# Chapter 16 - Internal combustion engines¶

## Example 1: pg 553¶

In [1]:
#pg 553
print('Example 16.1');

# aim : To determine
# (a) the net power output of the turbine plant if the turbine is coupled to the compresser
# (b) the thermal efficiency of the plant
# (c) the work ratio

# Given values
P1 = 100.;# inlet pressure of compressor, [kN/m^2]
T1 = 273.+18;# inlet temperature, [K]
P2 = 8*P1;# outlet pressure of compressor, [kN/m^2]
n_com = .85;# isentropic efficiency of compressor
T3 = 273.+1000;#inlet temperature of turbine, [K]
P3 = P2;# inlet pressure of turbine, [kN/m^2]
P4 = 100.;# outlet pressure of turbine, [kN/m^2]
n_tur = .88;# isentropic efficiency of turbine
m_dot = 4.5;# air mass flow rate, [kg/s]
cp = 1.006;# [kJ/kg K]
Gamma = 1.4;# heat capacity ratio

# (a)
# For the compressor
T2_prime = T1*(P2/P1)**((Gamma-1)/Gamma);# [K]
T2 = T1+(T2_prime-T1)/n_com;#  exit pressure of compressor, [K]

# for turbine
T4_prime = T3*(P4/P3)**((Gamma-1)/Gamma);# [K]
T4 = T3-(T3-T4_prime)*n_tur;# exit temperature of turbine, [K]

P_output = m_dot*cp*((T3-T4)-(T2-T1));# [kW]
print ' (a) The net power output is (kW) = ',round(P_output)

# (b)
n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;# thermal efficiency
print ' (b) The thermal efficiency of the plant is (percent) = ',round(n_the)

# (c)
P_pos = m_dot*cp*(T3-T4);# Positive cycle work, [kW]

W_ratio = P_output/P_pos;# work ratio
print ' (c) The work ratio is  =  ',round(W_ratio,3)

#  End

Example 16.1
(a) The net power output is (kW) =  1014.0
(b) The thermal efficiency of the plant is (percent) =  32.0
(c) The work ratio is  =   0.446


## Example 2: pg 554¶

In [2]:
#pg 554
print('Example 16.2');

# aim : To determine
# (a) the pressure ratiowhich will give the maximum net work output
# (b) the maximum net specific work output
# (c) the thermal efficiency at maximum work output
# (d) the work ratio at maximum work output
# (e) the carnot efficiency within the cycle temperature limits
from math import sqrt
# Given values
# taking the refrence as Fig.16.35
T3 = 273.+1080;# [K]
T1 = 273.+10;# [K]
cp = 1.007;# [kJ/kg K]
Gamma = 1.41;#  heat capacity ratio

# (a)
r_pmax = (T3/T1)**((Gamma)/(Gamma-1));# maximum pressure ratio
# for maximum net work output
r_p = sqrt(r_pmax);
print ' (a) The pressure ratio which give the maximum network output is  = ',round(r_p,2)

# (b)
T2 = T1*(r_p)**((Gamma-1)/Gamma);# [K]
# From equation [23]
T4 = T2;
W_max = cp*((T3-T4)-(T2-T1));# Maximum net specific work output, [kJ/kg]

print ' (b) The maximum net specific work output is (kJ/kg) = ',round(W_max)

# (c)
W = cp*(T3-T2);
n_the = W_max/W;# thermal efficiency
print ' (c) The thermal efficiency at maximum work output is (percent) =  ',round(n_the*100)

# (d)
# From the equation [26]
W_ratio = n_the;# Work ratio
print ' (d) The work ratio at maximum work output is  = ',round(W_ratio,2)

# (e)
n_carnot = (T3-T1)/T3*100;# carnot efficiency
print ' (e) The carnot efficiency within the cycle temperature limits is (percent) = ',round(n_carnot)

