# Chapter 12 : Heat Exchangers¶

## Example 12.1 Page No : 503¶

In [1]:
import math

#INPUT
T = 80.;		        	#Bulk Temperature of water in degrees C
Di = 0.0254;			#Inner diameter of steel pipe in m
Do = 0.0288;			#Outer diameter of steel pipe in m
k = 50.;			        #Thermal conductivity of steel in W/m.K
ho = 30800.;		    	#Average convection coefficient in W/m**2.K
v = 0.50;		    	#Velocity of water in m/s
# Variables FROM HEAT AND MASS TRANSFER DATA BOOK FOR WATER AT BULK TEMPERATURE OF 80 degree C
d = 974.;		    	#Density in kg/m**3
v1 = 0.000000364;		#Kinematic viscosity in m**2/s
k1 = 0.6687;			#Thermal conductivity in W/m.K
Pr = 2.2;			    #Prantl Number

# Calculations
Re = (v*Di)/v1;			#Reynold's number
Nu = (0.023*Re**0.8*Pr**0.4);			#Nusselts number
hi = Nu*(k1/Di);			#Heat transfer coefficient in W/m**2.K
ri = (Di/2);			#Inner radius of steel pipe in m
ro = (Do/2);			#Outer radius of steel pipe in m
U = (1./((1./ho)+((ro/k)*math.log(ro/ri))+(ro/(ri*hi))));			#Overall heat transfer coefficient in W/m**2.K

# Results
print 'Overall heat transfer coefficient is %3.1f W/m**2.K'%(U)

Overall heat transfer coefficient is 2591.8 W/m**2.K


## Example 12.2 Page No : 504¶

In [3]:
import math
# Variables
Do = 0.0254;			#Outer Diameter of heat exchanger tube in m
Di = 0.02286;			#Inner Diameter of heat exchanger tube in m
k = 102;			#Thermal conductivity of the tube in W/m.K
hi = 5500;			#Heat transfer coefficients at the inner side of tube in W/m**2.K
ho = 3800;			#Heat transfer coefficients at the outer side of tube in W/m**2.K
Rfi = 0.0002;			#Fouling factor in m**2.W.K
Rfo = 0.0002;			#Fouling factor in m**2.W.K

# Calculations
ro = (Do/2);			#Outer radius of heat exchanger tube in m
ri = (Di/2);			#Inner radius of heat exchanger tube in m
U = (1./((1./ho)+Rfo+((ro/k)*math.log(ro/ri))+((ro*Rfi)/ri)+(ro/(ri*hi))));			#Overall heat transfer coefficient in W/m**2.K

# Results
print 'Overall heat transfer coefficient is %i W/m**2.K'%(U)

Overall heat transfer coefficient is 1110 W/m**2.K


## Example 12.3 Page No : 509¶

In [4]:
import math
# Variables
mh = 10000.;			#Mass flow rate of oil in kg/h
ch = 2095.;			#Specific heat of oil J/kg.K
Thi = 80.;			#Inlet temperature of oil in degree C
Tho = 50.;			#Outlet temperature of oil in degree C
mc = 8000.;			#Mass flow rate of water in kg/h
Tci = 25.;			#Inlet temperature of water in degree C
U = 300.;			#Overall heat ransfer coefficient in W/m**2.K
cc = 4180.;			#Specific heat of water in J/kg.K

# Calculations
Q = (mh*ch*(Thi-Tho));			#Heat transfer rate in W
Tco = ((Q/(mc*cc))+Tci);			#Outlet temperature of water in degree C
T = (Thi-Tco);			#Temperature difference between oil inlet temperature and water outlet temperature in degree C
t = (Tho-Tci);			#Temperature difference between oil outlet temperature and water inlet temperature in degree C
A = (((Q/U)*math.log(t/T))/(3600*(t-T)));			#Area of heat exchanger in m**2

