A reheat Rankine cycle operates with water as the working fluid. Steam enters the first turbine at 8 MPa and 450°C and exits at 0.8 MPa. It

Question

A reheat Rankine cycle operates with water as the working fluid. Steam enters the first turbine at 8 MPa and 450°C and exits at 0.8 MPa. It is then reheated to 400°C before entering the second turbine, where it exits at 10 kPa. If the amount of work into the pump is 8.04 kJ/kg and the net work per cycle produced is 1410.25 kJ/kg, determine the thermal efficiency of the cycle

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Thu Thảo 3 weeks 2021-08-23T23:25:10+00:00 1 Answers 1 views 0

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    2021-08-23T23:27:04+00:00

    Answer:

    The thermal efficiency, \eta _{reheat}, of the Rankine cycle with reheat is 36.81%

    Explanation:

    p₁ = 8 MPa = 80 Bars

    T₁ = 450°C = 723.15 K

    From steam tables, we have;

    v₁ = 0.0381970 m³/kg

    h₁ = 3273.23 kJ/kg

    s₁ = 6.5577 kJ/(kg·K) = s₂

    The p₂ = 0.8 MPa

    T₂ = Saturation temperature at 0.8 MPa = 170.414°C = 443.564 K

    h₂ = 2768.30 kJ/kg

    T_{2'} = 400°C = 673.15 K

    h_{2'} = at 400°C and 0.8 MPa = 3480.6 kJ/kg

    p₃ = 10 kPa = 0.1 Bar

    T₃ = Saturation temperature at 10 kPa = 45.805 °C = 318.955 K

    h₃ = 2583.89 kJ/kg

    h₄ = h_{3f} = 191.812 kJ/kg

    The thermal efficiency, \eta _{reheat}, of a Rankine cycle with reheat is given as follows;

    \eta _{reheat} = \dfrac{\left (h_{1}-h_{2}  \right )+\left (h_{2'}-h_{3}  \right )-W_{p}}{h_{1}-\left (h_{4}+W_{p}  \right )+\left (h_{2'}-h_{2}  \right )}

    Therefore, we have;

    \eta _{reheat} = \dfrac{(3273.23 -2768.30 ) + (3480.6 -2583.89 ) - 8.04)}{(3273.23 -(191.812 + 8.04) + (3480.6 -2768.30 ) } = 0.3681

    Which in percentage is 36.81%.

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