# Chapter6-Heat transfer equipment¶

## Ex1-pg142¶

In [1]:
import math
## Example 6.1
print('Example 6.1');
print('Page No. 142');

## given
L = 2.5;## Length of tubes in metre
Do = 10*10**-3;## Internal diameter of tubes in metre
m = 3.46;## mass flow rate in kg/s
Th = 120.;## Temperature of condening steam in degree celcius
Tl_i = 20.;## Inlet temperature of liquid in degree celcius
Tl_o = 80.;## Outlet temperature of liquid in degree celcius
Cp = 2.35*10**3;## Specific heat capacity of liquid in J/kg-K
U = 950.;## Overall heat transfer coefficent in W/m**2-K

T1 = Th- Tl_i;## in degree celcius
T2 = Th- Tl_o;## in degree celcius
Tm = ((T2-T1)/math.log(T2/T1));## logarithmic mean temperature of pipe in  degree celcius
a = math.pi*Do*L;##Surface area per tube in m**2
A = ((m*Cp*(Tl_o - Tl_i))/(U*Tm));## in m**2
N = A/a;
print'%s %.2f %s'%('The number of tubes required is',N,'')

Example 6.1
Page No. 142
The number of tubes required is 99.85


## Ex2-pg142¶

In [2]:
import math
## Example 6.2
print('Example 6.2');
print('Page No. 142');

## given
v = 1.50;## velocity in m/s
N_t = 100.;## Number of tubes
Do = 10*10**-3;## Internal diameter of tubes in metre
m = 3.46;## mass flow rate in kg/s
p = 1180.;## density in kg/m**3

A = (N_t*math.pi*Do**2)/4.;## otal cross-sectional area in m**2
V = m/p;##Volumetric flow rate in m**3/s
Fv = V/A;## Fluid velocity in m/s
N_p = v/Fv;
print'%s %.2f %s'%('the number of passes is',N_p,'')

Example 6.2
Page No. 142
the number of passes is 4.02


## Ex3-pg144¶

In [3]:
import math
## Example 6.3
print('Example 6.3');
print('Page No. 144');

## given
Th_i = 130.;##Inlet temperature of hot liquid in degree celcius
Th_o = 90.;## Outlet temperature of hot liquid in degree celcius
Tc_i = 20.;## Inlet temperature of cold liquid in degree celcius
Tc_o = 50.;## Outlet temperature of cold liquid in degree celcius

##For Couter-current flow
T1 = Th_i - Tc_o;
T2 = Th_o - Tc_i;
Tm_1 = ((T2-T1)/math.log(T2/T1));
print'%s %.2f %s'%('The logarithmic mean temperature difference for counter-current flow ',Tm_1,' degree celcius')

##For Co-current flow
T3 = Th_i - Tc_i;
T4 = Th_o - Tc_o;
Tm_2 = ((T3-T4)/math.log(T3/T4));
print'%s %.2f %s'%('The logarithmic mean temperature difference for co-current flow is ',Tm_2,' degree celcius ')

Example 6.3
Page No. 144
The logarithmic mean temperature difference for counter-current flow  74.89  degree celcius
The logarithmic mean temperature difference for co-current flow is  69.20  degree celcius


## Ex4-pg147¶

In [4]:
import math
## Example 6.4
print('Example 6.4');
print('Page No. 147');

## given
F = 1.;## Fuel feed required in kg
##By ultimate analysis of feed
C = 0.86;## Carbon percentage - [%]
H2 = 0.05;## Hydrogen percentage - [%]
S = 0.001;## Sulphur percentage - [%]
O2 = 0.08;## Oxygen percentage - [%]

w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
##Calculation of excess air
C_req = mol_C*1.;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1.;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## in kmol
print'%s %.2f %s'%('Total number of kmol of O2 required per kg of fuel is',O2_req, 'kmol ')
m_O2 = O2_req*w_O2;## Mass of O2 required per kg of fuel
print'%s %.2f %s'%('Mass of O2 required per kg of fuel is ',m_O2,' kg ')
##Calculation of air
m_air = m_O2/0.232;## in kg
print'%s %.2f %s'%('Mass of air required per kg of fuel is ',m_air,' kg ')
##Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
R = 8310;##Universal gas constant in J/kmol-K
T = (273+20);## in K
P = 1.013*10**5;## in N/m**2
n = 1;## 1 kmol of air
V_kmol = (n*R*T)/P;## In m**3/kmol
M_air = 29;## Mol. weight of air
V_kg = V_kmol/M_air;## in m**3/kg
V_air = m_air*V_kg;## in m**3
print'%s %.2f %s'%('Volume of air required is ',V_air,' m^3 ')
##Deviation in answer is due to some approximation in calculation in the book

