Chapter 18: Other Heat Exchange Equipment

ILLUSTRATIVE EXAMPLE 18.2, Page number: 384

In [2]:
from __future__ import division

#Variable declaration:
T1 = 25                      #Temperature of H2SO4 (°C)
m = 50+200                   #Mass of H2SO4 (lb)
#From figure 18.2:
W1 = 50+100                  #Weight of H2SO4 (lb)
W2 = 100                     #Weight of H2O (lb)

#Calculation:
m = W1/(W1+W2)*100          #Percent weight of H2SO4 (%)
m2 = W1+W2                  #Mass of mixture (lb)
#From fgure 18.2:
T2 = 140                    #Final temperature between the 50% solution and pure H2SO4 at  25°C (°F)
h1 = -86                    #Specific heat capacity of H2O (Btu/lb)
h2 = -121.5                 #Specific heat capacity of H2SO4 (Btu/lb)
Q = m2*(h2-h1)              #Heat transferred (Btu)

#Result:
print "The final temperature between the 50% solution and pure H2SO4 at  25°C is :",round(T2)," °F ."
print "The heat transferred is :",round(Q)," Btu ."
The final temperature between the 50% solution and pure H2SO4 at  25°C is : 140.0  °F .
The heat transferred is : -8875.0  Btu .

ILLUSTRATIVE EXAMPLE 18.3, Page number: 386

In [3]:
from __future__ import division

#Variable declaration:
F = 10000                       #Mass flow rate of NaOH (lb/h)
C1 = 10                         #Old concentration of NaOH solution (%)
C2 = 75                         #New concentration of NaOH solution (%)
h1 = 1150                       #Enthalpy of saturated steam at 14.7 psia (Btu/lb)
U = 500                         #Overall heat transfer coefficient (Btu/h.ft^2.°F)
T1 = 212                        #Absolute temperature of evaporator (°F)
T2 = 340                        #Saturated steam temperature (°F)

#Calculation:
L = F*(C1/100)/(C2/100)         #Flow rate of steam leaving the evaporator (lb/h)
V = F-L                         #Overall material balance (lb/h)
#From figure 18.3:
hF = 81                         #Enthalpy of solution entering the unit (Btu/lb)
hL = 395                        #Enthalpy of the 75% NaOH solution (Btu/lb)
Q = round(V)*h1+round(L)*hL-F*hF    #Evaporator heat required (Btu/h)
A = Q/(U*(T2-T1))               #Area of the evaporaor (ft^2)

#Result:
print "The heat transfer rate required for the evaporator is :",round(Q,-2)," Btu/h ."
print "The area requirement in the evaporator is :",round(A,1)," ft^2 ."
The heat transfer rate required for the evaporator is : 9683600.0  Btu/h .
The area requirement in the evaporator is : 151.3  ft^2 .

ILLUSTRATIVE EXAMPLE 18.4, Page number: 388

In [4]:
from __future__ import division

#Variable declaration:
U1 = 240                    #Overall heat transfer coefficient for first effect (Btu/h.ft^2.°F)
U2 = 200                    #Overall heat transfer coefficient for second effect (Btu/h.ft^2.°F)
U3 = 125                    #Overall heat transfer coefficient for third effect (Btu/h.ft^2.°F)
A1 = 125                    #Heating surface area in first effect (ft^3)
A2 = 150                    #Heating surface area in second effect (ft^3)
A3 = 160                    #Heating surface area in third effect (ft^3)
T1 = 400                    #Condensation stream temperature in the first effect (°F)
T2 = 120                    #Vapor leaving temperature in the first effect (°F)

#Calculation:
R1 = 1/(U1*A1)              #Resistance across first effect
R2 = 1/(U2*A2)              #Resistance across second effect
R3 = 1/(U3*A3)              #Resistance across third effect
R = R1+R2+R3                #Total resistance
DT1 = (R1/R)*(T1-T2)        #Temperature drop across the heating surface in the first effect (°F)

#Result:
print "The temperature drop across the heating surface in the first effect is :",round(DT1)," °F ."
The temperature drop across the heating surface in the first effect is : 80.0  °F .