# Results
print 'Area of heat exchanger is %3.2f m**2'%(A)

Area of heat exchanger is 19.23 m**2


## Example 12.4 Page No : 510¶

In [6]:
import math
# Variables
Ch = 2500.;			#Capacity rate of hot oil in W/K
Thi = 360.;			#Temperature of hot fluid at the entrance of heat exchanger in degree C
Tho = 300.;			#Temperature of hot fluid at the exit of heat exchanger in degree C
Tci = 30.;			#Temperature of cold fluid at the entrance of heat exchanger in degree C
Tco = 200.;			#Temperature of hot fluid at the exit of heat exchanger in degree C
U = 800.;			#Overall heat transfer coefficient in W/m**2.K

# Calculations
Q = (Ch*(Thi-Tho));			#Heat transfer from the oil in W
#Parallel flow
T1 = Thi-Tci;			#Temperature difference between hot fluid inlet temperature and cold fluid inlet temperature in degree C
T2 = Tho-Tco;			#Temperature difference between hot fluid outlet temperature and cold fluid outlet temperature in degree C
Tlm1 = ((T1-T2)/math.log(T1/T2));			#LMTD for parallel flow arrangement in degree C
A1 = (Q/(U*Tlm1));			#Area of heat exchanger in m**2
#Counter flow
t1 = Thi-Tco;			#Temperature difference between hot fluid inlet temperature and cold fluid outlet temperature in degree C
t2 = Tho-Tci;			#Temperature difference between hot fluid outlet temperature and cold fluid inlet temperature in degree C
Tlm2 = ((t1-t2)/math.log(t1/t2));			#LMTD for counter flow arrangement in degree C
A2 = (Q/(U*Tlm2));			#Area of heat exchanger in m**2

# Results
print 'Area of heat exchanger in parallel flow arrangement is %3.3f m**2  \n \
Area of heat exchanger in counter flow arrangement is %3.3f m**2'%(A1,A2)

Area of heat exchanger in parallel flow arrangement is 0.973 m**2
Area of heat exchanger in counter flow arrangement is 0.892 m**2


## Example 12.5 Page No : 511¶

In [7]:
import math
# Variables
ch = 2130.;			#Specific heat of oil in J/kg.K
T1 = 160.;			#Temperature of hot fluid (oil) at the entrance of heat exchanger in degree C
T2 = 60.;			#Temperature of hot fluid (oil) at the exit of heat exchanger in degree C
t1 = 25.;			#Temperature of cold fluid (water) at the entrance of heat exchanger in degree C
d = 0.5;			#Inner diameter of the tube in m
mc = 2.;			#Mass flow rate of cooling water in kg/s
D = 0.7;			#outer annulus outer diameter in m
mh = 2.;			#Mass flow rate of hot oil in kg/s
U = 250.;			#Overall heat transfer coefficient in W/m**2.K
cc = 4186.;			#Specific heat of water in J/kg.K

# Calculations
Q = (mh*ch*(T1-T2));			# Required heat transfer rate in W
t2 = ((Q/(mc*cc))+t1);			#Outer water temperature in degree C
T = T1-t2;  	    		    #Change in temperature between inlet tmperature of hot fluid and outlet temperature of cold fluid in degree C
t = T2-t1;	    	        	#Change in temperature between outlet tmperature of hot fluid and inlet temperature of cold fluid in degree C
Tlm = ((T-t)/(math.log(T/t)));	#Value of LMTD in degree C
L = (Q/(U*3.14*d*Tlm));			#Length of the heat exchanger in m

# Results
print 'Length of the heat exchanger is %3.2f m'%(L)

Length of the heat exchanger is 19.38 m


## Example 12.6 Page No : 512¶

In [8]:
# Variables
T = 120.;			#Saturated steam temperature in degree C
U = 1800.;			#Heat transfer coefficient in W/m**2.K
m = 1000.;			#mass flow rate of water in kg/h
t1 = 20.;			#Inlet temperature of water in degree C
t2 = 90.;			#Outlet tmperature of water in degree C
hfg = 2200.;		#Enthalpy of steam in kJ/kg
c = 4186.;			#Specific het of water in J/kg.K

