# Chapter 3 : Heat transfer coefficient¶

## Example 3.1 Page No : 53¶

In :
# Variables
di = 0.06        			#m,initial diameter of iceball
T1 = 30.          			#C, room temp.
T2 = 0.           			#ice ball temp.
h = 11.4         			#W/m**2 C, heat transfer coefficient
x = 40.           			#% for reduction
rho = 929.        			#kg/m**3, density of ice
Lv = 3.35*10**5   			#j/kg, latent heat of fusion

# Calculations
# m = 4/3*math.pi*r**3      			#kg,mass of ice ball
#rate of melting = -dm/dt
#rate of heat adsorption  = -4*math.pi*r**2*rho*dr/dt*lamda
#at initial time t = 0
C1 = di/2        			#consmath.tant of integration
#if the volume of the ball is reduced by 40% of the original volume
r = ((1-x/100)*(di/2)**3)**(1./3)
#time required for melting umath.sing eq. 1
t = (di/2-r)/(h*(T1-T2)/(rho*Lv))

# Results
print "The time required for melting the ice is %.0f s"%(t)

# note : rounding off error.

The time required for melting the ice is 4274 s


## Example 3.2 Page No : 54¶

In :
import math
#calculate the time required for the  heating coil.

# Variables
P = 1.*10**3           			#W, electrical heating capacity
V = 220.              			#V, applied voltage
d = 0.574*10**-3      			#m, diameter of wire
R = 4.167            			#ohm, electrical resistance
Tr = 21.              			#C, room temp.
h = 100.              			#W/m**2 C, heat transfer coefficient
rho = 8920.           			#kg/m**3, density of wire
cp = 384.             			#j/kg C, specific heat of wire
percent = 63.         			#%, percent of the steady state

#Calculation
R_ = V**2/P           			#ohm, total electrical resistance
l = R_/R             			#m, length of wire
A = math.pi*d*l          			#m**2, area of wire
Tf = P/(h*A)+Tr      			#final temp.
dtf = Tf-Tr          			#C. steady state temp. rise
#temp. of wire after 63% rise
T = Tr+(percent/100)*dtf
#rate of heat accumulation on the wire
#d/dt(m*cp*T)                       (1)
#rate of heat loss
#h*A*(T-Tr).........................(2)
#heat balance eq.       (1) = (2)
m = math.pi*d**2*l*rho/4  			#kg. mass of wire
#integrating heat balance eq.

def f6(T):
return 1/((P/(m*cp))-((h*A)/(m*cp))*(T-Tr))

# Results
print "The time required for the heating coil is %.1f s"%(t)

The time required for the heating coil is 4.9 s


## Example 3.3 Page No : 56¶

In :
# Variables
t = 0.2           			#m, thickness of wall
k = 1.163         			#W/m C, thermal conductivity of material
Ta = 30           			#C, ambient temp

# Calculations and Results
#(a) at x = 0.2   let T = T1 at x = x1
x1 = 0.2
T1 = 250-2750*x1**2
#let     D = dT/dx
D = -5500*0.2     			#C/m, at x = 0.2
h = -k*D/(T1-Ta)
print " the heat transfer coefficient is %.2f W/m**2 C "%(h)

#(b)at other surface of wall, x = 0 = x2 (say)
x2 = 0
a = -5500*0
print "So there is no heat flow at other surface of the wall "

#(c)
A = 1            			#m**2, area
Vw = A*x1        			#m**3, volume of wall
HL = h*(T1-Ta)   			#W, heat loss from unit area
Vav = HL/x1
print "average volumetric rate of heat generation is %.0f W/m**3"%(Vav)

 the heat transfer coefficient is 11.63 W/m**2 C
So there is no heat flow at other surface of the wall
average volumetric rate of heat generation is 6396 W/m**3


## Example 3.4 Page No : 61¶

In :
from scipy.optimize import fsolve
import math
# Variables
id_ = 97.*10**-3         			#m,internal diameter of steam pipe
od = 114.*10**-3        			#m,outer diameter of steam pipe
pr = 30.               			#bar, absolute pressure os saturated steam
Ti = 234.                			#C, temp. at 30 bar absolute pressure
Ts = 55.               			#C, skin temp.
To = 30.               			#C, ambient temp.
kc = 0.1              			#W/m C, thermal conductivity of wool
kw = 43.               			#W/m C, thermal conductivity of pipe
h = 8.                 			#W/m**2 C, external air film coefficient
L = 1.                 			#m, assume length

