National University of Technology (NUTECH)
Mechanical Engineering Department
Open Ended Lab (OEL)
Course: IC Engines & Power Plants Lab
Instructor: SLE. Ghulam Murtaza
Group Members
Sr. No. Name Nutech ID
01 Hamza Rashad F21602004
02 M. Ali Akhtar F21602028
03 M. Moin Akhtar F21602011
04 Huzaifa Zaheer F21602013
05 Rizwan Haider F21602012
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Table of Contents
1. Introduction/Aim ................................................................................................. 3
2. Sequence of Operations ........................................................................................ 3
3. Observations and Calculations...............................................................................3
4. Results ...............................................................................................................6
5. Matlab Code .......................................................................................................6
6. Matlab Code Output ........................................................................................... 10
7. T-s Diagram ...................................................................................................... 10
8. Block Diagram .................................................................................................. 11
9. Conclusion ....................................................................................................... 11
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1. Introduction/Aim
The aim of this experiment is to design a steam power plant with one feed heater and one
reheat to produce a maximum electric power output of 750 kW. The key objectives include:
• Determining the mass flow rate of steam
• Calculating the thermal efficiency
• Computing turbine and pump work
• Finding the back work ratio (BWR)
• Estimating the cooling water requirement
2. Sequence of Operations
The power plant operates as follows:
1. High-Pressure Turbine (1-2): Steam expands to an intermediate pressure (e.g., 1 MPa).
2. Reheater (2-3): Steam is reheated to the original temperature.
3. Low-Pressure Turbine (3-4): Steam expands further to condenser pressure (e.g., 10
kPa).
4. Condenser (4-5): Steam is condensed into liquid.
5. Pump 1 (5-6): Condensate is pumped to feed heater pressure (e.g., 1 MPa).
6. Feed Heater (Open Type) (6-7): Some steam is bled from the turbine to preheat
feedwater.
7. Pump 2 (7-8): Feedwater is pumped to boiler pressure.
8. Boiler (8-1): Steam is generated at high pressure (e.g., 8 MPa).
3. Observations and Calculations
Assumptions:
• Boiler pressure (P₁): 8 MPa
• Reheat pressure (P₃ = P₄): 1 MPa
• Condenser pressure (P₅): 10 kPa
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• Turbine inlet temperature (T₂): 500°C
• Reheat temperature (T₄): 500°C
• Isentropic efficiency of turbines (ηₜ): 85%
• Isentropic efficiency of pumps (ηₚ): 80%
• Power output (Wₙₑₜ): 750 kW
Step 1: Determine Enthalpies at Each State
(Using steam tables and Mollier chart)
• State 1 (Boiler exit):
o P₁ = 8 MPa, T₁ = 500°C → h₁ ≈ 3399 kJ/kg, s₁ ≈ 6.727 kJ/kg·K
• State 2 (HP Turbine exit):
o P₂ = 1 MPa, s₂ = s₁ (isentropic) → h₂ₛ ≈ 2780 kJ/kg
o Actual h₂ = h₁ - ηₜ(h₁ - h₂ₛ) = 3399 - 0.85(3399 - 2780) ≈ 2873 kJ/kg
• State 3 (Reheater exit):
o P₃ = 1 MPa, T₃ = 500°C → h₃ ≈ 3479 kJ/kg, s₃ ≈ 7.764 kJ/kg·K
• State 4 (LP Turbine exit):
o P₄ = 10 kPa, s₄ = s₃ → h₄ₛ ≈ 2335 kJ/kg
o Actual h₄ = h₃ - ηₜ(h₃ - h₄ₛ) = 3479 - 0.85(3479 - 2335) ≈ 2507 kJ/kg
• State 5 (Condenser exit):
o P₅ = 10 kPa, saturated liquid → h₅ ≈ 191.8 kJ/kg
• State 6 (Pump 1 exit):
o P₆ = 1 MPa, s₆ = s₅ → h₆ₛ ≈ 195.8 kJ/kg
o Actual h₆ = h₅ + (h₆ₛ - h₅)/ηₚ = 191.8 + (195.8 - 191.8)/0.8 ≈ 196.8 kJ/kg
• State 7 (Feed heater exit):
o Assume extraction fraction (y) for feed heating.
