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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```

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 ."
```