# Chapter 19: Heat Transmission¶

## Exa 19.2¶

In [ ]:
#A large flat wall consists of two well bonded layers of material 8 in thick
#and 1 in thick. The 8 in thick layer is concrete having a k=0.50. and the 1- in
#k=0.02. The surface temp is  -20 F. and the surface temp. of concrete is 60 F.
#Find the rate of heat flow per unit area, the temperature at the interface
#between the two layers and the resistances per unit area of the two layers
import math
#initialisation of variables
t= 8. 							#in
t1= 1. 							#in
k= 0.50 						#Btu/hr ft F
k1= 0.02 						#Btu/hr ft F
A= 1 							#ft^2
T= 60 							#F
T1= -20 						#F
#CALCULATIONS
Rc= (t/12.)/(k*A)				#Resistance 1
Rf= (t1/12.)/(k1*A)				#Resistance 2
R= Rc+Rf 						#Total resistance
q= (T-T1)/R 					#Heat
T2= (T+(Rc/Rf)*T1)/(1+(Rc/Rf)) 	#temp of the interface
#RESULTS
print '%s %.2f' %('Rate of heat flow (Btu/hr) = ',q)
print '%s %.2f' %(' \n Temperature at the interface (F) = ',T2)
raw_input('press enter key to exit')

Rate of heat flow (Btu/hr) =  14.55

Temperature at the interface (F) =  40.61


## Exa 19.3¶

In :
#A steel pipe of 10 in inside dia. and 0.375 in. wall thickness carries steam
#at 500 f. The pipe is covered by 2 in of insulation to reduce heat losses to
#surroundings at 80 F. It is known from tests that the convection coefficients
#for inside and outside are 2500 and 1.6. to protect personnel it is desired
#that the outside shouldn't exceed 140 f. If ksteel=26. and insulation is 0.045.
#will the 2 in thickness of insulation meet the requirement?
import math
#initialisation of variables
h1= 2500. 							#Btu/sq ft hr F
r= 10. 								#in
t= 0.375 							#in
Ts= 500. 							#F
Ta= 80. 							#F
r2= 5.375 							#in
r1= 5. 								#in
r3= 7.375 							#in
kp= 26. 							#Btu ft/hr
ki= 0.045 							#Btu ft/hr
h1= 2500. 							#Btu/sq ft hr F
h3= 1.6 							#Btu/sq ft hr F
r4= 14.750
#CALCULATIONS
R1= 1/(h1*math.pi*(r/12.))			#Resistance 1
Rp= math.log(r2/r1)/(2*math.pi*kp) 	#Resistance 2
Ri= math.log(r3/r2)/(2*math.pi*ki) 	#Resistance 3
R3= 1/(h3*math.pi*(r4/12.)) 		#Resistance 4
R0= R1+Rp+Ri+R3 					#Total reistance
T3=Ta+ (Ts-Ta)*R3/R0 				#Interface temp.
#RESULTS
print '%s %.6f' %('Temperature at the interface (F) = ',T3)
raw_input('press enter key to exit')

Temperature at the interface (F) =  133.055520
press enter key to exit

Out:
''

## Exa 19.4¶

In :
#Lubricating oil is to be cooled from 170 F to 120 F by water ssupplied at 100F.
#Inorder to minimize the water requirement it is desired, if possible to have
#water rise to 140F, the highest permissible temp. The oil flow rate is 40000.
#the oil cp=0.5 and U=120. (b) Find the water flow rate for the counter flow
#operation. (c) Find the heat transfer surface required for the counter flow.
#(d) with the water flow rate found in (b), to what temp. could the oil be
#cooled in a parallel exchanger of unlimited area (e) In the exchanger of (d)
#how much water flow would be required to cool oil to 120F?
import math
#initialisation of variables
wh= 40000. 									#lb.hr
cph= 0.5 									#Btu/lb F
th1= 170. 									#F
th2= 120. 									#F
cpc= 1 										#Btu/lb F
tc2= 140. 									#F
tc1= 100. 									#F
t= 140 										#F
U= 120 										#Btu/sq ft hr F
#CALCULATIONS
dh= t-th2 									#Change in temp. for hot
dc= tc2-tc1 								#Change in temp. for cold
wc= (wh*cph*(th1-th2))/(cpc*dc) 			#weight of cold
dtm= (-(tc1-th2)-(th1-tc2))/math.log((-tc1+th2)/(th1-tc2)) #lmtd
q= wh*cph*(th1-th2) 						#heat
print(q)
A= q/(U*dtm) 								#area
th2= ((wh/wc)*(cph/cpc)*th1+tc1)/((wh/wc)*(cph/cpc)+1)#Hot final
wc1= (wh*cph*(th1-th2))/(cpc*(th2-tc1)) 	#weight of cold final
#RESULTS
print '%s %.2f' %('Water flow rate (lb/hr) = ',wc)
print '%s %.2f' %(' \n Area of heat transfer surface (sq ft) = ',A)
print '%s %.2f' %(' \n temperature of the oil (F) = ',th2)
print '%s %.2f' %(' \n flow rate (lb/hr) = ',wc1*2)
raw_input('press enter key to exit')

