Chapter 8 - Heat Exchanger

Example No : 8.1 - Page : 248

In [16]:
from __future__ import division
from math import log
# given data
t_hi=80   # in degree C
t_ci=30   # in degree C
t_ho=40   # in degree C
Mh=0.278   # in kg/s
Mc=0.278   # in kg/s
Cph=2.09# # in kJ/kg degree C
Cpc=4.18   #  in kJ/kg degree C
U=24   # in W/m**2 degree C
# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C
del_t1=t_hi-t_co   #in degree C
del_t2=t_ho-t_ci   #in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#
Cph=Cph*10**3   # in J/kg degree C
q=Mh*Cph*(t_hi-t_ho)#
#Formula q=U*A*del_tm
A=q/(U*del_tm)   # in m**2
print "Surface area of heat exchange = %0.1f square meter" %A

Surface area of heat exchange = 53.2 square meter

Example No : 8.2 - Page : 249

In [17]:
from numpy import pi
# given data
t_hi=160   # in degree C
t_ci=25   # in degree C
t_ho=60   # in degree C
Mh=2   # in kg/s
Mc=2   # in kg/s
Cph=2.035# # in kJ/kg degree C
Cpc=4.187   #  in kJ/kg degree C
U=250   # in W/m**2 K
d=0.5   # in m
# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C
del_t1=t_hi-t_co   #in degree C
del_t2=t_ho-t_ci   #in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#
Cph=Cph*10**3   # in J/kg degree C
q=Mh*Cph*(t_hi-t_ho)#
#Formula q=U*pi*d*l*del_tm
l=q/(U*pi*d*del_tm)#
print "Length of the heat exchanger = %0.2f meter" %l

Length of the heat exchanger = 18.22 meter

Example No : 8.3 - Page : 251

In [18]:
from math import log
from __future__ import division
# given data
t_hi=110   # in degree C
t_ci=35   # in degree C
t_co=75   # in degree C
Mh=2.5   # in kg/s
Mc=1   # in kg/s
Cph=1.9# # in kJ/kg K
Cpc=4.18   #  in kJ/kg K
U=300   # in W/m**2 K

# Energy balance Mc*Cpc*(t_co-t_ci) = Mh*Cph*(t_hi-t_ho)
t_ho=t_hi- Mc*Cpc*(t_co-t_ci)/(Mh*Cph)   # in degree C
del_t1=t_hi-t_co   #in degree C
del_t2=t_ho-t_ci   #in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#
Cph=Cph*10**3   # in J/kg degree C
q=Mh*Cph*(t_hi-t_ho)#
#Formula q=U*A*del_tm
A=q/(U*del_tm)#
print "Area of the heat exchanger = %0.2f square meter" %A

Area of the heat exchanger = 14.92 square meter

Example No : 8.4 - Page : 252

In [19]:
from math import log
# given data
Fi=0.00014   # in m**2 degree C/W
hi=2000   # in W/m**2degree C
Fo=0.00015   # in m**2 degree C/W
ho=1000   # in W/m**2degree C
di=3*10**-2   # in m
do=4*10**-2   #in m
ro=do/2#
ri=di/2#
k=53   # in W/m degree C
Uo=1/(do/di*1/hi+ do/(2*k)*log(ro/ri) + 1/ho + do*Fi/di + Fo)#
print "The overall heat transfer coefficient = %0.3f W/m**2 degree C" %Uo
# Note : Answer in the book is not accurate

The overall heat transfer coefficient = 473.509 W/m**2 degree C

Example No : 8.5 - Page : 252

In [20]:
import math
# given data
V=0.15   # in m/s
di=2.5*10**-2   # in m

delta=0.364*10**-6   # in m**2/s
k=0.668   # in W/m degree C
Pr=2.22#

Re=V*di/delta#
# Formula Nu= hi*di/k = 0.023*Re**0.8*Pr**0.3
hi=0.023*Re**0.8*Pr**0.3*k/di   # in W/m**2 degree C

