Design a steam power plant that can achieve an actual overall thermal efficiency of at least 46 percent under the conditions that all turbines have isentropic efficiencies of 90%, all pumps have...

Power_Plant/POWER PLANT DESIGN.docx
POWER PLANT DESIGN
Aim
To design a steam power plant with an actual overall thermal efficiency equal or more than or more than 46% with turbines have isentropic efficiencies of 90%, all pumps have isentropic efficiencies of 85%, all open feed-water heaters have effectiveness of 100%, and all closed feed-water heaters have effectiveness of 85%.
Methodology
1. Take a four component of simple steam power plant consisting of a boiler, pump and condenser
2. Take the best possible operating condition and calculate the thermal efficiency and make sure that it is more than or equal to 46%
3. Modified the power plant after adding the components and then again find the thermal efficiency and compared with Carnot efficiency
Analysis
Schematic Diagram
Figure 1: Schematic Diagram of Simple Steam Power Plant
Case 1
Pboiler = 22 MPa
Tboiler =100 0C
Pcondenser = 7 kPa
3-4 isentropic expansion
Turbine inlet operating condition
P3 = 22 MPa, T = 1000 0C
h3 = 4578.9 kJ/kg @ P = 22 MPa and T = 1000 0C
h4s = 2306 @ s2 = s3 and Pcondenser = 7 kPa
s2 = 7.447 kJ/kg.K
xs = 0.89
h1 = 163.35 kJ/kg
Wpump = v1 * (Pboiler - Pcondenser) = 0.0010075 * (2200 – 7) = 22.15 kJ/kg
Wactual.pump = Wpump/(isentropic efficiency of pump)
isentropic efficiency of pump = 85 %
Wactual.pump = Wpump/(0.85) = 26.06 kJ/kg
h2a = 163.35+ 26.06 = 189.41 kJ/kg
Wturbine = (h3 – h4s)*(isentropic efficiency of turbine) = ( 4578.9 – 2306)*0.9 = 2045.9 kJ/kg
Wnet = Wturbine - Wactual.pump = 2045.9 – 26 = 2029.9 kJ/kg
Qin = h3 – h2a = 4578.9 – 189.41 = 4389.5 kJ/kg
Actuall thermal efficiency = Wnet / Qin = 2029.9/4389.5 = 46.244%
Theoretical Thermal Efficiency
h2s = 163.35 + 22.15 = 175.5 kJ/kg
Wnet = ( 4578.9 – 2306) - 22.15 = 2250.75
Qin = h3 – h2s = 4578.9 – 175.5 = 4403.4 kJ/kg
Theoretical Thermal Efficiency ( With 100 % isentropic efficiency of turbine and pumps)
= Wnet / Qin = 2250.75/4403.4 = 51.11%
Case 2: Rankine Cycle with Reheating
Figure 2: Schematic Diagram of Rankine Cycle with Reheater
Addition of Components as Compare to previous Case
Reheater
A reheater is basically a superheater that superheats steam exiting the high-pressure stage of a turbine. The reheated steam is then sent to the low-pressure stage of the turbine. By reheating steam between high-pressure and low-pressure turbine it is possible to increase the electrical efficiency of the power plant cycle beyond 40%.The reheat cycle is used in large power boilers since it is feasible economically only in larger power plants. Reheater design is very much similar to superheater design because both operate at high temperature conditions
