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4vir4ik [10]
3 years ago
10

Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater heater. Superheated vapor enters the

turbine at 10 MPa, 480 C and the condenser pressure is 6 kPa. Steam expands through the first stage turbine where some is extracted and diverted to a closed feedwater heater at 0.7 MPa. Condensate drains from the feedwater heater as saturated liquid at 0.7 MPa and is trapped into the condenser. The feedwater leaves the heater at 10 MPa and a temperature equal to the saturation temperature at 0.7 MPa.
For the cycle of problem 8.49, reconsider the analysis assuming the pump and each turbine stage have isentropic efficiencies of 80%.

Determine:

a) the rate of heat transfer to the working fluid passing through the steam generator in kJ per kg of steam entering the first stage turbine.

b) the thermal efficiency

c) the rate of heat transfer from the working fluid passing through the condenser to the cooling water in kJ per kg of steam entering the first stage turbine.
Engineering
1 answer:
Oksanka [162]3 years ago
3 0

Answer:

<u>note: </u>

<u> solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment</u>

Download docx
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Answer:

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Explanation:

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∴ 1100 - t_{A2} = 1100/3

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\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

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t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

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Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

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Answer:

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