# Calculations
Tlm = (((T-t1)-(T-t2))/(math.log((T-t1)/(T-t2))));			#LMTD in a condenser in degree C
Q = ((m/3600)*c*(t2-t1));	    		#Rate of heat transfer in W
A = (Q/(U*Tlm));		               	#Surface area of heat exchanger in m**2
ms = ((Q*3600)/(hfg*1000));		    	#Rate of condensation of steam in kg/h

# Results
print 'Surface area of heat exchanger is %3.2f m**2  \n \
Rate of condensation of steam is %3.1f kg/h'%(A,ms)

Surface area of heat exchanger is 0.78 m**2
Rate of condensation of steam is 133.2 kg/h


## Example 12.7 Page No : 516¶

In [11]:
import math
# Variables
T = 100.;			#Temperature of saturated steam in degree C
t1 = 30.;			#Inlet temperature of water in degree C
t2 = 70.;			#Exit temperature of water in degree C

# Calculations
#COUNTER FLOW
Tc = (T-t2);			#Temperature difference between saturated steam and exit water temperature in degree C
tc = (T-t1);			#Temperature difference between saturated steam and inlet water temperature in degree C
Tlmc = ((Tc-tc)/math.log(Tc/tc));			#LMTD for counter flow in degree C

#PARALLEL FLOW
Tp = (T-t1);			#Temperature difference between saturated steam and inlet water temperature in degree C
tp = (T-t2);			#Temperature difference between saturated steam and exit water temperature in degree C
Tlmp = ((Tp-tp)/math.log(Tp/tp));			#LMTD for counter flow in degree C
#CROSS FLOW
R = ((T-T)/(t2-t1));			#R value for Correction factor F
P = ((t2-t1)/(T-t1));			#P value for Correction Factor F
F = 1;			#Referring to Fig.12.12 in page no 515
Tlmx = (F*Tlmc);			#LMTD for cross flow in degree C

# Results
print 'The effective math.log mean temperature difference for: \
\ni)COUNTER FLOW is %3.1f degree C  \
\nii)PARALLEL FLOW is %3.1f degree C  \niii)CROSS FLOW is %3.1f degree C'%(Tlmc,Tlmp,Tlmx)

The effective math.log mean temperature difference for:
i)COUNTER FLOW is 47.2 degree C
ii)PARALLEL FLOW is 47.2 degree C
iii)CROSS FLOW is 47.2 degree C


## Example 12.8 Page No : 516¶

In [12]:
# Variables
Ti = 18;			#Inlet temperature of Shell fluid in degree C
To = 6.5;			#Outlet temperature of Shell fluid in degree C
ti = -1.1;			#Inlet temperature of Tube fluid in degree C
to = 2.9;			#Outlet temperature of Tube fluid in degree C
U = 850;			#Overall heat transfer coefficient in W/m**2.K
Q = 6000;			#Design heat load in W

# Calculations
T = (Ti-to);			#Temperature difference between shell side inlet fluid and tube side outlet fluid in degree C
t = (To-ti);			#Temperature difference between shell side outlet fluid and tube side inlet fluid in degree C
Tlm = ((T-t)/math.log(T/t));			#LMTD for a counterflow arrangement in degree C
P = ((to-ti)/(Ti-ti));			#P value to calculate correction factor
R = ((Ti-To)/(to-ti));			#R value to calculate correction factor
F = 0.97			#Taking correction factor from fig. 12.9 on page no.514
A = (Q/(U*F*Tlm));			#Area of shell aand tube heat exchanger in m**2