ILLUSTRATIVE EXAMPLE 18.6, Page number: 389

In [6]:
from __future__ import division

#Variable declaration:
F = 5000                        #Mass of soltuion fed in the evaporator (lb)
xF = 2/100                      #Concentration of feed
xL = 5/100                      #Concentration of liquor
U = 280                         #Overall heat transfer coefficient (Btu/h.ft^2.°F)
#From figure 18.1 & 18.3:
TF = 100                        #Feed temperature (°F)
TS = 227                        #Steam temperature (°F)
TV = 212                        #Vapour temperature (°F)
TL = 212                        #Liquor temperature (°F)
TC = 227                        #Condensate temperature (°F)

#Calculation:
#From steam tables:
hF = 68                         #Enthalpy of feed (Btu/lb)
hL = 180                        #Enthalpy of liquor (Btu/lb)
hV = 1150                       #Enthalpy of vapour (Btu/lb)
hS = 1156                       #Enthalpy of steam (Btu/lb)
hC = 195                        #Enthalpy of condensate (Btu/lb)
s1 = F*xF                       #Total solids in feed (lb)
w = F-s1                        #Total water in feed (lb)
s2 = F*xF                       #Total solids in liquor (lb)
L = s2/xL                       #Total water in liquor (lb)
V = F-L                         #Overall balance (lb)
S = (V*hV+L*hL-F*hF)/(hS-hC)    #Mass of steam (lb)
Q = S*(hS-hC)                   #Total heat requirement (Btu)
A = Q/(U*(TS-TL))               #Required surface aea (ft^2)

#Result:
print "The mass of vapor produced is :",round(V)," lb ."
print "The total mass of steam required is :",round(S)," lb ."
print "The surface area required is :",round(A)," ft^2 ."
The mass of vapor produced is : 3000.0  lb .
The total mass of steam required is : 3611.0  lb .
The surface area required is : 826.0  ft^2 .

ILLUSTRATIVE EXAMPLE 18.7, Page number: 390

In [7]:
from __future__ import division

#Variable declaration:
F = 5000                        #Mass flow rate of NaOH (lb/h)
xF = 20/100                     #Old concentration of NaOH solution
TF = 100                        #Feed temperature (°F)
xL = 40/100                     #New concentration of NaOH solution
xv = 0                          #Vapour concentration at x
yv = 0                          #Vapour concentration at y
T1 = 198                        #Boiling temperature of solution in the evaporator (°F)
T2 = 125                        #Saturated steam temperature (°F)
U = 400                         #Overall heat transfer coefficient (Btu/h.ft^2.°F)
Ts = 228                        #Steam temperature (°F)

#Calculation:
#From steam tables at 228°F and 5 psig:
hS = 1156                       #Enthalpy of steam (Btu/lb)
hC = 196                        #Enthalpy of condensate (Btu/lb)
hV = hS-hC                      #Enthalpy of vapour (Btu/lb)
Tw = 125.4                      #Boiling point of water at 4 in Hg absolute (°F)
hS2 = 1116                      #Enthalpy of saturated steam at 125°F (Btu/lb)
hs = 0.46                       #Heat capacity of superheated steam (Btu/lb.°F)
#From figure 18.3:
hF = 55                         #Enthalpy of feed (Btu/lb)
hL = 177                        #Enthalpy of liquor (Btu/lb)
L = F*xF/xL                     #Mass of liquor (lb)
V = L                           #Mass of vapour (lb)
hV = hS2+hs*(T1-T2)             #Enthalpy of vapour leaving the solution (Btu/lb)
S = (V*hV+L*hL-F*hF)/(hS-hC)    #Mass flow rate of steam (lb/h)
Q = S*(hS-hC)                   #Total heat requirement (Btu)
A = Q/(U*(Ts-T1))               #Required heat transfer area (ft^2)

#Result:
print "The steam flow rate is :",round(S,-1)," lb/h ."
print "The required heat transfer area is :",round(A)," ft^2 ."
The steam flow rate is : 3170.0  lb/h .
The required heat transfer area is : 253.0  ft^2 .