# End

Example 16.2
(a) The pressure ratio which give the maximum network output is  =  14.74
(b) The maximum net specific work output is (kJ/kg) =  401.0
(c) The thermal efficiency at maximum work output is (percent) =   54.0
(d) The work ratio at maximum work output is  =  0.54
(e) The carnot efficiency within the cycle temperature limits is (percent) =  79.0


## Example 3: pg 558¶

In [3]:
#pg 558
print('Example 16.3');

# aim : To determine
# (a) the net power output of the plant
# (b) the exhaust temperature from the heat exchanger
# (c) the thermal efficiency of the plant
# (d) the thermal efficiency of the plant if there were no heat exchanger
# (e) the work ratio

# Given values
T1 = 273.+15;# temperature, [K]
P1 = 101.;# pressure, [kN/m^2]
P2 = 6*P1; # [kN/m^2]
eff = .65;# effectiveness of the heat exchanger,
T3 = 273.+870;# temperature, [K]
P4 = 101.;# [kN/m^2]
n_com = .85;# efficiency of compressor,
n_tur = .80;# efficiency of turbine
m_dot = 4.;# mass flow rate, [kg/s]
Gama = 1.4;# heat capacity ratio
cp = 1.005;# [kJ/kg K]

# solution
# (a)
# For compressor
T2_prim = T1*(P2/P1)**((Gama-1)/Gama);# [K]

# using n_com = (T2_prim-T1)/(T2-T1)')

T2 = T1+(T2_prim-T1)/n_com
# For turbine
P3 = P2;
T4_prim = T3*(P4/P3)**((Gama-1)/Gama);# [K]

T4=T3-n_tur*(T3-T4_prim); #  [K]
P_out = m_dot*cp*((T3-T4)-(T2-T1));#  net power output, [kW]
print ' (a) The net power output of the plant is (kW) = ',round(P_out)

# (b)
mtd = T4-T2;# maximum temperature drop for heat transfer, [K]
atd = eff*mtd;# actual temperature, [K]
et = T4-atd;# Exhaust temperature from heat exchanger, [K]
t6 = et-273;# [C]
print ' (b) The exhaust temperature from the heat exchanger is (C) = ',round(t6)

# (c)
T5 = T2+atd;# [K]
n_the = ((T3-T4)-(T2-T1))/(T3-T5)*100;# thermal effficiency
print ' (c) The thermal efficiency of the plant is (percent) = ',round(n_the,1)

# (d)
# with no heat exchanger
n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;# thermal efficiency without heat exchanger
print ' (d) The thermal efficiency of the plant if there wereno heat exchanger is (percent) = ',round(n_the,1)

# (e)
P_pos = m_dot*cp*(T3-T4);# positive cycle work;# [kW]
w_rat = P_out/P_pos;# work ratio
print ' (e) The work ratio is  =  ',round(w_rat,2)

#  End

Example 16.3
(a) The net power output of the plant is (kW) =  562.0
(b) The exhaust temperature from the heat exchanger is (C) =  333.0
(c) The thermal efficiency of the plant is (percent) =  30.5
(d) The thermal efficiency of the plant if there wereno heat exchanger is (percent) =  22.3
(e) The work ratio is  =   0.38


## Example 4: pg 562¶

In [4]:
#pg 562
print('Example 16.4');

# aim : To determine
# (a) the pressure and temperature as the air leaves the compressor turbine
# (b) the power output from the free power turbine
# (c) the thermal efficiency of the plant
# (d) the work ratio
# (e) the carnot efficiency within the cycle temperature limits

# Given values
T1 = 273.+19;# temperature, [K]
P1 = 100.;# pressure, [kN/m^2]
P2 = 8*P1; # [kN/m^2]
P3 = P2;# [kN/m^2]
T3 = 273.+980;# temperature, [K]
n_com = .85;# efficiency of rotary compressor
P5 = 100.;# [kN/m^2]
n_cum = .88;# isentropic efficiency of combustion chamber compressor,
n_tur = .86;# isentropic efficiency of turbine
m_dot = 7.;# mass flow rate of air, [kg/s]
Gama = 1.4;# heat capacity ratio
cp = 1.006;# [kJ/kg K]

