A steam power plant design consists of an ideal Rankine cycle with regeneration.
ID: 1842826 • Letter: A
Question
A steam power plant design consists of an ideal Rankine cycle with regeneration. Steam enters Turbine 1 at P1 and T1 at the rate of m1 and exits at P2. A fraction (y') of the steam exiting Turbine 1 is diverted to a closed feedwater heater while the remainder enters Turbine 2. A portion (y") of the steam exiting Turbine 2 at P3 is diverted to an open feedwater heater while the reaminder enters Turbine 3. The exit of Turbine 3 is fed into a condenser that operates at P4. Saturated liquid exits the condenser and is fed to Pump 1. The outlet of Pump 1 is fed into the open feedwater heater. Saturated liquid exits the open feedwater heater and is fed to Pump 2. The outlet of pump 2 is fed to the closed feedwater heater. Saturated liquid exits the low pressure output of the closed feedwater heater and is fed through a steam trap to the open feedwater heater. Both exits of the closed feedwater heater are at the same temperature. All turbines and pumps are isentropic.
Given Values: m1 = 45 kg/s, P1 = 140 bar, T1 = 520 C, P2 = 10 bar, P3 = 1 bar, P4 = 0.06 bar
Find :
The specific enthalpy at the inlet of turbine 1 inlet
The specific entropy at the inlet of turbine 1 inlet
The specific enthalpy at the exit of turbine 1
The specific enthalpy at the exit of turbine 2
The specific enthalpy at the exit of turbine 3
The specific enthalpy at the condenser exit
The specific enthalpy at the exit of the low pressure pump
The specific enthalpy at the exit of the open feedwater heater
The specific enthalpy at the exit of the high pressure pump
The specific enthalpy at hte low pressure exit of the closed feedwater heater
The specific enthalpy at the exit of the steam trap
The specific enthalpy at the inlt to the boiler
Determine the fraction of flow (y') diverted to the closed feedwater heater
The power (MW) produced by turbine 1
The power (MW) produced by turbine 2
Determine the fraction of flow (y") diverted to the open feedwater heater
The power (MW) produced by turbine 3
The power (kW) required bu the low pressure pump
The power (kW) required by the high pressre pump
The total rate of heat transfer (MW) supplied to the boiler
The thermal efficiency (%) of the pwer plant
Explanation / Answer
solution:
1)here given that steam entering turbine1 at T1=520 c and p1=140 bar has enthaphy as
h1=3379.8091 kj/kg
s1=6.4642 kj/kg k
2)this steam is exiting turbine 1 at point 2 with isentropic expansion so we can write
s1=s2
for P2=10 bar
sf=2.1384 and sg=6.5849 and cp=2.7149
s1=sf+x(sg-sf)
6.4642=2.1384+x(6.5849-2.1384)
x=.9728
h2=hf2+x(hg2-hf2)
hf2=762.68,hg2=2777.1195
h2=2722.43 kj/kg
as liquid at exit of closed heater is saturated liquid
hf2=h9=h10=762.68 kj/kg
,steam at exit of turbine 1
2)from that steam certain mass bleeded to closed heater and remaining at pressure P2=10 bar get enter in turbine 2 entry as
for turbine 2 exit,p3=1 bar
sf=1.3025,sg=7.3588
s1=s3
6.4642=1.3025+x(7.3588-1.3025)
x=.8522
h3=hf3+x(hg3-hf3)
hf3=417.43,hg3=2674.94
h3=417.43+.8522*(2674.94-417.43)
enthalphy at turbine 2 exit=h3=2341.47kj/kg
enthalphy at turbine 2 exit h3=2341.47 kj/kg
when this steam enters turbine 3 and exited at p4=.06 bar,so we get that
s1=s4
s1=sf4+x(sg4-sf4)
sf4=.5208,sg4=8.3291
6.4642=.5208+x(8.3291-.5208)
x=.7611
h4=hf4+x(hg4-hf4)
hf4=151.49.hg4=2566.66
enthalphy at turbine3 exit=h4=1989.83 kj/kg
enthalphy at condenser exit is nothing but saturated liquid at p4=.06 bar,so we get that
h5=hf4=151.49 kj/kg
where work done in pump 1 is
h6=h5+(p3-p4/10)=151.49+((1-.06)/10)=151.584 kj/kg
where enthalphy at open heater is saturated liquid at p3=1 bar sowe get that
h7=hf3=417.43 kj/kg
where enthalphy at closed heater again saturated liquid at p2=10 bar
enthalphy to steam trap=enthalphy to boiler=h9=h10=762.68 kj/kg
work done in high pressure pump is
enthalphy at high pressure pump exit=h8=h7+((p2-p3)/10)=417.43+((10-1)/10)=418.33 kj/kg
8)in this way we get all enthalpies and entropy as per question and for getting steam fraction y' and y'' we have equation at point 7 and 9 as
(1-y')y''m1h3+(1-y')(1-y'')m1h6+m1h10y'=m1h7
and
m1y'h2+m1h8=m1y'h10+m1h9
on putting value we get y' as
y'=.175711 of m1
and putting value of y' in first equation at point 7 we get
y''=.08779 of (1-y')m1
9)in this way we haev calculated all enthalpies and entropies and fraction of steam bleeded so from that we can easily calculated workdone in turbine and heat supplied in reheater.
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