In [1]:

```
import math
## Example 11.1
print('Example 11.1\n\n');
print('Page No. 308\n\n');
##given
V = 205.;## Flow rate in m^3
T1 = 74.;## in degree celcius
T2 = 10.;## in degree celcius
m = 1000.;## Steam in kg
p = 950.;## Density of steam in kg/m^3
C = 85.;## Cost in Pound per m^3
C_V = 43.3*10**6;## Calorific value in J/kg
Cp = 4.18*10**3;## heat capacity of water J/kg-K
h = 2.33*10**6;## Heat of the steam in J/kg
n = 0.65;## Average bolier efficiency
S_cost = ((m*h*C)/(C_V*p*n));## Steam cost in Pound per 1000 kg
E_save = V*m*(T1 - T2)*Cp;## Energy saving in J per day
S_save = E_save/h;## in kg per day
print'%s %.2f %s'%('the steam saving is ',S_save,' kg per day \n')
G_save = (S_cost*S_save)/m;## Pound per day
print'%s %.2f %s'%('The gross saving is ',G_save,' Pound per day per year')
```

In [2]:

```
import math
import numpy
#import tabulate
## Example 11.2
print('Example 11.2\n\n');
print('Page No. 313\n\n');
##given
p1 = 10.;##heat-sensitive liquor percen
p2 = 50.;##heat-sensitive liquor percent
m = 0.28;## mass rate in kg/s
t = 150.;## time in h per week
## This question does not contain any calculation part in it.
I = ['8250', '1150', '14850', '16500'];##Installation cost in Pound
A = ['69300' ,'36800' ,'23600' ,'24600'];## Annual steam cost in Pound
A_S =['0' ,'32500' ,'45700' ,'44700'];## Annual savings in Pound
for column in zip(I, A, A_S):
print ' '.join(column)
#from tabulate import tabulate
#print tabulate([['single effect', I[0], A[0],[A_S[0]]], ['double effect', I[1], A[1],[A_S[1]]] ,['double effect+vapour compression',I[2], A[2],[A_S[2]]] ,['Triple effect',I[3], A[3],[A_S[3]]]], headers=['Installation cost', 'Annual steam cost','Annual saving'])
#print'%s %.2f %s'%(' The results enable the return on investment to be assessed by one of the standard economic procedures and the final selsction made.')
```

In [3]:

```
import math
## Example 11.3
print('Example 11.3\n\n');
print('Page No. 314\n\n');
##given
f = 1.;## feed of sodium hydroxide in kg
v = 0.5;## produed vapour in kg
A = 30.;## in m^2
T1 = 95.;## Temperature of boiling solution in deg C
U = 3.*10**3;## heat transfer coefficent in W/m^2-K
m = 1.;## feed rate in kg/s
Tf = 70.;## Feed temperature in deg C
h_f = 260.*10**3.;## Enthalpy of feed in J/kg
h_b = 355.*10**3.;## Enthalpy of boiling solution in J/kg
h_v = 2.67*10**6.;## Enthalpy of vapour in J/kg
P1 = 0.6;## Pressure in vapour space in bar
Q = (v*h_b) + (v*h_v) -(f*h_f);## in W
print'%s %.2f %s'%('The total energy requirement is ',Q,' W \n')
## As Q = A*U*dT
dT = Q/(U*A);## in degree celcius
T2 = dT + T1;## in degree celcius
##The temperature of the heating steam T2 corresponds to a pressure of 1.4 bar. Dry saturated steam at 1.4 bar has a total enthalpy of 2.69*10^6 J/kg
##Assuming an isentropic compression of the vapour from 0.6 bar to 1.4 bar, the outlet enthalpy is 2.84*10^6 J/kg
## from steam table
P2 = 1.4## pressure in bar
h_s = 2.69*10**6;## enthalpy of dry saturated steam in J/kg
h_v2 = 2.84*10**6 ;## the outlet enthalpy of vapour in J/kg
W = v*(h_v2 - h_s);## Work in W
T_E = W + 60.*10**3;## in W
print'%s %.2f %s'%('The total energy consumption is ',T_E,' W')
```

In [4]:

```
import math
## Example 11.4
print('Example 11.4\n\n');
print('Page No. 316\n\n');
##given
Cm_S = 10000.;## Company saving in Pound per annum
S = Cm_S/12.;## Saving in Pound per months
Ca_C = 10500.;## Capital cost in Pound
Ins_C = 7500.;## Installation cost in Pound
T_C = Ca_C + Ins_C;## Total cost in Pound
T = T_C/S;## pay-back time in months
print'%s %.2f %s'%('The pay-back period was ',T,' months\n')
```

In [5]:

```
import math
## Example 11.5
print('Example 11.5\n\n');
print('Page No. 318\n\n');
##From the heat balance:-
##Heat recovered in the boiler = heat gained by the air = heat lost by the flue gases
##=> Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f
## As mass flow rate of air/flue gas is not given in the book
##Assuming m_a = m_f = 2.273 kg/s & Cp_a = 1*10^3 J/kg-K
m_a = 2.273;## in kg/s
m_f = m_a;## in kg/s
Cp_a = 1.*10**3;## Specific heat capacity of air in J/kg-K
T1_a = 20.;## Entrance temperature of air in degree celcius
T2_a = 130.;## Exit temperature of air in degree celcius
dT_a = T2_a - T1_a;##in K
T1_f = 260.;## Entrance temperature of flue gases in degree celcius
T2_f = 155.;## Entrance temperature of flue gases in degree celcius
dT_f = T1_f - T2_f;##in K
##From heat balance:- Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f
Cp_f = ((m_a*Cp_a*dT_a)/(m_f*dT_f));## in J/kg-K
Q = m_f*Cp_f*dT_f;## in W
print'%s %.2e %s'%('The total heat recovered at full load if ',Q,' W')
```