Example 6.4
Page No. 147
Total number of kmol of O2 required per kg of fuel is 0.08 kmol
Mass of O2 required per kg of fuel is  2.61  kg
Mass of air required per kg of fuel is  11.27  kg
Volume of air required is  9.34  m^3


## Ex5-pg148¶

In [5]:
import math
## Example 6.5
print('Example 6.5');
print('Page No. 148');

## given
F = 1.;## Weight of coal in kg
##By analysis of coal in weight basis
C = 0.74;## Carbon percentage - [%]
H2 = 0.05;## Hydrogen percentage - [%]
S = 0.01;## Sulphur percentage - [%]
N2 = 0.001;## Nitrogen percentage - [%]
O2 = 0.05;## Oxygen percentage - [%]
H20 = 0.09;## Moisture percentage - [%]
Ash = 0.05;## Ash percentage - [%]

w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## Total number of kmol of O2 required per kg of fuel in kmol
m_O2 = O2_req*w_O2;## Mass of O2 required per kg of fuel
print'%s %.2f %s'%('Mass of O2 required per kg of fuel is ',m_O2,' kg ')
##Calculation of air
m_air = m_O2/0.232;## in kg
print'%s %.2f %s'%('Mass of air required per kg of fuel is',m_air,' kg ')
##Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
R = 8310.;##Universal gas constant in J/kmol-K
T = (273.+0);## in K
P = 1.013*10**5;## in N/m**2
n = 1;## 1 kmol of air
V_kmol = (n*R*T)/P;## In m**3/kmol
M_air = 29.;## Mol. weight of air
V_kg = V_kmol/M_air;## in m**3/kg
V_air = m_air*V_kg;## in m**3
print'%s %.2f %s'%('Volume of air required is ',V_air,' m^3')

Example 6.5
Page No. 148
Mass of O2 required per kg of fuel is  2.33  kg
Mass of air required per kg of fuel is 10.06  kg
Volume of air required is  7.77  m^3


## Ex6-pg149¶

In [1]:
import math
## Example 6.9
print('Example 6.9');
print('Page No. 157');

## given
P = 10.;## Boiler pressure in bar
Ts = 180.;## Steam temperature in degree celcius
Tf = 80.;## Feed water temperature in degree celcius
X = 0.95;## Steam dryness fraction
m_s = 4100.;## steam rate in kg/h
m_f = 238.;## Gas rate in kg/h
G_CV = 53.5*10**6.;## In J/kg
N_CV = 48*10**6.;##in J/kg

##from steam table,AT 10 bar and at temperature T = Ts
h2 = (763.+(X*2013.))*10**3;##Specific enthalpy of steam in J/kg
##At temperature T = Tf
h1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg

E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));##
print'%s %.2f %s'%('The gross efficiency  percentage is',E_G,'')

E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));##
print'%s %.2f %s'%('The net efficiency  percentage is ',E_N,'')

Example 6.9
Page No. 157
The gross efficiency  percentage is 75.36
The net efficiency  percentage is  83.99


## Ex7-pg150¶

In [30]:
import math
## Example 6.7
print('Example 6.7');
print('Page No. 150');

## given
F = 1;## Weight of fuel in kg
e = 0.5;## excess air percentage
C = 0.74;## Mass of Carbon in kg
H2 = 0.05;## Mass of Hydrogen in kg
S = 0.01;## Mass of Sulphur in kg
N2 = 0.001;##Mass of Nitrogen in kg
O2 = 0.05;## Mass of Oxygen in kg
H2O = 0.09;## Mass of Moisture in kg
Ash = 0.05;## Mass of Ash in kg

w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
w_N2 = 28.;## mol. weight of N2
w_H20 = 18.;## mol. weight of H2O
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
mol_N2 = N2 /w_N2;##kmol of N2
mol_H2O = H2O /w_H20;##kmol of H20

##By kmol of product
CO2 = mol_C*1.;## CO2 formed by the reaction C + O2 -> CO2
H2O_air = mol_H2*1.;## H2O formed by the reaction H2 + (1/2)O2 -> H2O
SO2 = mol_S*1.;## SO2 formed by the reaction S + O2 -> SO2
Pdt = CO2 + H2O_air + SO2 + mol_N2 + mol_H2O;## Total kmol of combustion products in kmol
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## Total number of kmol of O2 required per kg of fuel in kmol

Ex_O2 = O2_req*e;##  Amount of excess oxygen in kmol

N2_air = (O2_req*(1+e)*79.)/21.;## in kmol (considering air consists of 79% N2 and 21% O2 by moles)
N2_flue = mol_N2 + N2_air;## Total N2 in flue gas in kmol
H2O_flue =  mol_H2O+ H2O_air;## Total H2O in flue gas in kmol

