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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...

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Design a steam power plant that can achieve an actual overall thermal efficiency of at
least 46 percent under the conditions that all tu
ines 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%.
Start with a simple 4-component steam power plant and then consider modifying it by
adding additional components such as re-heaters, re-generators (open and closed feed-water
heaters), a combination of re-heater(s) and re-generator(s), as well as changing operating
conditions, etc. to achieve an actual overall cycle thermal efficiency of 46% or higher for
practical operating conditions.
Prepare an engineering report describing your design criteria, methodology, analysis,
and results. Justify your design criteria and any assumptions made. Your design report must
include, but is not limited to, the following parts:

a)​ A schematic diagram of each plant (design) with numbering and nomenclatures as
specified in your textbook.

)​ Design criteria based on consideration of the recommended practices (operating
conditions of pressure, temperature, steam quality, etc.; materials selection; cost,
safety and environmental factors).

c)​ Detailed calculations of each cycle investigated. Specify all input data for each cycle
at the beginning of your calculations.

d)​ A schematic T-s diagram with labeled thermodynamic states of the working fluid at
different states in the cycle for each case.

e)​ Operating conditions for each component of the plant and related thermodynamic
properties of the working fluid (water).

f)​ Comparison of actual and ideal cycle for the same operating conditions.

g)​ Tables summarizing calculations and results for various cases considered.

h)​ Discussion of various cycles attempted to meet the goal, as well as the positive and
negative aspects of your design.

i)​ A summary of your design project (the second page of your report, after the title
page).

j)​ Conclusions and recommendations.
o You need to use Steam Table to find all properties.
o No need for code, do calculations by hand and show details of calculations for EACH cycle
o However, you need to type the introduction, summary, discussion, and conclusions.
Answered Same Day May 10, 2021

Solution

Rahul answered on May 14 2021
141 Votes
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 tu
ines 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 condense
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
Tu
ine 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
Wtu
ine = (h3 – h4s)*(isentropic efficiency of tu
ine) = ( 4578.9 – 2306)*0.9 = 2045.9 kJ/kg
Wnet = Wtu
ine - 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 tu
ine and pumps)
= Wnet / Qin = 2250.75/4403.4 = 51.11%
Case 2: Rankine Cycle with Reheating
Figure 2: Schematic Diagram of Rankine Cycle with Reheate
Addition of Components as Compare to previous Case
Reheate
A reheater is basically a superheater that superheats steam exiting the high-pressure stage of a tu
ine. The reheated steam is then sent to the low-pressure stage of the tu
ine. By reheating steam between high-pressure and low-pressure tu
ine 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
Wtu
ine = ((h3 – h4) + (h5 – h6))*(isentropic efficiency of tu
ine)
= ( (4578.9 – 4483.6) + (4585.7 - 2331.06 ))*0.9 = 2114.96 kJ/kg
Wnet = Wtu
ine - 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 tu
ine)*(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 tu
ine)*(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
alances 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
alances 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 heate
Figure 4: Open Feed Water Heate
Figure 5: Rankine Cycle with Open Feed Wate
An open feedwater heater is basically a mixing chamber, where the steam extracted from the tu
ine 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 tu
ine 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 tu
ine continues to expand to the condenser pressure (state 7). Saturated...
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