#Calculation
ri = id_/2             			#m,
r1 = (114.*10**-3)/2        			#m,outer radius of steam pipe

#thermal resistance of insulation
#Ri = math.log(ro/r1)/(2*math.pi*L*kc)
#Thermal resistance of pipe wall
Rp = math.log(r1/ri)/(2*math.pi*L*kw)
#RT = Ri+Rp
DF = Ti-Ts            			#C, driving force
#At steady state the rate of heat flow through the insulation
# and the outer air film are equ

#by trial and error method :
def f(ro):
return (Ti-Ts)/(math.log(ro/r1)/kc+math.log(r1/ri)/kw)-(h*ro*(Ts-To))
ro = fsolve(f,0.1)
th = ro-r1        			#m, required thickness of insulation
Q = 2*math.pi*ro*h*L*(Ts-To)

# Results
print "The rate of heat loss is %.1f W"%(Q)

The rate of heat loss is 150.9 W


## Example 3.5 Page No : 62¶

In :
import math
# Variables
w1 = 8.            			#%, solubility of alcohol
w2 = 92.           			#%, solubility of water
k1 = 0.155        			#W/m C, thermal conductivity of alcohol
k2 = 0.67         			#W/m C thermal conductivity of water
ka = 0.0263       			#W/m C thermal conductivity of air
kw = 45.           			#W/m Cthermal conductivity of pipe wall
ki = 0.068        			#W/m C , thermal cond. of glass
id_ = 53.*10**-3     			#m, internal diameter of pipe
od = 60.*10**-3     			#m, outer  diameter of pipe
t = 0.04          			#m, thickness of insulation
hi = 800.          			#W/m**2 C, liquid film coefficient
ho = 10.           			#W/m**2 C, air film coefficient
L = 1.             			#m, length of pipe
T1 = 75.           			#C, initial temp.
T2 = 28.           			#C, ambient air temp.

# Calculations and Results
#(a)
km = (w1/100)*k1+(w2/100)*k2-0.72*(w1/100)*(w2/100)*(-(k1-k2))
deli = km/hi     			#m, effective thickness of liquid film
delo = ka/ho     			#m, effective thickness of air film
print "effective thickness of air  is %.2f mm"%(deli*10**3)
print "effective thickness of liquid films is %.1f mm."%(delo*10**3)

#(b)
Ai = 2*math.pi*id_/2*L      			#m**2, inside area
ri = id_/2                    			#m,inside radius of pipe
r_ = od/2                      			#m, outside radius of pipe
ro = r_+t              	        		#m, outer radius of insulation
Ao = 2*math.pi*ro*L        		    	#m**2, outer area
#from eq. 3.11, overall heat transfer coefficient
Ui = 1/(1/hi+(Ai*math.log(r_/ri))/(2*math.pi*L*kw)+(Ai*math.log(ro/r_))/(2*math.pi*L*ki)+Ai/(Ao*ho))
print "the overall heat transfer coefficient based on i.d of pipe is %.3f W/m**2 C"%(Ui)

#(c)
#frim eq. 3.14
Uo = Ui*Ai/Ao
print "the overall heat transfer coefficient based on od of pipe is %.3f W/m**2 C"%(Uo)

#(d)
R = 1/(Ui*Ai)          			#C/W, total heat transfer resistance
Rair = 1/(Ao*ho)       			#C/W, heat transfer resistance of air film
p = Rair/R
print "the percentage of total resistance  offered by air film. is %.2f percent"%(p*100)

#(e)
Q = Ui*Ai*(T1-T2)
print "Rate of heat loss is %.1f W"%(Q)

