Chapter 5 : Steam Boilers

Example 5.1 pg : 45

In [1]:
# Variables
Mf = 1300.;                #in kg
Ma = 13000.;               #in kg
P = 7.;                    #in bar
Cpw = 4.187;              #in kJ/kg K
CV = 30000.;               #in kJ/kg
x = 0.95;                 #Dryness Fraction
Tfw = 40.;                 #in C

Hfw = Tfw*Cpw;

#At 7 bar
Hf = 697.2;               #in kJ/kg
Hfg = 2066.3;             #in kJ/kg

# Calculations and Results
H = Hf+(x*Hfg);
Ms = Ma/Mf;

Me = (Ms*(H-Hfw))/(2257);
print 'Equivalent evaporation: %3.2f kg/kg of coal'%(Me);


Eff = 100*(Ma*(H-Hfw))/(Mf*CV);
print 'Boiler Efficiency: %3.1f Percent'%(Eff);

Equivalent evaporation: 11.04 kg/kg of coal
Boiler Efficiency: 83.1 Percent

Example 5.2 pg : 46

In [2]:
# Variables
Ma = 5400.;                #in kg/hr
Tfw = 42.;                 #in C
P = 7.6;                  #in bar
Mf = 670.;                 #in kg/hr
x = 0.98;                 #Dryness Fraction
CV = 31000.;               #kJ/kg
Ms = Ma/Mf;
Hf = 175.81;              #in kJ/kg
Hfw = Hf;

#Now at 7.6 bar pressure
Hf = 711.8;               #in kJ/kg
Hfg = 2055.2;             #in kJ/kg

# Calculations and Results
H = Hf+(x*Hfg);
Eff = 100*(Ma*(H-Hfw))/(Mf*CV);
print 'Boiler Efficiency %3.1f percent'%(Eff);


Me = (Ms*(H-Hfw))/(2257);
print 'Equivalent evaporation: %3.2f kg/kg of coal'%(Me);

Boiler Efficiency 66.3 percent
Equivalent evaporation: 9.11 kg/kg of coal

Example 5.3 pg : 47

In [4]:
# Variables
P = 12;                   #in bar
CV = 34000;               #in kJ/kg
T = 250;                  #in C
Ms = 10;                  #in kg/kg of coal
Tfw = 36;                 #in C
Hfw = 150.74;             #in kJ/kg
Hg = 2784.8;              #in kJ/kg
Tsup = T;
Tsat = 188;               #in C
Cps = 2.1;                #in kJ/kg K
H = Hg+(Cps*(Tsup-Tsat));

# Calculations and Results
Me = (Ms*(H-Hfw))/2257;
print 'Equivalent evaporation: %3.2f kg/kg of coal'%(Me);


Eff = (Me*250)/21.296;
print 'Boiler Power: %3.2f kW'%(Eff);

# rounding off error.

Equivalent evaporation: 12.25 kg/kg of coal
Boiler Power: 143.78 kW

Example 5.4 pg : 48

In [6]:
# Variables
Ma = 35500;           #kg of steam
Mf = 3460;
CV = 39500;
Ms = Ma/Mf;


Hfw2 = 313.9;             #in kJ/kg
Hfw1 = 71.4;              #in kJ/kg

Q = Ma*(Hfw2-Hfw1);           #Heat added in economizer
H = 2915.0;           #in kJ/kg

# Calculations and Results
Me = (Ms*(H-Hfw2))/2257;
print 'Equivalent evaporation: %3.2f kg/kg of Oil'%(Me);


Eff1 = (Ma*100*(H-Hfw2))/(Mf*CV);
print 'Thermal Efficiency of boiler: %3.1f Percent'%(Eff1);


Eff2 = (Ma*100*(H-Hfw1))/(Mf*CV);
print 'Thermal Efficiency of Boiler plant: %3.1f Percent'%(Eff2);


HU = 860875000/(Mf*CV);
print 'Heat Utilized by Economizer: %3.1f Percent'%(HU);

# rounding off error.

Equivalent evaporation: 11.52 kg/kg of Oil
Thermal Efficiency of boiler: 67.6 Percent
Thermal Efficiency of Boiler plant: 73.9 Percent
Heat Utilized by Economizer: 6.0 Percent