# Results
print 'Area of shell-and-tube heat exchanger is %3.2f m**2'%(A)

Area of shell-and-tube heat exchanger is 0.67 m**2


## Example 12.9 Page No : 517¶

In [13]:
# Variables
Q = 6000;			#Taking design heat load value in W from Example no. 12.8 on page no.516
U = 850;			#Taking overall heat transfer coefficient value in W/m**2.K from Example no. 12.8 on page no.516
Tlm = 10.92			#Taking LMTD for a counterflow arrangement in degree C from Example no. 12.8 on page no.517
R = 2.875;			#Taking R value from Example no. 12.8 on page no.517
P = 0.209;			#Taking P value from Example no. 12.8 on page no.517
F = 0.985;			#Taking correction factor from Fig. 12.10 on page no.514

# Calculations
A = (Q/(U*F*Tlm));			#Area of shell-and-tube heat exchanger in m**2

# Results
print 'Area of shell aand tube heat exchanger is %3.3f m**2'%(A)

Area of shell aand tube heat exchanger is 0.656 m**2


## Example 12.10 Page No : 517¶

In [14]:
# Variables
Ti = 360.;			#Inlet temperature of hot fluid in degree C taken from Example no. 12.4 on page no. 510
To = 300.;			#Outlet temperature of hot fluid in degree C taken from Example no. 12.4 on page no. 510
ti = 30.;			#Inlet temperature of cold fluid in degree C taken from Example no. 12.4 on page no. 510
to = 200.;			#Outlet temperature of cold fluid in degree C taken from Example no. 12.4 on page no. 510
U = 800.;			#Overall heat transfer coefficient in W/m**2.K taken from Example no. 12.4 on page no. 510
Q = 150000.;			#Calculated heat transfer rate in W from Example no. 12.4 on page no. 510
Tlm = 210.22			#Calculated LMTD for counterflow arrangement in degree C taken from Example no. 12.4 on page no. 511

# Calculations
P = ((to-ti)/(Ti-ti));			#P value for calculation of correction factor
R = ((Ti-To)/(to-ti));			#R value for calculation of correction factor
F = 0.98;			#Correction Factor value taken from Fig.12.11 on page no.515
A = (Q/(U*F*Tlm));			#Required surface area in a cross flow heat exchanger in m**2

# Results
print 'The required surface area in a cross flow heat exchanger is %3.2f m**2'%(A)

The required surface area in a cross flow heat exchanger is 0.91 m**2


## Example 12.11 Page No : 518¶

In [4]:
import math
# Variables
mc = 4.;			#Mass flow rate of cold water in kg/s
Tci = 38.;			#Inlet Temperature of cold water in degree C
Tco = 55.;			#Outlet Temperature of cold water in degree C
D = 0.02;			#Diameter of the tube in m
v = 0.35;			#Velocity of water in m/s
Thi = 95.;			#Inlet Temperature of hot water in degree C
mh = 2.;			#Mass flow rate of hot water in kg/s
L = 2.;			    #Length of the tube in m
U = 1500.;			#Overall heat transfer coefficient in W/m**2.K
c = 4186.;			#Specific heat of water in J/kg.K
d = 1000.;			#Density of water in kg/m**3

# Calculations
Q = (mc*c*(Tco-Tci));			#Heat transfer rate for cold fluid in W
Tho = (Thi-(Q/(mh*c)));			#Outlet temperature of hot fluid in degree C
T = Thi-Tco;			#Difference of temperature between hot water inlet and cold water outlet in degree C
t = Tho-Tci;			#Difference of temperature between hot water outlet and cold water inlet in degree C
Tlm = ((T-t)/math.log(T/t));			#LMTD for counterflow heat exchanger
A = (Q/(U*Tlm));			#Area of heat exchanger in m**2
A1 = (mc/(d*v));			#Total flow area in m**2
n = ((A1*4)/(3.14*D**2));			#Number of tubes
L = (A/(36*3.14*D));			#Length of each tube taking n = 36 in m
N = 2;			#Since this length is greater than the permitted length of 2m, we must use more than one tube pass. Let us try 2 tube passes
P = ((Tco-Tci)/(Thi-Tci));			#P value for calculation of correction factor
R = ((Thi-Tho)/(Tco-Tci));			#R value for calculation of correction factor
F = 0.9;			#Corrction Factor from Fig.12.9 on page no. 514
A2 = (Q/(U*F*Tlm));			#Total area required for one shall pass,2 tube pass exchanger in m**2
L1 = (A2/(2*36*3.14*D));			#Length of tube per pass taking n = 36 in m

