# Chapter 22: Design Principles and Industrial Applications¶

### ILLUSTRATIVE EXAMPLE 22.6, Page number: 471¶

In [6]:
from __future__ import division
from sympy import symbols,solve

#Variable declaration:
#From steam tables:
h1 = 1572                           #Enthalpy for super heated steam at (P = 40 atm, T = 1000°F) (Btu/lb)
h2 = 1316                           #Enthalpy for super heated steam at (P = 20 atm, T = 600°F) (Btu/lb)
h3 = 1151                           #Enthalpy for saturated steam (Btu/lb)
h4 = 28.1                           #Enthalpy for saturated water (Btu/lb)
m1 = 1000                           #Mass flowrate of  steam (lb/h)
m = symbols('m')                    #Mass flow rate of steam (lb/h)

#Calculation:
Dh1 = m1*(h3-h4)                    #The change in enthalpy for the vaporization of the water stream (Btu/h)
Dh2 = m*(h1-h2)                     #The change in enthalpy for the cooling of the water stream (Btu/h)
x = solve(Dh1-Dh2,m)                #Mass flowrate of steam (lb/h)
m2 = x[0];                          #Mass flowrate of steam (lb/h)

#Result:
print "The mass flowrate of the utility steam required is :",round(m2)," lb/h."

The mass flowrate of the utility steam required is : 4386.0  lb/h.


### ILLUSTRATIVE EXAMPLE 22.7, Page number: 473¶

In [7]:
from __future__ import division

#Variable declaration:
#From table 22.1:
QH1 = 12*10**6                  #Heat duty for process unit 1 (Btu/h)
QH2 = 6*10**6                   #Heat duty for process unit 2 (Btu/h)
QH3 = 23.5*10**6                #Heat duty for process unit 3 (Btu/h)
QH4 = 17*10**6                  #Heat duty for process unit 4 (Btu/h)
QH5 = 31*10**6                  #Heat duty for process unit 5 (Btu/h)
T1 = 90                         #Supply water temperature (°F)
T2 = 115                        #Return water temperature (°F)
cP = 1                          #Cooling water heat capacity (Btu/(lb.°F))
p = 62*0.1337                   #Density of water (lb/gal)
BDR = 5/100                     #Blow-down rate

#Calculation:
QHL = (QH1+QH2+QH3+QH4+QH5)/60  #Heat load (Btu/min)
DT = T2-T1                      #Change in temperature (°F)
qCW = round(QHL,-5)/(DT*cP*p)   #Required cooling water flowrate (gpm)
qBD = BDR*qCW                   #Blow-down flow (gpm)

#Result:
print "The total flowrate of cooling water required for the services is :",round(qCW,-1)," gpm."
print "The required blow-down flow is :",round(qBD)," gpm."

The total flowrate of cooling water required for the services is : 7240.0  gpm.
The required blow-down flow is : 362.0  gpm.


### ILLUSTRATIVE EXAMPLE 22.8, Page number: 474¶

In [8]:
from __future__ import division

#Variable declaration:
Q1 = 10*10**6                       #Unit heat duty for process unit 1 (Btu/h)
Q2 = 8*10**6                        #Unit heat duty for process unit 2 (Btu/h)
Q3 = 12*10**6                       #Unit heat duty for process unit 3 (Btu/h)
Q4 = 20*10**6                       #Unit heat duty for process unit 4 (Btu/h)
hv = 751                            #Enthalpy of vaporization for pressure 500 psig (Btu/lb)

#Calculation:
mB1 = Q1/hv                         #Mass flowrate of 500 psig steam through unit 1 (lb/h)
mB2 = Q2/hv                         #Mass flowrate of 500 psig steam through unit 2 (lb/h)
mB3 = Q3/hv                         #Mass flowrate of 500 psig steam through unit 3 (lb/h)
mB4 = Q4/hv                         #Mass flowrate of 500 psig steam through unit 4 (lb/h)
mBT = mB1+mB2+mB3+mB4               #Total steam required (lb/h)

#Result:
print "The total steam required is :", round(mBT,-1)," lb/h."

The total steam required is : 66580.0  lb/h.


