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Lumped Capacitance Analysis Using Python
When an object at very high temperature is suddenly dropped in a cooler liquid and if it is assumed that the conductive resistance of the solid is very small in comparison to the surrounding convective resistance then the heat transfer analysis is called as lumped capacitance analysis (as shown in the figure given below). Here, we treat the system as a lump. In that case, we can assume that the rate of change of internal energy of lump will be equal to the heat interaction with the surrounding fluid.
Mathematically, this can be written as
\mathrm{pcV\frac{\partial T}{\partial t} \: = \: hA(T \: \: T_{\infty}) \: \dotso\dotso \: (1)}
\mathrm{\frac{\partial T}{\partial t} \: = \: \frac{hA}{pcV}(T \: \: T_{\infty}) \: \dotso\dotso \: (2)}
Now the Eq 2 has to be solved to obtain the variation of temperature with time.
If we integrate Eq. 2 with the initial temperature of object \mathrm{T_{0}} to some arbitrary temperature T in time 0 to t then the final answer will be
\mathrm{\frac{T \: \: T_{\infty}}{T_{0} \: \: T_{\infty}} \: = \: exp( \frac{hA}{pcV}t) \: \dotso\dotso \: (3)}
Now this equation can be used in the following way
If time is required to arrive at some temperature \mathrm{T_{f}}
\mathrm{t \: = \: \frac{pcV}{hA}In(\frac{T_{f} \: \: T_{\infty}}{T_{0} \: \: T_{\infty}}) \: \dotso\dotso \: (4)}
If temperature is required at some time \mathrm{t_{f}}
\mathrm{T \: = \: T_{\infty} \: + \: (T_{0} \: \: T_{\infty}) \: \times \: exp (\frac{hA}{pcV}t_{f}) \: \dotso\dotso \: (5)}
Let us demonstrate this with the help of following examples
Example 1
A steel ball of radius 1 mm is at temperature \mathrm{1200^{\circ}C} is placed in open air having temperature \mathrm{25^{\circ}C}. Calculate the time required to cool the ball to arrive at \mathrm{100^{\circ}C}. \mathrm{(k_{b} \: = \: 50 \: W/mK, \: c_{b} \: = \: 500kJ/kgK, \: p_{b} \: = \: 8000kg/m^{3}, \: and \: h_{air} \: = \: 10000W/m^{2}K)}.
Solution
As time is expected, so we will be using Equation 4.
# lumped capacitance analysis from pylab import * #Input Data T_inf=25 r=1.E-3 T0=1200 Tf=100 k=50 c=500 ?=8000 h=10000 #Evaluating Volume and Area A=(4*pi*r**2) V=(4/3)*pi*r**3 ?0=?*c*V/(h*A) t=-?0*log((Tf-T_inf)/(T0-T_inf)) print(f't = {round(t,3)} s')
Output
The program output will be
t = 0.367 s
Example 2
In Example 1, what is the temperature of the ball after 0.1 sec.
Solution
In this case, Eq. 5 has to be solved. Everything will be same except the formula.
# lumped capacitance analysis from pylab import * # Input Data T_inf=25 r=1.E-3 T0=1200 Tf=100 t=0.1 k=50 c=500 ?=8000 h=10000 # Evaluating Volume and Area A=(4*pi*r**2) V=(4/3)*pi*r**3 ?0=?*c*V/(h*A) T=T_inf+(T0-T_inf)*exp(-t/?0) print(f'T = {round(T,3)} deg. C')
Output
The output will be
T = 580.031 deg. C
To understand the picture, one should plot the variation of temperature. For that case, Eq. 3 will be used as follows
Example 3
from pylab import * # Input Data T_inf=25 r=1.E-3 T0=1200 k=50 c=500 ?=8000 h=10000 # Evaluating Volume and Area A=(4*pi*r**2) V=(4/3)*pi*r**3 # Evaluating time constant ?0=?*c*V/(h*A) t= linspace(0,1.0,50) T=T_inf+(T0-T_inf)*exp(-t/?0) figure(1,dpi=300) plot(t,T,'r-o') xlabel('t (s)') ylabel('T ($^\circ$C)') savefig('plot1.jpg') show()
Output
Here is the plot that shows the variation in temperature
Numerical Approach
If one wants to follow the numerical approach then you can also do that. Let us use explicit method to solve Eq. 2.
