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Numerical Analysis of Heat Transfer within a Cooling Fin
Numerical Analysis of Heat Transfer within a Cooling Fin
Objective
The goal of this analysis was to practice using numerical methods as solutions to differential equations, in order to prepare for situations where the analytical solutions are not known.
Outcome
Finite-difference schemes and assumed boundary conditions were applied to convective temperature and heat transfer equations, then directly compared to analytical solutions. A table of errors was generated for interpreting accuracy.
Findings
Findings
This analysis includes four figures and two tables. Below the figures and tables is the python code used to calculate the figures.
It is clear that the temperature errors are much smaller than the heat transfer errors, because calculating the heat transfer from the temperature slope numerically will cause the errors to greatly expound. To increase accuracy, fourth-order finite difference schemes could be implemented. Increasing the number of nodes and modifying the step size would also cause results to converge better.
Figure 1: Shown is the temperature vs the position of a cooling fin modeled by an analytical solution and a second-order finite difference method. This model assumes that the ends of the fin are held at fixed temperatures.
Figure 2: The heat transfer vs the position of a cooling fin, also modeled by an analytical solution and a second-order finite difference method. The ends of the cooling fin are also held at fixed temperatures.
Figure 3: Shown is the temperature vs the position of a cooling fin modeled by an analytical solution and a second-order finite difference method. This time, the front end is held at a constant while the end of the fin is assumed to be adiabatic (zero heat transfer).
Figure 4: The Heat Transfer vs the position of a cooling fin modeled by an analytical solution and a second-order finite difference method. This model is additionally modeled with an adiabatic tip and a constant-temperature beginning
Python Code
Python Code
'''
Heat Transfer of a Cooling Fins; Numerical vs
Analytical Models
Dallin Carter
The purpose of this code is to compare the difference
of temperature and heat transfer values across cooling
fins between numerical and analytical solutions.
'''
#Libraries
import numpy as np
import matplotlib.pyplot as plt
from scipy.linalg import solve
#Physical Conditions
length_fin = 0.15 #m
area_fin = 1.67E-4 #m^2
length_perimeter = 0.066 #m
k = 16 #W/m-K
h = 25 #W/m^2-K
temp_base = 500 #K
temp_freestream = 300 #K
temp_tip = 350 #K
#Numerical Conditions
dx = 0.03 #m
m = np.sqrt(h*length_perimeter/(k*area_fin))
'''
Analytical Functions:
The input is distance (in meters) across the fin, and returns the value
of Temperature (in Kelvin) or Heat Transfer (in Watts). Also included is
a linspace of xx, which has calculates the temperature/heat for many values
across the fin.
'''
def T1_ana(x):
return ((temp_tip-temp_freestream)*np.sinh(m*x)+(temp_base-temp_freestream)*np.sinh(m*(length_fin-x)))/np.sinh(m*length_fin) + temp_freestream
def q1_ana(x):