### ILLUSTRATIVE EXAMPLE 22.9, Page number: 474¶

In [9]:
from __future__ import division
from sympy import symbols,solve
from math import log,pi

#Variable declaration:
po = 53*16.0185                     #Density of oil (kg/m^3)
co = 0.46*4186.7                    #Heat capacity of oil (J/kg.°C)
muo = 150/1000                      #Dynamic viscosity of oil (kg/m.s)
ko = 0.11*1.7303                    #Thermal conductivity of oil (W/m.°C)
qo = 28830*4.381*10**-8             #Volumetric flowrate of oil (m^3/s)
pw = 964                            #Density of water (kg/m^3)
cw = 4204                           #Heat capacity of water (J/kg.°C)
muw = 0.7/3600*1.4881               #Dynamic viscosity of water (kg/m.s)
kw = 0.678                          #Thermal conductivity of water (W/m.°C)
qw = 8406*4.381*10**-8              #Volumetric flowrate of water (m^3/s)
t1 = 23.5                           #Initial temperature of oil (°C)
t2 = 27                             #Final temperature of oil (°C)
T1 = 93                             #Water heating temperature of water (°C)
T2 = symbols('T2')                  #Minimum temperature of heating water (°C)
A = symbols('A')                    #Heat transfer area (m^2)
Uc = 35.4                           #Clean heat transfer coefficient (W/m^2.K)
Rf = 0.0007                         #Thermal resistance (m^2.K/W)
D = 6*0.0254                        #Inside diameter of pipe (m)

#Calculation:
vo = muo/po                         #Kinematic viscosity of oil (m^2/s)
mo = po*qo                          #Mass flowrate of oil (kg/s)
vw = muw/pw                         #Kinematic viscosity of  (m^2/s)
mw = pw*qw                          #Masss flow rate of water (kg/s)
Q1 = mo*co*(t2-t1)                  #Duty of exchanger of oil (W)
T2m = t1                            #Lowest possible temperature of the water (°C) (part 1)
Qmw = mw*cw*(T1-T2m)                #Maximum duty of exchanger of water (W) (part 2)
Q2 = mw*cw*(T1-T2)                  #Duty of exchanger of water in terms of T2 (W)
x = solve(Q1-Q2,T2)                 #Solving value for T2 (°C)
T3 = x[0];                          #Minimum temperature of heating water (°C)
DT1 = T3-t1                         #Inlet temperature difference (°C)
DT2 = T1-t2                         #Outlet temperature difference (°C)
DTlm = (DT1-DT2)/log(DT1/DT2)       #Log mean temperature difference (°C)
Ud1 = 1/Uc+Rf                       #Dirty heat transfer coefficient (W/m^2.K) (part 3)
Ud2 = 34.6                          #Dirty heat transfer coefficient (W/m^2.°C)
Q3 = Ud2*A*DTlm                     #Duty of exchanger (W) (part 4)
y = solve(Q1-Q3,A)                  #Heat transfer area (m^2)
A1 = y[0];                          #Required heat transfer area (m^2)
L = A1/(pi*D)                       #Required heat transfer length (m)
Qmo = mo*co*(T1-t1)                 #Maximum duty of exchanger of oil (W) (part 5)
Qm = Qmw                            #Maximum duty of exchanger (W)
E = Q1/Qm*100                       #Effectiveness (%)
NTU = Ud2*A1/(mw*cw)                #Number of transfer units

#Result:
print "1. The lowest possible temperature of the water is :",T2m," °C ."
print "2. The log mean temperature difference is :",round(DTlm,2)," °C ."
print "3. The overall heat transfer coefficient for the new clean exchanger is :",round(Ud2,1)," (W/m^2.°C ."
print "4. The length of the double pipe heat exchanger is :",round(L,2)," m ."
print "5. The effectiveness of the exchanger is :",round(E,2)," % ."
print "   The NTU of the exchanger is :",round(NTU,4)," ."

1. The lowest possible temperature of the water is : 23.5  °C .
2. The log mean temperature difference is : 65.33  °C .
3. The overall heat transfer coefficient for the new clean exchanger is : 34.6  (W/m^2.°C .
4. The length of the double pipe heat exchanger is : 6.68  m .
5. The effectiveness of the exchanger is : 6.97  % .
The NTU of the exchanger is : 0.0741  .