Given:
T3 = T5 = 1000 0C
P3 = 22 MPa
P6 = P5 = 7 kPa
x6 = 0.9
s6 = sf + x * (sfg)
sf = 0.5593 kJ/(kg.K) @ Pcondenser = 7 kPa
sfg = 7.7154 kJ/(kg.K) @ Pcondenser = 7 kPa
s6 = 0.5593 + 0.9*7.7154 = 7.50316 kJ/(kg.K)
h6 = 163.5 + (0.9*(2408.4)) = 2331.06 kJ/kg.K
s5 = s6 ( Due to isentropic process)
s5 = 7.50316 kJ/(kg.K)
P5 = 19.5 MPa @ s5 = 7.50316 kJ/(kg.K) & T5 = 1000 oC
h5 = 4585.7 kJ/kg
s3 = 7.4470 kJ/(kg.K)
s3 = 7.4470 kJ/(kg.K)
s3 = s4 ( Due to isentropic process)
s4 = 7.4470 kJ/kg.K
h4 = 4483.6 kJ/(kg.K)
Wpump = v1 * (Pboiler - Pcondenser) = 0.0010075 * (22000 – 7) = 22.15 kJ/kg
Wactual.pump = Wpump/(isentropic efficiency of pump)
isentropic efficiency of pump = 85 %
Wactual.pump = Wpump/(0.85) = 26.06 kJ/kg
h2 = 163.35+ 26.06 = 189.41 kJ/kg
Wturbine = ((h3 – h4) + (h5 – h6))*(isentropic efficiency of turbine)
= ( (4578.9 – 4483.6) + (4585.7 - 2331.06 ))*0.9 = 2114.96 kJ/kg
Wnet = Wturbine - Wactual.pump = 2114.96 – 26 = 2088.9 kJ/kg
Qin = h3 – h2 = 4578.9 – 189.41 = 4389.5 kJ/kg
Thermal Efficiency = Wnet / Qin = 2088.9 / 4389.5 = 47.6%
Theoretical Thermal Efficiency
h2s = 163.35 + 22.15 = 175.5 kJ/kg
Wnet = ((h3 – h4) + (h5 – h6))
Wnet ( (4578.9 – 4483.6) + (4585.7 - 2331.06 )) = 2349.95 kJ/(kg.K)
Qin = h3 – h2 = 4578.9 – 175.5 = 4403.4 kJ/kg
Theoretical Ideal Thermal Efficiency = = Wnet / Qin = 2349.95.75/4403.4 = 53.36%
Case 3
Rankine Cycle with Reheat - Regenerative
Figure 3: Schemetic & T-s disgram of Rankine Cycle with Reheat - Regenerative
P9 = 15 MPa
T9 = 600 oC
h9 = 3583.0 kJ/kg
P10 = 4 MPa
h10 = 3155.0 kJ/kg
hact_10 = 3583 – (isentropic efficiency of turbine)*(3583 – 3155)
hact_10 = 3583 – (0.9)*(3583 – 3155) = 3197.8 kJ/kg
h11 = 3674.9 kJ/kg @ P11 = 4 MPa & T11 = 600 oC
h12 = 3014.8 kJ/kg @ P12 = 0.5 MPa & s11 = s12
h13 = 2335.7 kJ/kg
hact_13 = 3674.9 – (isentropic efficiency of turbine)*(3674.9 – 2335.7)
hact_13 = 3674.9 – (0.9)*( 3674.9 – 2335.7) = 2469.62 kJ/kg
w1-2 = 0.00101 * (500 -10)/(isentropic efficiency) = 0.49/0.85 = 0.5764 kJ/kg
h1 = 191.81 kJ/kg
h2 = h1 + 0.5764 = 191.81 + 0.5764 = 192.38 kJ/kg
w3-4 = 3.83/(isentropic efficiency) = 3.83/0.85 = 4.505 kJ/kg
h3 = 640.09 kJ/kg
h4 = h1 + 0.5764 = 640.09 + 4.505 = 645.40 kJ/kg
w6-7 = 13.77/(isentropic efficiency) = 13.77/0.85 = 16.2 kJ/kg
h6 = 1087.4 kJ/kg
h7 = h6 + 16.2 = 1087.4 + 16.2 = 1103.6 kJ/kg
The fractions of steam extracted are determined from the mass and energy
balances of the feedwater heaters:
Closed feedwater heater:
Ein = Eout
y* hact_10 + (1-y)*h4 = (1-y)*h5 + y*h6
