# Chapter 9:Heat Transfer¶

## Example no:9.1,Page no:386¶

In [11]:
#Variable declaration
M_dot1=20  #rate of mass to be cooled
M_dot2=25  #rate of cooling water
Cp=4.18e3  #Heat capacity
T1=360  #Initial temp.
T2=340  #Final temp.
theta_1=300  #Temperature of cooing water entering
U=2e3  #Overall heat transfer coefficient

#Calculation
import sympy
import math
x=sympy.Symbol('x')
theta_2=sympy.solve(Q-(M_dot2*Cp*(x-300)))
theta_m1=((T1-theta_2[0])-(T2-theta_1))/(math.log((T1-theta_2[0])/(T2-theta_1)))
A1=Q/(U*theta_m1)
theta_m2=((T1-theta_1)-(T2-theta_2[0]))/(math.log((T1-theta_1)/(T2-theta_2[0])))
A2=Q/(U*theta_m2)

#Result
print"(a).In counter flow,The surface area required",round(A1,2),"m**2"
print"(b).In cocurrent flow,The surface area required",round(A2,2),"m**2"

(a).In counter flow,The surface area required 19.92 m**2
(b).In cocurrent flow,The surface area required 21.28 m**2


## Example no:9.2,Page no:390¶

In [12]:
#Variable declaration
dx=0.5  #Thickness of wall
T1=400  #Temperartue of inner surface
T2=300  #Temperature of outer surface
K=0.7  #Thermal conductivity
A=1  #Area of heat transfer

#Calculation
#From equation 9.12:
Q=K*A*(T1-T2)/dx

#Result
print"The heat loss per square metre of surface =",Q,"W/m**2"

The heat loss per square metre of surface = 140.0 W/m**2


## Example no:9.3,Page no:391¶

In [16]:
#Variable declaration
dx1=0.20  #thickness of firebrick
dx2=0.10  #thickness of insulating brick
dx3=0.20  #thickness of building brick
k1=1.4  #Thermal conductivity of firebrick
k2=0.21  #Thermal conductivity of insulating brick
k3=0.7  #Thermal conductivity of building brick
T1=1200  #Temperature at junction 1
T4=330  #Temperature at junction 4

#Calculation
Q=(T1-T4)/((dx1/k1)+(dx2/k2)+(dx3/k3))
#The ratio (Temperature drop over firebrick)/(Total temperature drop)
R=(dx1/k1)/((dx1/k1)+(dx2/k2)+(dx3/k3))
#Temperature drop over firebrick
dT=(T1-T4)*R
T2=(T1-dT)

#Result
print"\n Heat loss per unit area =%d"%Q,"W/m**2"
print"\n Temperature drop over firebrick =%d"%dT,"K"
print"\n The temperature at the firebrick-insulating brick interface =",round(T2),"K"

 Heat loss per unit area =961 W/m**2

Temperature drop over firebrick =137 K

The temperature at the firebrick-insulating brick interface = 1063.0 K


## Example no:9.4,Page no:398¶

In [19]:
#Variable declaration
T=295  #initial temperature of surfaces
T2f=375  #Final temperature of far surface
dT1=900  #Temperature of near face raised

#Calculation
import sympy
R=(T2f-T)/(2*(dT1-T))  #ratio of theta to twice of theta dash
x=sympy.Symbol('x')
t=sympy.solve((1.30**2*x)-346**2)

#Result
print"Time taken to rise from 295 to 375 K =",round(t[0]/3600,1),"h"

Time taken to rise from 295 to 375 K = 19.7 h


## Example no:9.5,Page no:400¶

In [22]:
#Variable declaration
T=295  #initial temperature of surfaces
T2f=375  #Final temperature of far surface
dT1=900  #Temperature of near face raised
DH=4.2e-7  #Thermal diffusivity
#The development of the temperature profile is shown in Figure 9.12
#The problem will be solved by taking relatively large intervals for dx.
#Choosing dx = 50 mm, the construction shown in Figure 9.12
dx=50e-3
#Because the second face is perfectly insulated, the temperature gradient must
# be zero at this point.
#It is seen that the temperature is
#less than 375 K after time 23dt and greater than 375 K after time 25dt
#Thus:
#t=24*dt
#from equation 9.43
dt=dx**2/(2*DH)
t=24*dt

#Result
print"\n The time taken to rise from 295 to 375 K =",round(t/3600.0,1),"h"

 The time taken to rise from 295 to 375 K = 19.8 h


## Example no:9.6,Page no:403¶

In [28]:
#Variable declaration
d=25e-3  #Diameter of copper sphere
l=25e-3  #Side length of a copper cube
h=75  #External heat transfer coefficient
rho_cu=8950  #Density of copper at mean temperature
Cp=0.38e3  #Heat capacity of copper at mean temperature
k=385  # Thermal conductivity of copper at mean temperature
Tf=923  #Temperature of the furnace
Ta=368  #Temperature at which they are annealed
t=5*60  # time taken

#Calculation
import math
import sympy
V_Ae_S=(d/6.0)  #V/Ae tor the sphere
V_Ae_C=(l/6.0)  #V/Ae tor the cube
Bi=h*(V_Ae_S)/k
#The use of a lumped capacity method is therefore justified
tao=rho_cu*Cp*V_Ae_S/h
x=sympy.Symbol('x')
T=sympy.solve(((x-Ta)/(Tf-Ta))-math.exp(-t/tao))

