#pg 292
# determine the equivalent evaporation
# Given
P = 1.4;# [MN/m^2]
m = 8.;# mass of water,[kg]
T1 = 39.;# entering temperature,[C]
T2 = 100.;# [C]
x = .95;#dryness fraction
# solution
hf = 830.1;# [kJ/kg]
hfg = 1957.7;# [kJ/kg]
# steam is wet so specific enthalpy of steam is
h = hf+x*hfg;# [kJ/kg]
# at 39 C
h1 = 163.4;# [kJ/kg]
# hence
q = h-h1;# [kJ/kg]
Q = m*q;# [kJ]
evap = Q/2256.9;# equivalent evaporation[kg steam/(kg coal)]
#results
print 'The equivalent evaporation, from and at 100 C is (kg steam/kg coal) = ',round(evap,2)
# End
#pg 292
#aim : To determine the mass of oil used per hour and the fraction of enthalpy drop through the turbine
# heat transfer available per kilogram of exhaust steam
# Given values
ms_dot = 5000.;# generation of steam, [kg/h]
P1 = 1.8;# generated steam pressure, [MN/m^2]
T1 = 273.+325;# generated steam temperature, [K]
Tf = 273+49.4;# feed temperature, [K]
neta = .8;# efficiency of boiler plant
c = 45500.;# calorific value, [kJ/kg]
P = 500.;# turbine generated power, [kW]
Pt = .18;# turbine exhaust pressure, [MN/m^2]
x = .98;# dryness farction of steam
# solution
# using steam table at 1.8 MN/m^2
hf1 = 3106.;# [kJ/kg]
hg1 = 3080.;# [kJ/kg]
# so
h1 = hf1-neta*(hf1-hg1);# [kJ/kg]
# again using steam table specific enthalpy of feed water is
hwf = 206.9;# [kJ/kg]
h_rais = ms_dot*(h1-hwf);# energy to raise steam, [kJ]
h_fue = h_rais/neta;# energy from fuel per hour, [kJ]
m_oil = h_fue/c;# mass of fuel per hour, [kg]
# from steam table at exhaust
hf = 490.7;# [kJ/kg]
hfg = 2210.8;# [kJ/kg]
# hence
h = hf+x*hfg;# [kJ/kg]
# now
h_drop = (h1-h)*ms_dot/3600;# specific enthalpy drop in turbine [kJ]
f = P/h_drop;# fraction ofenthalpy drop converted into work
# heat transfer available in exhaust is
Q = h-hwf;# [kJ/kg]
#results
print ' The mass of oil used per hour is (kg) = ',round(m_oil,1)
print ' The fraction of the enthalpy drop through the turbine that is converted into useful work is = ',round(f,3)
print ' The heat transfer available in exhaust steam above 49.4 C is (kJ/kg) = ',round(Q,1)
# End
#pg 293
print('Example 10.3');
# aim : To determine
# (a) the thermal efficiency of the boiler
# (b) the equivalent evaporation of the boiler
# (c) the new coal consumption
# given values
ms_dot = 5400.;# steam feed rate, [kg/h]
P = 750;# steam pressure, [kN/m**2]
x = .98;# steam dryness fraction
Tf1 = 41.5;# feed water temperature, [C]
CV = 31000.;# calorific value of coal used in the boiler, [kJ/kg]
mc1 = 670.;# rate of burning of coal/h, [kg]
Tf2 = 100.;# increased water temperature, [C]
# solution
# (a)
SRC = ms_dot/mc1;# steam raised/kg coal, [kg]
hf = 709.3;# [kJ/kg]
hfg = 2055.5;# [kJ/kg]
h1 = hf+x*hfg;# specific enthalpy of steam raised, [kJ/kg]
# from steam table
hfw = 173.9;# specific enthalpy of feed water, [kJ/kg]
EOB = SRC*(h1-hfw)/CV;# efficiency of boiler
print ' (a) The thermal efficiency of the boiler is (percent) = ',round(EOB*100,1)
# (b)
he = 2256.9;# specific enthalpy of evaporation, [kJ/kg]
Ee = SRC*(h1-hfw)/he;# equivalent evaporation[kg/kg coal]
print ' (b) The equivalent evaporation of boiler is (kg/kg coal) = ',round(Ee,2)
# (c)
hw = 419.1;# specific enthalpy of feed water at 100 C, [kJ/kg]
Eos = ms_dot*(h1-hw);# energy of steam under new condition, [kJ/h]
neb = EOB+.05;# given condition new efficiency of boiler if 5%more than previous
Ec = Eos/neb;# energy from coal, [kJ/h]
mc2 = Ec/CV;# mass of coal used per hour in new condition, [kg]
print ' (c) Mass of coal used in new condition is (kg) = ',round(mc2)
print ' The saving in coal per hour is (kg) = ',round(mc1-mc2)
# End
#pg 294
print('Example 10.4');
# aim : To determine the
# (a) Heat transfer in the boiler
# (b) Heat transfer in the superheater
# (c) Gas used
# given values
P = 100.;# boiler operating pressure, [bar]
Tf = 256.;# feed water temperature, [C]
x = .9;# steam dryness fraction.