ILLUSTRATIVE EXAMPLE 18.10, Page number: 398

In [10]:
from __future__ import division

#Variable declaration:
T1 = 2000                       #Hot gas temperature (°F)
T2 = 550                        #Cool gas temperature (°F)
T3 = 330                        #Steam temperature (°F)
T4 = 140                        #Water temperature (°F)
m = 30000                       #Mass flow rate of steam (lb/h)
cp = 0.279                      #Average heat capacity of gas (Btu/lb.°F)
N = 800                         #Number of boiler tubes

#Calculation:
DT = (T1-T3)/(T2-T3)            #Temperature difference ratio
Tav = (T1+T2)/2                 #Average gas temperature (°F)
#From steam tables (Appendix):
hs = 1187.7                     #Steam enthalpy (Btu/lb)
hw = 107.89                     #Water enthalpy (Btu/lb)
Q = m*(hs-hw)                   #Heat duty (Btu/h)
mh = Q/cp*(T1-T2)               #Mass flow rate of gas (lb/h)
x = mh/N                        #Gas mass flow rate per tube (lb/h)
#From figure 18.5:
L = 15                          #Length of boiler tubes (ft)

#Result:
print "The length of boiler tubes is :",L," ft ."
The length of boiler tubes is : 15  ft .

ILLUSTRATIVE EXAMPLE 18.12, Page number: 399

In [12]:
from __future__ import division

#Variable declaration:
T1 = 1800                       #Hot gas temperature (°F)
T2 = 500                        #Cool gas temperature (°F)
#From steam tables:
Tw = 312                        #Boiling point of water at 80 psia (°F)
m1 = 120000                     #Mass flow rate of flue gas (lb/h)
D = 2/12                        #Inside diameter of tube (ft)
cp = 0.26                       #Average heat capacity of flue gas (Btu/lb.°F)

#Calculation:
DT = (T1-Tw)/(T2-Tw)            #Temperature difference ratio
Tav = (T1+T2)/2                 #Average gas temperature (°F)
#From figure 18.4:
x = 150                         #Gas mass flow rate per tube (m/N) (lb/h)
N = m1/x                        #Number of tubes
L = 21.5                        #Length of tubes (ft)
A = N*L*D                       #Total heat transfer area (ft^2)
Q = m1*cp*(T1-T2)               #Heat duty (Btu/h)
#From steam tables (Appendix):
hs = 1183.1                     #Steam enthalpy at 80 psia (Btu/lb)
hw = 168.1                      #Water enthalpy at 200°F (Btu/lb)
m2 = Q/(hs-hw)                  #Mass flow rate of water (lb/h)

#Result:
print "The required heat transfer area is :",round(A)," ft^2 ."
print "The tube length is :",L," ft ."
print "The heat duty is :",round(Q/10**7,2)," x 10^7 ."
print "The water mass flow rate is :",round(m2,-4)," lb/h ."
The required heat transfer area is : 2867.0  ft^2 .
The tube length is : 21.5  ft .
The heat duty is : 4.06  x 10^7 .
The water mass flow rate is : 40000.0  lb/h .

ILLUSTRATIVE EXAMPLE 18.18, Page number: 407

In [18]:
#Variable declaration:
m1 = 144206                     #Mass flow rate of flue gas (lb/h)
cp = 0.3                        #Average flue gas heat capacity (Btu/lb.°F)
T1 = 2050                       #Initial temperature of gas (°F)
T2 = 560                        #Final temperature of gas (°F)
T3 = 70                         #Ambient air temperature (°F)

#Calculation:
Q = m1*cp*(T1-T2)               #Duty rate (Btu/h)
#From appendix:
cpa = 0.243                     #Average ambient air heat capacity 70°F (Btu/lb.°F)
MW = 29                         #Molecular weight of air at 70°F
ma = round(Q,-5)/(cpa*(T2-T3))  #Mass of air required (lb/h)
m2 = round(ma)/MW               #Moles of air required (lb mol/h)
m3 = round(ma)*13.32            #Volume of air required (ft^3/h)

#Result:
print "The mass of air required is :",round(ma,-2)," lb/h ."
print "The moles of air required is :",round(m2,-1),"lb mol/h ."
print "The volume of air required is :",round(m3,-3)," ft^3/h ."
The mass of air required is : 541700.0  lb/h .
The moles of air required is : 18680.0 lb mol/h .
The volume of air required is : 7215000.0  ft^3/h .