#Result
print"Temperature of the sphere and of the cube at the end of 5 minutes =%d"%(T[0]-273),"C"

Temperature of the sphere and of the cube at the end of 5 minutes =208 C


## Example no:9.7,Page no:409¶

In [32]:
#Variable declaration
k=2.5  #Thermal conductivity
DH=2e-7 #Thermal diffusivity of the surrounding fluid
h=100  #External heat transfer coefficient
To=293  #Initial Temperature
T_dash=373  #Oven Temperture
Tc=353  #temperature throughout the whole of the sheet reaches a minimum
l=10e-3  #thickness of sheet
L=l/2

#Calculation
#For the given process, the Biot number
Bi=h*L/k
Bi_1=1/Bi
lim_val=(T_dash-Tc)/(T_dash-To)
#From Figure 9.17, the Fourier number
Fo=7.7
t=Fo*L**2/DH

#Result
print"The minimum time for which the sheet must be heated =%d"%(t/60.0),"min"

The minimum time for which the sheet must be heated =16 min


## Example no:9.8,Page no:413¶

In [37]:
#Variable declaration
l=5 #Length of the channel of uranium reactor
Q=.25e6  #Heat release from uranium reactor
k=33  #Thermal conductivity of the uranium

#Calculation
import math
Q_m=Q/l  #Heat release rate
#Thus, from equation 9.52:
dT=Q_m/(4*math.pi*k)

#Result
print"The temperature difference between the surface and the centre of the uranium element =",round(dT),"K"

The temperature difference between the surface and the centre of the uranium element = 121.0 K


## Example no:9.9,Page no:429¶

In [39]:
#Variable declaration
Cp=2380 #specific heat capacity of nitrobenzene
k=0.15
Meu=0.70e-3 #Viscosity of nitrobenzene
d_i=15e-3 #internal diameter of tube
d_o=19e-3 #external diameter of the tube
d_s=0.44 #shell diameter
b_s=0.150 #baffle spacing
p=0.025 #pitch
c=0.006 #clearance
#(i)Tube side coefficient
h_i=1000 #based on inside area

#Calculation
import math
h_io=1000*d_i/d_o #based on outside area
#(ii) Shell side coefficient.
A=d_s*b_s*c/p #Area for flow
G_s_=4/A
#Taking Meu/Meu_s=1 in equation 9.91
d_e=4*((25e-3**2-(math.pi*d_o**2/4))/(math.pi/d_o))
h_o=0.36*k/d_e*(d_e*G_s_/Meu)**0.55*(Cp*Meu/k)**0.33
#(iii) Overall coefficient
#The math.logarithmic mean temperature difference is given by:
Tm=(((400.0-345.0)-(315.0-305.0))/math.log((400.0-345.0)/(315.0-305.0)))
#The corrected mean temperature difference is
Tm_c=Tm*0.8
Q=4*Cp*(400.0-315.0)
#The surface area of each tube
A_t=0.0598
U_o=Q/(2*166*5*A_t*Tm_c)
#(iv) Scale resistance.
R_d=(1/U_o)-(1.0/750.0)-(1.0/1000.0)

#Result
print"\n Value of scale resistance that could be allowed =",round(R_d,5),"m**2 K/W"

 Value of scale resistance that could be allowed = 0.00026 m**2 K/W


## Example no:9.10,Page no:432¶

In [47]:
#Variable declaration
G=15.0 #Mass flow rate of benzene
d_s=1.0  #Internal diameter of Heat Exchanger
l=5.0  #Length of tubes
od=19e-3  #Outer diameter of tubes
C=6e-3  #Clearance
l_b=0.25  #Baffle spacing
Meu=.5e-3
Y=25e-3  #dimension of square pitch
N=19.0  #no. of Baffles

#Calculation
import math
As=d_s*l_b*C/Y  #Cross-flow area
G_dash_s=G/As  #Mass flow
d_e=4*(Y**2-(math.pi*od**2.0/4.0))/(math.pi*od) #Equivalent Diameter
Re=G_dash_s*d_e/Meu
#From Figure 9.29:
f_dash=0.280
rho_b=881 #density of benzene
DPf=f_dash*G_dash_s**2*(N+1)*d_s/(2*rho_b*d_e)

#Result
print"The pressure drop over the tube bundle =%d"%DPf,"N/m**2=",round(DPf/(rho_b*9.81)),"m of Benzene"
print"NOTE:Approx value of pressure drop is given in book"

The pressure drop over the tube bundle =8680 N/m**2= 1.0 m of Benzene
NOTE:Approx value of pressure drop is given in book


## Example no:9.11,Page no:437¶

In [48]:
#Variable declaration
d=0.15  #Diameter of pipe
Ts=400  #Surface temperature
Ta=294  #Air temperture
k=0.0310  #Thermal conductivity ---Table 6, Appendix A1

#Calculation
X=36/k**4
#From Equation 9.102:
GrPr=X*(Ts-Ta)*d**3
#From Table 9.5:
n=0.25
C_dd=1.32
#Thus, in Equation 9.104:
h=C_dd*(Ts-Ta)**n*d**(3*n-1)