Th = 450.;# superheater exit temperature, [C]
m = 1200.;# steam generation/h, [tonne]
TE = .92;# thermal efficiency
CV = 38.;# calorific value of fuel, [MJ/m^3]
# solution
# (a)
# from steam table
hw = 1115.4;# specific enthalpy of feed water, [kJ/kg]
# for wet steam
hf = 1408.;# specific enthalpy, [kJ/kg]
hg = 2727.7;# specific enthalpy, [kJ/kg]
# so
h = hf+x*(hg-hf);# total specific enthalpy of wet steam, [kJ/kg]
# hence
Qb = m*(h-hw);# heat transfer/h for wet steam, [MJ]
print ' (a) The heat transfer/h in producing wet steam in the boiler is (MJ) = ',Qb
# (b)
# again from steam table
# specific enthalpy of superheated stem at given condition is,
hs = 3244;# [kJ/kg]
Qs = m*(hs-h);# heat transfer/h in superheater, [MJ]
print ' (b) The heat transfer/h in superheater is (MJ) = ',Qs
# (c)
V = (Qb+Qs)/(TE*CV);# volume of gs used/h, [m^3]
print ' (c) The volume of gas used/h is (m^3) = ',round(V)
print 'There is calculation mistake in the book so our answer is not matching'
# End
#pg 300
print('Example 10.5');
#aim : To determine
# the flow rate of cooling water
#Given values
P=24;#pressure, [kN/m^2]
ms_dot=1.8;#steam condense rate,[tonne/h]
x=.98;#dryness fraction
T1=21.;#entrance temperature of cooling water,[C]
T2=57.;#outlet temperature of cooling water,[C]
#solution
#at 24 kN/m^2, for steam
hfg=2616.8;#[kJ/kg]
hf1=268.2;#[kJ/kg]
#hence
h1=hf1+x*(hfg-hf1);#[kJ/kg]
#for cooling water
hf3=238.6;#[kJ/kg]
hf2=88.1;#[kJ/kg]
#using equation [3]
#ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so
mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);#[tonne/h]
#results
print 'The flow rate of the cooling water is =',round(mw_dot,1),'tonne/h'
#End
#pg 306
print('Example 10.6');
# aim : To determine
# (a) the energy supplied in the boiler
# (b) the dryness fraction of the steam entering the condenser
# (c) the rankine efficiency
# given values
P1 = 3.5;# steam entering pressure, [MN/m^2]
T1 = 273+350;# entering temperature, [K]
P2 = 10;#steam exhaust pressure, [kN/m^2]
# solution
# (a)
# from steam table, at P1 is,
hf1 = 3139;# [kJ/kg]
hg1 = 3095;# [kJ/kg]
h1 = hf1-1.5/2*(hf1-hg1);
# at Point 3
h3 = 191.8;# [kJ/kg]
Es = h1-h3;# energy supplied, [kJ/kg]
print ' (a) The energy supplied in boiler/kg steam is (kJ/kg) = ',Es
# (b)
# at P1
sf1 = 6.960;# [kJ/kg K]
sg1 = 6.587;# [kJ/kg K]
s1 = sf1-1.5/2*(sf1-sg1);# [kJ/kg K]
# at P2
sf2 = .649;# [kJ/kg K]
sg2 = 8.151;# [kJ/kg K]
# s2=sf2+x2(sg2-sf2)
# theoretically expansion through turbine is isentropic so s1=s2
# hence
s2 = s1;
x2 = (s2-sf2)/(sg2-sf2);# dryness fraction