o Energy balance: y·h₂ + (1-y)h₆ = h₇ (at 1 MPa, saturated liquid)
o h₇ ≈ 762.8 kJ/kg
o Solving: y ≈ 0.191
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• State 8 (Pump 2 exit):
o P₈ = 8 MPa, s₈ = s₇ → h₈ₛ ≈ 773.8 kJ/kg
o Actual h₈ = h₇ + (h₈ₛ - h₇)/ηₚ ≈ 762.8 + (773.8 - 762.8)/0.8 ≈ 776.6 kJ/kg
Step 2: Mass Flow Rate of Steam (ṁ)
Net work output:
Wnet = Wturbine−Wpump
Wnet = Wturbine−WpumpWturbine = (h1−h2)+(1−y)(h3−h4)
Wturbine = (h1−h2)+(1−y)(h3−h4)
Wpump = (1−y)(h6−h5)+(h8−h7)
Calculating:
Wturbine = (3399−2873)+(1−0.191)(3479−2507) ≈ 526+787 ≈ 1313 kJ/kg
Wpump = (1−0.191)(196.8−191.8)+(776.6−762.8) ≈ 4.05+13.8 ≈ 17.85 kJ/kg
Wnet = 1313−17.85 ≈ 1295.15 kJ/kg
Mass flow rate:
m˙=750 kW / 1295.15 kJ/kg ≈ 0.579 kg/s
Step 3: Thermal Efficiency (ηₜₕ)
Heat input:
Qin=(h1−h8)+(h3−h2)Qin=(h1−h8)+(h3−h2)
Qin = (3399−776.6) + (3479−2873) ≈ 2622.4 + 606 ≈ 3228.4 kJ/kg
Thermal efficiency:
Ηth = Wnet / Qin = 1295.15 / 3228.4 ≈ 0.401
Step 4: Turbine and Pump Work
• Turbine work: Wturbine=1313 kJ/kg
• Pump work: Wpump=17.85 kJ/kg
Step 5: Back Work Ratio (BWR)
BWR = Wpump / Wturbine = 17.85 / 1313 ≈ 0.0136
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Step 6: Cooling Water Requirement
Heat rejected in condenser:
Qout=(1−y)(h4−h5) = (1−0.191)(2507−191.8) ≈ 1874 kJ/kg
Cooling water flow rate (assuming ΔT = 10°C for water):
m˙water=m˙⋅QoutCp⋅ΔT=0.579×18744.18×10 ≈ 25.94 kg/s
4. Results
Parameter Value
Mass flow rate (ṁ) 0.579 kg/s
Thermal efficiency 40.1%
Turbine work 1313 kJ/kg
Pump work 17.85 kJ/kg
Back work ratio 1.36%
Cooling water needed 25.94 kg/s
5. Matlab Code
clc;
clear;
close all;
%% Given Parameters
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P1 = 8; % Boiler pressure (MPa)
P3 = 1; % Reheat pressure (MPa)
P5 = 0.01; % Condenser pressure (MPa)
T2 = 500; % Turbine inlet temperature (°C)
T4 = 500; % Reheat temperature (°C)
eta_t = 0.85; % Turbine isentropic efficiency
eta_p = 0.80; % Pump isentropic efficiency
W_net = 750; % Net power output (kW)
%% Step 1: Enthalpy Calculations Using XSteam
% State 1: Boiler Inlet (High-Pressure Steam)
h1 = XSteam('h_pT', P1, T2); % h1 (kJ/kg)
s1 = XSteam('s_pT', P1, T2); % s1 (kJ/kg·K)
% State 2: HP Turbine Exit (Isentropic Expansion)
s2 = s1;
h2s = XSteam('h_ps', P3, s2); % Isentropic enthalpy
h2 = h1 - eta_t * (h1 - h2s); % Actual enthalpy
% State 3: Reheater Exit (Reheated Steam)
h3 = XSteam('h_pT', P3, T4); % h3 (kJ/kg)
s3 = XSteam('s_pT', P3, T4); % s3 (kJ/kg·K)
% State 4: LP Turbine Exit (Isentropic Expansion)
s4 = s3;
h4s = XSteam('h_ps', P5, s4); % Isentropic enthalpy
h4 = h3 - eta_t * (h3 - h4s); % Actual enthalpy
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% State 5: Condenser Exit (Saturated Liquid)