#result
print"\n The heat transfer coefficient =",round(h,2),"W/m**2 K"

 The heat transfer coefficient = 6.81 W/m**2 K


## Example no:9.12,Page no:439¶

In [52]:
#Variable declaration
lamda=1e-6 #Wavelength
E_l_b=1e9  #Emissive power at given lambda
C2=1.439e-2
C1=3.742e-16

#Calculation
import math
T=C2/lamda/math.log(C1/(E_l_b*lamda**5))
#With an error of +2 per cent, the correct value is given by:
E_l_b_n=(100-2)*E_l_b/100
#In equation 9.108:1
T_n=C2/lamda/math.log(C1/(E_l_b_n*lamda**5))

#Result
print"\n The temperature of surface =%d"%T,"K"
print"\n The temperature of surface with +2 per cent error=",round(T_n),"K"

 The temperature of surface =1121 K

The temperature of surface with +2 per cent error= 1120.0 K


## Example no:9.13,Page no:441¶

In [55]:
#Variable declaration
d=10e-3  #Diameter of carbide elements
l=0.5  #Length of carbide elements
Ts=1750  #Maximun surface temperature of carbide
P=500e3  #Thermal power output required
sigma=5.67e-8

#Calculation
Eb=sigma*Ts**4
A=math.pi*d*l
P1=Eb*A #Power dissipated by one element
n=P/P1  #Number of elements required

#Result
print"Number of elements required =%d"%round(n)

Number of elements required =60


## Example no:9.14,Page no:444¶

In [59]:
#Variable declaration
A=10.0  #Area of the surface
T1=1500.0  #First Temperature
T2=1600.0  #Second Temperatue
sigma=5.67e-8

#Calculation
E=P_r/A  #The emissive Power
#From equation 9,118:
e=E/(sigma*T1**4)
E2=e*sigma*T2**4

#Result
print"\n Emissivity when T=1500 K =",round(e,3)
print"\n The Emissive power when T=1600 K =",round(E2*1e-2),"kW"

 Emissivity when T=1500 K = 0.348

The Emissive power when T=1600 K = 1295.0 kW


## Example no:9.15,Page no:448¶

In [64]:
import math
#Variable declaration
A1=2.0  #Area of rectangle(Surface 1)
A2=math.pi*1**2.0/4.0  #Area of disc (Surface 2)
T1=1500.0  #Temperature of Surface 1
T2=750.0  #Temperature of Surface 2
F12=0.25  #View factor
sigma=5.67e-8

#Calculation
#From equation 9. 1 26:
F21=A1*F12/A2
Q12=sigma*A1*F12*(T1**4-T2**4)

#Result
print"View factor, F12 =",round(F21,3)
print"NOTE:Calculation mistake in book"

View factor, F12 = 0.637
The net radiation transfer = 134.6 kW
NOTE:Calculation mistake in book


## Example no:9.16,Page no:449¶

In [68]:
#Variable declaration
X=4.0 #width of horizontal plate and length vertical plate
Y=6.0 #length of horizontal plate
Z=3.0 #height of verical plate

#Calculation
W=Y/X
H=Z/X
A1=Z*X #Area of plate 1
A2=X*Y #Area of plate 2
F12=0.12
#From equation 9.126:
F21=A1*F12/A2
#For the two spheres
r1=1.0  #Diameter of sphere 1
r2=2.0  #Diameter of sphere 2
F12b=1.0
F21b=(r1/r2)**2
F22b=1.0-F21b

#Result
print"For vertical plate"
print"View Factor, F12=",F12
print"View Factor, F21=",F21
print"\nFor sphere:"
print"View Factor, F21=",F21b
print"View Factor, F22=",F22b

For vertical plate
View Factor, F12= 0.12
View Factor, F21= 0.06

For sphere:
View Factor, F21= 0.25
View Factor, F22= 0.75


## Example no:9.17,Page no:454¶

In [73]:
#Variable declaration
ri_u=0.2 # Inner radius of the upper ring
ro_u=0.3 # Outer radius of the upper ring
ri_l=0.3 # Inner radius of the lower ring
ro_l=0.4 # Outer radius of the lower ring
F12_34=0.4
F12_4=0.22
F1_34=0.55
F14=0.30

#Calculation
A12_A2=ro_l**2/(ro_l**2-ri_l**2)
A1_A2=ro_u**2/(ro_l**2-ri_l**2)
F23=((A12_A2)*(F12_34-F12_4))+((A1_A2)*(F1_34-F14))

#Result
print"\n F23 =",round(F23,2)

 F23 = 0.73


## Example no:9.18,Page no:455¶

In [75]:
#Variable declaration
d=1  #Diameter of plate
r1=0.5
r4=r1  #Radius of the imaginary disc sealing the hemisphere
L=r1  #The distance between the plate and the bottom of the dome

