 3 Step 4 We need to choose three repeating parameters since j = 3. We cannot choose φ since it is dimensionless. We choose a length (D) and a density (ρ). We’d rather have gravitational constant g than surface tension σs in our Πs. So, we choose ρ, g, D Repeating parameters: Step 5 The dependent Π is generated. Since h has the same dimensions as D, we immediately write Π 1: Π1 = h D The first independent Π is generated by combining σs with the repeating parameters, Π 2 = σ s ρ a2 g b2 D c2 mass: time: length: {m } = {m m } 0 1 {t } = { t 0 {L } = {L 0 } −3a2 b2 c2 L L The first independent Π is thus Π 2: a2 −2 −2 b2 t {Π 2 } = } {( m t 1 −2 )( m L ) ( L t ) ( L ) 1 −3 a2 1 −2 b2 1 c2 } 0 = 1 + a2 a2 = −1 0 = −2 − 2b2 b2 = −1 0 = −3a2 + b2 + c2 c2 = −2 c2 = 3a2 − b2 Π2 = σs ρ gD 2 Finally, the third Π (second independent Π) is simply angle φ itself since it is dimensionless, 7-119 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Π3 = φ Π 3: Step 6 We write the final functional relationship as Relationship between Πs: ⎛ σs ⎞ h = f⎜ ,φ ⎟ 2 D ⎝ ρ gD ⎠ (b) From Chap. 2 we see that the exact analytical solution is Exact relationship: Chapter 7 Dimensional Analysis and Modeling h= 4σ s cos φ ρ gD (2) (3) Comparing Eqs. 2 and 3, we see that they are indeed of the same form. In fact, Functional relationship: Π1 = constant × Π 2 × cos Π 3 (4) Discussion We cannot determine the constant in Eq. 4 by dimensional analysis. However, one experiment is enough to establish the constant. Or, in this case we can find the constant exactly. Viscosity is not relevant in this problem since there is no fluid motion. 7-120 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-120 Solution tube. Assumptions parameter. We are to find a functional relationship for the time scale required for the liquid to climb up the capillary 1 trise is a function of the same parameters listed in the previous problem, but there is another relevant Analysis Since this is an unsteady problem, the rise time will surely depend also on fluid viscosity μ. The list of parameters now involves seven parameters, List of relevant parameters: trise = f ( ρ , g , σ s , D, φ , μ ) n=7 (1) and we expect four Πs. We choose the same repeating parameters and the algebra is similar to that of the previous problem. It turns out that Π 1: Π1 = trise g D The second and third Π are the same as those of the previous problem. Finally, the fourth Π is formed by combining μ with the repeating parameters. We expect some kind of Reynolds number. We can do the algebra in our head. Specifically, a velocity scale can be formed as gD . Thus, Π 4: Π 4 = Re = The final functional relationship is Relationship between Πs: Discussion trise ρ D gD μ ⎛ σs ⎞ g , φ , Re ⎟ = f⎜ 2 D gD ρ ⎝ ⎠ If we would have defined a time scale as (2) D / g , we could have written Π1 by inspection as well, saving ourselves some algebra. 7-121 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-121 Solution We are to use dimensional analysis to find the functional relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the I = f ( P, c, ρ ) Step 1 There are four parameters in this problem; n = 4, List of relevant parameters: n=4 (1) Step 2 The dimensions of I are those of power per area. The primary dimensions of each parameter are listed, ρ {m t } {m L t } {L t } { m L } I P 1 −3 c 1 −1 −2 1 −1 1 −3 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−3 =1 Step 4 We need to choose three repeating parameters since j = 3. The problem is that the three independent parameters form a Π all by themselves (c2ρ/P is dimensionless). Let’s see what happens if we don’t notice this, and we pick all three independent parameters as repeating variables, P, ρ , and c Repeating parameters: Step 5 The Π is generated: Π1 = I × P a ρ b c c mass: time: length: {m } = { m m m } 0 {t } = {t 0 1 t −a t {( m t 1 −3 } L−3b Lc } )( m L 1 −1 −2 t ) (m L ) (L t ) } a 1 −3 b 1 −1 c 0 = 1+ a + b a = −1 − b 0 = −3 − 2a − c c = −1 + 2b 0 = − a − 3b + c c = −1 + 2b b −3 −2 a − c {L } = { L 0 a {Π1} = c = −3 − 2a c = a + 3b This is a situation in which two of the equations agree, but we cannot solve for unique exponents. If we knew b, we could get a and c. The problem is that any value of b we choose will make the Π dimensionless. For example, if we choose b = 1, we find that a = –2 and c = 1, yielding Π1 for the case with b = 1: Since there is only one Π, we write Functional relationship for the case with b = 1: Π1 = I ρc P2 I = constant × P2 ρc (2) However, if we choose a different value of b, say b = –1, then a = 0 and c = –3, yielding 7-122 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling Π1 for the case with b = –1: Π1 = Since there is only one Π, we write I ρ c3 I = constant × ρ c 3 Functional relationship for the case with b = –1: (3) Similarly, you can come up with a whole family of possible answers, depending on your choice of b. We double check our algebra and realize that any value of b works. Hence the problem is indeterminate with three repeating variables. We go back now and realize that something is wrong. As stated previously, the problem is that the three independent parameters can form a dimensionless group all by themselves. This is another case where we have to reduce j by 1. Setting j = 3 – 1 = 2, we choose two repeating parameters, ρ and c Repeating parameters: We jump to Step 5 of the method of repeating variables, Step 5 The first Π is generated: Π1 = I ρ a c b {m } = { m m } mass: 0 1 {t } = {t time: 0 length: t −3 a 0 a −3 − b {L } = { L {Π1} = } Lb } {( m t 1 −3 )( m L ) ( L t ) } 1 −3 1 −1 a b 0 = 1+ a a = −1 0 = −3 − b b = −3 0 = −3a + b b = −3 b = 3a Fortunately, the results for time and length agree. The dependent Π is thus Π 1: Π1 = We form the second Π with sound pressure P, {Π 2 } = Π2 = Pρ ec f mass: time: length: {m } = { m m } 0 1 {t } = { t 0 {L } = { L 0 −2 − f t −1 −3 e e } L Lf The second Π is thus } {( m L t 1 −1 −2 )( m L ) ( L t ) 1 −3 e 1 −1 f } 0 = 1+ e e = −1 0 = −2 − f f = −2 0 = −1 − 3e + f f = 1 + 3e f = −2 Π 2: Step 6 We write the final functional relationship as Relationship between Πs: I ρ c3 Π2 = P ρ c2 ⎛ P ⎞ I = f⎜ 2⎟ 3 ρc ⎝ ρc ⎠ (4) 7-123 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling (b) We try the force-based primary dimension system instead. I = f ( P, c, ρ ) Step 1 There are four parameters in this problem; n = 4, List of relevant parameters: n=4 (5) Step 2 The dimensions of I are those of power per area. The primary dimensions of each parameter are listed, ρ {F L t } {F L } {L t } {F t L } I P 1 −1 −1 c 1 −2 1 −1 1 2 −4 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (F, L, and t). Again, however, the three independent parameters form a dimensionless group all by themselves. Thus we lower j by 1. j=2 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 4−2 = 2 Step 4 We need to choose two repeating parameters since j = 2. We pick the same two parameters as in Part (a), ρ and c Repeating parameters: Step 5 The first Π is generated: Π1 = I × ρ b c c {F } = { F F } force: 0 {t } = {t time: 0 length: {( F L t −1 2 b − c t t } −1 −4 b L Lc } 1 −1 −1 )( F t L ) ( L t ) } 1 2 −4 1 −1 b c 0 = 1+ b b = −1 0 = −1 + 2b − c c = −3 0 = −1 − 4b + c c = −3 1 b {L } = { L 0 {Π1} = c = −1 + 2b c = 1 + 4b Again the two results for length and time agree. The dependent Π is thus Π 1: Π1 = We form the second Π with sound pressure P, Π2 = P × ρ ec f force: time: {F } = { F F } 0 {t } = {t 0 1 e −2 e − f t } {Π 2 } = {( F L 1 −2 ρ c3 I )( F t L ) ( L t ) 1 2 −4 e 1 −1 f } 0 = 1+ e e = −1 0 = −2e − f f = −2e f = −2 7-124 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. length: {L } = { L 0 −2 L−4 e Lf The second Π is thus } 0 = −2 − 4e + f f = −2 f = 2 + 4e Π 2: Step 6 We write the final functional relationship as Relationship between Πs: Chapter 7 Dimensional Analysis and Modeling Π2 = ρ c2 P ⎛ P ⎞ I = f⎜ 2⎟ 3 ρc ⎝ ρc ⎠ (6) Discussion Equations 4 and 6 are the same. This exercise shows that you should get the same results using mass-based or force-based primary dimensions. 