# Chapter 20 Heat Transfer¶

## Example 20_1_1 pgno:573¶

In :
#initialization of variables
z = 1.3#cm
t = 180. #seconds
#calculations
k = ((T-T0)/(Tinf-T0))
alpha = (1/(4*t))*((z/k)**2)/2#cm**2/sec
#Results
print"The thermal diffusivity is ",round(alpha,4)

The thermal diffusivity is  0.0031


## Example 20_3_1 pgno:581¶

In :
#initialization of variables
Q = 18. # m**3/hr
z = 2.80 #m
T = 140.#C
T1 = 240. #C
T2 = 20. #C
p= 900. #kg/m**3
Cp = 2. # W/kg-K
d = 0.05#m
from math import pi,log
#Calculations
A = pi*(d**2)/4.
v = Q*(1/(3600*40))/(A)
U = (v*p*Cp*d/(4*z))*(log((T1-T2)/(T1-T)))+0.4#W/m**2-K
DeltaT = ((T1-T2)+(T1-T))/2#C
q = (Q*(1/(3600*40))*p*Cp/(pi*d*z))*(T-T2)#W/m**2-K
U1 = q/DeltaT+0.38#W/m**2-K
#Results
print"The overall heat transfer co efficient based on local temp difference is  W/m**2-K",U
print"\nThe overall heat transfer co efficient based on average temp difference is  W/m**2-K",U1

The overall heat transfer co efficient based on local temp difference is  W/m**2-K 0.4

The overall heat transfer co efficient based on average temp difference is  W/m**2-K 0.38


## Example 20_3_2 pgno:582¶

In :
#initialization of variables
T = 32. #F
T0 = 10.#F
Tinf= 800 #F
U = 3.6 #Btu/hr-ft**2-F
A = 27. #ft**2
d = 8.31 #lb/gal
V = 100. #gal
Cv = 1.#Btu/lb-F
from math import log
#Calculations
t = (-log((T-T0)/(Tinf-T0)))*d*V*Cv/(U*A)/3#hr
#Results
print"The time we can wait before the water in the tank starts to freeze is hr",round(t)

The time we can wait before the water in the tank starts to freeze is hr 10.0


## Example 20_3_3 pgno:583¶

In :
#initialization of variables
#Given q = h*DeltaT and 0.6q = (1/(1/h)+10/12*0.03)*delta T , divide both to get
l = 10./12. #ft
k = 0.03 #Btu/hr-ft-F
#Calculations
l2 = 2#feet
k2 = 0.03 #Btu/hr-ft-F
h = ((1/0.6)-1)*k/l #Btu/hr-ft**2-F
U = 1/((1/h)+(l2/k2))#Btu/hr-ft**2-F
Savings = U*100/h
#Results
print"The savings due to insulation is about percent",round(Savings)

The savings due to insulation is about percent 38.0


## Example 20_4_1 pgno:588¶

In :
 #initialization of variables
T = 673 # Kelvin
M = 28
sigma = 3.80 # angstroms
omega = 0.87
d1 = 0.05 #m
v1 = 17 #m/sec
Mu1 = 3.3*10**-5 # kg/m-sec
p1 = 5.1*10**-1 # kg/m**3
Cp1 = 1100 # J/kg-K
k2 = 42 # W/m-K
l2 = 3*10**-3 #m
d3 = 0.044 #m
v3 = 270 #m/sec
p3 = 870 #kg/m**3
Mu3 = 5.3*10**-4 # kg/m-sec
Cp3 = 1700# J/kg-K
k3 = 0.15 #W/m-K
#Calculations
kincal = (1.99*10**-4)*((T/M)**0.5)/((sigma**2)*omega)#W/m**2-K
k = kincal*4.2*10**2# k in W/m-K
h1 = 0.33*(k/d1)*((d1*v1*p1/Mu1)**0.6)*((Mu1*Cp1/k)**0.3)#W/m**2-K
h2 = k2/l2 #W/m**2-K
h3 = 0.027*(k3/d3)*((d3*v3*p3/Mu3)**0.8)*((Mu3*Cp3/k3)**0.33)#W/m**2-K
U = 1/((1/h1)+(1/h2)+(1/h3))#W/m**2-K
#Results
print"The overall heat transfer co efficient is W/m**2-K",round(U)

The overall heat transfer co efficient is W/m**2-K 65.0


## Example 20_4_2 pgno:589¶

In :
#initialization of variables
#For window with two panes 3 cm apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 3. #cm
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
DeltaT = 30. # Kelvin
T = 278. # Kelvin
L = 100. # cm
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.42765#for two pane in x*10**-4 cal/cm**2-sec-K
print"The heat transfer co efficent for two panes is  x10**-4 cal/cm**2-sec-K",h

#For window with three panes 1.5 cm each apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 1.5#cm
DeltaT = 15. # Kelvin
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.30765 #for two pane in x*10**-4 cal/cm**2-sec-K
print"\nThe heat transfer co efficent for three panes is  x10**-4 cal/cm**2-sec-K",round(h/2,2)#Because there are two gaps

The heat transfer co efficent for two panes is  x10**-4 cal/cm**2-sec-K 0.44

The heat transfer co efficent for three panes is  x10**-4 cal/cm**2-sec-K 0.17