#Calculation
import math
A1=math.pi*d**2/4.0  #Area of the plate
A2=2*math.pi*d**2/4.0  #Area of the underside of the Hemisphere
A4=math.pi*r4**2/4.0 #Area of an imaginary disc sealing the hemisphere and parallel
#to the plate
T1=750 #Temperature of the plate
T2=1200 #Temperature of hemispherical cone
T3=290 #Temperature of the surroundings
sigma=5.67e-8
R1=r1/L
R4=r4/L
S=1+(1+R4**2)/(R1**2)
F14=0.5*(S-(S**2-4*(r4/r1)**2)**0.5)
F12=F14
F13=1-F12
Q1=sigma*A1*F12*(T2**4-T1**4)+sigma*A1*F13*(T3**4-T1**4)

#Result
print"The net rate of heat transfer by radiation to the plate =",round(Q1*1e-3,1),"kW"

The net rate of heat transfer by radiation to the plate = 21.4 kW


## Example no:9.19,Page no:457¶

In [79]:
#Variable declaration
d=2  #Diameter of the cylinder
h=1  #Depth of insulated cylinder
T1=1500
T2=373
#From Figure 9.40ii, with i = 1, j = 2
r1=1
r2=1
L=1

#Calculation
A2=A1  #Under-Surface of the vessel
A_R=math.pi*d*h
#The view factor may also be obtained from Figure 9.39ii as follows:
R1=r1/L
R2=r2/L
S=1+(1+R2**2)/(R1**2)
F12=0.5*(S-(S**2-4*(r2/r1)**2)**0.5)
sigma=5.67e-8
#Using the summation rule
#F11=0
F1R=1-F12
F2R=F1R
Q2=(A1*F12+((1/(A1*F1R)+(1/(A2*F2R))))**-1)*sigma*(T1**4-T2**4)
#If the surroundings without insulation are surface 3 at
T3=290
F23=F2R
#from equation 9.135
Q2_d=sigma*A1*F12*(T1**4-T2**4)+sigma*A2*F23*(T3**4-T2**4)
red=(Q2-Q2_d)/Q2*100  #Percentage Reduction

#Result
print"The rate of radiant heat transfer to the vessel =%d"%(Q2*1e-3),"kW"
print"If the insulation were removed ,The rate of radiant heat transfer to the vessel=",round(Q2_d*1e-3),"kW"
print"Reduction percentage =",round(red),"%"

The rate of radiant heat transfer to the vessel =620 kW
If the insulation were removed ,The rate of radiant heat transfer to the vessel= 342.0 kW
Reduction percentage = 45.0 %


## Example no:9.20,Page no:459¶

In [84]:
#Variable declaration
e=0.75  #Emissivity of grey surface
r=1-e  #reflectivity of surface
Ts=400  #Temperature of surface
T_amb=295
sigma=5.67e-8
q1=3e3  #Rate of radiation arriving at grey surface

#Calculation
#From equation 9.118
Eb=sigma*Ts**4
#From equation 9.138
qo=e*Eb+r*q1
#From equation 9.140
Q_A=e/r*(Eb-qo)
q=Q_A
#For convective heat transfer from the surface
qc=-1*q
hc=qc/(Ts-T_amb)

#Result
print"\n The net rate of radiation trasfer = %d"%q,"W/m**2"
print"\n Coefficient of heat transfer =",round(hc,1),"W/m**2 K"

 Radiosity =1839 W/m**2

The net rate of radiation trasfer = -1161 W/m**2

Coefficient of heat transfer = 11.1 W/m**2 K


## Example no:9.21,Page no:464¶

In [5]:
#Variable declaration
sigma=5.67e-8
T=[1000,500,300] #tempertaure of surfaces
A=[1.07,1.07,0.628] #Array of area of surfaces
e=[0.75,0.50,1.0] #Array of emissivity of the surfaces
r=[0.250,0.50] # Array of radius of two surfaces
L=0.2  #distance between two discs

X=[0]*3
Y=[0]*3
R=[0]*2

#Calculation
from scipy.optimize import fsolve
for i in range(0,2):
X[i]=A[i]/r[i]
Y[i]=A[i]*e[i]/r[i]
R[i]=r[i]/L
F11=0
F22=0
S=1+(1+R[1]**2)/(R[1]**2)
F12=0.5*(S-(S**2-4*(r[1]/(2*r[0]))**2)**0.5)
A1_F11=0
A2_F22=0
A1_F12=A[0]*F12
A1_F13=A[0]-(A[0]*F11+A[1]*F12)
#for surface 2:
A2_F21=A1_F12
A2_F23=A1_F13
#for surface 3:
#By reciprocity rule
A3_F31=A1_F13
A3_F32=A2_F23
A3_F33=A[2]-(A3_F31+A3_F32)

#From equation 9.112:
E_b=[0]*3
for i in range(0,3):
E_b[i]=sigma*T[i]**4/1000.0

#since surface 3 is a black body
q_o3=E_b[2]
#From equations 9.157 and 9.158:
#we get
f=[0,0]
def F(p):

x,y=p
f[0]=(A1_F11-A[0]/r[0])*x+A2_F21*y+A3_F31*q_o3+E_b[0]*A[0]*e[0]/r[0]
f[1]=(A1_F12*x)+((A2_F22-A[1]/r[1])*y)+E_b[1]*A[1]*e[1]/r[1]
return(f[0],f[1])
q_o=fsolve(F,(0,0))
#From equation 9.140:
Q1=(A[0]*e[0]/r[0])*(E_b[0]-q_o[0])
Q2=(A[1]*e[1]/r[1])*(E_b[1]-q_o[1])