y = (h5 – h4)/((hact_10 – h6) + (h5 – h4)) = (1087.4 - 645.40)/(( 3197.8 - 1087.4) + (1087.4 - 645.40))
y = 0.173
Open feedwater heater:
Ein = Eout
z*h12 + (1-y-z)*h2 = (1-y)*h3
z = (h3 – h2)*(1 – y)/(h12 – h2) = (1 – 0.173)*(640.09 – 192.38)/(3014.8 – 192.38)
z = 0.131
h8 = (1-y)*h5 + y*h7 = (1 – 0.173)*(1087.4) + (0.173 *1103.6) = 1090.2 kJ/kg
qin = (h9 – h8) + (1 – y)*(h11 – hact_10)
qin = (3583.1 – 1090.2) + (1 – 0.173)*(3674.9 – 3197.8)
qin = 2887.46 kJ/kg
qout = (1-y-z)*(hact_13 – h1)
qout = (1-0.173-0.131)*(2469.62 – 191.81) = 1500.35
thermal efficiency = 1 – (1535.5/2887.46) = 46.82 %
Ideal Thermal Efficiency
P9 = 15 MPa
T9 = 600 oC
h9 = 3583.0 kJ/kg
P10 = 4 MPa
h10 = 3155.0 kJ/kg
h11 = 3674.9 kJ/kg @ P11 = 4 MPa & T11 = 600 oC
h12 = 3014.8 kJ/kg @ P12 = 0.5 MPa & s11 = s12
h13 = 2335.7 kJ/kg
w1-2 = 0.49
h1 = 191.81 kJ/kg
h2 = h1 + 0.49= 191.81 + 0.49 = 192.3 kJ/kg
w3-4 = 3.83
h3 = 640.09 kJ/kg
h4 = h1 + 3.83 = 640.09 + 3.83 = 643.92 kJ/kg
w6-7 = 13.77
h6 = 1087.4 kJ/kg
h7 = h6 + 13.77 = 1087.4 + 13.77 = 1101.2 kJ/kg
The fractions of steam extracted are determined from the mass and energy
balances of the feedwater heaters:
Closed feedwater heater:
Ein = Eout
y* hact_10 + (1-y)*h4 = (1-y)*h5 + y*h6
y = (h5 – h4)/((h10 – h6) + (h5 – h4)) = (1087.4 - 643.92)/(( 3155 - 1087.4) + (1087.4 - 643.92))
y = 0.1766
Open feedwater heater:
Ein = Eout
z*h12 + (1-y-z)*h2 = (1-y)*h3
z = (h3 – h2)*(1 – y)/(h12 – h2) = (1 – 0.1766)*(640.09 – 192.3)/(3014.8 – 192.3)
z = 0.1306
h8 = (1-y)*h5 + y*h7 = (1 – 0.1766)*(1087.4) + (0.1766 *1101.2) = 1089.8 kJ/kg
qin = (h9 – h8) + (1 – y)*(h11 – h10)
qin = (3583.1 – 1089.8) + (1 – 0.1766)*(3674.9 – 3155)
qin = 2921.4 kJ/kg
qout = (1-y-z)*(hac13 – h1)
qout = (1-0.1766-0.1306)*(2335.7 – 191.81) = 1485.3
Theoretical thermal efficiency = 1 – (1485.5/2921.4) = 49.2 %
Additional Components as compare to previous case
Open feed Water heater
Figure 4: Open Feed Water Heater
Figure 5: Rankine Cycle with Open Feed Water
An open feedwater heater is basically a mixing chamber, where the steam extracted from the turbine mixes with the water exiting the pump. In an ideal condition, the water leaves the heater as a saturated liquid at the heater pressure. The schematic of a steam power plant with one open feedwater heater is shown on the left. In an ideal regenerative Rankine cycle with an open feedwater heater, steam from the boiler (state 5) expands in the turbine to an intermediate pressure (state 6). At this state, some of the steam is extracted and sent to the feedwater heater, while the remaining steam in the turbine continues to expand to the condenser pressure (state 7). Saturated...