# Now, Reynold number for flow of air across the tube
delta=18.22*10**-6   # in m**2/s
k=0.0281   # in W/m degree C
Pr=0.703#
d=2.5*10**-2   # in m
u=10   # in m/s
Re=u*d/delta#
Re=math.floor(Re)#
#The Nusselt number for this case
Nu=(0.04*Re**0.5+ 0.006*Re**(2/3))*Pr**0.4
# Formula Nu= ho*do/k
do=di#
ho=Nu*k/do   # in W/m**2 degree C
print "Heat transfer coefficient = %0.3f W/m**2 degree C" %ho
U=1/(1/hi+1/ho)#
print "The overall heat transfer coefficient neglecting the wall resistance = %0.3f W/m**2 degree C" %U
l=1   # in m
Ti=90   # in degree C
To=10   # in degree C
q=U*pi*d*l*(Ti-To) #
print "Heat loss per meter length of the tube = %0.3f W/m" %q
# Note: Answer in the book is wrong

Heat transfer coefficient = 7.931 W/m**2 degree C
The overall heat transfer coefficient neglecting the wall resistance = 7.882 W/m**2 degree C
Heat loss per meter length of the tube = 49.523 W/m

Example No : 8.6 - Page : 253

In [21]:
# given data
t_hi=83   # in degree C
t_ho=45   # in degree C
t_ci=25   # in degree C
Mh=5   # in kg/min
Mc=9   # in kg/min
Cph=4.18# # in kJ/kg K
Cpc=2.85   #  in kJ/kg K
# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci) 
t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C
print "t_co = %0.2f degree C" %t_co
if(t_co>t_ho) :
    print "since t_co > t_ho, hence counter flow arrangment will be suitable"

t_co = 55.96 degree C
since t_co > t_ho, hence counter flow arrangment will be suitable

Example No : 8.7 - Page : 254

In [22]:
from math import log
# given data
# (a) For parallel flow arrangment
del_t1=60-10   # in degree C
del_t2=40-30   # in degree C
del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C
q=100*10**3   # in W
U=75   # in W/m**2 degree C
# Formula q=U*A*del_tm#
A=q/(U*del_tm)#
print "Area for paraller flow arrangment = %0.1f square meter" %A
# (b) For counter flow heat exchange
del_t1=60-30   # in degree C
del_t2=40-10   # in degree C
# In this case
del_tm=(del_t1+del_t2)/2   # in degree C
A=q/(U*del_tm)#
print "Area For counter flow heat exchange = %0.2f square meter" %A

Area for paraller flow arrangment = 53.6 square meter
Area For counter flow heat exchange = 44.44 square meter

Example No : 8.8 - Page : 255

In [23]:
from numpy import pi
# given data
Cp=4180   # in J/kg degree C
miu=0.86*10**-3   # in kg/m-s
Pr=60#
k=0.60   # in W/m degree C
h_fg=2372400   # in W
ho=6000   # in W/m**2 degree C
di=2*10**-2   # in m
d_o=3*10**-2   # in m
t_co=35   # in degree C
t_ci=15   # in degree C
M=0.9#
Re=4*M/(pi*di*miu)#
# since Re > 2300, hence flow inside tube is a turbulent flow.
# Hence  Nu= hi*di/k = 0.023*Re**0.8*Pr**0.4
hi= 0.023*Re**0.8*Pr**0.4*k/di#
Uo= 1/(1/10213.6*(d_o/di)+1/ho)#
del_t1=50-15   # in degree C
del_t2=50-35   # in degree C
del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C
# Formula q= Uo*pi*d_i*L*del_tm = M*Cp*(t_co-t_ci)
L= M*Cp*(t_co-t_ci)/(Uo*pi*d_o*del_tm)#
print "Length of tube = %0.3f meter" %L
q=M*Cp*(t_co-t_ci)   # in watt
m=q/h_fg#
print "Rate of condensation = %0.3f kg/sec" %m