#Result
print"\n Power input to the heater ,Q1 =",round(Q1,2),"kW"
print"\n The rate of heat transfer to the plate ,Q2 =",round(Q2,2),"kW"

 Power input to the heater ,Q1 = 36.19 kW

The rate of heat transfer to the plate ,Q2 = -14.44 kW


## Example no:9.22,Page no:470¶

In [14]:
#Variable declaration
d=0.5 #diameter of chamber
l=2.0 #Length of chamber
e=0.5 #Emissivity
T_s=750.0 #Temperature at which the chamber is maintained
P=150e3
T_g=1250.0
sigma=5.67e-8
P_c=0.1*P
P_w=P_c

#Calculation
import math
V=math.pi/4.0*d**2*l #Volume of the chamber
A_s=(2*math.pi/4.0*d**2)+(math.pi*d*l) #total surface are of chamber
L_e=3.6*(V/A_s)
C_w=1.4
e_w1=C_w*0.075
C_c1=1.2
e_c1=(C_c1*0.037)
A=(P_w+P_c)*L_e
B=P_c/(P_c+P_w)
De=0.001
e_g=e_w1+0.044-De
e_w2=(0.12*C_w)
a_w=e_w2*(T_g/T_s)**0.65
C_c2=1.02
e_c2=(0.08*C_c2)
a_c=e_c2*(T_g/T_s)**0.65
a_g=a_w+a_c
#If the surrounding surface is black, then:
Q=sigma*A_s*(e_g*T_g**4-a_g*T_s**4)
#For grey walls, the correction factor allowing for multiple reflection of
C_g=0.5/(1.0-(1.0-0.326)*(1.0-0.5))
Q_w=(Q*C_g)

#Result
print"\n Radiation to the walls if the surface is black =%.1f"%(Q*1e-3),"kW"
print"\n Net radiation to the walls =%.2f"%(Q_w*1e-3),"kW"

 Radiation to the walls if the surface is black =50.3 kW

Net radiation to the walls =37.97 kW


## Example no:9.23,Page no:479¶

In [5]:
#Variable declaration:
def mole(w,m):
n = w/m
return(n)
def partial(n1):
p = 308*(n1/total)
return(p)
w_steam = 0.57       #mass flow rate of steam entering in [kg/sec]
w_CO2 = 0.20         #mass flow rate of CO2 entering in [kg/sec]
m_water = 18         #molecular mass of water in kg
m_CO2 = 44           #molecular mass of CO2 in kg
n_steam = mole(w_steam,m_water)      #number of moles in kmol
n_CO2 = mole(w_CO2,m_CO2)            #number of moles in kmol

#Calculation
import math
total = n_steam + n_CO2
p_steam = partial(n_steam)
p_CO2 = partial(n_CO2)
mean_mol = (0.57 + 0.20)/total      #mean molecular weight of the mixture in kg/kmol
outlet_steam = 11.7             #partial pressure of water in kN/m**2
outlet_CO2 = 308 - outlet_steam  #partial pressure of water in kN/m**2
n_s = n_CO2*outlet_steam/outlet_CO2
steam_condensed = n_steam - n_s
p_steam_401K = 252.2    #[kN/m**2]
p_CO2_401K = 308 - 252.2 #[kN/m**2]
steam_remaining = 0.0045*p_steam_401K/p_CO2_401K
s_c = n_steam - steam_remaining    #[kmol]
Heat_cond = s_c*18*(2180 + 1.93*(404-401))          #[kW]
Heat_uncondensed_steam = 0.0203*18*1.93*(404-401)   #[kW]
Heat_CO2 = 0.020*0.92*(404-401)
total_heat = Heat_cond + Heat_uncondensed_steam + Heat_CO2
flow_water = 1407.3/(4.187*(319-300))       #[kg/sec]
hi = 6.36        #[kW/m**2 K]Based on flow velocity of 1425 kg/m**2 sec
ho = 5.25        #[kW/m**2 K]Based on outside area
Cp = (0.20*0.92 + 0.57*1.93)/0.77   #[kJ/kg K]
k_mean = 0.025    #[kW/m K]
a = 0.0411        #[m**2]
mass_velocity = (0.20+0.57)/0.0411      #[kg/m**2 sec]
hg = 107          #[W/m**2 K] at Re = 29,800 at equivalent diameter = 0.024m
u_pD = 0.62       #(u/pD)**0.67 = 0.62
Cpu_k = 1.01      #(Cp*u/k)**0.67
Psf = (122.6 - 38)/math.log(122.6/38)
kG = hg*(Cpu_k)/(1000*Cp*Psf*u_pD)

#Result
print"\n Assuming no scale resistance, the overall coefficient =",round(1407.3/(34.8*74.2),3) ,"W/m** K"
print"\n Actual coeficient =",round(1407.3/(53.9*74.2),3) ,"kW/m**2 K"
print"\n Dirt factor =",round((0.545-0.352)/(0.545*0.352),2),"K/kW"