return (m*k*area_fin/np.sinh(m*length_fin))*((temp_base-temp_freestream)*np.cosh(m*(length_fin-x))-(temp_tip-temp_freestream)*np.cosh(m*x))
def T2_ana(x):
return (np.cosh(m*(length_fin-x))/np.cosh(m*length_fin))*(temp_base-temp_freestream)+temp_freestream
def q2_ana(x):
return m*k*area_fin*(np.sinh(m*(length_fin-x))/np.cosh(m*length_fin))*(temp_base-temp_freestream)
xx = np.linspace(0,length_fin,100)
'''First Boundary Conditions'''
n = 6
A = np.zeros([n,n])
b = np.zeros(n)
x = np.linspace(0,length_fin,n)
#Boundary Conditions for Temperature
A[0,0] = 1
A[-1,-1] = 1
b[0] = temp_base
b[-1] = temp_tip
# Fill in matrix
for i in range(1,n-1):
A[i,i-1] = 1
A[i,i] = -2-dx**2*m**2
A[i,i+1] = 1
b[i] = -dx**2*m**2*temp_freestream
# Solve for Temperature
T1 = solve(A,b)
#Plotting Temp vs Position
plt.plot(x,T1,marker='o', linestyle = ':', label = 'Numerical Solution')
plt.plot(xx,T1_ana(xx), label = 'Analytical Solution', color='k')
plt.xlabel('Position (m)')
plt.ylabel('Temperature (K)')
plt.legend()
plt.savefig('HeatTransfer1')
plt.show()
#Solve for Heat Transfer
q1 = np.zeros([n])
q1[0] = -k*area_fin*(-T1[2]+4*T1[1]-3*T1[0])/(2*dx)
q1[n-1] = -k*area_fin*(T1[-3]-4*T1[-2]+3*T1[-1])/(2*dx)
for i in range(1, n-1):
q1[i] = -k*area_fin*(-T1[i-1]+T1[i+1])/(2*dx)
#Plotting Heat Transfer vs Position
plt.plot(x,q1,marker='o', linestyle = ':', label = 'Numerical Solution')
plt.plot(xx,q1_ana(xx), label = 'Analytical Solution', color='k')
plt.xlabel('Position (m)')
plt.ylabel('Heat Transfer (W)')
plt.legend()
plt.savefig('HeatTransfer2')
plt.show()
'''Second Boundary Conditions'''
#Boundary Conditions for Temperature
A[0,0] = 1
A[-1,-3] = 1
A[-1,-2] = -4
A[-1,-1] = 3
b[0] = temp_base
b[-2] = 0
b[-1] = 0
# Fill in matrix
for i in range(1,n-1):
A[i,i-1] = 1
A[i,i] = -2-dx**2*m**2
A[i,i+1] = 1
b[i] = -dx**2*m**2*temp_freestream
# Solve for Temperature
T2 = solve(A,b)
#Plot Temp vs Position
plt.plot(x,T2,marker='o', linestyle = ':', label = 'Numerical Solution')
plt.plot(xx,T2_ana(xx), label = 'Analytical Solution', color='k')
plt.xlabel('Position (m)')
plt.ylabel('Temperature (K)')
plt.legend()
plt.savefig('HeatTransfer3')
plt.show()
#Solve for Heat Transfer
q2 = np.zeros([n])
q2[0] = -k*area_fin*(-T2[2]+4*T2[1]-3*T2[0])/(2*dx)
q2[n-1] = -k*area_fin*(T2[-3]-4*T2[-2]+3*T2[-1])/(2*dx)
for i in range(1, n-1):
q2[i] = -k*area_fin*(-T2[i-1]+T2[i+1])/(2*dx)
#Plot of Heat Transfer vs Position
plt.plot(x,q2,marker='o', linestyle = ':', label = 'Numerical Solution')
plt.plot(xx,q2_ana(xx), label = 'Analytical Solution',color='k')
plt.xlabel('Position (m)')
plt.ylabel('Heat Transfer (W)')
plt.legend()
plt.savefig('HeatTransfer4')
plt.show()
'''
Calculating Error and Printing the Solutions
'''
#Calculating the Errors
ErrorT1 = abs(T1 - T1_ana(x))/T1_ana(x)*100
ErrorT2 = abs(T2 - T2_ana(x))/T2_ana(x)*100
Errorq1 = abs(q1 - q1_ana(x))/q1_ana(x)*100
Errorq2 = abs(q2 - q2_ana(x))/q2_ana(x)*100
#Printing Solutions at all Nodes
#First LaTeX Table Print
for i in range(n):
print(f"{x[i]:.3f} & {T1[i]:.2f} & {ErrorT1[i]:.2f} & {q1[i]:.2f} & {Errorq1[i]:.2f} \\\\")
print()
#Second LaTeX Table Print
for i in range(n):
print(f"{x[i]:.3f} & {T2[i]:.2f} & {ErrorT2[i]:.2f} & {q2[i]:.2f} & {Errorq2[i]:.2f} \\\\")
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