T_wet = CO2 + H2O_flue + SO2 + Ex_O2 + N2_flue;##Total components of flue gas on a wet basis in kmol
T_dry = CO2 + SO2 + Ex_O2 + N2_flue;##Total components of flue gas on a dry basis in kmol
H2O_dry = 0;
C_wet = ((CO2 / T_wet)*100.);## in percentage
H_wet = ((H2O_flue/T_wet)*100.);## in percentage
S_wet = ((SO2/T_wet)*100.);## in percentage
N_wet = ((N2_flue/T_wet)*100.);## in percentage
O_wet = ((Ex_O2/T_wet)*100);## in percentage

C_dry = ((CO2 / T_dry)*100.);## in percentage
H_dry = ((H2O_dry/T_dry)*100.);## in percentage
S_dry = ((SO2/T_dry)*100.);## inpercentage
N_dry = ((N2_flue/T_dry)*100.);## in percentage
O_dry =((Ex_O2/T_dry)*100.);## in percentage
T1 = C_wet + H_wet + S_wet + N_wet +O_wet;## in percentage
T2 = C_dry + S_dry + N_dry + O_dry;## in percentage

print('components\t\t\t\t\t\t kmol \t\t\t\t\t\t % Composition by volume' '\n\n\n')
print('\t wet \t\t dry \t\t\t\t wet \t\t dry ')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and  '',CO2,'' '\t\t'  '',CO2,'' '\t\t\t\t'   '',C_wet,'' and  '\t\t\t\t' '\t\t\t\t',C_dry,' \t\t\t\t ')#and '',CO2,'' and '',C_wet,'' and '',C_dry,'')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and '',H2O_flue,'' '\t\t' '',H2O_dry,''  '\t\t\t\t'   '',H_wet,'' '\t\t\t\t'   '',H_dry,'')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and  '',SO2,'' '\t\t' '',SO2,'' '\t\t\t\t'  '',S_wet,'' '\t\t\t\t'  '',S_dry,'')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and  '',N2_flue,' ''\t\t'  '',N2_flue,'' '\t\t\t\t'   '',N_wet,'' '\t\t\t\t'  '',N_dry,'')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and '',Ex_O2,'' '\t\t' '',Ex_O2,'' '\t\t\t\t' '',O_wet,'' '\t\t\t\t' '',O_dry,'')
print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and  '',T_wet,''  '\t\t' '',T_dry,'' '\t\t\t\t'   '',T1,'' '\t\t\t\t'   '',T2,'')

#printf('\t\t      kmol \t\t   percent composition by volume\n Component \t Wet \t     Dry \t\t    Wet \t  Dry \n    CO2 \t %.4f    %.4f  \t\t   %.1f \t  %.1f \n    H2O \t %.4f    %.0f  \t\t\t   %.1f \t\t  %.1f \n    SO2 \t %.4f    %.4f  \t\t   %.1f \t\t  %.1f \n    N2 \t\t %.4f    %.4f  \t\t   %.1f \t  %.1f \n    O2 \t\t %.4f    %.4f  \t\t   %.1f \t\t  %.1f \n    TOTAL \t %.4f    %.4f  \t\t   %.0f \t\t  %.0f'
#	,CO2,CO2,C_wet,C_dry,H2O_flue, H2O_dry,H_wet,H_dry,
#SO2,SO2,S_wet,S_dry,N2_flue, N2_flue,N_wet,N_dry,Ex_O2,Ex_O2,O_wet,O_dry,T_wet,T_dry,T1,T2)
#//Deviation in answes is due to some calculation approxiamation in the book.

Example 6.7
Page No. 150
components						 kmol 						 % Composition by volume

wet 		 dry 				 wet 		 dry
0.06 		 0.06 				  11.42  12.09
0.03 		 0.00 				  5.56 				 0.00
0.00 		 0.00 				  0.06 				 0.06
0.41  		 0.41 				  76.21 				 80.70
0.04 		 0.04 				  6.75 				 7.15
0.54 		 0.51 				  100.00 				 100.00


## Ex8-pg156¶

In [36]:
import math
## Example 6.8
print('Example 6.8');
print('Page No. 156');

## given
H = 0.05;## Hydrogen percentage - [%]
O = 0.08;## Oxygen percentage - [%]
C = 0.86;## Carbon percentage - [%]
S = 0.001;## Sulphur percentage - [%]

G_CV = ((33.9*C)+143*(H-(O/8.))+(9.1*S))*10**6;
print'%s %.2f %s'%('The gross calorific value is ',G_CV,' J/kg ')

N_CV = ((33.9*C)+121*(H-(O/8.))+(9.1*S))*10**6.;
print'%s %.2f %s'%('The net calorific value is ',N_CV,' J/kg')

Example 6.8
Page No. 156
The gross calorific value is  34883100.00  J/kg
The net calorific value is  34003100.00  J/kg