 Assuming no scale resistance, the overall coefficient = 0.545 W/m** K

Actual coeficient = 0.352 kW/m**2 K

Dirt factor = 1.01 K/kW


## Example no:9.24,Page no:498¶

In [23]:
#Variable declaration
d_v=1.0 #diameter of the vessel
L=0.3 #diameter of propeller agitator
N=2.5 #rotating speed of propeller agitator
T=310.0 #Temperature
G=0.5 #circulation speed of cooling water
d_o=25e-3 #outer diameter of stainless steel coil
d=22e-3 #inner diameter of stainless steel coil
d_w=(d_o+d)/2
d_c=0.8 #diameter of helix
T_m=290.0 #mean temperature
k1=0.59
Meu1=1.08e-3
C_p1=4.18e3
x_w=1.5e-3

#Calculation
h_i=(k1/d)*(1+3.5*(d/d_c))*0.023*(d*1315.0/Meu1)**0.8*(C_p1*Meu1/k1)**0.4
#The external film coefficient is given by equation 9.204:
C_p2=1.88e3 #Specefic heat capacity
Meu2=6.5e-3 #viscosity
k2=0.40
rho=1666
Meu_s=8.6e-3
h_o=0.87*(C_p2*Meu2/k2)**(1.0/3.0)*(L**2*N*rho/Meu2)**0.62*(Meu2/Meu_s)**0.14*k2/d_v

k_w=15.9
R_o=0.0004
R_i=0.0002
U_o=((1/h_o)+(x_w*d_o/(k_w*d_w))+(d_o/(h_i*d))+(R_o)+(R_i*d_o/d))**-1

#Result
print"The overall coeffecient of heat transfer =",round(U_o),"W/m**2.K"

The overall coeffecient of heat transfer = 498.0 W/m**2.K


## Example no:9.25,Page no:501¶

In [31]:
#Variable declaration
C_p=4e3

#Calculation
import sympy
x=sympy.Symbol('x')
T_max=sympy.solve((600*0.5)*(393-x)-(10*6)*(x-293))
#solving the equation finally we get
def f(T):
return(11111*(1/(376.3-T)))
#The steam is turned off for 7200 s and during this time a heat balance gives:
#on solving as given in book we get
T=346.9
#The time taken to reheat the liquid to 353 K is then given by:

#Result
print"\n\n Maximum temperature to which it can be heated =",round(T_max[0],1),"K"
print"\n Time taken to heat the liquid from 293 K to 353 K =%d"%t1[0],"s"
print"\n Time taken to reheat the liquid to 353 K =%d"%round(t2[0]),"s"


Maximum temperature to which it can be heated = 376.3 K

Time taken to heat the liquid from 293 K to 353 K =14155 s

Time taken to reheat the liquid to 353 K =2584 s


## Example no:9.26,Page no:516¶

In [33]:
#Variable declaration
#As in Example 9.1, the heat load = 1672 kW
Q=1672
#With reference to Figure 9.71:
T1=360
T2=340
theta1=300 #Temperature of cooling water entering
theta2=316

#Calculation
X=(theta2-theta1)/(T1-theta1)
Y=(T1-T2)/(theta2-theta1)
#from Figure 9.58
F=0.97
theta_m=41.9
#and hence:
A=Q/(2*F*theta_m) #the heat transfer area

#Result
print"\n The heat transfer area is =",round(A,1),"m**2"

 The heat transfer area is = 20.6 m**2


## Example no:9.27,Page no:521¶

In [37]:
#Variable declaration
#As in Example 9.1, the heat load = 1672 kW
Q=1672.0
#With reference to Figure 9.71:
T1=360.0
T2=340.0
theta1=300.0 #Temperature of cooling water entering
theta2=316.0
F_theta_m=40.6 #corrected mean temperature difference

#Calculation
T=(T1+T2)/2.0
d=1.9e-3 #Tube diameter
u=1 #Water velocity
#then, in equation 9,221:
h_i=4.28*(0.00488*T-1)*u**0.8/d**0.2
#From Table 9.18, an estimate of the shell-side film coefficient is:
h_o=(1700.0+11000.0)/2000.0
#For steel tubes of a wall thickness of 1.6 mm, the thermal resistance of the wall, from Table 9.15 is:
xw_kw=0.025
#the thermal resistance for treated water, from Table 9.16, is 0.26 m2K/kW
Ri=0.26
Ro=Ri
U=((1.0/h_o)+xw_kw+Ri+Ro+(1.0/h_i))**-1
A=Q/(F_theta_m*U)

#Result
print"\n The heat transfer area =",round(A,1),"m**2"

 The heat transfer area = 32.8 m**2


## Example no:9.28,Page no:531¶

In [1]:
#Variable declaration
mh = 30.0             #[kg/s] Hot fluid flow rate
Thi = 370.0          #[K] Hot Fluid Inlet Temperature
Tho = 315.0          #[K] Hot Fluid outlet Temperature
Tci = 300.0          #[K] Cold Fluid Inlet Temperature
Tco = 315.0          #[K] Cold Fluid Outlet Temperature
cpc = 4.18*10**3         #[J/kg.K] Thermal Conductivity of Cold Fluid
#From table A1.3 at mean temperature 343 K
cph = 2.9*10**3             #[J/kg.K] Thermal Capacity of Hot fluid

