2 years ago

Can someone please answer this, it will be very appreciative

Engineering

1 answer:

Lisa [10]2 years ago

7 0

Answer:

sorry the image isn't clear

Explanation:

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In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W

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Answer:

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Explanation:

Given:

Temperature $T = 1273.15$ K

Initial Pressure $P_{1} = 1.8$ MPa

Final pressure $P_{2} = 0.1$ MPa

From the table superheated,

$h_{i} = 4635$ $\frac{K J}{Kg}$ and $h_{f} = 2706.54$ $\frac{K J}{Kg}$

Work done by shaft is,

$W = h_{f} - h_{i}$

$W = 2706.54 - 4635$

$W = -1928.46 \frac{kJ}{kg}$

But here efficiency is 0.56,

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Work = $0.56 \times(- 1928.46)$

Work = $-1.3$ $\frac{MJ}{kg}$

Therefore, the shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

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