Rankine cycle Simulator
Rankine cycle Simulator
AIM: Creating a Rankine Cycle Simulator & Calculating the state points of the Rankine Cycle based on user inputs, Plotting the corresponding T-S and H-S plots for the given set of inputs
Theory:
Rankine cycle is the most basic thermodynamic cycle for any steam powerplant. It starts with heating of water in boiler at constant pressure to produce steam. The steam produced is then isentropically expanded in turbine to produce work. After turbine the steam in goes into the condenser where it is cooled and converted into water at constant pressure. The water is then again pumped back to boiler by a pump and cylce continues.
Turbine Inlet temperature = 400° C
Turbine Inlet Pressure = 30 bars
Turbine Outlet Pressure = 0.05 bars = Condenser Inlet Pressure
Process 1-2 : Reversible adiabatic expansion in the turbine (or steam engine).
Process 2-3 : Constant-pressure transfer of heat in the condenser.
Process 3-4 : Reversible adiabatic pumping process in the feed pump.
Process 4-1 : Constant-pressure transfer of heat in the boiler.
PROGRAM EXPLANATION:
1. Clearing workspace, command window & closing all plots
2. General calculation for Rankine Cycle Simulator
3. Printing Introduction
4. User Inputs, Dryness fraction,State point 1(Enthalpy, Entropy, Enthalpy of saturated liquid, Enthalpy of saturated vapor, Saturation Temperature, Entropy of saturated liquid, Entropy of saturated vapor)
State point 2 (Dryness fraction), State point 3, State point 4.
5. Temp space for plotting saturation curve
6. Calculation & Plotting saturation Liquid and Vapor lines.
7. Plotting T-s Plot
8. Plotting h-s Plot
9. Calculation of Work done, heat, SSC and efficiency
PROGRAM: Rankine Cycle Simulator
% General calculation for Rankine Cycle Simulator
clear all
close all
clc
% Printing Introduction
fprintf('RANKINE CYCLE SIMULATOR \n')
fprintf('\n1-2 is isentropic expansion in the Turbine \n')
fprintf('2-3 is Constant Pressure Heat Rejection by the Condenser \n')
fprintf('3-4 is isentropic Compression in pump \n')
fprintf('4-1 is Constant Pressure Heat Addition by the Boiler \n')
% User Inputs
P1 = input('\nEnter the Pressure at the Turbine Inlet (in bar): ');
T1 = input('Enter the Temprature at the Turbine Inlet (in Degree Celcius): ');
P2 = input('Enter the Pressure at the Turbine Outlet (in bar): ');
% State point 1
H1 = XSteam('h_pt',P1,T1); %Enthalpy
S1 = XSteam('s_pt',P1,T1); %Entropy
Hf1 = XSteam('hL_p',P1); %Enthalpy of saturated liquid
Hg1 = XSteam('hV_p',P1); %Enthaply of saturated vapour
Ts1 = XSteam('Tsat_p',P1); %Saturation Temprature
Sf1 = XSteam('sL_p',P1); %Entropy of saturated liquid
Sg1 = XSteam('sV_p',P1); %Entropy of saturated vapour
% State point 2
S2 = S1;
Sf2 = XSteam('sL_p',P2);
Sg2 = XSteam('sV_p',P2);
x = (S2-Sf2)/(Sg2-Sf2); % Dryness fraction
Hf2 = XSteam('hL_p',P2);
Hg2 = XSteam('hV_p',P2);
H2 = Hf2 + x*(Hg2-Hf2);
T2 = XSteam('T_ph',P2,H2);
Ts2 = XSteam('Tsat_p',P2);
% State point 3
H3 = Hf2;
S3 = Sf2;
T3 = T2;
P3 = P2;
% State point 4
P4 = P1;
S4 = S3;
H4 = XSteam('h_ps',P4,S4);
T4 = XSteam('T_ps',P4,S4);
t = linspace(0,374,500); % temp space for plotting saturation cutves
%Plot T-S Diagram
figure(1)
hold on
% Plot Saturation Curve
for i = 1 : length(t)