#Calculation
import math
q = mh*cph*(Thi-Tho)         #[kW] Heat load
mc = q/(cpc*(Tco-Tci))       #[kg/s] Flow of cooling water
Tln = ((Thi-Tho)-(Tco-Tci))/(math.log((Thi-Tho)/(Tco-Tci)))     #[K] logarithm mean temperature difference
X = (Thi - Tho)/(Tco-Tci)
Y = (Tco-Tci)/(Thi - Tci)
F = .85
U = 500.0        #[W/m**2.K]
A = q/(F*Tln * U)
od = .02        #[m] outer dia
id = .016       #[m] inner dia
l = 4.83        #[m] effective tube length
s = math.pi*od*l
N = A/s
db = (1210/.249)**(2.207)**-1*20/1000.0             #[m]
dc = .068        #[m] diametric clearance between shell and tubes
ds = db+dc            #[m] Shell dia
Ac = math.pi/4*id**2            #[m**2] Cross sectional area
Ntp = N/2.0
Af = N/2.0*Ac        #[m**2] Tube side flow area
mw = 76.3/Af            #[kg/m**2.s] Mass velocity of water
rho = 995        #[kg/m**3] mas density of water
u = mw/rho            #[m/s] water velocity
vu = .8*10**-3            #[N.s/m**2] viscosity
k = .59                #[W/m.K]
Re = id*u*rho/vu
Pr = cpc*vu/k
ld = l/id
jh = 3.7*10**-3
hi = jh*Re*Pr**.3334*.59/id             #[W/m**2.K]
dbf = .20*ds         #[m] Baffle Dia
tb = 1.25*20*10**-3        #[mm] Tube Pitch
As = (25.0-20.0)/25.0*10**3*(ds*Ac)            #[m**2]
Gs = 30.0/As                         #[kg/m**2.s]
de = 1.1*(.025**2-.917*od**2)/od             #[m]
rho2 = 780.0            #[kg/m**3] density
vu2 = .8*10**-3            #[N.s/m**2] viscosity
Cp2 = 3.1*10**3                #[J/kg.K] Heat capacity
k2 = .16                #[W/m.K]
Re2 = Gs*de/vu2
Pr2 = Cp2*vu2/k2
jh2 = 5*10**-3
hs = jh2*Re2*Pr2**.334*k2/de
k3 = 50.0                #[W/m.K]        Thermal Conductivity
Rw = .00020            #[m**2.K/W]      Scale Resistances
Ro = .00015            #[m**2.K/W]      Resistance for organic
U = (1.0/hs + Rw + 0.5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1
#From figure 9.78
jf = 4.5*10**-3
n = 2.0
delP = n*(4*jf*(4.830/id) + 1.25)*(rho*u**2)
u2 = Gs/rho2
jf2 = 4.6*10**-2
N2 = 1.0
delP2 = N2*(4*jf2*(4.830/od)*(1005/14.2))*(rho2*u2**2)

#Increasing the baffle spacing pressure drop is reduced one-fourth
delPs = delP2/4.0
ho = hs*(.5)**.8             #[W/m**2.K]
U2 = (1.0/ho + Rw + .5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1

#REsult
print"Overall Coefficient of %d"%U2," W/m**2.K \n Number of tubes/pass =",round(Ntp),"\n Number of tubes required = %d"%N
print"NOTE:Very approximate values are used in textbook for Calculation\nThat's why answer is different"

Overall Coefficient of 563  W/m**2.K
Number of tubes/pass = 603.0
Number of tubes required = 1205
NOTE:Very approximate values are used in textbook for Calculation


## Example no:9.29,Page no:535¶

In [45]:
#Variable declaration
G=1.0  #Flow rate of organic liquid
Cp=2e3#Heat capacity of organic liquid
T1=350.0
T2=330.0
theta1=290.0
theta2=320.0

#Calculation
G_cool=Q/(4187*(theta1-theta2)) #flow of water
GCp_hot=(G*Cp) #for organic
GCp_cold=(G_cool*4187)
#From equation 9.235:
eta=GCp_hot*(T1-T2)/(GCp_cold*(T1-theta1))

#Result
print" Effectiveness of the given double pipe heat exchanger =",round(eta,2)

 Effectiveness of the given double pipe heat exchanger = 0.5


## Example no:9.30,Page no:538¶

In [60]:
#Variable declaration
Tci = 320.0        #[K] Cold Fluid Initial Temperature
Tce = 340.0        #[K] Cold Fluid Final Temperature
mc = 4.0            #[kg/s] Flow rate of cold fluid
mh = 8.0            #[kg/s] Flow rate of hot fluid
import numpy as np
Thi = np.array([380,370,360,350])    #[K] Hot fluid initial temperature
Cp = 4.18            #[kJ/kg.K] mean heat capacity
U = 1.5        #[W/m**2.K] Overall heat transfer coefficient

#Calculation
GCpu= mh*Cp             #[kW/K]
GCpp= mc*Cp             #[kW/K]
if(GCpu<GCpp):
GCpmin = GCpu        #[kW/K]
ratio = GCpmin/GCpp
else:
GCpmin = GCpp       #[kW/K]
ratio = GCpmin/GCpu