Length of tube = 10.604 meter
Rate of condensation = 0.032 kg/sec

Example No : 8.9 - Page : 257

In [24]:
from math import log
# given data
Cph=3850# # in J/kg degree C
t_hi=100   # in degree C
t_ci=20   # in degree C
t_ho=50   # in degree C
Mh=8   # in kg/s
Mc=10   # in kg/s
Cpc=4.18*10**3   #  in J/kg degree C
U=400   # in W/m**2 degree C
# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C
# Heat load
q=Mh*Cph*(t_hi-t_ho)   # in W
# (a) Parallel flow
del_t1=90   # in degree C
del_t2=3.16   # in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#
A=q/(U*del_tm)#
print "Surface area for parallel flow = %0.3f meter square" %A
# (b) Counter flow heat exchanger
del_t1=53.16   # in degree C
del_t2=40   # in degree C
del_tm_counter= (del_t1-del_t2)/log(del_t1/del_t2)#
A=q/(U*del_tm_counter)#
print "Surface area for counter flow heat exchanger = %0.1f meter square" %A
#(c) One shell pass and two tube pass.
#here
t1=10   # in degree C
t2=46.84   # in degree C
T1=100   # in degree C
T2=50   # in degree C
P=(t2-t1)/(T1-t1)#
R=(T1-T2)/(t2-t1)#
F=0.88#
del_tm=F*del_tm_counter   # in degree C
A=q/(U*del_tm)#
print "Surface area for one shell pass and two tube pass = %0.2f meter square" %A
# (d) For cross flow, correction factor 
F=0.9#
del_tm=F*del_tm_counter#
A=q/(U*del_tm)#
print "Surface area for cross flow = %0.3f meter square" %A

Surface area for parallel flow = 148.486 meter square
Surface area for counter flow heat exchanger = 83.2 meter square
Surface area for one shell pass and two tube pass = 94.56 meter square
Surface area for cross flow = 92.456 meter square

Example No : 8.10 Page No : 258

In [25]:
from math import exp
# given data
Cpc=4.18*10**3   #  in J/kg degree C
Mc=1   # in kg/s
Mh=2.4   # in kg/s
Cph=2050# # in J/kg degree C
t_hi=100   # in degree C
t_ci=20   # in degree C
C_c=Mc*Cpc   # in W/degree C
C_h=Mh*Cph   # in W/degree C
U=300   # in W/m**2 degree C
A=10   # in m**2
C_min=C_c#
C_max=C_h#
N= A*U/C_min#
C=C_min/C_max#
# Effectiveness for counter flow heat exchanger
epsilon= (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#
# Total heat transfer
q=epsilon*C_min*(t_hi-t_ci)   # in watt
print "Total heat transfer = %0.3f kW" %(q*10**-3) 
t_co=t_ci+epsilon*C*(t_hi-t_ci) #
print "Exit temperature of water = %0.3f degree C" %t_co

Total heat transfer = 144.170 kW
Exit temperature of water = 49.303 degree C

Example No : 8.11 Page No : 259

In [26]:
from math import log
# given data
t_hi=135   # in degree C
t_ci=20   # in degree C
t_ho=65   # in degree C
t_co=50   # in degree C
# Energy balance  Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
# C =C_min/C_max = Mh*Cph/( Mc*Cpc)
C= (t_co-t_ci)/(t_hi-t_ho)#
epsilon=(t_hi-t_ho)/(t_hi-t_ci)#
# Also epsilon = epsilon_parallel = (1-exp(-NTU*(1+C)))/(1+C)
NTU= -log(1-epsilon*(1+C))/(1+C)#
# if the existing heat exchanger is to be used as counter flow mode, its NTU will not change, i.e.
epsilon_c= (1-exp(-NTU*(1-C)))/((1-C*exp(-NTU*(1-C))))#
# Exit temperature
# (i) Hot fluid
t_ho=t_hi-epsilon_c*(t_hi-t_ci)   # in degree C
print "Exit temperature for hot fluid = %0.3f degree C" %t_ho

#(ii) Cold fluid
t_co= t_ci+epsilon_c*C*(t_hi-t_ci)#
print "Exit temperature for cold fluid = %0.1f degree C" %t_co