#(f)
Ts = Uo*Ao*(T1-T2)/(ho*Ao)+T2
print "insulation skin temp.is %.1f C"%(Ts)

effective thickness of air  is 0.75 mm
effective thickness of liquid films is 2.6 mm.
the overall heat transfer coefficient based on i.d of pipe is 2.707 W/m**2 C
the overall heat transfer coefficient based on od of pipe is 1.025 W/m**2 C
the percentage of total resistance  offered by air film. is 10.25 percent
Rate of heat loss is 21.2 W
insulation skin temp.is 32.8 C


## Example 3.6 Page No : 64¶

In :
import math
# Variables
id_ = 1.5                			#m, internal diameter of math.tank
h = 2.5                 			#m, height of math.tank
t1 = 0.006              			#m, thickness of wall
t2 = 0.04               			#m, thickness of insulation
Ta = 25.                 			#C, ambient temp.
T1 = 80.                 			#C, outlet temp. of liquid
cp = 2000.               			#j/kg C, specific heat of liquid
FR = 700./3600           			#KG/s, Liquid flow rate

# Calculations and Results
ri = id_/2+t1            			#m, inner radius of insulation
ro = ri+t2              			#m, outer radius of insulation
ki = 0.05               			#W/m C, thermal conductivity of insulation
hc = 4                  			#W/m**2 C, heat transfer coefficient at cylindrical surface
ht = 5.5                			#W/m**2 C, heat transfer coefficient at flat surface
l = h+t1+t2             			#m, height of the top of insulation
#fromm eq. 3.10
#heat transfer resistance of cylindrical wall
Rc = math.log(ro/ri)/(2*math.pi*l*ki)+1/(2*math.pi*ro*l*hc)
#heat transfer resistance of flat insulated top surface
Ri = (1/(math.pi*ro**2))*((ro-ri)/ki+1/ht)
tdf = T1-Ta             			#C, temp. driving force
Q = tdf/Rc + tdf/Ri       			#W, total rate of heat loss
Tt = Q/(FR*cp)+T1        			#C, inlet temp. of liquid
print "Inlet liquid temp. should be %.0f C "%(Tt)
Q1 = tdf/Ri   			#W, rate of heat loss from flat surface
T1 = Q1/(math.pi*ro**2*ht)+Ta
print " the insulation skin temp. at the flat top surface is %.0f C "%(T1)
#similarly
T2 = 38
print "similarly  the insulation skin temp at cylindrical surface is %.0f C"%(T2)

Inlet liquid temp. should be 82 C
the insulation skin temp. at the flat top surface is 35 C
similarly  the insulation skin temp at cylindrical surface is 38 C


## Example 3.7 Page No : 66¶

In :
import math
# Variables
id_ = 2.5*10**-2              			#m, internal diameter of glass tube
t = 0.3*10**-2               			#m, thickness of wall
l = 2.5                     			#m, length of nichrome wire
L = 0.12                    			#m, length of steel covered with heating coil
Re = 16.7                   			#ohm, electrical resistance
ti = 2.5*10**-2              			#m, thickness of layer of insulation
kg = 1.4                    			#W/m C, thermal conductivity of glass
ki = 0.041                  			#W/m C, thermal conductivity of insulation
T1 = 91.                     			#C, boiling temp. of liquid
T2 = 27.                     			#C, ambient temp.
ho = 5.8                    			#W/m **2 C outside air film coefficient
V = 90.                      			#V,  voltage

#Calculation
Rc = Re*l                   			#ohm, resistance of heating coil
Q = V**2/Rc                  			#W, rate of heat generation
ri = id_/2                   			#m, inner radius of glass tube
r_ = ri+t                   			#m, outer radius of glass tube
ro = r_+ti                   			#m,outer radius of insulation
#heat transfer resistance of glass wall
Rg = math.log(r_/ri)/(2*math.pi*L*kg)
#combined resistance of insulation and outer air film
Rt = math.log(ro/r_)/(2*math.pi*L*ki)+1/(2*math.pi*ro*L*ho)
#Rate of heat input to the boiling liquid in steel = Q1 = (Ts-T1)/Rg
#Rate of heat loss through insulation ,Q2 = (Ts-To)/(Rt)
#Q1+Q2 = Q
Ts = (Q+ T1/Rg +T2/Rt)/(1/Rg +1/Rt)
Q1 = (Ts-T1)/Rg
Q2 = Q-Q1