### ILLUSTRATIVE EXAMPLE 22.10, Page number: 477¶

In [10]:
from __future__ import division
from math import log,pi

#Variable declaration:
#From example 22.9:
t1 = 23.5                           #Initial temperature of oil (°C)
t2 = 27                             #Final temperature of oil (°C)
T1 = 93                             #Water heating temperature of water (°C)
T2 = 88.16                          #Minimum temperature of heating water (°C)
U = 34.6                            #Overall heat transfer coefficient (W/m^2.°C)
Q = 7227.2                          #Duty of exchanger (W)
D = 6*0.0254                        #Inside diameter of pipe (m)
l = 6.68                            #Previous heat transfer length (m)

#Calculation:
DT1 = T1-t1                         #Inlet temperature difference (°C)
DT2 = T2-t2                         #Outlet temperature difference (°C)
DTlm = (DT1-DT2)/log(DT1/DT2)       #Log mean temperature difference (°C)
A = Q/(U*DTlm)                      #Required heat transfer area (m^2)
L = A/(pi*D)                        #Required heat transfer length (m)

#Result:
print "The length of the parallel pipe heat exchanger is :",round(L,2)," m ."
if L>l:
print "The tube length would increase slightly."
elif L<l:
print "The tube length would decrease slightly."

The length of the parallel pipe heat exchanger is : 6.69  m .
The tube length would increase slightly.


### ILLUSTRATIVE EXAMPLE 22.12, Page number: 478¶

In [1]:
from __future__ import division
from scipy.optimize import fsolve
from math import log

#Variable declaration:
T = 80                                #Pipe surface temperature (°F)
t1 = 10                               #Inlet temperature of brine solution (°F)
m = 1200                              #mass flowrate of solution (kg/s)
c = 0.99                              #Heat capacity of brine solution (Btu/lb.°F)
A = 2.5                               #Heat transfer area (ft^2)
U1 = 150                              #Overall heat transfer coefficient at temperature approach (Btu/h.ft^2.°F)
U2 = 140                              #Overall heat transfer coefficient at inlet brine temperature (Btu/h.ft^2.°F)

#Calculation:
DT1 = T-t1                            #Temperature approach at the pipe entrance (°F)

def equation(DT2):
Q1 = m*c*(DT1-DT2)               #Energy balance to the brine solution across the full length of the pipe (Btu/h)
DTlm = (DT1-DT2)*log(DT2/DT1)    #Log mean temperature difference (°F)
Q2 = A*(U2*DT1-U1*DT2)/log((U2*DT1)/(U1*DT2)) #Heat transfer rate (Btu/h)
return Q2-Q1
t2 = round(T-fsolve(equation,1),1)	   #The temperature of the brine solution (°F)

#Results:
print 'The temperature of brine solution is: ',(t2-32)/1.8,'°C'

The temperature of brine solution is:  -2.0 °C


### ILLUSTRATIVE EXAMPLE 22.13, Page number: 479¶

In [13]:
from __future__ import division
from math import log

#Variable declaration:
m = 1200                            #mass flowrate of solution (kg/s)
c = 0.99                            #Heat capacity of brine solution (Btu/lb.°F)
DT1 = 70                            #Temperature approach at the pipe entrance (°F)
DT2 = 51.6                          #Temperature difference at the pipe exit (°F)

#Calculation:
Q = m*c*(DT1-DT2)                   #Heat transfer rate (Btu/h)
DTlm = (DT1-DT2)/log(DT1/DT2)       #Log mean temperature difference (°F)

#Result:
print "1. The rate of heat transfer is :",round(Q,-1)," Btu/h."
print "Or, the rate of heat transfer is :",round(Q/3.412)," W."
print "2. The log mean temperature difference is :",round(DTlm,1)," °F."
print "Or, the log mean temperature difference is :",round(DTlm/1.8,1)," °C."

1. The rate of heat transfer is : 21860.0  Btu/h.
Or, the rate of heat transfer is : 6407.0  W.
2. The log mean temperature difference is : 60.3  °F.
Or, the log mean temperature difference is : 33.5  °C.