# (iii) # If the parallel flow heat exchanger is too long, then body fluid will have common outlet temperature (t)
# From MCp_h*(t_hi-t) = MCp_c*(t-t_ci)

t=(C*t_hi+t_ci)/(1+C)#
print "The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,"
print "in degree C = %0.1f" %t

Exit temperature for hot fluid = 55.909 degree C
Exit temperature for cold fluid = 53.9 degree C
The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,
in degree C = 54.5

Example No : 8.12 - Page : 261

In [27]:
from math import log
# given data
t_hi=78   # in degree C
t_ci=23   # in degree C
t_ho=65   # in degree C
t_co=36   # in degree C
# Energy balance  Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
# C =C_min/C_max = Mh*Cph/( Mc*Cpc)
C= (t_co-t_ci)/(t_hi-t_ho)#
epsilon=(t_hi-t_ho)/(t_hi-t_ci)#
# Formula epsilon = (1-exp(-N*(1+C)))/(1+C)
N= -log(1-epsilon*(1+C))/(1+C)#
# When flow rates of both fluids are doubled , the deat capacity ratio will not change, i.e.
# C=1
#  MCp_new =2* MCp_old
# N=U*A/C_min=N/2
N=N/2#
epsilon=(1-exp(-N*(1+C)))/(1+C)#
# exit temperature
t_ho=t_hi-epsilon*(t_hi-t_ci)   # in degree C
t_co= t_ci+epsilon*(t_hi-t_ci)#
print "Exit temperature  in degree C is : ",round(t_ho,2),"and",round(t_co,2)
# Note: Answer in the book is wrong due to put wrong value of t_ci in second last line

Exit temperature  in degree C is :  70.47 and 30.53

Example No : 8.13 - Page : 262

In [28]:
from numpy import pi
# given data
t_hi=125   # in degree C
t_ci=22   # in degree C
Mh=21   # in kg/s
Mc=5   # in kg/s
C_ph=2100   # in J/kg K
C_pc=4100   # in J/kg K
Ch=Mh*C_ph   # in Js/kg
Cc=Mc*C_pc   # in Js/kg
C_min=Cc   # in Js/kg
C_max=Ch   # in Js/kg
U=325   # in W/m**2 K
d=2.2*10**-2   # in m
l=5   # in m
total_tube=195   # number of total tubes
A=pi*d*l*total_tube
NTU=U*A/C_min#
C=C_min/C_max#
epsilon = (1-exp(-NTU*(1-C)))/(1-C*exp(-NTU*(1-C)))#
t_co= t_ci+epsilon*(t_hi-t_ci)#
t_ho= t_hi-epsilon*Cc/Ch*(t_hi-t_ci)#
print "Exit temperature  in degree C :",round(t_co,1),"and",round(t_ho,2)
# Total heat transfer
q=epsilon*C_min*(t_hi-t_ci)#
print "Total heat transfer = %0.3f kW" %(q*10**-3)

Exit temperature  in degree C : 82.8 and 96.73
Total heat transfer = 1246.589 kW

Example No : 8.14 - Page : 263

In [29]:
from math import exp
# given data
t_hi=94   # in degree C
t_ci=15   # in degree C
Mw=0.36   # in kg/s
Mo=0.153   # in kg/s
C_po=2*10**3   # in J/kg K
C_pw=4.186*10**3   # in J/kg K
U=10.75*10**2   # in W/m**2 K
A=1   # in m**2
Ch=Mo*C_po   # in kW/K
Cc=Mw*C_pw   # in kW/K
C_min=Ch   # in W/K
C_max=Cc   # in W/K
C=C_min/C_max#
NTU=U*A/C_min#
# Effectiveness
N=NTU#
epsilon = (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#
mCp_min=C_min#
q_max= mCp_min*(t_hi-t_ci)   # in W
q_actual= epsilon*q_max   # in W
print "Total heat transfer = %0.3f watt" %q_actual
# Outlet temp. of water
t_co= q_actual/Cc+t_ci   # in degree C
print "Outlet temperature of water = %0.3f degree C" %t_co
# Outlet temp. of oil
t_ho=t_hi-q_actual/Ch   #in degree C
print "Outlet temperature of oil = %0.3f degree C" %t_ho
#Note: Evaluation of Cc and Ch in the book is wrong so the Answer in the book is wrong