# Results
print "the heat imput to the boiling.is %.1f W"%(Q1)

the heat imput to the boiling.is 191.2 W


## Example 3.8 Page No : 68¶

In :
import math

# Variables
ri = 1.3*10**-3            			#m, radius of 10 gauge wire
t = 1.3*10**-3             			#m, thickness of rubber insulation
Ti = 90.                   			#C, temp. 0f insulation
To = 30.                   			#C, ambient temp.
h = 15.                    			#W/m**2 C, air film coefficient
km = 380.                  			#W/m C, thermal cond. of copper
kc = 0.14                 			#W/m C, thermal cond. of rubber(insulation)
Rc = 0.422/100            			#ohm/m, eletrical resistance of copper wire

# Calculations and Results
Tcmax = 90.                			#X, the maximum temp. in insulation
ro = ri+t                 			#m, outside radius of 10 gauge wire
Sv = ((Tcmax-To)*(2*kc/ri**2))/(math.log(ro/ri)+kc/(h*ro))
I = (math.pi*ri**2*Sv/Rc)**0.5      			#A, Current strength
print "maximum allowable current is %.2f A"%(I)

#(b) at r = 0
Tm = To+(ri**2*Sv/2)*(1/km+(math.log(ro/ri))/kc+1/(h*ro))
print "remp. at the centre of wire is %.3f C"%(Tm)

#at r = ro
Tc = 30+(ri**2*Sv/(2*kc))*(kc/(h*ro))
print "The temprature at the outer surface of insulation is %.1f C"%(Tc)

maximum allowable current is 54.04 A
remp. at the centre of wire is 90.005 C
The temprature at the outer surface of insulation is 80.3 C


## Example 3.9 Page No : 72¶

In :
# Variables
tA = 0.25          			#m, thickness of slab A
tB = 0.1           			#m, thickness of slab B
tC = 0.15          			#m, thickness of slab C
kA = 15.            			#W/m C, thermal comductivity of slab A
kB = 10.            			#W/m C, thermal comductivity of slab B
kC = 30.            			#W/m C, thermal comductivity of slab C
#Temprature distribution in slab A
T1 = 40.            			#C, fluid temp.
T2 = 35.            			#C, medium temp.

# Calculations and Results
#(a)
x1 = tB
TA1 = 90.+4500*x1-11000*x1**2
#similarly at the right surface
x2 = tA+tB
TA2 = 90.+4500*x2-11000*x2**2
#let dTA/dx = D
D = 0              			#for maximum temp.
x3 = 4500./22000
TAmax = 90.+4500*x3-11000*x3**2
print "At x = 0.1 the temp. at the surface of slab A  is %.0f C"%(TA1)
print "At x = 0.35 the temp. at the surface of slab A  is %.0f C"%(TA2)
print " the maximum Temp. in A occurs at  %.4f m"%(x3)
print " the maximum Temp. in A is %.1f TAmax "%(TAmax)

#(b)
#At the interface 2
D1 = 4500-2.*11000*x1       			#C/W, D1 = dTA/dx, at x = 0.1
#At the interface 3
D2 = 4500-2.*11000*x2       			#D12 = dTA/dx, at x = 0.35
#Temprature gradient in slab B and C
#by umath.sing the continuity of heat flux at interface (2)
D3 = -kA*D1/(-kB)          			#D3 = dTB/dx,  at x = 0.1
#at interface (1)
D4 = D3                    			#D4 = dTB/dx  at x = 0
#similarly
D5 = -1600.                 			#C/W, dTB/dx, x = 0.35
D6 = D5                    			#at interface 4
print "temp. gradient at interface 2 of the slabs A is %.0f C/W"%(D1)
print "temp. gradient at interface 3 of the slabs A is %.0f C/W"%(D2)
print "temp. gradient at interface 2 of the slabs B is %.0f C/W"%(D3)
print "temp. gradient at interface 1 of the slabs B is %.0f C/W"%(D4)
print "temp. gradient at interface 3 of the slabs C is %.0f C/W"%(D5)
print "temp. gradient at interface 4 of the slabs C is %.0f C/W"%(D6)