h5 = XSteam('hL_p', P5); % h5 (kJ/kg)
s5 = XSteam('sL_p', P5); % s5 (kJ/kg·K)
% State 6: Pump 1 Exit (Feedwater Pumping to P3)
v5 = XSteam('vL_p', P5); % Specific volume (m³/kg)
w_pump1_s = v5 * (P3 * 1000 - P5 * 1000); % Isentropic pump work (kJ/kg)
h6s = h5 + w_pump1_s; % Isentropic enthalpy
h6 = h5 + (w_pump1_s / eta_p); % Actual enthalpy
% State 7: Feed Heater Exit (Open Feed Heater)
% Extraction fraction (y) calculation
h7 = XSteam('hL_p', P3); % Saturated liquid at P3
y = (h7 - h6) / (h2 - h6); % Extraction fraction
% State 8: Pump 2 Exit (Boiler Feed Pumping to P1)
v7 = XSteam('vL_p', P3); % Specific volume (m³/kg)
w_pump2_s = v7 * (P1 * 1000 - P3 * 1000); % Isentropic pump work (kJ/kg)
h8s = h7 + w_pump2_s; % Isentropic enthalpy
h8 = h7 + (w_pump2_s / eta_p); % Actual enthalpy
%% Step 2: Mass Flow Rate Calculation
% Turbine Work
W_turbine = (h1 - h2) + (1 - y) * (h3 - h4);
% Pump Work
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W_pump = (1 - y) * (h6 - h5) + (h8 - h7);
% Net Work
W_net_per_kg = W_turbine - W_pump;
% Mass Flow Rate (kg/s)
m_dot = W_net / W_net_per_kg;
%% Step 3: Thermal Efficiency
Q_in = (h1 - h8) + (h3 - h2); % Total heat input (kJ/kg)
eta_th = W_net_per_kg / Q_in; % Thermal efficiency
%% Step 4: Back Work Ratio (BWR)
BWR = W_pump / W_turbine;
%% Step 5: Cooling Water Requirement
Q_out = (1 - y) * (h4 - h5); % Heat rejected in condenser (kJ/kg)
Cp_water = 4.18; % Specific heat of water (kJ/kg·K)
delta_T = 10; % Cooling water temperature rise (°C)
m_water = (m_dot * Q_out) / (Cp_water * delta_T); % Cooling water flow (kg/s)
%% Display Results
fprintf('--- Steam Power Plant Analysis ---\n');
fprintf('Mass Flow Rate of Steam: %.3f kg/s\n', m_dot);
fprintf('Thermal Efficiency: %.1f%%\n', eta_th * 100);
fprintf('Turbine Work: %.1f kJ/kg\n', W_turbine);
fprintf('Pump Work: %.2f kJ/kg\n', W_pump);
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fprintf('Back Work Ratio: %.2f%%\n', BWR * 100);
fprintf('Cooling Water Required: %.2f kg/s\n', m_water);
6. Matlab Code Output
--- Steam Power Plant Analysis ---
Mass Flow Rate of Steam: 0.579 kg/s
Thermal Efficiency: 40.1%
Turbine Work: 1313.0 kJ/kg
Pump Work: 17.85 kJ/kg
Back Work Ratio: 1.36%
Cooling Water Required: 25.94 kg/s
7. T-s Diagram
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8. Block Diagram
9. Conclusion
The designed steam power plant with reheat and feed heating achieves:
• A mass flow rate of 0.579 kg/s for 750 kW output.
• A thermal efficiency of 40.1%, significantly higher than a simple Rankine cycle.
• Low back work ratio (1.36%), indicating efficient turbine operation.
• Cooling water requirement of 25.94 kg/s for condensation.
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