print ' (b) The dryness fraction of steam entering the condenser is = ',round(x2,3)
# (c)
# at point 2
hf2 = 191.8;# [kJ/kg]
hfg2 = 2392.9;# [kJ/kg]
h2 = hf2+x2*hfg2;# [kJ/kg]
Re = (h1-h2)/(h1-h3);# rankine efficiency
print ' (c) The Rankine efficiency is (percent) = ',round(Re*100,1)
# End
#pg 307
print('Example 10.7');
# aim : To determine
# the specific work done and compare this with that obtained when determining the rankine effficiency
# given values
P1 = 1000;# steam entering pressure, [kN/m^2]
x1 = .97;# steam entering dryness fraction
P2 = 15;#steam exhaust pressure, [kN/m^2]
n = 1.135;# polytropic index
# solution
# (a)
# from steam table, at P1 is
hf1 = 762.6;# [kJ/kg]
hfg1 = 2013.6;# [kJ/kg]
h1 = hf1+hfg1; # [kJ/kg]
sf1 = 2.138;# [kJ/kg K]
sg1 = 6.583;# [kJ/kg K]
s1 = sf1+x1*(sg1-sf1);# [kJ/kg K]
# at P2
sf2 = .755;# [kJ/kg K]
sg2 = 8.009;# [kJ/kg K]
# s2 = sf2+x2(sg2-sf2)
# since expansion through turbine is isentropic so s1=s2
# hence
s2 = s1;
x2 = (s2-sf2)/(sg2-sf2);# dryness fraction
# at point 2
hf2 = 226.0;# [kJ/kg]
hfg2 = 2373.2;# [kJ/kg]
h2 = hf2+x2*hfg2;# [kJ/kg]
# at Point 3
h3 = 226.0;# [kJ/kg]
# (a)
Re = (h1-h2)/(h1-h3);# rankine efficiency
print ' (a) The Rankine efficiency is (percent) = ',round(Re*100,1)
# (b)
vg1 = .1943;# specific volume at P1, [m^3/kg]
vg2 = 10.02;# specific volume at P2, [m^3/kg]
V1 = x1*vg1;# [m^3/kg]
V2 = x2*vg2;# [m^3/kg]
W1 = n/(n-1)*(P1*V1-P2*V2);# specific work done, [kJ/kg]
# from rankine cycle
W2 = h1-h2;# [kJ/kg]
print ' (b) The specific work done is (kJ/kg) = ',round(W1,1)
print ' The specific work done (from rankine) is (kJ/kg) = ',round(W2,1)
print 'there is calculation mistake in the book so our answer is not matching'
# End
#pg 309
print('Example 10.8');
# aim : To determine
# (a) the rankine fficiency
# (b) the specific steam consumption
# (c) the carnot efficiency of the cycle
# given values
P1 = 1100.;# steam entering pressure, [kN/m^2]
T1 = 273.+250;# steam entering temperature, [K]
P2 = 280.;# pressure at point 2, [kN/m^2]
P3 = 35.;# pressure at point 3, [kN/m^2]
# solution
# (a)
# from steam table, at P1 and T1 is
hf1 = 2943.;# [kJ/kg]
hg1 = 2902.;# [kJ/kg]
h1 = hf1-.1*(hf1-hg1); # [kJ/kg]
sf1 = 6.926;# [kJ/kg K]
sg1 = 6.545;# [kJ/kg K]
s1 = sf1-.1*(sf1-sg1);# [kJ/kg K]
# at P2
sf2 = 1.647;# [kJ/kg K]
sg2 = 7.014;# [kJ/kg K]
# s2=sf2+x2(sg2-sf2)
# since expansion through turbine is isentropic so s1=s2
# hence
s2 = s1;
x2 = (s2-sf2)/(sg2-sf2);# dryness fraction
# at point 2
hf2 = 551.4;# [kJ/kg]
hfg2 = 2170.1;# [kJ/kg]
h2 = hf2+x2*hfg2;# [kJ/kg]
vg2 = .646;# [m^3/kg]
v2 = x2*vg2;# [m^3/kg]
