Chapter 14: Heat Transfer

Example 14.1, page no. 279

In [2]:
#initialization
T = 50.0                        #F
Q = 3.9                         #B/hr-ft^2

#calculation
#From table 14.1
k = 0.026                       #B/hr-ft-F
dx = k*T/Q

#result
print "Required thickness  =  ft", round(dx, 3), "lb/in^2"

Required thickness  =  ft 0.333 lb/in^2

Example 14.2, page no. 281

In [3]:
#initialization
x1 = 1.0                        #in
x2 = 4.0                        #in
T1 = 85.0                       #F
T2 = 30.0                       #F

#calculation
QbyA = 12*(T1-T2)/(x1/0.3 + x2/0.026)

#result
print "Rate of heat flow  = ", round(QbyA, 1), "B/r-ft^2-F"

Rate of heat flow  =  4.2 B/r-ft^2-F

Example 14.3, page no. 283

In [4]:
import math

#initialization
L = 6.5                         #in
thick = 1.0                     #in
k = 0.06                        #B/hr-ft-F
T1 = 350.0                      #F
T2 = 110.0                      #F

#calculation
QbyL = 2*math.pi*k*(T1-T2)/math.log(1+2/L)

#result
print "heat flow  =  %d B/hr-ft" %QbyL

heat flow  =  337 B/hr-ft

Example 14.4, page no. 283

In [6]:
import math

#initialization
t = 0.25                        #in
dia = 5.5                       #in
t2 = 0.6                        #in
To = 100.0                      #F
kp = 34.5                       #B/hr-ft-F
ki = 0.05                       #B/hr-ft-F
l = 10.0                        #ft
Q = 2000.0                      #B/hr

#calculation
dT = Q*(1/kp *math.log(1+ 2*t/dia) + 1/ki *math.log(1 + 4*t/(dia+2*t)))/(2*math.pi*l)
T1 = dT+To

#result
print "Inner surface temperature of the pipe  = ", round(T1, 1), "F"

Inner surface temperature of the pipe  =  198.2 F

Example 14.5, page no. 286

In [7]:
#initialization
Tsurr = 90.0                    #F
T = 85.0                        #F

#calculation
H = 4.2/(Tsurr-T)

#result
print "Film coefficient  =  %.2f B/hr-ft^2-F" %H

Film coefficient  =  0.84 B/hr-ft^2-F

Example 14.6, page no. 288

In [8]:
#initialization
k = 0.04                        #B/hr-ft-F
thick = 1.0                     #in
T1 = 90.0                       #F
T2 = 30.0                       #F
Air = 2.5                       #B/hr-ft^2-F
film = 2.0                      #B/hr-ft^2-F

#calculation
U = 1.0/(1.0/Air + thick/12.0/k + 1/film)
Q = U*(T1-T2)

#result
print "Rate of heat transfer per unit square area  =  %.1f B/hr-ft^2" %Q

Rate of heat transfer per unit square area  =  20.1 B/hr-ft^2

Example 14.7, page no. 291

In [14]:
import math

#initialization
U = 115.0                   #B/hr-ft^2-F
T1 = 190.0                  #F
T2 = 160.0                  #F
Tc1 = 65.0                  #F
Tc2 = 100.0                 #F
w = 140.0                   #lbm/min
c = 0.8                     #B/lbm F

#calculation
Q = w*60.0*c*(T1-T2) #answer in textbook for Q is wrong. Hence, the difference in final answer
dT = ((T1-Tc2) - (T2-Tc1))/math.log((T1-Tc2)/(T2-Tc1))
A = Q/(U*dT)

#result
print "Required Area  =  %.1f ft^2" %A

Required Area  =  19.0 ft^2

Example 14.8, page no. 294

In [15]:
#initialization
e = 0.8
T1 = 100+460.0                  #R
T2 = 300+460.0                  #R

#calculation
Qdot = 0.173*10**-8 *(T2**4 - T1**4)/(1/e +1/e -1)

#result
print "Radiant heat transfer per sq. foot  =  %d B/hr-ft^2" %(Qdot+1)

Radiant heat transfer per sq. foot  =  272 B/hr-ft^2

Example 14.9, page no. 294

In [19]:
#initialization
T1 = 400+460.0                  #R
A = 40.0                        #in^2
e = 0.1
T2 = 70+460.0                   #R

#calculation
Q = A*e*0.173*10**-8 *(T1**4 - T2**4)/144.0

#result
print "Rate of heat transfer  =  %.2f B/hr" %Q
#answer in the book is wrong

Rate of heat transfer  =  22.49 B/hr