ILLUSTRATIVE EXAMPLE 18.19, Page number: 407

In [19]:
#Variable declaration:
#From example 18.19:
m1 = 144200                     #Mass flow rate of flue gas (lb/h)
m2 = 541700                     #Mass flow rate of air (lb/h)
R = 0.73                        #Universal gas constant (psia.ft^3/lbmol.°R)
P = 1                           #Absolute pressure (psia)
T = 1020                        #Absolute temperature (°R)
MW = 29                         #Molecular weight of air
t = 1.5                         #Residence time (s)

#Calculation:
m = m1+m2                       #Total mass flow rate of the gas (lb/h)
q = m*R*T/(P*MW)                #Volumetric flow at 560°F (ft^3/h)
V = q*t/3600                    #Volume of tank (ft^3)

#Result:
print "The total mass flow rate of the gas is :",round(m,-2)," lb/h ."
print "The volumetric flow at 560°F is :",round(q/10**7,2)," x 10^7 ft^3/h ."
print "The volume of tank is :",round(V)," ft^3 ."
The total mass flow rate of the gas is : 685900.0  lb/h .
The volumetric flow at 560°F is : 1.76  x 10^7 ft^3/h .
The volume of tank is : 7338.0  ft^3 .

ILLUSTRATIVE EXAMPLE 18.20, Page number: 408

In [21]:
from __future__ import division
from math import pi

#Variable declaration:
#Fro example 18.20:
V = 7335                    #Volume of tank (ft^3)

#Calculation:
D = (4*V/pi)**(1/3)         #Diameter of tank (ft)
H = D                       #Height of tube (ft)

#Result:
print "The diameter of tank is :",round(H,2)," ft ."
print "The height of tube is :",round(D,2)," ft ."
The diameter of tank is : 21.06  ft .
The height of tube is : 21.06  ft .

ILLUSTRATIVE EXAMPLE 18.21, Page number: 408

In [22]:
from __future__ import division

#Variable declaration:
m1 = 144206                     #Mass flow rate of flue gas (lb/h)
cp1 = 0.3                       #Average heat capacities of the flue gas (Btu/lb°F)
cp2 = 0.88                      #Average heat capacities of the solid (Btu/lb°F)
#From example 18.18:
T1 = 550                        #Initial temperature of gas (°F)
T2 = 2050                       #Final temperature of gas (°F)
T3 = 70                         #Initial temperature of solid (°F)
T4 = 550-40                     #Final temperature of solid (°F)

#Calculation:
Dhf = m1*cp1*(T2-T1)            #For the flue gas, the enthalpy change for one hour of operation (Btu)
Dhs = round(Dhf,-4)             #For the solids, the enthalpy change for one hour of operation (Btu)
m2 = Dhs/(cp2*(T4-T3))          #Mass of solid (lb)

#Result:
print "The mass of solid is :",round(m2)," lb ."
The mass of solid is : 167588.0  lb .

ILLUSTRATIVE EXAMPLE 18.22, Page number: 409

In [23]:
from __future__ import division
from math import log,sqrt,pi

#Variable declaration:
#From example 18.21:
m = 144206                      #Mass flow rate of flue gas (lb/h)
cp = 0.3                        #Average heat capacities of the flue gas (Btu/lb°F)
T1 = 2050                       #Initial temperature of gas (°F)
T2 = 180                        #Final temperature of gas (°F)
T3 = 60                         #Ambient air temperature (°F)
U = 1.5                         #Overall heat transfer coefficient for cooler (Btu/h.ft^2.°F)
MW = 28.27                      #Molecular weight of gas
R = 379                         #Universal gas constant (psia.ft^3/lbmol.°R)
v = 60                          #Duct or pipe velcity at inlet (2050°F) (ft/s)

#Calculation:
Q = m*cp*(T1-T2)                #Heat duty (Btu/h)
DTlm = ((T1-T3)-(T2-T3))/log((T1-T3)/(T2-T3))   #Log-mean temperature difference (°F)
A1 = round(Q,-5)/(U*round(DTlm))                 #Radiative surface area (ft^2)
q = m*R*(T1+460)/(T3+460)/MW    #Volumetric flow at inlet (ft^3/h)
A2 = q/(v*3600)                 #Duct area (ft^2)
D = sqrt(A2*4/pi)               #Duct diameter (ft)
L = A1/(pi*D)                   #Length of required heat exchange ducting (ft)

#Result:
print " The radiative surface area required is :",round(A1,-1)," ft^2 ."
print " The length of required heat exchange ducting is :",round(L)," ft ."
 The radiative surface area required is : 80980.0  ft^2 .
 The length of required heat exchange ducting is : 3476.0  ft .