Example 5.5 pg : 49

In [7]:
# Variables
Ma = 10000;           #in kg/hr
P = 7;                #in bar

Tfw = 40;             #in C
Hfw = 167.6;          #in kJ/kg
H = 2763.5;           #in kJ/kg

# Calculations and Results
Q = Ma*(H-Hfw)/60;            #Heat per minute
SA = Q/2720;                  #Heating surface area required
print 'Heating surface area required: %3.1f m**2'%(SA);


GA = SA/25;
print 'Grate area required: %3.1f m**2'%(GA);

Heating surface area required: 159.1 m**2
Grate area required: 6.4 m**2

Example 5.6 pg : 50

In [9]:
# Variables
Ma = 2400;            #in kg
Mf = 240;             #in kg
P = 12;               #in bar
CV = 33500;           #in kJ/kg
Tfw = 120;            #in C
Cpw = 4.187;
Hfw = Cpw*Tfw;
H = 2784.8;           #in kJ/kg
Mfa = Mf-(0.1*Mf);

# Calculations and Results
Eff = (Ma*100*(H-Hfw))/(Mfa*CV);
print 'Thermal Efficiency: %3.1f percent'%(Eff);


Eff1 = (Ma*100*(H-Hfw))/(Mf*CV);
print 'Thermal Efficiency of boiler and grate: %3.1f percent'%(Eff1);

Thermal Efficiency: 75.7 percent
Thermal Efficiency of boiler and grate: 68.1 percent

Example 5.7 pg : 50

In [11]:
# Variables
Mf = 255;         #in kg
x = 0.94;         #Dryness Fraction
CV = 30100;       #in kJ/kg
P = 11.5;         #in bar
Ma = 2100;        #in kg
Tfw = 25;         #in C
Ms = Ma/Mf;

Hfw = 104.9;          #in kJ/kg
Hf = 790.1;           #in kJ/kg
Hfg = 1993.2;         #in kJ/kg
H = Hf+(x*Hfg);

# Calculations and Results
Me = (Ms*(H-Hfw))/2257;
Eff = (Ma*100*(H-Hfw))/(Mf*CV);
print 'Equivalent Evaporation: %3.2f kg/kg of coal '%(Me)
print 'Thermal Efficiency: %3.1f percent'%(Eff);

Equivalent Evaporation: 9.07 kg/kg of coal 
Thermal Efficiency: 70.0 percent

Example 5.8 pg : 51

In [12]:
# Variables
Hf = 762.8;               #in kJ/kg
Hfg = 2015.3;             #in kJ/kg
x = 0.95;                 #Dryness Fraction
Ma = 1000;
Eff = 0.75;
CV = 31000;

# Calculations
H = Hf+(x*Hfg);
Cpw = 4.187;
T = 50;

Hfw = Cpw*T;
Q = Ma*(H-Hfw);

Mf = Q/(Eff*CV);
y = Mf/0.9;

Eff1 = (Q*100)/(y*CV);

# Results
print 'Efficiency of Boiler and grate: %3.1f percent'%(Eff1);

Efficiency of Boiler and grate: 67.5 percent

Example 5.9 pg : 52

In [13]:
# Variables
#At 10 bar
Hg = 2778.1;              #in kJ/kg
Cp = 2.1;                 #in kJ/kg K
T = 50;
CV = 30000;           #in kJ/kg

# Calculations and Results
H = Hg+(Cp*T);
C = 4.187;
Tf = 30;
Hfw = C*Tf;

Ms = 800./100;

Me = (Ms*(H-Hfw))/2257;
print 'Equivalent Evaporation: %3.2f kg/kg of coal'%(Me);


Eff = (Ms*100*(H-Hfw))/CV;
print 'Efficiency of Boiler and grate: %3.1f percent'%(Eff);

Equivalent Evaporation: 9.77 kg/kg of coal
Efficiency of Boiler and grate: 73.5 percent

Example 5.10 pg : 53

In [14]:
# Variables
#At 10 bar pressure
Tsat = 179.9;
Tsup = 250;
Cps = 2.1;                #in kJ/kg K

Hg = 2778.1;              #in kJ/kg
Ms = 10;                  #in kg/kg of coal
Hsup = Hg+(Cps*(Tsup-Tsat));

Hfw = 155;
Me = (Ms*(Hsup-Hfw))/2257;

