Is Your Rankine Cycle Configuration Optimal for Maximum Efficiency?

[julia.julia]

A steam power plant operates on an ideal reheat–regenerative Rankine cycle with one reheater and two open feedwater heaters. Steam enters the high-pressure turbine at 10 MPa and 600°C and leaves the low-pressure turbine at 7.5 kPa. Steam is extracted from the turbine at 1.8 and 0.3 MPa, and it is reheated to 550 °C at a pressure of 1 MPa. Water leaves both feedwater heaters as a saturated liquid. Heat is transferred to the steam in the boiler at a rate of 400 MW. Show the cycle on a T-s diagram with respect to saturation lines, and determine(a) the mass flow rate of steam through the boiler, (b) the net power output of the plant,(c) the thermal efficiency of the cycle.



answer:

Pressure of the closed feed water heaterP4=1MPa
Take the extracted pressure from the second turbine P11=0.3MPa
Operating pressure of open feed water heater=0.3MPa
State7: from the superheated water
P7=10MPa and t7=600degree C
H7=3625.8kJ/kg
S7=6.9045kJ/kg-k
Since state 2 is fixed so we know that
S7=s8=6.9045kJ/kg.K and P8=1.8Mpa
H8=3047.77kJ/kg
The state exit of the first turbine is the same as a exit of the first turbine.
State9:from the saturtrated water pressure table
P9=1MPa and s9=s7=6.9045kJ/kg-K
S9=sf+x3(sfx)
6.9045=2.1381+(x9)(4.4470)
X9=0.9895
Then
H9=hf=x9(hfg)
=762.51+(0.98)(2014.6)
=2736.818kJ/kg
State 10:
Steam is superheated state so
From the superheated water
P10 =1MPa and T10=550 degree C
H10=3588.85Kj/kg
S10=7.89765kJ/kg-K
State11:
Steam is superheated state so
From the superheated water
P11=0.3MPa and s11=s10=7.89765kJ/kg-K
H11=3093.325kJ/kg
We know that
From the saturated water table
P1=P12=7.5KPa and S12=S10=7.89765kJ/kg-K
H11=3093.325kJ/kg
We know that
From the saturated water table
P1=P12=7.5KPa and s12=s10=7.89765kJ/kg-K
The quality at state 6 is found to be X12=0.95
H12=hf+x12(hfx)
=168.75+(0.95)(2405.3)
=2453.785kJ/kg
And
State1:
H1@7.5KPa=hf=168.75kJ/kg
V1=0.001008M cube/kg
State2:
H2=h1+v1(P2-P1)
=168.75=(0.001008)(0.3-0.0075)
=168.75kJ/kg
State 3:
Since liquid leaving the open feed water heater at state 9is saturated liquid at 0.3MPa.
Thenh3=561047kJ/kg
The specific enthalpy at the exit of the second pump is h4=h3=v3(P4-p3)
=561.47+(0.001073)(10-0.3)
561.48kj/kg
The specific enthalpy of the feed water heater existing of the closed fed water heater at 10MPa and 550 degree C
Then
H5=hf=vf(P5-P sat @550 degree C)
=884.233kJ/kg
By applying the mass balance equation closed feed water heater
Y1=h3-h4/h5-h4
=884.233-561.48/3047.77-561.48
=0.1298
Similarly for open feed water heater
Yii=(1-y)(h2)+y1(h3-h2)/(h2-h11)
=(0.8702)(168.75)+(0.1298)(561.48-168.25)/168.75-3093.325
=0.0675
Work developed by the turbine per unit mass entering
Wt1=(h7-hg)=(1-y1)(hg-h9)
=(3625.8-3047.77)+(0.8702)(3093.325-2453.785)
=944.564kJ/kg
First pump work
Wp1=(1-y1-yii)(h1-h3)
=(0.8027)(168.75-561.47)
=315kJ/kg
Second pump work is
Wp2=h4-h3
=561.48-561.47
=0.01kJ/kg
Heat input is
Qtn=(h7-h6)+(1-y1)(h10-h9)
=(3625.8-884.233)+(0.8702)(3588.85-2736.818)
=3483.00kJ/kg
So here we needed 400 MW power so that vary the mass flow rate according to the need of the demand,
So here we required 400 MW so mass flow rate is
M=W/Wn+Wt2-Wp1-Wp2
=400MW(3600s/h)/822.521+944.527-315-0.01
=991.700k.kg/h
Required power output is
P=W*m
=1452.038kJ/kg*991.2*10 cubic kg/h/3600
=399.99MW
=400MW.

 

[vacuum]

Thermal efficiency of the cycle is
Ht=(Wn+wt2-Wp1-Wp2)/Qtn
=(822.521+944.527-315-0.01)/3483.00
=0.839
Or
83.9%