# solution
# (a)
# For compressor
T2_prim = T1*(P2/P1)**((Gama-1)/Gama);# [K]

T2 = T1+(T2_prim-T1)/n_com;# temperature, [K]

# for compressor turbine
# T3-T4 = T2-T1,because compressor turbine power=compressor power so
T4 = T3-(T2-T1);#turbine exit temperature, [K]
T4_prim = T3-(T3-T4)/n_cum;# [K]

# For turbine
# T4_prim = T3*(P4/P3)^((Gama-1)/Gama)
P4 = P3*(T4_prim/T3)**(Gama/(Gama-1));# exit air pressure of air, [kN/m^2]

print ' (a) The temperature as the air leaves the compressor turbine is (C) = ',round(T4-273)
print '      The pressure as the air leaves the compressor turbine is (kN/m^2) = ',round(P4)

# (b)
T5_prim = T4*(P5/P4)**((Gama-1)/Gama);# [K]

T5 = T4-n_tur*(T4-T5_prim);# temperature, [K]

PO = m_dot*cp*(T4-T5);# power output
print ' (b) The power output from the free power turbine is (kW) = ',round(PO)

# (c)

n_the = (T4-T5)/(T3-T2)*100;# thermal effficiency
print ' (c) The thermal efficiency of the plant is (percent) = ',round(n_the)

# (d)

WR = (T4-T5)/(T3-T5);# work ratio
print ' (d) The work ratio is  = ',round(WR,2)

# (e)
CE = (T3-T1)/T3;# carnot efficiency
print ' (e) The carnot efficiency is (percent) = ',round(CE*100)

print 'The answers are a bit different due to rounding off error in textbook'
#  End

Example 16.4
(a) The temperature as the air leaves the compressor turbine is (C) =  701.0
The pressure as the air leaves the compressor turbine is (kN/m^2) =  288.0
(b) The power output from the free power turbine is (kW) =  1541.0
(c) The thermal efficiency of the plant is (percent) =  32.0
(d) The work ratio is  =  0.44
(e) The carnot efficiency is (percent) =  77.0
The answers are a bit different due to rounding off error in textbook


## Example 5: pg 564¶

In [5]:
#pg 564
print('Example 16.5');

# aim : To determine
# (a) the pressure and temperature of the air compression
# (b) the power developed by the gas turbine
# (c) the temperature and pressure of the airentering the exhaust jet as it leaves the gas turbine
from math import log
# Given values
T1 = 273-22.4;# temperature, [K]
P1 = 470.;# pressure, [bar]
P2 = 30*P1; # [kN/m**2]
P3 = P2;# [kN/m**2]
T3 = 273.+960;# temperature, [K]
r = 1.25;# ratio of turbine power to compressor power
n_tur = .86;# isentropic efficiency of turbine
m_dot = 80.;# mass flow rate of air, [kg/s]
Gama = 1.41;# heat capacity ratio
cp = 1.05;# [kJ/kg K]

# solution
# (a)
# For compressor
T2_prim = T1*(P2/P1)**((Gama-1)/Gama);# [K]
# using n_tur=(T2_prim-T1)/(T2-T1)
T2 = T1+(T2_prim-T1)/n_tur;# temperature, [K]

print ' (a) The pressure of the air after compression is (bar) = ',P2

print '      The temperature of the air after compression is (C) = ',round(T2-273,1)

# (b)
Td  = r*(T2-T1);# temperature drop in turbine, [K]
PO = m_dot*cp*Td;# power output, [kW]
print ' (b) The power developed by the gas turbine is (MW) = ',round(PO*10**-3,2)