# Calculations
FOE = Me/Ms;          #Factor of Evaporation
BP = (Me*370)/21.296;

# Results
print 'Equivalent Evaporation: %3.1f kg/kg of coal'%(Me);
print 'Boiler Power: %3.1f kW'%(BP);

Equivalent Evaporation: 12.3 kg/kg of coal
Boiler Power: 213.3 kW

Example 5.11 pg: 53

In [15]:
# Variables
Ma = 1100;            #in kg/hr
CV = 33000;               #in kJ/kg
Tfw = 46;             #in C
P = 10;           #in bar
x = 0.9;          #Dryness Fraction
Eff = 0.81;           #Efficiency

# Calculations
Hf = 762.8;
Hfg = 2015.3;
H = Hf+(x*Hfg);
Hfw = 192.6;

Mf = (Ma*(H-Hfw))/(CV*Eff);

# Results
print 'Amount of Coal Consumed per hour: %3.1f kg'%(Mf);

Amount of Coal Consumed per hour: 98.1 kg

Example 5.12 pg : 54

In [16]:
# Variables
Ms = 7.3;             #kg/kg of fuel
Tfw = 46;             #in C
P = 10;               #in bar
FOE = 1.17;           #Factor of Evaporation
Eff = 0.79;

# Calculations and Results
Me = FOE*Ms;
print 'Equivalent Evaporation: %3.2f kg/kg of coal'%(Me);


Hfw = 192.6;              #in kJ/kg
Hg = 2778.1;              #in kJ/kg
Tsat = 179.9;             #in C
Cps = 2.1;                #in kJ/kg K
H = (2257*FOE)+Hfw;
Tsup = ((H-Hg)/Cps)+Tsat;
print 'Temperature of Superheated Steam: %3.1f C'%(Tsup);


CV = (Ms*(H-Hfw))/Eff;
print 'Calorific Value: %3.1f kJ/kg'%(CV);

Equivalent Evaporation: 8.54 kg/kg of coal
Temperature of Superheated Steam: 206.2 C
Calorific Value: 24401.3 kJ/kg

Example 5.13 pg : 54

In [18]:
# Variables
Ma = 18000;           #in kg/hr
P = 10;               #in bar
x = 0.97;             #Dryness Fraction
Tfw = 40;             #in C
Mf = 2050;            #in kg/hr
CV = 28000;           #kJ/kg

#At 10 bar
Hf1 = 762.8;
Hfg1 = 2015.3;
H = Hf1+(x*Hfg1);
Hfw = 167.6;

# Calculations and Results
Eff = (Ma*100*(H-Hfw))/(Mf*CV);
print 'Boiler efficiency: %3.2f Percent'%(Eff);

EA = ((Ma/Mf)*(H-Hfw))/2257;
print 'Equivalent Evaporation: %3.2f kg/kg of coal'%(EA);

# rounding off error.

Boiler efficiency: 79.97 Percent
Equivalent Evaporation: 9.04 kg/kg of coal

Example 5.14 pg : 55

In [20]:
# Variables
Ma = 18000;           #in kg/hr
P = 12;           #in bar
x = 0.97;         #Dryness Fraction
CV = 27400;       #in kJ/kg
Mf = 2050;            #in kg>hr

# Calculations and Results
Qs = Mf*CV;
print 'Heat Supplied per hour: %3.1f kJ/hr'%(Qs);


#At 12 bar 
Hf = 798.6;           #in kJ/kg
Hfg = 1986.2;         #in kJ/kg
H1 = Hf+(x*Hfg);

#At 105 C
Hfw = 438.9;          #in kJ/kg
Eff = (Ma*100*(H1-Hfw))/Qs;
print 'Thermal Efficiency: %3.2f Percent'%(Eff);


Ms = Ma/Mf;
print 'Factor of Evaporation: %3.2f kg of steam '%(Ms);

Heat Supplied per hour: 56170000.0 kJ/hr
Thermal Efficiency: 73.27 Percent
Factor of Evaporation: 8.00 kg of steam

Example 5.15 pg : 56

In [21]:
# Variables
Ms = 7.5;         #kg/kg of coal
P = 11;           #in bar
Tf = 70;          #in C
Eff = 0.75;       #Efficiency
FOE = 1.15;       #Factor of Evaporation
Cps = 2.1;        #in kJ/kg K
Hfw = 293;        #in kJ/kg
H = (FOE*2257)+Hfw;