## Ex9-pg157¶

In [31]:
import math
## Example 6.9
print('Example 6.9');
print('Page No. 157');

## given
P = 10.;## Boiler pressure in bar
Ts = 180.;## Steam temperature in degree celcius
Tf = 80.;## Feed water temperature in degree celcius
X = 0.95;## Steam dryness fraction
m_s = 4100.;## steam rate in kg/h
m_f = 238.;## Gas rate in kg/h
G_CV = 53.5*10**6.;## In J/kg
N_CV = 48*10**6.;##in J/kg

##from steam table,AT 10 bar and at temperature T = Ts
h2 = (763.+(X*2013.))*10**3;##Specific enthalpy of steam in J/kg
##At temperature T = Tf
h1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg

E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));##
print'%s %.2f %s'%('The gross efficiency  percentage is',E_G,'')

E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));##
print'%s %.2f %s'%('The net efficiency  percentage is ',E_N,'')

Example 6.9
Page No. 157
The gross efficiency  percentage is 75.36
The net efficiency  percentage is  83.99


## Ex10-pg 158¶

In [32]:
import math
## Example 6.10
print('Example 6.10');
print('Page No. 158');

## given
##for Boiler-1
P_1 = 15.;## Boiler pressure in bar
Ts_1 = 300.;## Steam temperature in degree celcius
Tf_1 = 80.;## Feed water temperature in degree celcius
X_1 = 0.;## Steam dryness fraction
m_s1 = 9000.;## steam rate in kg/h
m_f1 = 700.;## Gas rate in kg/h
G_CV1 = 43.0*10**6;## In J/kg
##from steam table,at P = 15 bar and at given temperatures
h2_1 = 3039.*10**3;##Specific enthalpy of steam in J/kg
h1_1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg

E_G1 = ((m_s1*(h2_1-h1_1)*100.)/(m_f1*G_CV1));##
print'%s %.2f %s'%('The gross efficiency  percentage is ',E_G1,'')
Ee_1 = ((m_s1/m_f1)*(h2_1-h1_1))/(2257*10**3);
print'%s %.2f %s'%('the  equivalent evaporation for boiler-1 is',Ee_1,' kg ')

##for Boiler-2
P_2 = 10.;## Boiler pressure in bar
Ts_2 = 180.;## Steam temperature in degree celcius
Tf_2 = 60.;## Feed water temperature in degree celcius
X_2 = 0.96;## Steam dryness fraction
m_s2 = 7000.;## steam rate in kg/h
m_f2 = 510.;## Gas rate in kg/h
G_CV2 = 43.0*10**6.;## In J/kg
##from steam table,AT 10 bar and at temperature T = Ts_2
h2 = (763.+(X_2*2013.))*10**3.;##Specific enthalpy of steam in J/kg
##At temperature T = Tf_2
h1 = 251.*10**3.;##Specific enthalpy of feed steam in J/kg

E_G2 = ((m_s2*(h2-h1)*100)/(m_f2*G_CV2));##
print'%s %.2f %s'%('The gross efficiency  percentage is',E_G2,'')
Ee_2 = ((m_s2/m_f2)*(h2-h1))/(2257*10**3);
print'%s %.2f %s'%('the  equivalent evaporation for boiler-2 is ',Ee_2,' kg')

Example 6.10
Page No. 158
The gross efficiency  percentage is  80.85
the  equivalent evaporation for boiler-1 is 15.40  kg
The gross efficiency  percentage is 78.03
the  equivalent evaporation for boiler-2 is  14.87  kg


## Ex11-pg167¶

In [33]:
import math
## Example 6.11
print('Example 6.11\n\n');
print('Page No. 167\n\n');

## given
m = 10.*10**3;## Production of boiler in kg/h
X = 0.95;##Dryness fraction
P = 10.;##Pressure ib bar
T_fw = 95.;## Feed water temperature in degree celcius
T_mf = 230.;## Mean flue gae temperature in degree celcius
T_mb = 25.;## Mean boiler house temperature in degree celcius
Coal_c = 900.;## Coal consumption in kg/h
A = 0.08;## Ash content in coal
C_c = 0.15;##carbon content in coal
CV_coal = 33.50*10**6;## Calorific value of coal in J
M = 28.;## Mass of flue gas per kg coal in kg
Cp = 1.05*10**3;## Mean Specific heat capacity of the flue gas in J/kg-K
CV_c = 34.*10**6;## Calorific value of carbon in J/kg