7-125 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-122 Solution We are to find the dimensionless relationship between the given parameters. Assumptions 1 The given parameters are the only relevant ones in the problem. Analysis Πs). The step-by-step method of repeating variables is employed to obtain the nondimensional parameters (the I = f ( P, c, ρ , r ) Step 1 There are now five parameters in this problem; n = 5, List of relevant parameters: n=5 (1) Step 2 The dimensions of I are those of power per area. The primary dimensions of each parameter are listed, ρ {m t } {m L t } {L t } {m L } I P 1 −3 c 1 −1 −2 1 −1 1 −3 {L } r 1 Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). j=3 Reduction: If this value of j is correct, the expected number of Πs is Number of expected Πs: k = n− j = 5−3 = 2 Step 4 We need to choose three repeating parameters since j = 3. We pick the three simplest independent parameters (r instead of P), r , ρ , and c Repeating parameters: Step 5 The first Π is generated: Π1 = I × r a ρ b c c mass: time: length: {m } = { m m } 0 1 {t } = { t 0 {L } = { L L 0 a −3b {( m t b −3 − c t {Π1} = } Lc The first Π is thus } 1 −3 )( L ) ( m L ) ( L t ) } 1 −3 a 1 b = −1 0 = −3 − c c = −3 0 = a − 3b + c a=0 a = 3b − c Π1 = {( m L t We form the second Π with sound pressure P, mass: {m } = { m m } 0 1 e c 0 = 1+ b Π 1: Π2 = P × r d ρ ec f 1 −1 b {Π 2 } = 1 −1 −2 I ρ c3 )( L ) ( m L ) ( L t ) 1 d 1 −3 e 1 −1 f } e = −1 0 = 1+ e 7-126 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. time: length: {t } = { t 0 {L } = { L 0 −2 − f t −1 d } L L−3e Lf The second Π is thus f = −2 0 = −2 − f } 0 = −1 + d − 3e + f d =0 d = 1 + 3e − f Π 2: Step 6 We write the final functional relationship as Relationship between Πs: Chapter 7 Dimensional Analysis and Modeling Π2 = ρ c2 P ⎛ P ⎞ I = f⎜ 2⎟ 3 ρc ⎝ ρc ⎠ (2) Discussion This is an interesting case in which we added another independent parameter (r), yet this new parameter does not even appear in the final functional relationship! The list of independent parameters is thus over specified. (It turns out that P is a function of r, so r is not needed in the problem.) The result here is identical to the result of the previous problem. It turns out that the function in Eq. 2 is a constant times Π22, which yields the correct analytical equation for I, namely Analytical result: I = constant × P2 ρc (3) 7-127 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 7 Dimensional Analysis and Modeling 7-123 Solution We are to calculate the speed of a robotic tuna to match the Reynolds number of a real tuna. Assumptions 1 The fluid in the model and prototype tests is the same (seawater) so that fluid properties such as density and viscosity do not change from model to prototype. Analysis We match Reynolds number between the model and prototype, Re m = Reynolds number matching: ρ pVp Lp ρ mVm Lm = Re p = μm μp from which we solve for the required speed of the model, Required model speed: ⎞ ⎛ ρ p ⎞ ⎛ Lp ⎞ ⎟⎜ ⎟ ρ m ⎟ ⎜ Lm ⎟ ⎠⎝ ⎠ ⎠⎝ ⎛ 2.5 m ⎞ = (12.5 m/s )(1)(1) ⎜ ⎟ = 31.25 m/s ≅ 31 m/s ⎝ 1.0 m ⎠ ⎛μ Vm = Vp ⎜ m ⎜ μp ⎝ Discussion Note that both the increased size and increased speed influence the required speed of the model. We round the final answer to two significant digits in keeping with the precision of the given information. KJ 7-128 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.