Total heat transfer = 22987.488 watt
Outlet temperature of water = 30.254 degree C
Outlet temperature of oil = 18.877 degree C

Example No : 8.15 - Page : 265

In [30]:
from math import log
# given data
U=1800   # in W/m**2 degree C
h_fg=2200*10**3   # in J/kg
t_ci=20   # in degree C
t_co=90   # in degree C
del_t1=120-20   # in degree C
del_t2=120-90   # in degree C
del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C
Mc=1000/3600   # in kg/s
Cc=4180   # in kg/s
# Rate of heat transfer
q=Mc*Cc*(t_co-t_ci)   # in watt
# Formula q=U*A*del_tm
A=q/(U*del_tm)#
print "Surface area = %0.3f square meter" %A
#Rate of condensation of steam
ms=q/h_fg   # in kg/sec
print "Rate of condensation of steam = %0.4f kg/sec" %ms

Surface area = 0.777 square meter
Rate of condensation of steam = 0.0369 kg/sec

Example No : 8.16 - Page : 265

In [31]:
from math import log
# given data
Mh=10000/3600   # in kg/sec
Mc=8000/3600   # in kg/sec
Cph=2095   # in J/kg K
Cpc=4180   # in J/kg K
t_hi=80   # in degree C
t_ci=25   # in degree C
t_ho=50   # in degree C
U=300   # in W/m**2 K
# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)
t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C
del_t1=t_hi-t_co   #in degree C
del_t2=t_ho-t_ci   #in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#
q=Mh*Cph*(t_hi-t_ho)#
#Formula q=U*A*del_tm
A=q/(U*del_tm)   # in m**2
print "Surface area of heat exchange = %0.2f square meter" %A

Surface area of heat exchange = 19.23 square meter

Example No : 8.17 - Page : 266

In [32]:
from math import log
from numpy import pi
# given data
ho=5000   # in W/m**2 degree C
rho=988.1   # in kg/m**3
K=0.6474#
D=555*10**-9   # in m**2/s
Pr=3.54#
n=100#
d_i=2.5*10**-2   # in m
ri=d_i/2#
d_o=2.9*10**-2   # in m
ro=d_o/2#
Cp=4174   # in J/kg degree C
Mc=8.333   # in kg/s
Mw=Mc#
t_c1=30   # in degree C
t_c2=70   # in degree C
t_n1=100   # in degree C
t_n2=t_n1   # in degree C
R_fi=0.0002   # in m**2 degree C/W (In the book, there is miss print in this line,they took here R_fi = .002)
# Heat gain by water
Q=Mc*Cp*(t_c2-t_c1)#
# Also Q= U*A*del_tm
del_t1=t_n1-t_c1   #in degree C
del_t2=t_n2-t_c2   #in degree C
del_tm= (del_t1-del_t2)/log(del_t1/del_t2)# 
# Mw= 1/4*pi*d_i**2*V*rho*N, here
N=n#
V=4*Mw/(pi*d_i**2*rho*N)#
# Formula Re=V*d_i/v, here
v=D#
Re=V*d_i/v#
# Formula Nu= hi*d_i/K = 0.023*Re**0.8*Pr**0.33
hi= 0.023*Re**0.8*Pr**0.33*K/d_i#
# Formula 1/Vi= 1/hi + R_fi + ri/ro*1/ho
Vi= 1/(1/hi + R_fi + ri/ro*1/ho)   # in W/m**2 degree C
#Formula Q = Vi*(N*pi*d_i*L)*del_tm
L= Q/(Vi*(N*pi*d_i)*del_tm)#
print "Length of the tube bundle = %0.1f m" %L

Length of the tube bundle = 4.6 m