# (c)
t3 = T3-273;# [C]
t4 = t3-Td;# temeprerature of air leaving turbine,[K]
Tdi = Td/n_tur;# isentropic temperature drop, [K]
T4_prim = t3-Tdi+273;# temperature, [K]
# using T4_prim=T3*(P4/P3)**((Gama-1)/Gama)
P4 = P3*(T4_prim/T3)**(Gama/(Gama-1));# exit air pressure of air, [kN/m**2]

print ' (c) The air pressure as it leaves the gas turbine is (bar) = ',round(P4,0)

print 'Result in the book is not matching because they have taken pressure in mbar  but in in question it is given in bar'

#   End

Example 16.5
(a) The pressure of the air after compression is (bar) =  14100.0
The temperature of the air after compression is (C) =  469.6
(b) The power developed by the gas turbine is (MW) =  51.66
(c) The air pressure as it leaves the gas turbine is (bar) =  714.0
Result in the book is not matching because they have taken pressure in mbar  but in in question it is given in bar


## Example 6: pg 566¶

In [6]:
#pg 566
print('Example 16.6');

# aim : To determine
# (a) the mass of fuel oil used by the gas turbine
# (b) the mass flow of steam from the boiler
# (c) the theoretical output from the steam turbine
# (d) the overall theoretical thermal efficiency of the plant

# given values
Po = 150.;# generating plant output, [MW]
n_the1 = .35;# thermal efficiency
CV = 43.;# calorific value of fuel, [MJ]
me = 400.;# flow rate of exhaust gas, [kg/s]
T = 90.;# boiler exit temperature, [C]
T1 = 550.;# exhaust gas temperature, [C]
P2 = 10.;# steam generation pressure, [MN/m**2]
T2 = 450.;# boiler exit temperature, [C]
Tf = 140.;# feed water temperature, [C]
n_tur = .86;# turbine efficiency
P3 = .5;# exhaust temperature, [MN/m**2]
n_boi = .92;# boiler thermal efficiency
cp = 1.1;# heat capacity, [kJ/kg]

#  solution
# (a)
ER = Po*3600/n_the1;# energy requirement from the fuel, [MJ/h]
mf = ER/CV*10**-3;# fuel required, [tonne/h]
print ' (a) The mass of fuel oil used by the gas is (tonne/h) = ',round(mf,1)

# (b)

ET = me*cp*(T1-T)*3600*n_boi;# energy transferred to steam,[kJ/h]
# from steam table
h1 = 3244;# specific enthalpy, [kJ/kg]
hf = 588.5;# specific enthalpy, [kJ/kg]
ERR = h1-hf;# energy required to raise steam, [kJ/kg]
ms = ET/ERR*10**-3;# mass flow of steam, [tonne/h]
print ' (b) The mass flow  rate of steam from the boiler is (tonne/h) = ',round(ms,1)

# again from steam table
s1 = 6.424;# specific entropy, [kJ/kg K]
sf2 = 1.86;# specific entropy, [kJ/kg K
sg2 = 6.819;# specific entropy, [kJ/kg K]

hf2 = 640.1;# specific enthalpy,[kJ/kg]
hg2 = 2747.5;# specific enthalpy, [kJ/kg]
# for ths process s1=s2=sf2+x2*(sg2-sf2)
s2 = s1;
# hence
x2 = (s2-sf2)/(sg2-sf2);# dryness fraction

h2_prim = hf2+x2*(hg2-hf2);# specific enthalpy of steam, [kJ/kg]

TO = n_tur*(h1-h2_prim);#theoretical steam turbine output, [kJ/kg]
TOt = TO*ms/3600.;# total theoretical steam turbine output, [MW]

print ' (c) The theoretical output from the steam turbine is (MW) = ',round(TOt,2)

# (d)
n_tho = (Po+TOt)*n_the1/Po;# overall theoretical thermal efficiency
print ' (d) The overall thermal efficiency is (percent) = ',round(n_tho*100,1)

#  End

Example 16.6
(a) The mass of fuel oil used by the gas is (tonne/h) =  35.9
(b) The mass flow  rate of steam from the boiler is (tonne/h) =  252.4
(c) The theoretical output from the steam turbine is (MW) =  40.06
(d) The overall thermal efficiency is (percent) =  44.3