# Is Your Rankine Cycle Configuration Optimal for Maximum Efficiency?

• julia.julia
In summary, a steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two open feedwater heaters. The steam enters the high-pressure turbine at 10 MPa and 600°C and leaves the low-pressure turbine at 7.5 kPa. It is extracted from the turbine at 1.8 and 0.3 MPa and reheated to 550 °C at a pressure of 1 MPa. The water leaves both feedwater heaters as a saturated liquid. Heat is transferred to the steam in the boiler at a rate of 400 MW. The mass flow rate of steam through the boiler is 991.700 kg/h, the net power output of the plant is
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.

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.

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%

## 1. What is the specific problem that needs to be checked in this thermodynamic problem?

The specific problem that needs to be checked in this thermodynamic problem could vary, as there are many different types of thermodynamic problems that could arise. It would be helpful to provide more information or context about the problem in question.

## 2. Is there a specific equation or formula that needs to be used to solve this thermodynamic problem?

Yes, there are many equations and formulas in thermodynamics that can be used to solve different types of problems. It would be helpful to provide which specific equation or formula is being used in the problem, as well as any given values or variables.

## 3. Are there any given conditions or assumptions that need to be considered for this thermodynamic problem?

Yes, thermodynamic problems often involve certain conditions or assumptions, such as constant temperature or pressure. It's important to clarify these conditions or assumptions in order to accurately solve the problem.

## 4. Can you provide a diagram or illustration to better understand this thermodynamic problem?

Yes, providing a diagram or illustration can be very helpful in understanding a thermodynamic problem. It can also aid in visualizing the problem and finding a solution.

## 5. How do you approach solving a thermodynamic problem?

The approach to solving a thermodynamic problem can vary, but typically involves identifying known and unknown variables, choosing the appropriate equation or formula, and following a step-by-step process to solve for the unknown variable. It may also involve using graphs or diagrams to aid in the solution process.

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