# Results
print 'Number of tubes per pass is %.f  \n \
Number of passes is %i  \n \
Length of tube per pass is %3.3f m'%(n,N,L1)

Number of tubes per pass is 36
Number of passes is 2
Length of tube per pass is 1.518 m


## Example 12.12 Page No : 524¶

In [18]:
import math
# Variables
mh = 250.;			#Mass flow rate of hot liquid in kg/h
ch = 3350.;			#Specific heat of hot liquid in J/kg.K
Thi = 120.;			#Inlet temperature of hot liquid in degree C
mc = 1000.;			#Mass flow rate of cold liquid in kg/h
Tci = 10.;			#Inlet temperature of cold liquid in degree C
U = 1160.;			#Overall heat transfer coefficient in W/m**2.K
A = 0.25;			#Surface area of heat exchanger in m**2
cc = 4186.;			#Specific heat of cold liquid in J/kg.K

# Calculations
Cc = ((mc*cc)/3600);			#Heat capacity rate for cold liquid in W/K
Ch = ((mh*ch)/3600);			#Heat capacity rate for hot liquid in W/K
Cmin = min(Cc,Ch);			#Minimum heat capacity rate in W/K
Cmax = max(Cc,Ch);			#Maximum heat capacity rate in W/K
r = (Cmin/Cmax);			#Ratio of min amd max heat capacity rates
NTU = ((U*A)/Cmin);			#Number of transfer units
e = ((1-math.exp(-NTU*(1+r)))/(1+r));			#Effectiveness for a parallel flow heat exchanger
Qmax = (Cmin*(Thi-Tci));			#Maximum possible heat transfer rate in W
Q = (e*Qmax);			#Actual rate of heat transfer in W
Tco = ((Q/Cc)+Tci);			#Outlet temperature of cold liquid in degree C
Tho = (Thi-(Q/Ch));			#Outlet temperature of hot liquid in degree C

# Results
print 'Effectiveness for a parallel flow heat exchanger is %3.3f \n \
Outlet temperature of water is %3.2f degree C  \n \
Outlet temperature of cooled liquid is %3.2f degree C'%(e,Tco,Tho)

Effectiveness for a parallel flow heat exchanger is 0.647
Outlet temperature of water is 24.23 degree C
Outlet temperature of cooled liquid is 48.87 degree C


## Example 12.13 Page No : 527¶

In [5]:
# Variables
Tci = 15.;			#Inlet temperature of water in degree C
mc = 1300.;			#Mass flow rate of water in kg/h
ch = 2000.;			#Specific heat of oil in J/kg.K
mh = 550.;			#Mass flow rate of oil in kg/h
Thi = 94.;			#Inlet temperature of oil in degree C
A = 1.;			    #Area of heat exchanger in m**2
U = 1075.;			#Overall heat transfer coefficient in W/m**2.K
cc = 4186.;			#Specific heat of water in J/kg.K

