How Does Efficiency Impact Calculations in a Regenerative Rankine Cycle?

• Engineering
• Kamuna
In summary, the high-pressure turbine stage efficiency of a steam power plant is 90%, the low-pressure turbine stage efficiency is 85%, and the pump efficiency is 75%.
Kamuna
Homework Statement
Regenerative rankine cycle
Relevant Equations
regenerative rankine cycle related formulas
I tried to solve and get the answer but i know this is wrong since i didnt use the efficiencies. If you can help me pls indicate where did you use the given efficiency so that i can learn and understand from it. I am still new to this topic. Thank you

In a steam power plant operating on regenerative Rankine cycle with one contact feedwater heater, steam enters the turbine at 8.0 MPa, 350 C and condensed in the condenser with a pressure at saturated temperature of 40 C. Expansion of steam at high-pressure stage turbine up to saturation at 3.5 MPa then reheated to 350 C. After partial expansion at low-pressure stage turbine to 700 kPa, some steam is bled at this pressure for feedwater heating while the remaining steam is expanded to condenser pressure. Consider high-pressure turbine stage efficiency of 90%, low-pressure turbine stage of 85 %, and pump efficiency of 75 %. There are 27.5 kg/s of steam generated from the boiler, determine:

• extracted steam flow per second
• turbine work output in kW
• thermal efficiency of the cycle

Solution(no efficiency used)
State 1 (superheated steam)

@ P= 8.00 MPa & t = 350 C

h1 = 2988.1 kJ/kg

s1 = 6.1321 kJ/kg K

State 2 (saturated steam)

@ P2 = 3.5 MPa

For S = C expansion

s2 = s1 = sg = 6.1321 kJ/kg K

h2 = hg = 2802.7 kJ/kg

State 3 (superheated steam)

@ P3 = P2 = 3.5 MPa & t3 = 350 C

h3 = 3104.9 kJ/kg

s3 = 6.6601 kJ/kg K

State 4 (wet steam)

@ P2 = 700 kPa ( tsat = 165 C)

For S = C expansion

S4= s3 = 6.6601 KJ/kg

sf = 1.9918

sfg = 4.7153

hf = 697.00

hfg = 2065.8

vf = 0.00118 m3/kg

h4 = hf + X4 hfg

= 697.00 + X2 (2065.8)

= 697.00 + 0.878 (2065.8)

h4 = 2510.89 kJ/kg

where:

s4 = s3= sf + X4 (sfg)

6.6601 = 1.9918 + X4 (4.7153)

X4 = 0.878

State 5 (wet steam)

@ P5 = 7.38 kPa (tsat = 40 C)

For S = C expansion

s5= s3 = 6.6601 kJ/kg

sf = 0.5724

sfg = 7.6832

hf = 167.53

hfg = 2406.0

vf = 0.001008 m3/kg
h5= hf + X5 hfg

= 167.53 + X3 (2406.0)

= 167.53 + 0.724 (2406.0)

h5 = 1909.47 kJ/kg

where:

s5 = s3= sf + X5 (sfg)

6.6601 = 0.5724 + X5 (7.6832)​

X5 = 0.724
State 6 (saturated liquid)

@ P6 = P5 = 7.38 kPa

h6 = hf = 167.53 kJ/kg

v6 = vf = 0.001008 m3/kg

State 7 (subcooled liquid)

@ P6 = 700 kPa

WpA = m (h7-h6) kW or wpA = (h7-h6) kJ/kg

wpA ≈ vf (P7-P6) kJ/kg approx

= 0.001008 (700 – 7.38)​

wpA = 0.698 kJ/kg
h7 = h6 + wpA

= 167.53 + 0.698

h7 = 168.23 kJ/kg

State 8 (saturated liquid)

@ P8 = P4 = 700 kPa

h8 = hf= 697.0 kJ/kg

v8 = vf =0.00118 m3/kg

State 9 (subcooled liquid)

@ P9 = P1 = 8.0 MPa

WpB = m (h9-h8) kW or wpB = (h9-h8) kJ/kg

wpB ≈ vf (P9-P8) kJ/kg approx

= 0.00118 (8000 -700)​

wpB = 8.614 kJ/kg

h9 = h8 + wpB

= 697.0 + 8.614

h9 = 705.614 kJ/kg
from the 1st Law of Thermo on OPEN FWHΣ Energy in = Σ Energy out

m1 h4 + (1 – m1) h7 = 1 (h8)m1 = (h8 – h7) / (h4 – h7)

= (697.0 – 168.23) / (2510.89 – 168.23)

= 0.2257 kg extracted / kg total

m1 = 0.2257 kg extracted /kg total (ms) kg total steam /s

= 0.2257 (27.5)m1 = 6.21 kg /s

WT = ms [(h1-h2) +(h3 – h4) + (1- m1) (h4 – h5)]WT = 27.5 kg/s [(2988.1 – 2802.7) + (3104.9 – 2802.7) + (1- 0.2257) (2802.7 – 1909.47)] kJ/kgWT = 32,428.77 KW = 32.4 MWQA = QBoiler + QReheater

= ms [(h1 – h9) + (h3 – h2)]

= 27.5 [(2988.1 – 705.614) + (3104.9 –2802.7)]

QA = 71,078.86 kWΣWP = WPA + WPB

WpA = wpA (1 – m1) (ms)

= 0.698 kJ/kg (1 – 0.2257) (27.5) kg/s

= 14.89 kWalsoWPB = wpB kJ/kg (ms) kg/s

= 8.614 kJ/kg (27.5) kg/s

= 236.9 kWthen ΣWP = WPA + WPB = 236.9 + 14.86 = 251.76 kWCycle TE = (WT- ΣWp) / QA

Cycle TE = (32,428.77 – 251.76) / 71,078.86

= 0.453

Cycle TE = 45.3 %

Answersextracted steam flow per second = 6.21 kg/sturbine work output in kW = 32.4 MWthermal efficiency of the cycle = 45.3%

1. What is a Regenerative Rankine cycle?

A Regenerative Rankine cycle is a thermodynamic cycle used in power plants to convert heat into mechanical work. It is a modification of the traditional Rankine cycle, where the feedwater is heated by the steam extracted from the turbine, resulting in improved efficiency.

2. How does a Regenerative Rankine cycle work?

In a Regenerative Rankine cycle, the steam from the turbine is extracted at various stages and used to heat the feedwater before it enters the boiler. This preheating of the feedwater reduces the amount of energy needed to heat it in the boiler, resulting in increased efficiency.

3. What are the advantages of a Regenerative Rankine cycle?

The main advantage of a Regenerative Rankine cycle is its increased efficiency compared to a traditional Rankine cycle. This is due to the preheating of the feedwater, which reduces the amount of energy needed to heat it in the boiler. Additionally, the use of multiple extraction points allows for better heat utilization and reduces the temperature difference between the steam and the feedwater, resulting in less energy loss.

4. What are the applications of a Regenerative Rankine cycle?

The Regenerative Rankine cycle is commonly used in power plants to generate electricity. It is also used in other industries, such as in geothermal power plants and waste heat recovery systems, to convert heat into mechanical work.

5. What are the limitations of a Regenerative Rankine cycle?

One of the limitations of a Regenerative Rankine cycle is the complexity of its design and operation. It requires more equipment and control systems compared to a traditional Rankine cycle, which can increase the cost of implementation. Additionally, the efficiency gains are limited by the temperature difference between the steam and the feedwater, which can be a constraint in certain applications.

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