M_s = m/Coal_c;## Mass of steam produced per kg coal in kg
H_w = (M_s*(763.+(X*2013.) - 398.)*10**3)/10**6;## Heat absorbed by water per kg coal in 10^6 J(from steam table at given pressure and dryness fraction)
H_f = (M*Cp*(T_mf - T_mb))/10**6;## Heat in flue gas in 10^6 J
H_uc = (A*C_c*CV_c)/10**6;##Heat in unburnt carbon in 10^6 J
h_sup = (CV_coal)/10**6;## Heat supplied by coal in 10^6 J
un_acc = (h_sup - (H_w + H_f + H_uc));## unaccounted heat losses in 10^6 J
a = (h_sup/h_sup)*100.;
b = (H_w/h_sup)*100.;
c = (H_f/h_sup)*100.;
d = (H_uc/h_sup)*100.;
e = (un_acc/h_sup)*100.;
T = b + c + d + e;
print(' THERMAL BALANCE SHEET :\n\t\t\t\t\t\t 10^6 J')
print'%s %.2f %s %.2f %s '%('percentage',a,'\t'   'Heat supplied by coal   ',h_sup,'')
print'%s %.2f %s %.2f %s '%('percentage',b,'\t'   'Heat absorbed by water  ',H_w,''  )
print'%s %.2f %s %.2f %s '%('percentage',c,'\t'   'Heat in flue gas',H_f,'')
print'%s %.2f %s %.2f %s '%('percentage',d,'\t'   'Heat in unburnt carbon',H_uc,'')
print'%s %.2f %s %.2f %s '%('percentage',e,'\t'   'unaccounted heat losses',un_acc,'')
print'%s %.2f %s %.2f %s '%('TOTALpercentage ',T,'\t'  'TOTAL heat supplied',h_sup,'')

Example 6.11

Page No. 167

THERMAL BALANCE SHEET :
10^6 J
percentage 100.00 	Heat supplied by coal    33.50
percentage 75.53 	Heat absorbed by water   25.30
percentage 17.99 	Heat in flue gas 6.03
percentage 1.22 	Heat in unburnt carbon 0.41
percentage 5.26 	unaccounted heat losses 1.76
TOTALpercentage  100.00 	TOTAL heat supplied 33.50


## Ex12-pg168¶

In [34]:
import math
## Example 6.12
print('Example 6.12');
print('Page No. 168');

## given
C_Rate = 2920.;## Coal consumption rate in kg/h
S_Rate = 22.5*10**3;## Steam consumption rate in kg/h
Ps = 20.;## Steam pressure in bar
Ts = 350.;## Steam Temperature in degree celcius
Tf_in = 70.;## Feed water temperature inlet economiser in degree celcius
Tf_out = 110.;## Feed water temperature outlet economiser in degree celcius
Tm_b = 25.;## Mean Boiler house temperature in degree celcius
Tm_f = 260.;## Mean exit flue gas temperature in degree celcius
CO2_f = 15.8;## CO2 content of dry exit flue gas by volume
CO_f = 0.;## CO content of dry exit flue gas by volume
C_ash = 0.025;## Carbon in ash in [%]
G = 0.005;## Grit produced in [%]
##Analysis of coal(as fired)
M = 0.105;## Moisture [%]
VM = 0.308;##Volatile matter [%]
FC = 0.497;## FIxed carbon [%]
Ash =0.09;## ASh [%]
C = 0.66;## Carbon percentage - [%]
H2 = 0.042;## Hydrogen percentage - [%]
S = 0.015;## Sulphur percentage - [%]
N2 = 0.012;## Nitrogen percentage - [%]
O2 = 0.076;## Oxygen percentage - [%]
H20 = 0.105;## Moisture percentage - [%]
G_CV = 26.90;## Gross Calorific Value in 10**6 J/kg
CV_C = 33.8*10**6;## Calorif Value of carbon in J/kg
CV_G = 33.8*10**6;## Calorif Value of Grit in J/kg
Ps_l = 20.;## Pressure of steam leaving the boiler in bar

##(a) Calculation of excess air usage
##(a.1) Theoretical oxygen requirement
F = 1.;## Fuel feed required in kg
w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_S = 32.; ##mol. weight of S
w_N2 = 28.; ## mol. weight of N2
w_O2 = 32.; ## mol. weight of O2
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_S = S /w_S;##kmol of S
mol_N2 = N2 /w_N2;##kmol of N2
mol_O2 = O2 /w_O2;##kmol of O2
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## in kmol
N2_air = (O2_req*76.8)/23.2;## in kmol (considering air consists of 76.8% N2 and 23.2% O2 )
print('(a.1) ')
print'%s %.2f %s'%('Total number of kmol of O2 required per kg of fuel is ',O2_req,' kmol ')
print'%s %.2f %s'%('N2 associated with O2 is ',N2_air,' kmol ')

##(a.2) Theoretical CO2 content of dry flue gas
T = C_req + S_req + mol_N2 + N2_air;## Total flue gas in kmol
CO2 = (C_req/T)*100.;## in [%]
print('(a.2) ')
print'%s %.2f %s'%('Theoretical CO2 content of dry flue gas in percentage is ',CO2,' ')