# Calculations and Results
#At 11 bar
Hg = 2781.7;      #in kJ/kg
Tsat = 184.1;     #in C
Tsup = ((H-Hg)/Cps)+Tsat;
DOS = Tsup-Tsat;      #Degree of Superheat
print 'Degree of Superheat: %3.1f C'%(DOS);


Me = (Ms*(H-Hfw))/2257;
print 'Equivalent evaporation: %3.2f kg/kg of coal'%(Me);


CV = (Ms*(H-Hfw))/Eff;
print 'Calorific value of Boiler: %3.2f kJ/kg '%(CV);

Degree of Superheat: 50.9 C
Equivalent evaporation: 8.62 kg/kg of coal
Calorific value of Boiler: 25955.50 kJ/kg

Example 5.16 pg: 57

In [22]:
# Variables
Ma = 17000;       #in kg/hr
P = 14;           #in bar
x = 0.95;         #Dryness Fraction
Tfw = 102;            #in C
Mf = 2050;            #in kg/hr
CV = 27400;       #Calorific Value

# Calculations and Results
HS = Mf*CV;
print 'Heat Supplied per hour: %3.2f kJ'%(HS);


Hf = 830.3;           #in kJ/kg
Hfg = 1959.7;         #in kJ/kg
Hfw = 427.5;          #in kJ/kg

H = Hf+(x*Hfg);
Eff = (Ma*100*(H-Hfw))/(Mf*CV);
print 'Efficiency of Boiler: %3.2f Percent'%(Eff);


Ms = Ma/Mf;
Me = (Ms*(H-Hfw))/2257;
print 'Equivalent evaporation: %3.2f kg/kg of coal'%(Me);

Heat Supplied per hour: 56170000.00 kJ
Efficiency of Boiler: 68.54 Percent
Equivalent evaporation: 8.03 kg/kg of coal

Example 5.17 pg: 58

In [23]:
# Variables
Ma = 1800;        #kg/hr
P = 12;           #in bar
x = 0.97;         #Dryness Fraction
Tfw = 105;        #in C
Mf = 2050;        #in kg/hr
CV = 27400;       #in kJ/kg

# Calculations and Results
Q = Mf*CV;
print 'Heat Supplied: %3.2f kJ'%(Q);

#At 12 bar pressure
Hf = 798.6;           #in kJ/kg
Hfg = 1986.2;         #in kJ/kg
H = Hf+(x*Hfg);
Hfw = 4.187*Tfw;

Me = (Ma*(H-Hfw))/(2257*Mf);
print 'Equivalent Evaporation: %3.2f kg/kg of coal'%(Me);

Eff = (Ma*100*(H-Hfw))/(CV*Mf);
print 'Efficiency of boiler: %3.2f Percent'%(Eff);

Heat Supplied: 56170000.00 kJ
Equivalent Evaporation: 0.89 kg/kg of coal
Efficiency of boiler: 7.32 Percent

Example 5.18 pg : 59

In [24]:
# Variables
Me = 10;   		   #kg/kg
CV = 34000.        #kJ/kg

# Calculations
x = Me*2257;
Eff = 100*x/CV;

# Results
print 'Efficiency of Boiler: %3.2f Percent'%(Eff);

Efficiency of Boiler: 66.38 Percent

Example 5.19 pg : 60

In [26]:
# Variables
Ma = 5500;        #kg/hr
P = 1;            #bar
x = 0.94;         #Dryness Fraction
Tfw = 40;         #in C
Mf = 600;         #kg/hr
CV = 32000;       #kJ/kg
Hfw = Tfw*4.187;

#At 1 bar pressure
Hf = 417.5;       #kJ/kg
Hfg = 2258;       #kJ/kg
H = Hf+(x*Hfg);
Ms = Ma/Mf;

# Calculations and Results
Me = (Ms*(H-Hfw))/2257;
print 'Equivalent Evaporation: %3.3f kg/kg of coal'%(Me);

Eff = (Ms*100*(H-Hfw))/CV;
print 'Efficiency: %3.2f percent'%(Eff);

# rounding off error.

Equivalent Evaporation: 9.461 kg/kg of coal
Efficiency: 66.73 percent