# Calculations
Cc = ((mc*cc)/3600);			#Heat capacity of water in W/K
Ch = ((mh*ch)/3600);			#Heat capacity of oil in W/K
Cmin = min(Cc,Ch);			#Minimum heat capacity in W/K
Cmax = max(Cc,Ch);			#Maximum heat capacity in W/K
r = (Cmin/Cmax);			#Ratio of min and max heat capacity
NTU = ((U*A)/Cmin);			#Number of transfer Units
e = 0.94			#Effectiveness of heat exchanger from Fig. 12.15 on page no.524
Qmax = (Cmin*(Thi-Tci));			#Maximum possible heat transfer rate in W
Q = (e*Qmax);			#Actual heat transfer rate in W
Tco = ((Q/Cc)+Tci);			#Outlet Temperature of water in degree C
Tho = (Thi-(Q/Ch));			#Outlet Temperature of oil in degree C

# Results
print 'The total heat transfer is %3.1f W  \n \
Outlet Temperature of water is %i degree C  \n \
Outlet Temperature of oil is %3.2f degree C'%(Q,Tco,Tho)

The total heat transfer is 22690.6 W
Outlet Temperature of water is 30 degree C
Outlet Temperature of oil is 19.74 degree C


## Example 12.14 Page No : 528¶

In [20]:
# Variables
N = 3000.;			#Number of brass tubes
D = 0.02;			#Diameter of brass tube in m
Tci = 20.;			#Inlet temperature of cooling water in degree C
mc = 3000.;			#Mass flow rate of cooling water in kg/s
ho = 15500.;			#Heat transfer coefficient for condensation in W/m**2.K
Q = (2.3*10**8);			#Heat load of the condenser in W
Thi = 50.;			#Temperature at which steam condenses in degree C
hfg = 2380.			#Enthalpy of liquid vapour mixture in kJ/kg
m = 1.;			#Flow rate of each tube in kg/s
Cc = 4180.;			#Specific heat of water in J/kg.K
#Properties of water at 300K from data book
Cc = 4186.;			#Specific heat in J/kg.K
mu = (855.*10**-6);			#Dynamic vismath.cosity in Ns/m**2
k = 0.613;			#Thermal Conductivity in W/mK
Pr = 5.83			#Prantl number

# Calculations
Tco = ((Q/(mc*Cc))+Tci);			#Outlet temperature of cooling water in degree C
Re = ((4*m)/(3.1415*D*mu));			#Reynold's number
Nu = (0.023*Re**(4./5)*Pr**(2./5));			#Nusselts number
hi = (Nu*(k/D));			#Heat transfer coefficient in W/m**2.K
U = (1./((1./ho)+(1./hi)));			#Overall heat transfer coefficient in W/m**2.K
Cmin = (mc*Cc);			#Minimum heat capacity in W/K
Qmax = (Cmin*(Thi-Tci));			#Maximum heat transfer rate in W
e = (Q/Qmax);			#Effectiveness of heat transfer
NTU = 0.8;			#Number of transfer units from Fig. 12.16 on page no.525
A = ((NTU*Cmin)/U);			#Area of heat exchanger in m**2
L = (A/(2*N*3.1415*D));			#Length of tube per pass in m
ms = (Q/(hfg*1000));			#Amount of steam condensed in kg/s

# Results
print 'The outlet temperature of the cooling water is %3.2f degree C  \n \
The overall heat transfer coefficient is %3.1f W/m**2.K  \n \
Tube length per pass using NTU method is %3.2f m  \n \
The rate of condensation of steam is %3.0f kg/s'%(Tco,U,L,ms)

The outlet temperature of the cooling water is 38.32 degree C
The overall heat transfer coefficient is 6525.8 W/m**2.K
Tube length per pass using NTU method is 4.08 m
The rate of condensation of steam is  97 kg/s


## Example 12.15 Page No : 530¶

In [21]:
# Variables
Tci = 5.;			#Inlet temperature of water in degree C
mc = 4600.;			#Mass flow rate of water in kg/h
mh = 4000.;			#Mass flow rate of air in kg/h
Thi = 40.;			#Inlet temperature of air in degree C
U = 150.;			#Overall heat transfer coefficient in W/m**2.K
A = 25.;			#Area of heat exchanger in m**2
Cc = 4180.;			#Specific heat of water in J/kg.K
Ch = 1010.;			#Specific heat of air in J/kg.K