##(a.3)Excess air based on CO2 content
Ex_air = ((CO2 - CO2_f)/CO2_f)*100.;##  in [%]
print('(a.3) ')
print'%s %.2f %s'%('Excess air based on CO2 content in percentage is ',math.floor(Ex_air),'')

##(b) Fuel gas components
##(b.1) Composition per kg fuel
w_CO2 = 44.;## mol. weight of CO2
w_SO2 = 64.;## mol. weight of SO2
## FOR DRY GAS
CO2_d = C_req * w_CO2;## In kg/kg
SO2_d = S_req * w_SO2;## In kg/kg
N2_d = mol_N2 * w_N2;##  N2 from fuel In kg/kg
N2_air_d = N2_air * w_N2;##  N2 from air In kg/kg
T_N2 = N2_d + N2_air_d;## In kg/kg
T_dry = CO2_d + SO2_d + T_N2;## In kg/kg
print('(b.1) ')
print('Composition of dry gas ')
print'%s %.2f %s'%('CO2        ',CO2_d,'')
print'%s %.2f %s'%('SO2      ',SO2_d,'')
print'%s %.2f %s'%('N2 from fuel         ',N2_d,'')
print'%s %.2f %s'%('N2 from air         ',N2_air_d,'')
print'%s %.2f %s'%('Total dry air         ',T_dry,'kg/kg')

##FOR WET GAS
w_H2O = 18.;## mol. weight of H2O
H2O_f = M;##  H2O from fuel
H2O_H2 = mol_H2 * w_H2O;## H2O from H2
T_H2O = H2O_f + H2O_H2;## in kg/kg
print('Composition of wet gas ')
print'%s %.2f %s'%('H2O from fuel         ',H2O_f,'')
print'%s %.2f %s'%('H2O from H2         ',H2O_H2,'')
print'%s %.2f %s'%('Total H2O in wet gas         ',T_H2O,'kg/kg')

##FOR DRY EXCESS AIR
O2_dry_ex = O2_req * w_O2 *0.3;##in kg/kg
N2_dry_ex = N2_air * w_N2 *0.3;##in kg/kg
T_dry_ex = O2_dry_ex + N2_dry_ex;## in kg/kg
print('Composition of dry excess air ')
print'%s %.2f %s'%('O2      ',O2_dry_ex,'')
print'%s %.2f %s'%('N2     ',N2_dry_ex,'')
print'%s %.2f %s'%('Total dry excess air       ',T_dry_ex,'kg/kg')

##(b.2) Enthalpy
## From steam table or from the appendix C.2; at the given pressure and temperatures, the following specific heat capacity for different gases are obtained
Cp_CO2_T1 = 1.04*10**3;## Specific heat Capacity of CO2 at temperature Tm_f in J/kg-K
Cp_CO2_T2 = 0.85*10**3;## Specific heat Capacity of CO2 at temperature Tm_b in J/kg-K
Cp_SO2_T1 = 0.73*10**3;## Specific heat Capacity of SO2 at temperature Tm_f in J/kg-K
Cp_SO2_T2 = 0.62*10**3;## Specific heat Capacity of SO2 at temperature Tm_b in J/kg-K
Cp_N2_T1 = 1.07*10**3;## Specific heat Capacity of N2 at temperature Tm_f in J/kg-K
Cp_N2_T2 = 1.06*10**3;## Specific heat Capacity of N2 at temperature Tm_b in J/kg-K
Cp_O2_T1 = 0.99*10**3;## Specific heat Capacity of O2 at temperature Tm_f in J/kg-K
Cp_O2_T2 = 0.91*10**3;## Specific heat Capacity of O2 at temperature Tm_b in J/kg-K

Cp_dry_T1 = ((CO2_d * Cp_CO2_T1) + (SO2_d * Cp_SO2_T1) + (T_N2 * Cp_N2_T1))/T_dry;## in J/kg-K
Cp_dry_T2 = ((CO2_d * Cp_CO2_T2) + (SO2_d * Cp_SO2_T2) + (T_N2 * Cp_N2_T2))/T_dry;## in J/kg-K
Cp_air_T1 = ((O2_dry_ex * Cp_O2_T1) + (N2_dry_ex * Cp_N2_T1))/T_dry_ex;## in J/kg-K
Cp_air_T2 = ((O2_dry_ex * Cp_O2_T2) + (N2_dry_ex * Cp_N2_T2))/T_dry_ex;## in J/kg-K
print('(b.2) ')
print'%s %.2f %s'%('Specific heat Capacity of dry gas at 260 deg C is ',Cp_dry_T1,' J/kg-K ')
print'%s %.2f %s'%('Specific heat Capacity of dry gas at 25 deg C is ',Cp_dry_T2,' J/kg-K')
print'%s %.2f %s'%('Specific heat Capacity of dry excess air at 260 deg C is ',Cp_air_T1,' J/kg-K ')
print'%s %.2f %s'%('Specific heat Capacity of dry excess air at 25 deg C is ',Cp_air_T2,' J/kg-K ')