For the grid shown above the Eq. 2 can be discretized (using forward difference) as
\mathrm{\frac{T_{i} \: \: T_{i 1}}{\Delta t} \: = \: \tau_{0}(T_{i} \: \: T_{\infty})}
\mathrm{T_{i} \: = \: T_{i 1} \: \: \Delta t \: \times \: (T_{i} \: \: T_{\infty})/\tau_{0} \: \dotso\dotso \: (6)}
subscript represent the time step.
We will march in time but in every time step, we will iterate till the solution gets converged.
As at the initial time temperature is known, i.e., 1200 so will start with second time step. First, we will guess the value of temperature at this time step (say 0) then we will solve Eq. 6 and then compare the guess value and the value obtained. If the absolute difference becomes less than the convergence criterion \mathrm{10^{ 5}} then we will move to the next time step. Else, we will set the obtained value of temperature as the guess and again solve equation 6. The process is repeated till the last time step. The program for this is shown below. Where we have also compared the result with the exact analytical value (Eq. 3).
Example
from pylab import * # Input Data T_inf=25 r=1.E-3 T0=1200 k=50 c=500 ?=8000 h=10000 # Evaluating Volume and Area A=(4*pi*r**2) V=(4/3)*pi*r**3 ?0=?*c*V/(h*A) n=35 # Number of time steps t = linspace(0,1.0,n) ?t=1/(n-1) # initial guess for all time steps T=zeros(n) # initial condition T[0]=T0 # Array of guess Tg=T.copy() # Time marching for i in range(1,n): # Iteration loop error=1 while error>1.E-5: T[i]=T[i-1]-?t*(T[i]-T_inf)/?0 error=abs(T[i]-Tg[i]) Tg=T.copy() # Exact solution T_exact=T_inf+(T0-T_inf)*exp(-t/?0) # Data Plotting figure(2,dpi=300) plot(t,T,'r-o',label='Numerical') plot(t,T_exact,'b-o',label='Analytical') xlabel('t (s)') ylabel('T ($^\circ$C)') legend() savefig('plot2.jpg') show()
Output
The program output will be as follows
Data on the basis of which the above plot is generated is shown below
time T(analytical) T(numerical) 0.0 1200.0 1200.0 0.0294 987.6506 967.4051 0.0588 813.6776 780.853 0.0882 671.1455 631.2296 0.1176 554.3722 511.2245 0.1471 458.7025 414.9749 0.1765 380.3226 337.7781 0.2059 316.1076 275.8627 0.2353 263.4978 226.2036 0.2647 220.3958 186.3748 0.2941 185.0833 154.4301 0.3235 156.1526 128.809 0.3529 132.4503 108.2597 0.3824 113.0316 91.7782 0.4118 97.1223 78.5592 0.4412 84.0881 67.957 0.4706 73.4095 59.4535 0.5 64.6608 52.6334 0.5294 57.4932 47.1632 0.5588 51.6209 42.776 0.5882 46.8099 39.2572 0.6176 42.8684 36.4349 0.6471 39.6391 34.1713 0.6765 36.9935 32.3558 0.7059 34.826 30.8997 0.7353 33.0502 29.7319 0.7647 31.5954 28.7952 0.7941 30.4034 28.0439 0.8235 29.4269 27.4414 0.8529 28.6269 26.9581 0.8824 27.9714 26.5705 0.9118 27.4344 26.2596 0.9412 26.9945 26.0103 0.9706 26.634 25.8103 1.0 26.3387 25.6499
Conclusion
In this tutorial, lumped capacitance analysis has been explained and modelled using Python. Both analytical and numerical approaches were discussed.
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