# Calculations
C1 = ((mh*Ch)/3600);			#Heat capacity of air in W/K
C2 = ((mc*Cc)/3600);			#Heat capacity of water in W/K
Cmin = min(C1,C2);			#Minimum value of heat capacity in W/K
Cmax = max(C1,C2);			#Maximum value of heat capacity in W/K
r = (Cmin/Cmax);			#Ratio of min and max heat capacity in W/K
NTU = ((U*A)/Cmin);			#Number of heat transfer units
e = 0.92;			#Effectiveness of heat exchanger from Fig. 12.18 on page no.526
Q = (e*Cmin*(Thi-Tci));			#Heat transfer rate in W
Tco = ((Q/C2)+Tci);			#Outlet temperature of water in degree C
Tho = (Thi-(Q/C1));			#Outlet temperature of air in degree C

# Results
print 'The exit temperature of water is %3.1f degree C  \n \
The exit temperature of air is %3.1f degree C'%(Tco,Tho)

The exit temperature of water is 11.8 degree C
The exit temperature of air is 7.8 degree C


## Example 12.16 Page No : 532¶

In [23]:
# Variables
A = 15.82;			#Total outside area of heat exchanger in m**2
Thi = 110;			#Inlet temperature of oil in degree C
Ch = 1900;			#Specific heat of oil in J/kg.K
mh = 170.9;			#Mass flow rate of oil in kg/min
mc = 68;			#Mass flow rate of water in kg/min
Tci = 35;			#Inlet temperature of water in degree C
U = 320;			#Overall heat transfer coefficient in W/m**2.K
Cc = 4186;			#Specific heat of water in J/kg.K

# Calculations
C1 = ((mh*Ch)/60);			#Heat capacity of oil in W/K
C2 = ((mc*Cc)/60);			#Heat capacity of water in W/K
D = (U*A*((1./C1)-(1./C2)));			#Constant
r = (C1/C2);			#Ratio of heat capacity of oil and water
Tho = Thi-(((Thi-Tci)*(1-math.exp(D)))/(r-math.exp(D)));			#Outlet temperature of oil in degree C
Tco = Tci+(r*(Thi-Tho));			#Outlet temperature of water in degree C

# Results
print 'The exit temperature of oil is %3.2f degree C \n \
The exit temperature of water is %3.1f degree C'%(Tho,Tco)

The exit temperature of oil is 74.97 degree C
The exit temperature of water is 75.0 degree C


## Example 12.17 Page No : 533¶

In [24]:
# Variables
Tci = 20.;			#Inlet temperature of water in degree C
Tco = 50.;			#Outlet temperature of water in degree C
Th = 120.;			#Temperature at which steam condenses in degree C
newTci = 15.;			#New Inlet temperature of water in degree C

# Calculations
newTco = (((Tco-Tci)*(Th-newTci))/(Th-Tci))+newTci;			#New outlet temperature of water in degree C

# Results
print 'New outlet temperature of water is %3.1f degree C'%(newTco)

New outlet temperature of water is 46.5 degree C


## Example 12.18 Page No : 534¶

In [8]:
# Variables
T = 100;			#Total length of tubes in m

# Calculations
D = ((3.14*4000)/(3.14*30000))**0.5;			#Diameter of the exchanger in m
L = (2./(3.1415*D**2));			#Length of the exchanger in m
Cost = (10000+(15000*D**3*L)+(2000*D*L));			#Optimal math.cost in Rs

# Results
print 'The diameter of the exchanger is %3.3f m  \n \
The Length of the exchanger is %3.2f m  \n \
Optimal cost is %3.0f Rs'%(D,L,Cost)

# note : rounding error.

The diameter of the exchanger is 0.365 m
The Length of the exchanger is 4.77 m
Optimal cost is 16974 Rs