## From Steam table or Appendix B.3, Enthalpy of superheated steam is obtained at 260 deg C and 1 bar
E_s = 2995.*10**3;##in J/kg-K

##(c) Heat transferred to water
E_w = S_Rate / C_Rate;## Evaporation of water per kg of fuel in kg
E = (E_w*(461. - 293.)*10**3)/10**6;## in 10**6 J
B = (E_w*(2797. - 461.)*10**3)/10**6;## in 10**6 J
S = (E_w*(3139. - 2797.)*10**3)/10**6;## in 10**6 J
print('(c) ')
print'%s %.2f %s'%('Heat to water in Economiser is ',E,' *10^6 J ')
print'%s %.2f %s'%('Heat to water in Boiler is ',B,' *10^6 J ')
print'%s %.2f %s'%('Heat to water in Superheater is ',S,' *10^6 J ')

##(d) Heat loss in flue gas
hl = 1056*10**3;## Enthalpy of steam at 25 deg C (from steam table) in J/kg-K
loss_dry = T_dry*((Tm_f*Cp_dry_T1) - (Tm_b*Cp_dry_T2))/10**6;## in  10**6 J
loss_wet = T_H2O*(E_s - hl)/10**6;## in 10**6 J
loss_ex_air = T_dry_ex*((Tm_f*Cp_air_T1) - (Tm_b*Cp_air_T2))/10**6;## in 10**6 J
print('(d) ')
print'%s %.2f %s'%('Heat loss in dry flue gas is ',loss_dry,' *10^6 J ')
print'%s %.2f %s'%('Heat loss in wet flue gas is ',loss_wet,' *10^6 J ')
print'%s %.2f %s'%('Heat loss in dry excess air is ',loss_ex_air,' *10^6 J ')

##(e) Heat loss in combustile matter in ash
loss_ash = (Ash * C_ash * CV_C)/10**6;## in 10**6 J
print'%s %.2f %s'%('(e) Heat loss in combustile matter in ash is ',loss_ash,' *10^6 J ')

##(f) Heat loss in grit
loss_grit = (G * CV_G)/10**6;## in  10**6 J
print'%s %.2f %s'%('(f) Heat loss in grit is ',loss_grit,' *10^6 J ')

##(g) Radiation and unaccounted heat loss
h_sup = G_CV;## Heat supplied by the coal in 10**6 J
loss_rad = (h_sup - (E + B + S + loss_dry + loss_wet + loss_ex_air + loss_ash + loss_grit));## Radiation and unaccounted loss in 10**6 J
a = (h_sup/h_sup)*100.;
b = (E/h_sup)*100.;
c = (B/h_sup)*100.;
d = (S/h_sup)*100.;
e = (loss_dry/h_sup)*100.;
f = (loss_wet/h_sup)*100.;
g = (loss_ex_air/h_sup)*100.;
h = (loss_ash/h_sup)*100.;
i = (loss_grit/h_sup)*100.;
T = b + c + d + e + f + g + h + i + j;
print('(g) THERMAL BALANCE SHEET :\t percentage \t\t\t 10**6 J ')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',a,'\t'   'Heat supplied by coal   ',h_sup,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',b,'\t'   'Heat absorbed to loss in economiser  ',E,''  )
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',c,'\t'   'boiler',B,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',d,'\t'   'superheater',S,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',e,'\t'   'heat loss in :dry flue gas',loss_dry,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',f,'\t'   'wet flue gas   ',loss_wet,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',g,'\t'   'dry eecess air  ',loss_ex_air,''  )
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',h,'\t'   'heat loss in ash',loss_ash,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',i,'\t'   'heat loss in grit',loss_grit,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',j,'\t'   'radiation and unaccounted heat losses',loss_rad,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\tTOTAL  percentage ',T,'\t'  'TOTAL heat supplied',h_sup,'')