# by Fig10.20.
A6125 = h1-h2;# area of 6125, [kJ/kg]
A5234 = v2*(P2-P3);# area 5234, [kJ/kg]
W = A6125+A5234;# work done
hf = 304.3;# specific enthalpy of water at condenser pressuer, [kJ/kg]
ER = h1-hf;# energy received, [kJ/kg]
Re = W/ER;# rankine efficiency
print ' (a) The rankine efficiency is (percent) = ',round(Re*100)
# (b)
kWh = 3600;# [kJ]
SSC = kWh/W;# specific steam consumption, [kJ/kWh]
print ' (b) The specific steam consumption is (kJ/kWh) = ',round(SSC,2)
# (c)
# from steam table
T3 = 273+72.7;# temperature at point 3
CE = (T1-T3)/T1;# carnot efficiency
print ' (c) The carnot efficiency of the cycle is (percent) = ',round(CE*100,1)
# End
#pg 311
print('Example 10.9');
# aim : To determine
# (a) the theoretical power of steam passing through the turbine
# (b) the thermal efficiency of the cycle
# (c) the thermal efficiency of the cycle assuming there is no reheat
# given values
P1 = 6;# initial pressure, [MN/m^2]
T1 = 450;# initial temperature, [C]
P2 = 1;# pressure at stage 1, [MN/m^2]
P3 = 1;# pressure at stage 2, [MN/m^2]
T3 = 370;# temperature, [C]
P4 = .02;# pressure at stage 3, [MN/m^2]
P5 = .02;# pressure at stage 4, [MN/m^2]
T5 = 320;# temperature, [C]
P6 = .02;# pressure at stage 5, [MN/m^2]
P7 = .02;# final pressure , [MN/m^2]
# solution
# (a)
# using Fig 10.21
h1 = 3305.;# specific enthalpy, [kJ/kg]
h2 = 2850.;# specific enthalpy, [kJ/kg]
h3 = 3202.;# specific enthalpy, [kJ/kg]
h4 = 2810.;# specific enthalpy, [kJ/kg]
h5 = 3115.;# specific enthalpy, [kJ/kg]
h6 = 2630.;# specific enthalpy, [kJ/kg]
h7 = 2215.;# specific enthalpy, [kJ/kg]
W = (h1-h2)+(h3-h4)+(h5-h6);# specific work through the turbine, [kJ/kg]
print ' (a) The theoretical power/kg steam/s is (kW) = ',W
# (b)
# from steam table
hf6 = 251.5;# [kJ/kg]
TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));# thermal efficiency
print ' (b) The thermal efficiency of the cycle is (percent) = ',round(TE1*100,1)
# (c)
# if there is no heat
hf7 = hf6;
TE2 = (h1-h7)/(h1-hf7);# thermal efficiency
print ' (c) The thermal efficiency of the cycle if there is no heat is (percent) = ',round(TE2*100,1)
# End
#pg 313
print('Example 10.10');
# aim : To determine
# (a) the mass of steam bled to each feed heater in kg/kg of supply steam
# (b) the thermal efficiency of the arrangement
# given values
P1 = 7.;# steam initial pressure, [MN/m^2]
T1 = 273.+500;# steam initil temperature, [K]
P2 = 2.;# pressure at stage 1, [MN/m^2]
P3 = .5;# pressure at stage 2, [MN/m^2]
P4 = .05;# condenser pressure,[MN/m^2]
SE = .82;# stage efficiency of turbine
# solution
# from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are
h1 = 3410.;# [kJ/kg]
h2_prim = 3045.;# [kJ/kg]
# h1-h2=SE*(h1-h2_prim), so
h2 = h1-SE*(h1-h2_prim);# [kJ/kg]
h3_prim = 2790.;# [kJ/kg]
# h2-h3=SE*(h2-h3_prim), so
h3 = h2-SE*(h2-h3_prim);# [kJ/kg]
h4_prim = 2450;# [kJ/kg]
# h3-h4 = SE*(h3-h4_prim), so
h4 = h3-SE*(h3-h4_prim);# [kJ/kg]
# from steam table
# @ 2 MN/m^2
hf2 = 908.6;# [kJ/kg]
# @ .5 MN/m^2
hf3 = 640.1;# [kJ/kg]
# @ .05 MN/m^2
hf4 = 340.6;# [kJ/kg]
# (a)
# for feed heater1
m1 = (hf2-hf3)/(h2-hf3);# mass of bled steam, [kg/kg supplied steam]
# for feed heater2
m2 = (1-m1)*(hf3-hf4)/(h3-hf4);#
print ' (a) The mass of steam bled in feed heater 1 is (kg/kg supply steam) = ',round(m1,3)
print ' The mass of steam bled in feed heater 2 is (kg/kg supply steam) = ',round(m2,3)
# (b)
W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);# theoretical work done, [kJ/kg]
Eb = h1-hf2;# energy input in the boiler, [kJ/kg]
TE1 = W/Eb;# thermal efficiency
print ' (b) The thermal efficiency of the arrangement is (percent) = ',round(TE1*100,1)
# If there is no feed heating
hf5 = hf4;
h5_prim = 2370;# [kJ/kg]
# h1-h5 = SE*(h1-h5_prim), so
h5 = h1-SE*(h1-h5_prim);# [kJ/kg]
Ei = h1-hf5;#energy input, [kJ/kg]
W = h1-h5;# theoretical work, [kJ/kg]
TE2 = W/Ei;# thermal efficiency
print ' The thermal efficiency if there is no feed heating is (percent) = ',round(TE2*100,1)
# End