Chapter11-Heat recovery

Ex1-pg308

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')
Example 11.1


Page No. 308


the steam saving is  23537.17  kg per day 

The gross saving is  174.34  Pound per day per year

Ex2-pg313

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.')
Example 11.2


Page No. 313


8250 69300 0
1150 36800 32500
14850 23600 45700
16500 24600 44700

Ex3-pg314

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')
Example 11.3


Page No. 314


The total energy requirement is  1252500.00  W 

The total energy consumption is  135000.00  W

Ex4-pg316

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')
Example 11.4


Page No. 316


The pay-back period was  21.60  months

Ex5-pg318

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')
Example 11.5


Page No. 318


The total heat recovered at full load if  2.50e+05  W

Ex6-pg320

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)')
Example 11.6


Page No. 320


The pay back time is  2.86  year


In a heat-pump system the work input to drive the compressor,W, produces a heat absorption capacity,Q2,
and to balance the energy flow, a quantity of heat, Q1, must be dissipated.
Thus the energy equation is
 -> Q1 = W + Q2
and the coeffient of performance is 
C.O.P. = Q1/W = Q1/(Q1 - Q2)
   Consequently the C.O.P. is always greater than unity.
The maximum theoretical value of the C.O.P. is that predicted by the Carnot in chapter 2,namely :
 -> (C.O.P.)max = T1/(T1 - T2)

Ex7-pg320

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')
Example 11.7


Page No. 320


C.O.P. is  7.83  

C.O.P. when secondary fluid is used is  5.55  

Ex8-pg323

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')
Example 11.8


Page No. 323


No calculation is required as not in shown in book

Ex9-pg324

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.')
Example 11.9


Page No. 324


The heat transfer, Q is  6.57  *10^6 W 

The exit tempearture of the flue gas is  622.39  K 

The number of channels required is  800.00 

Consequently for this regenerator a square layout could be achieved with 40 channels arranged horizontally and 20 channels vertically.

Ex10-pg324

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
Example 11.10


Page No. 324


The steam production is  3.05  kg/s