Example 6.12
Page No. 168
(a.1)
Total number of kmol of O2 required per kg of fuel is  0.06  kmol
N2 associated with O2 is  0.21  kmol
(a.2)
Theoretical CO2 content of dry flue gas in percentage is  20.64
(a.3)
Excess air based on CO2 content in percentage is  30.00
(b.1)
Composition of dry gas
CO2         2.42
SO2       0.03
N2 from fuel          0.01
N2 from air          5.89
Total dry air          8.36 kg/kg
Composition of wet gas
H2O from fuel          0.10
H2O from H2          0.38
Total H2O in wet gas          0.48 kg/kg
Composition of dry excess air
O2       0.61
N2      1.77
Total dry excess air        2.38 kg/kg
(b.2)
Specific heat Capacity of dry gas at 260 deg C is  1060.09  J/kg-K
Specific heat Capacity of dry gas at 25 deg C is  997.61  J/kg-K
Specific heat Capacity of dry excess air at 260 deg C is  1049.47  J/kg-K
Specific heat Capacity of dry excess air at 25 deg C is  1021.50  J/kg-K
(c)
Heat to water in Economiser is  1.29  *10^6 J
Heat to water in Boiler is  18.00  *10^6 J
Heat to water in Superheater is  2.64  *10^6 J
(d)
Heat loss in dry flue gas is  2.09  *10^6 J
Heat loss in wet flue gas is  0.94  *10^6 J
Heat loss in dry excess air is  0.59  *10^6 J
(e) Heat loss in combustile matter in ash is  0.08  *10^6 J
(f) Heat loss in grit is  0.17  *10^6 J
(g) THERMAL BALANCE SHEET :	 percentage 			 10**6 J
percentage 100.00 	Heat supplied by coal    26.90
percentage 4.81 	Heat absorbed to loss in economiser   1.29
percentage 66.91 	boiler 18.00
percentage 9.80 	superheater 2.64
percentage 7.79 	heat loss in :dry flue gas 2.09
percentage 3.48 	wet flue gas    0.94
percentage 2.19 	dry eecess air   0.59
percentage 0.28 	heat loss in ash 0.08
percentage 0.63 	heat loss in grit 0.17
percentage 4.11 	radiation and unaccounted heat losses 1.11
TOTAL  percentage  100.00 	TOTAL heat supplied 26.90


## Ex13-pg188¶

In [35]:
import math
## Example 6.13
print('Example 6.13\n\n');
print('Page No. 188\n\n');

## given
P = 1.5;## Pressure in bar
T = 111.;## Temperature in degree celcius
m = 2.;## mass flow rate of process liquid in kg/s
Cp = 4.01*10**3;## Mean Specific heat capacity in J/kg_K
Tl_i = 20.;## Inlet temperature of liquid in degree celcius
Tl_o = 90.;## Outlet temperature of liquid in degree celcius
Ps = 15.;## Pressure of steam in bar
X = 0.97;## Dryness fraction of steam
Pa = 1.5;##Pressure after adiabatic expansion in bar
Ta = 80.;## Temperature of injecting condensate in degree celcius

##(a)
Q = m*Cp*(Tl_o - Tl_i);## in W
L = 2227.*10**3;## Latent heat of 1.5 bar steam in J/kg
m_a = Q/L;
print'%s %.2f %s'%('(a) Mass flow rate of 1.5 bar steam is ',m_a,' kg/s \n')

##(b)
##from steam table, Specific enthalpy of 0.97 dry 15 bar absolute steam
h = ((843.+(X*1946.))*10**3);## in J/kg
##the balance for the desuperheater,when y is the mass flow rate(kg/s) of condensate at 80 deg C is,on the basis of 1kg/s of superheated steam: => (1*2731*10^3)+(335*10^3*y)=(1+y)*2693*10^3
y = (((2731.-2693.)*10**3)/((2693.-335.)*10**3))## in kg/s
m_b = m_a/(1.+y);## in kg/s
print'%s %.2f %s'%('(b) Mass flow rate of 15 bar steam is ',m_b,' kg/s \n')

##(c)
m_c = y*m_b;##in kg/s
print'%s %.2f %s'%('(c) Mass flow rate of condensateis ',m_c,' kg/s\n')

##(d)
v = 30.;## steam velocity in m/s
##from steam table
V = 1.16;## Specific volum of 1.5 bar saturated steam in m^3/kg
V_d = V*m_a;## in m^3/s
d = ((V_d*4.)/(v*math.pi))**0.5;## im m
print'%s %.2f %s'%('(d) The vapour main diameter is ',d,' m \n')

##(e)
l = 2.5;## Length of tubes in m
d_i = 10*10**-3;## Internal Diameter of tube in m
U = 1500.;##  Overall heat transfer coefficent in W/m^2-K

a = math.pi*d_i*l;## in m^2
T1 = T - Tl_i;## in degree celcius
T2 = T - Tl_o;## in degree celcius
Tm = ((T2-T1)/math.log(T2/T1));## logarithmic mean temperature of pipe in  degree celcius
A = Q/(U*Tm);## in m^2
N = A/a;
print'%s %.2f %s'%('(e) The number of tubes required is ',N,'\n')

Example 6.13

Page No. 188

(a) Mass flow rate of 1.5 bar steam is  0.25  kg/s

(b) Mass flow rate of 15 bar steam is  0.25  kg/s

(c) Mass flow rate of condensateis  0.00  kg/s

(d) The vapour main diameter is  0.11  m

(e) The number of tubes required is  99.82