In [6]:

```
import math
## Example 11.6
print('Example 11.6\n\n');
print('Page No. 320\n\n');
C = 10000.;## Installation cost of the pump in Pound
S = 3500.;## Saving in Pound per annum
T = C/S;## in year
print'%s %.2f %s'%('The pay back time is ',T,' year\n\n')
## This question further does not contain any calculation part in it.
print('In a heat-pump system the work input to drive the compressor,W, produces a heat absorption capacity,Q2,\nand to balance the energy flow, a quantity of heat, Q1, must be dissipated.\nThus the energy equation is\n -> Q1 = W + Q2\nand the coeffient of performance is \nC.O.P. = Q1/W = Q1/(Q1 - Q2)\n Consequently the C.O.P. is always greater than unity.\nThe maximum theoretical value of the C.O.P. is that predicted by the Carnot in chapter 2,namely :\n -> (C.O.P.)max = T1/(T1 - T2)')
```

In [7]:

```
import math
## Example 11.7
print('Example 11.7\n\n');
print('Page No. 320\n\n');
##given
T1 = 40.;## in degree
T2 = 0.;## in degree celcius
##As from carnot cycle, C.O.P = (T1/(T1 - T2)), where temperature are in degree celcius
C_O_P1 = ((T1+273.)/((T1+273.) - (T2+273.)));
print'%s %.2f %s'%('C.O.P. is ',C_O_P1,' \n')
## A secondary fluid as hot water at 60 deg C is used
T3 = 60;## Temperature of hot water in degree celcius
C_O_P2 = ((T3+273.)/((T3+273.) - (T2+273.)));
print'%s %.2f %s'%('C.O.P. when secondary fluid is used is ',C_O_P2,' \n')
```

In [8]:

```
import math
## Example 11.8
print('Example 11.8\n\n');
print('Page No. 323\n\n');
## This question does not contain any calculation part in it.
print('No calculation is required as not in shown in book')
```

In [9]:

```
import math
## Example 11.9
print('Example 11.9\n\n');
print('Page No. 324\n\n');
##given
T1 = 273.;## Measured temperature In degree celcius
P = 1.;## Measured pressure in bar
T2 = 290.;## initial temperature In degree celcius
T3 = 1000.;## Final temperature In degree celcius
T4 = 1150.;## Entering tempearture In degree celcius
v1 = 7.;## in m^3/s
v2 = 8.;## in m^s
M = 22.7;## in kmol/m^3
d = 0.1;## Diameter in m
A = 0.01;## Surface area per regenerator channel in m^2
u = 1.;## maximum velocity in m/s
Cp_1 = 34.*10**3;## Heat capacity at T4 temperature in J/kmol-K
Cp_2 = 32.*10**3;## Heat capacity at outlet temperature in J/kmol-K
Cp_m = 30.*10**3;## Heat capacity at mean temperature in J/kmol-K
m_c = v1/M;## Molal air flow rate in kmol/s
H_c1 = Cp_m*(T3 - T1);## Enthalpy of air at 1000K in J/mol
H_c2 = Cp_m*(T2 - T1);## Enthalpy of air at 290 in J/mol
Q = (m_c*(H_c1 - H_c2))/10**6;## in 10^6 W
print'%s %.2f %s'%('The heat transfer, Q is ',Q,' *10^6 W \n')
m_F = v2/M;## Molal flow rate of flue gas in kmol/s
dH = (Q/m_F)*10**6;## enthaply chnage of the flue gas in J/kmol
H_F1 = Cp_1*(T4 - T1);## Enthalpy of the flue gas at 1150 K in J/kmol
H_F2 =H_F1 - dH;## Enthalpy at the exit temperature in J/kmol
T_F2 = (H_F2/Cp_2) + T1;## in K
print'%s %.2f %s'%('The exit tempearture of the flue gas is ',T_F2,' K \n')
S_R = v2/u;##cross sectional area of the regenerator in m^2
N = S_R/A;
print'%s %.2f %s'%('The number of channels required is ',N,'\n')
print('Consequently for this regenerator a square layout could be achieved with 40 channels arranged horizontally and 20 channels vertically.')
```

In [10]:

```
import math
## Example 11.10
print('Example 11.10\n\n');
print('Page No. 324\n\n');
##given
Pr = 100.;## Production in tonnes per day
p = 10.2;## percentage of sulphur dioxide
T1 = 900.;##Burner temperature in degree celcius
T2 = 425.;##Required temperature in degree celcius
P = 10.;## Dry saturated steam pressure in bar
T = 120.;## Dry saturated steam temperature in degree celcius
##At the given Temperature =T and Pressure P, the required heat Qr to geberate steam from feed water is calculated from the steam table.
Qr = 2.27*10**6;## in J/kg
Sp_1 = 1.14*10**3;## Specific heat of the inlet gas in J/kmol-K
Sp_2 = 1.03*10**3;## Specific heat of the outlet gas in J/kmol-K
pr_rate = 1.2;## production rate in kmol/s
##In the calculation part, the book has taken percentage of sulphur dioxide p = 10.6 in the place of p = 10.2, so there exists a deviation in answer
Q_in = ((Pr*pr_rate)/p) * Sp_1 * T1;## Heat content of the inlet gas in J/s
Q_out = ((Pr*pr_rate)/p) * Sp_2 * T2;## Heat content of the outlet gas in J/s
Qa = Q_in - Q_out;## Heat available for steam
S = Qa/Qr;## in kg/s
print'%s %.2f %s'%('The steam production is ',S,' kg/s')##Deviation in answer is due to some wrong value substition as discussed above
```