#(c)
#from D3 = 3450  and TB = beeta1*x+beeta2
beeta1 = 3450.
beeta2 = 85.
x = 0.
TB = beeta1*x+beeta2
#similary
TC = 877.5-1600*x
h1 = -kB*D4/(T1-TB)
#similarly
h2 = 1129.
print "The  heat transfer coefficient at one surface of solid fluid interface is %.1f W/m**2 C"%(h1)
print "The  heat transfer coefficient at other surface of solid fluid interface is %.0f W/m**2 C"%(h2)

At x = 0.1 the temp. at the surface of slab A  is 430 C
At x = 0.35 the temp. at the surface of slab A  is 318 C
the maximum Temp. in A occurs at  0.2045 m
the maximum Temp. in A is 550.2 TAmax
temp. gradient at interface 2 of the slabs A is 2300 C/W
temp. gradient at interface 3 of the slabs A is -3200 C/W
temp. gradient at interface 2 of the slabs B is 3450 C/W
temp. gradient at interface 1 of the slabs B is 3450 C/W
temp. gradient at interface 3 of the slabs C is -1600 C/W
temp. gradient at interface 4 of the slabs C is -1600 C/W
The  heat transfer coefficient at one surface of solid fluid interface is 766.7 W/m**2 C
The  heat transfer coefficient at other surface of solid fluid interface is 1129 W/m**2 C


## Example 3.10 Page No : 79¶

In :
import math

# Variables
id_ = 78.*10**-3       			#m, actual internal dia of pipe
tw = 5.5*10**-3      			#m, wall thickness
nl = 8.              			#no. of longitudinal fins
tf = 1.5*10**-3      			#m, thickness of fin
w = 30.*10**-3        			#m,breadth of fin
kf = 45.             			#W/m C, thermal conductivity of fin
Tw = 150.            			#C, wall temp.
To = 28.             			#C, ambient temp.
h = 75.              			#W/m**2C, surface heat transfer coefficient

#Calculation
#from eq. 3.27
e = math.sqrt(2*h/(kf*tf))
n = (1./(e*w))*math.tanh(e*w)  			#efficiency of fin
L = 1.              			#m, length of fin
Af = 2.*L*w         			#m**2, area of  math.single fin
Atf = nl*Af          			#m**2 total area of fin
Qmax = h*Atf*(Tw-To)   			#W, maximum rate of heat transfer
Qa = n*Qmax           			#W, actual rate of heat transfer
Afw = L*tf            			#m**2, area of contact of fin with pipe wall
Atfw = Afw*nl         			#m**2 , area of contact of all fin with pipe wall
ro = id_/2+tw          			#m, outer  pipe radius
A = 2*math.pi*L*ro        			#m**2  area per meter
Afree = A-Atfw        			#m**2, free outside area of finned pipe
#Rate of heat transfer from free area of pipe wall
Q1 = h*Afree*(Tw-To)  			#W,
#total rate of hewat gtransfer from finned pipe
Qtotal = Qa+Q1        			#W
#Rate of heat transfer fromm unfinned pipe
Q2 = h*A*(Tw-To)
per = (Qtotal-Q2)/Q2

# Results
print "the percentage increase in the rate of heat transfer is %.1f percent "%(per*100)

the percentage increase in the rate of heat transfer is 103.6 percent


## Example 3.11 Page No : 80¶

In :
import math

# Variables
id_ = 90.*10**-2       			#m, internal diameter of steel
od = 110.*10**-2      			#m, outer diameter of steel
Ti = 180.            			#C, inside temp. of steel
To = 170.            			#C, outside temp. of steel
k = 37.             			#W/m C, thermal conductivity of alloy
Q = 5.18*10**3       			#W, Rate of heat loss

# Calculations and Results
ri = id_/2           			#m, inside radius of shell
ro = od/2           			#m, outside radius of shell
r_ = 0.5            			#m, boundary between the layers
L = 1               			#m, length of shell
#Rate of heat transfer in the absence of contact resistance
Q1 = 2*math.pi*L*k*(Ti-To)/(math.log(ro/ri))
print "Rate of heat transfer in the absence of contact resistance is %.3f KW"%(Q1/1000)
print "The actual rate of heat loss is 5.18kW is much less than this value\
. So there is a thermal contact resistance at the interface between the layers "