### ILLUSTRATIVE EXAMPLE 22.23, Page number: 484¶

In [25]:
from __future__ import division
from math import log

#Variable declaration:
Too = 100                           #Steam temperature (°C)
Ti = 18                             #Initial temperature of liquid TCA (°C)
Tf = 74                             #Final temperature of liquid TCA (°C)
t = 180                             #Heating time (s)
p = 87.4                            #Density of TCA (lb/ft^3)
V = 18                              #Kinematic viscosity of TCA (m^2/s)
cp = 0.23                           #Heat capacity of TCA (Btu/lb.°F)
U = 200                             #Overall heat transfer coefficient (Btu/h.ft^2.°F)

#Calculation:
ui = Too-Ti                         #Initial excess temperature (°C)
uf = Too-Tf                         #Final excess temperature (°C)
R = log(ui/uf)                      #Ratio t/r
r = t/R                             #Thermal time constant (s)
A = p*V*cp/(3600*U*r)               #Required heating area (ft^3)
Ti_F = Ti*9/5+32                    #Initial temperature in fahrenheit scale (°F)
Tf_F = Tf*9/5+32                    #Final temperature in fahrenheit scale (°F)
Q = p*V*cp*(Tf_F-Ti_F)              #Total amount of heat added (Btu)

#Result:
print "1. The required surface area of the heating coil is :",(A,1)," ft^3 ."
print "2. The total heat added to the liquid TCA is :",round(Q)," Btu ."

1. The required surface area of the heating coil is : (3.2068907918330814e-06, 1)  ft^3 .
2. The total heat added to the liquid TCA is : 36473.0  Btu .


### ILLUSTRATIVE EXAMPLE 22.24, Page number: 486¶

In [27]:
#Variable declaration:
m1 = 62000                      #Mass flowrate of alcohol (lb/h)
h1 = 365                        #Enthalpy of vapour (Btu/lb)
cp = 1                          #Heat capacity of water (Btu/lb.°F)
T1 = 85                         #Entering temperature of water (°F)
T2 = 120                        #Exit temperature of water (°F)
a1 = 2.11                       #Flow area for the shell side (ft^2)
N = 700                         #Total number of tubes
a2 = 0.546                      #Flow area per tube (in^2/tube)
n = 4                           #Number of tube passes
p = 62.5                        #Density of water (lb/ft^3)
L = 16                          #Length of condenser (ft)
hio = 862.4                     #Cooling water inside film coefficient (Btu/h.ft^2.°F)
g = 9.8                         #Gravitational accleration (m^2/s)
Rf = 0.003                      #Fouling factor (Btu/h.ft^2.°F)

#Calculation:
Q1 = m1*h1                      #Heat loss from alcohol (Btu/h)
Q2 = Q1                         #Heat gained by water (Btu/h)
DT = T2-T1                      #Temperature difference (°F)
m2 = Q2/(cp*DT)                 #Water mass flow rate (lb/h)
LMTD = ((T2-32)-(T1-32))/log((T2-32)/(T1-32))   #Log mean temperature difference (°F)
at = (N*a2)/(144*n)             #Total flow area for tube side (ft^2)
G1 = m1/a1                      #Mass velocity of flow in shell side (lb/h.ft^2)
G2 = m2/at                      #Mass velocity of flow in tube side (lb/h.ft^2)
V = G2/(3600*p)                 #Velocity of water (ft/s)
#For alcohol:
kf = 0.105                      #Thermal conductivity (Btu/h.ft.°F)
muf = 0.55*2.42                 #Dynamic viscosity (lb/ft.h)
sf = 0.79                       #
pf = sf*p                       #Density (lb/ft^3)
h = 151*(((kf**3)*(pf**2)*g*muf)/((muf**2)*n*G3))**(1/3)   #Heat transfer coefficient for the shell side (Btu/h.ft^2.°F)
ho = h                          #Outside heat transfer coefficient of the tube bundle (Btu/h.ft^2.°F)
Uc = (hio*ho)/(hio+ho)          #Overall heat transfer coefficient for a new (clean) heat exchanger (Btu/h.ft^2.°F)
A = N*L*0.2618                  #Area for heat transfer (ft^2)
Ud = Q1/(A*DT)                  #Design (D) overall heat transfer coefficient (Btu/h.ft^2.°F)
Rd = (Uc-Ud)/(Uc*Ud)            #Dirt (d) factor (Btu/h.ft^2.°F)

#Result:
print "The dirt (d) factor is :",round(Rd,4)," Btu/h.ft^2.°F ."
if (Rd>Rd):
print "Therefore, the exchanger as specified is unsuitable for these process conditions since the fouling factor is above the recommended value. Cleaning is recommended."
else:
print "Therefore, the exchanger as specified is suitable for these process conditions since the fouling factor is below the recommended value. Cleaning is not recommended."

The dirt (d) factor is : -0.0157  Btu/h.ft^2.°F .
Therefore, the exchanger as specified is suitable for these process conditions since the fouling factor is below the recommended value. Cleaning is not recommended.
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