#Equation 9.235
n = mc*Cp*(Tce-Tci)*(mc*Cp*(Thi - Tci))**-1
#From Figure 9.85b Number of transfer Units
N = np.array([.45,.6,.9,1.7])     #[NTU]
A = N*GCpmin/U         #Area of required [m**2]

#Result
print"Thi(K)   n      N     A (m**2)"
for i in range(0,4):
print Thi[i],"\t",round(n[i],2),"\t",N[i],"\t",round(A[i],1)

Thi(K)   n      N     A (m**2)
380 	0.33 	0.45 	5.0
370 	0.4 	0.6 	6.7
360 	0.5 	0.9 	10.0
350 	0.67 	1.7 	18.9


## Example no:9.31,Page no:544¶

In [64]:
#Variable declaration
o_d=10e-3 #outer diameter of the tube
i_d=8.2e-3  #inner diameter of the tube
h=140 #coeffecient of heat transfer between gas and copper tube
k=350 #Thermal conductivity of copper tube
L=0.075

#Calculation
import math
b=math.pi*o_d #perimeter of tube
A=math.pi/4*(o_d**2-i_d**2) #cross sectional area of the metal
m=((h*b)/(k*A))**0.5
T_g=((475*math.cosh(m*L))-365)/(math.cosh(m*L)-1)

#Result
print"The gas temperature is =%d"%T_g,"K"

The gas temperature is =539 K


## Example no:9.32,Page no:545¶

In [67]:
#Variable declaration
d2=54e-3  #outer diameter of the tube
d1=70e-3  # fin diameter
w=2e-3  #fin thickness
n=230.0 # number of fins per metre run
T_s=370.0  #Surface temperature
T=280.0  #Temperature of surroundings
h=30.0  #Heat transfer coeffecient between gas and fin
k=43.0  #Thermal conductivity of steel
L=(d1-d2)/2.0

#Calculation
import math
theta1=T_s-T
#Assuming that the height of the fin is small compared with its circumference
#and that it may be treated as a straight fin of length
l=(math.pi/2.0)*(d1+d2)
b=2*l #perimeter
A=l*w #the average area at right-angles to the heat flow
m=((h*b)/(k*A))**0.5
#From equation 9.254, the heat flow is given for case (b) as:
Qf=m*k*A*theta1*(math.exp(2*m*L)-1)/(1+math.exp(2*m*L))
Q=Qf*n #Heat loss per meter run of tube

#Result
print"The heat loss per metre run of tube =",round(Q*1e-3,2),"kW/m"

The heat loss per metre run of tube = 1.91 kW/m


## Example no:9.33,Page no:556¶

In [71]:
#Variable declaration
d=150e-3 #Internal diameter of tube
d_o=168e-3 #outer diameter of tube
d_w=159e-3
d_s=268e-3
d_m=(d_s-d_o)/math.log(d_s/d_o) #log mean of d_o and d_s
h_i=8500 #The coefficient for condensing steam together with that for any scale
k_w=45
k_l=0.073
x_l=50e-3
x_w=9e-3
DT=444-294
sigma=5.67e-8

#Calculation
import math
#The temperature on the outside of the lagging is estimated at 314 K and (hr + hc) will be taken as 10 W/m2 K.
#total thermal resisitance
R=(h_i*math.pi*d)**-1+(10*math.pi*d_s)**-1+(k_w*math.pi*d_w/x_w)**-1+(k_l*math.pi*d_m/x_l)**-1
Q_l=DT/R  #The heat loss per metre of length(from eq 9.261)
DT_lagging=((k_l*math.pi*d_m/x_l)**-1/R)*DT
#Taking an emissivity of 0.9, from equation 9.119:

h_r=(0.9*sigma*(310**4-294**4))/(310-294)
C=1.32
#Substituting in equation 9.105 (putting l = diameter = 0.268 m):
h_c=C*((310.0-294.0)/d_s)**0.25
#If the pipe were unlagged,(hc+hr)for DT=150 K would be about 20 W/m2 K and the heat loss would then be:
Q_l=20*math.pi*d_o*150

#Result
print"The heat loss to the air =",round(Q_l*1e-3,2),"kW/m"

The heat loss to the air = 1.58 kW/m


## Example no:9.34,Page no:560¶

In [76]:
#Variable declaration
T1=420 #temperature of steam
k=0.1 #Thermal conductivity
T2=285  #Ambient temperature
h=10 #the coefficient of heat transfer from the outside of the lagging to
#the surroundings

#Calculation

#determining Q/l from equation 9.21 and equating it to heat loss from the
#outside of the lagging we get
#(Q/l)=84.82/(math.log(d_o/0.1)+(0.02/d_o)) W/m
#using various equations we finally get an equation in terms of d_o and we
# will solve it by using fsolve
def F(d_o):
return((1/(math.log(d_o/0.1)+(0.02/d_o))**2)-(2.35*(d_o**3)/(d_o-0.02)))
ans=fsolve(F,1)
E_t=(ans-0.1)/2

#Result
print"\n Economic thickness of lagging =",round(E_t[0]*1e3),"mm"

 Economic thickness of lagging = 163.0 mm