#(b)
Ri = (math.log(r_/ri)/(2*math.pi*L*k))  			#C/W, resistance of inner layer
Ro = (math.log(ro/r_)/(2*math.pi*L*k))  			#C/W, resistance of outer layer
Rc = ((Ti-To)/(Q))-(Ri+Ro)     			#C/W, contact resistance
print "The contact resistance is %f C/W "%(Rc)
Ac = 2*math.pi*L*r_                			#m**2, area of contact surface of shell
hc = 1/(Ac*Rc)                 			    #W/m**2 c, contact heat transfer coefficient
print "contact heat transfer coefficient is %.1f W/m**2 C "%(hc)

#(c)
dt = Q/(hc*Ac)
print "The temprature jump is %.1f C"%(dt)

Rate of heat transfer in the absence of contact resistance is 11.585 KW
The actual rate of heat loss is 5.18kW is much less than this value. So there is a thermal contact resistance at the interface between the layers
The contact resistance is 0.001067 C/W
contact heat transfer coefficient is 298.2 W/m**2 C
The temprature jump is 5.5 C


## Example 3.12 Page No : 84¶

In :
# Variables
d = 5.2*10**-3       			#m, diameter of copper wire
ri = d/2            			#inner radius of insulation
kc = 0.43           			#W/m C, thermal conductivity of PVC
Tw = 60.             			#C, temp. 0f wire
h = 11.35           			#W/m**2 C, film coefficient
To = 21.             			#C, ambient temp.

#calculation
Ro = kc/h           			#m,critical outer radius of insulation
t = Ro-ri

# Results
print "the critical thickness is %.2f mm"%(t*10**3)

the critical thickness is 35.29 mm


## Example 3.13 Page No : 85¶

In :
# calculate the critical  insulation thickness.

# Variables
d = 15.*10**-2        			#m, length of steam main
t = 10.*10**-2        			#m, thickness  of insulation
ki = 0.035          			#W/m C, thermal conductivity of insulation
h = 10.              			#W/m**2 C, heat transfer coefficient

#calculation
#from eq. 3.29
ro = ki/h

# Results
print "ro =  %.1f cm "%(ro*10**3)
print "Radius of bare pipe is larger than outer radius of insulation  So critical \
insulation thickness does not exist "

ro =  3.5 cm
Radius of bare pipe is larger than outer radius of insulation  So critical   insulation thickness does not exist


## Example 3.14 Page No : 87¶

In :
from scipy.optimize import fsolve
import math

# Variables
Ti = 172.           			#C, saturation temp.
To = 20.            			#C, ambient temp.
Cs = 700.           			#per ton, math.cost of steam
Lv = 487.           			#kcal/kg, latent heat of steam
ho = 10.32           			#kcal/h m**2 C, outer heat transfer coefficient
kc = 0.031             			#W/m C, thermal conductivity of insulation
n = 5.              			#yr, service life of insulation
i = 0.18            			#Re/(yr)(Re), interest rate

#Calculation
di = 0.168           			#m, inner diameter of insulation
#Cost of insulation
Ci = 17360.-(1.91*10**4)*di         			#Rs/m**3
Ch = Cs/(1000*Lv)                 			#Rs/cal, math.cost of heat energy in steam
sm = 1./(1+i)+1/(1+i)**2+1/(1+i)**3+1/(1+i)**4+1/(1+i)**n
#from eq. 3.33
ri = di/2         			#m  inner radius of insulation
L = 1             			#m, length of pipe
#Pt = Ch*sm*2*math.pi*ri*L*( 1/(((ri/kc)*('math.log(ro/ri)'))+ri/(ho*ro)))*7.2*10**3*(Ti-To)+math.pi*(ro**2-ri**2)*L*Ci
#On differentiating , dpt/dro = -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2
def f(ro):
return -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2+98960*ro
ro = fsolve(f,0.1)
t = ro-ri

# Results
print "The optimum insulation thickness is %.0f mm"%(t*1000)

The optimum insulation thickness is 71 mm