1000000.0
Water flow rate (lb/hr) =  25000.00

Area of heat transfer surface (sq ft) =  337.89

temperature of the oil (F) =  131.11

flow rate (lb/hr) =  50000.00
press enter key to exit

Out:
''

## Exa 19.5¶

In :
#Find the film coefficient for airr flowing at 100 fps thru a tube of 1 in outside
#diameter and 18 gage thickness , if the avg. bulk air temp. is 600F, the pressure
#is 1 atm,a nd the tube wall is 200 F?
import math
#initialisation of variables
Tw= 200. 			#F
Ta= 600. 			#F
V= 100 				#fps
Di= 0.902 			#in
d= 0.0375 			#lb/cu ft
u= 0.000020 		#lbm/sec
cp= 0.25 			#Btu/lb F
k= 0.027 			#Btu/sq ft hr
#CALCULATIONS
NRe= (Di*V*d)/(u*12)#Reynolds number
Npr= 0.66 			#prandtl number
h= k*0.023*math.pow(NRe,0.8)*math.pow(Npr,0.4)*12/Di
#RESULTS
print '%s %.2f' %('Film coefficient (Btu/sq ft hr F) = ',h)
raw_input('press enter key to exit')

Film coefficient (Btu/sq ft hr F) =  14.59
press enter key to exit

Out:
''

## Exa 19.6¶

In :
#Find the film coefficient for the conditions of example 5, using eq. 17
import math
#initialisation of variables
Tw= 200 					#F
Ta= 600 					#F
cpb= 0.25 					# Btu/lb F
tf= 0.68
uf= 0.000017 				#lbm/sec ft
D= 0.902 					#in
V= 100. 					#fps
d= 0.0375 					#lb/cu ft
#CALCULATIONS
Nre= (D/12.)*V*d/uf			#reynolds number
Npr= 0.68 					#prandtl number
h= cpb*V*3600*d*0.023/(math.pow(Nre,0.2)*math.pow(Npr,(2./3.)))
#RESULTS
print '%s %.2f' %('Film coefficient (Btu/sq ft hr F) = ',h)
raw_input('press enter key to exit')

Film coefficient (Btu/sq ft hr F) =  14.38
press enter key to exit

Out:
''

## Exa 19.7¶

In :
#An uninsulated 3 in steam pipe passes through a room in which the air and all
#solid surfaces are at an average temp. of 70 F. If the surface temp. of the
#steam pipe is 200F, estimate the heat loss per foot of pipe by radiation and
# compare the relative magnitudes of losses by radiation and by free convection.
import math
#initialisation of variables
A1= 0.916 			#ft^2
e1= 0.8
s= 0.173 			#BTU s^-1 in^-2 R^-4
T= 200 				#F
T1= 70 				#F
D= 0.292
#CALCULATIONS
q= (A1/math.pow(10,6))*e1*s*((math.pow((T+460),4)/100.)-(math.pow((T1+460),4)/100.))
hr= q/(A1*(T-T1))	#Coefficient of convection
hc= 0.27*math.pow(((T-T1)/D),0.25) #free convection coefficient
#RESULTS
print '%s %.2f' %('Heat loss (Btu/hr) = ',q)
print '%s %.2f' %(' \n hr (Btu/sq ft hr F) = ',hr)
print '%s %.2f' %(' \n hc (Btu/sq ft hr F) = ',hc)
raw_input('press enter key to exit')

Heat loss (Btu/hr) =  140.52

hr (Btu/sq ft hr F) =  1.18

hc (Btu/sq ft hr F) =  1.24
press enter key to exit

Out:
''

## Exa 19.8¶

In :
#A peep hole in the furnace wall of 13.5 in thickness is 8 in. square. the inside
#of the furnace is at uniform temp. of 1500 F and the external surroundings are at
#120 F. estimate the rate of heat loss by radiation through the open peep hole?
import math
#initialisation of variables
T= 120. 					#F
T1= 1500. 					#F
A= 64./144.
F= 0.86
Fe= 1
s= 0.173 					#BTU s^-1 in^-2 R^-4
#CALCULATIONS
q= (A/math.pow(10,6))*F*Fe*s*((math.pow((T1+460),4)/100.)-(math.pow((T+460),4)/100.))
#RESULTS
print '%s %.2f' %('Heat loss (Btu/hr) = ',q)
raw_input('press enter key to exit')

Heat loss (Btu/hr) =  9683.74
press enter key to exit

Out:
''