plot(XSteam('sL_T',t(i)),t(i),'.','color','b')
plot(XSteam('sV_T',t(i)),t(i),'.','color','b')
end
% Plot Cycle
plot([S1 S2],[T1 T2],'linewidth',2,'color','r')
if x>1
T3 = Ts2;
plot([S2 Sg2 S3 S4],[T2 Ts2 T3 T4],'linewidth',2,'color','r')
T2 = Ts2;
else
plot([S2 S3 S4],[T2 T3 T4],'linewidth',2,'color','r')
end
l = linspace(T1,T2,1000);
for i = 1 : length(l)
plot(XSteam('s_pT',P1,l(i)),l(i),'.','color','r')
end
plot([Sf1 Sg1],[Ts1 Ts1],'linewidth',2,'color','r')
xlabel('Entropy[KJ/Kg-K]')
ylabel('Temprature[K]')
title('T-S Diagram')
% Plot H-S Diagram
figure(2)
hold on
% Plot Saturation Curve
for i = 1 : length(t)
plot(XSteam('sL_T',t(i)),XSteam('hL_T',t(i)),'.','color','b')
plot(XSteam('sV_T',t(i)),XSteam('hV_T',t(i)),'.','color','b')
end
% Plot Cycle
plot([S1 S2 S3 S4 S1],[H1 H2 H3 H4 H1],'linewidth',2,'color','r')
xlabel('Entropy[KJ/Kg-K]')
ylabel('Enthalpy[KJ/Kg]')
title('H-S Diagram')
�lculations
% Turbine Work
Wt = H1-H2;
% Pump Work
Wp = H4-H3;
% Heat Addition
Qin = H1-H4;
% Heat Rejection
Qout = H2-H3;
% Net Work
Wnet = Wt-Wp;
�ficiency
Ntherm = (Wnet/Qin)*100;
% Specific steam consumption
SSC = 3600/Wnet;
�ck Work Ratio
BWR = Wp/Wt;
% Print Results
fprintf('\n\nResults\n')
fprintf('At State Point 1:\n')
fprintf('P1 is : %.2f bar\n',P1)
fprintf('T1 is : %.2f Deg Celcius\n',T1)
fprintf('H1 is : %.2f KJ/Kg\n',H1)
fprintf('S1 is : %.2f KJ/Kg-K\n',S1)
fprintf('\nAt State Point 2:\n')
fprintf('P2 is : %.2f bar\n',P2)
fprintf('T2 is : %.2f Deg Celcius\n',T2)
fprintf('H2 is : %.2f KJ/Kg\n',H2)
fprintf('S2 is : %.2f KJ/Kg-K\n',S2)
fprintf('\nAt State Point 3:\n')
fprintf('P3 is : %.2f bar\n',P3)
fprintf('T3 is : %.2f Deg Celcius\n',T3)
fprintf('H3 is : %.2f KJ/Kg\n',H3)
fprintf('S3 is : %.2f KJ/Kg-K\n',S3)
fprintf('\nAt State Point 4:\n')
fprintf('P4 is : %.2f bar\n',P4)
fprintf('T4 is : %.2f Deg Celcius\n',T4)
fprintf('H4 is : %.2f KJ/Kg\n',H4)
fprintf('S4 is : %.2f KJ/Kg-K\n',S4)
fprintf('\nWt is : %.2f KJ/Kg\n',Wt)
fprintf('Wp is : %.2f KJ/Kg\n',Wp)
fprintf('Wnet is : %.2f KJ/Kg\n',Wnet)
fprintf('Thermal Efficiency is : %.2f percent\n',Ntherm)
fprintf('S.S.C. is : %.2f Kg/KWh\n',SSC)
fprintf('Back Work Ratio is : %.3f \n',BWR)
OUTPUT:
COMMAND WINDOW:
RANKINE CYCLE SIMULATOR
1-2 is isentropic expansion in the Turbine
2-3 is Constant Pressure Heat Rejection by the Condenser
3-4 is isentropic Compression in pump
4-1 is Constant Pressure Heat Addition by the Boiler
Enter the Pressure at the Turbine Inlet (in bar): 30
Enter the Temperature at the Turbine Inlet (in Degree Celsius): 400
Enter the Pressure at the Turbine Outlet (in bar): 0.05
Results
At State Point 1:
P1 is : 30.00 bar
T1 is : 400.00 Deg Celsius
H1 is : 3231.57 KJ/Kg
S1 is : 6.92 KJ/Kg-K
At State Point 2:
P2 is : 0.05 bar
T2 is : 32.88 Deg Celsius
H2 is : 2110.71 KJ/Kg
S2 is : 6.92 KJ/Kg-K
At State Point 3:
P3 is : 0.05 bar
T3 is : 32.88 Deg Celsius
H3 is : 137.77 KJ/Kg
S3 is : 0.48 KJ/Kg-K
At State Point 4:
P4 is : 30.00 bar
T4 is : 32.95 Deg Celsius
H4 is : 140.76 KJ/Kg
S4 is : 0.48 KJ/Kg-K
Wt is : 1120.86 KJ/Kg
Wp is : 3.00 KJ/Kg
Wnet is : 1117.87 KJ/Kg
Thermal Efficiency is : 36.17 percent
S.S.C. is : 3.22 Kg/KWh
Back Work Ratio is : 0.003
ERROR: There is some text error i check again and again and solve them.