Regenerative Rankine cycle with one contact feedwater heater

In summary, the required steam generated from the boiler per hour is not explicitly given. The extracted steam per hour for feedwater heater can be found by solving for the mass flow rate using the given enthalpies and assuming steady-state conditions. The thermal efficiency of the cycle can also be solved using the given enthalpies and the formula for thermal efficiency. The key to solving for A and B is the given 50 MW power output.
  • #1
9
1
Homework Statement
a) required steam generated from the boiler per hour
b) extracted steam per hour for feedwater heater
c) thermal efficiency of the cycle
Relevant Equations
rankine cycle equations
My question is how do i get :
A. required steam generated from the boiler per hour
B. extracted steam per hour for feedwater heater
C. thermal efficiency of the cycle
I can solve the thermal efficiency but A and B is giving me difficulties, i know when solving the extraction of steam needs mass flow rate but in the given there isnt. I am assuming the 50 MW is the key to solving these two ? I still don't know the exact formula in finding A and B but I attached my solutions here although its still on finding the 9 enthalpies. I just need to know the correct formula in solving for A and B. Thanks in advance

Heres my problem set:
In a 50 MW steam power plant operating on regenerative Rankine cycle with one contact feedwater heater, steam enters the turbine at 10 MPa, 300 C and condensed in the condenser with a pressure at 10 kPa. Expansion of steam at high-pressure stage turbine up to saturation at 3.0 MPa then reheated to 300 C. After partial expansion at low-pressure stage turbine to 750 kPa, some steam is bled at this pressure for feedwater heating while the remaining steam is expanded to condenser pressure. Determine:

  • required steam generated from the boiler per hour
  • extracted steam per hour for feedwater heater
  • thermal efficiency of the cycle
IMG_20220408_142032.jpg
IMG_20220408_142039.jpg
 
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  • #2
My solution:Enthalpy at state 1 (Boiler): h1 = 3,466.3 kJ/kgEnthalpy at state 2 (High-pressure turbine inlet): h2 = 3,080.1 kJ/kgEnthalpy at state 3 (High-pressure turbine outlet): h3 = 2,510.5 kJ/kgEnthalpy at state 4 (Reheater): h4 = 3,080.1 kJ/kgEnthalpy at state 5 (Low-pressure turbine inlet): h5 = 2,785.7 kJ/kgEnthalpy at state 6 (Low-pressure turbine outlet): h6 = 2,385.6 kJ/kgEnthalpy at state 7 (Feedwater heater inlet): h7 = 2,510.5 kJ/kgEnthalpy at state 8 (Feedwater heater outlet): h8 = 2,660.3 kJ/kgEnthalpy at state 9 (Condenser): h9 = 1,180.7 kJ/kgThermal efficiency of the cycle:η = (h1 - h9) / (h1 - h2) = (3,466.3 - 1,180.7) / (3,466.3 - 3,080.1) = 0.228 = 22.8%
 

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