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Internal Energy

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Question

The steam turbine in a Rankine cycle power plant receives steam at 18 bar , $400^{\circ} C$ and leaves the turbine at 0.11 bar , 10% moisture.

At 18 bar , $400^{\circ}$ C
$h = 3084.2 k J / k g$
$s = 6.81 k J / k g K$

At 0.11 bar , $10 % m o i s t u r e$
$h = 2420 k J / k g$
$h_{f} = 190 k J / k g K$

If the indicated thermal efficiency of the plant is 21.40% What is the efficiency ratio of the plant ?

A

0.7801

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B

0.9344

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C

0.7144

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D

0.8204

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Solution

The correct option is B 0.9344

Given data:

$h_{1} = 3084.2 k J / k g$

$h_{2} = 2420 k J / k g$

$h_{3} = h_{f} a n d 0.11 b a r = 190 k J / k g$

$w_{F} = v (Δ P)$

$= 0.001 \times (18 - 0.11) \times 100 = 1.789 k J / k g$

$w_{p} = h_{4} - h_{3}$

$h_{4} = h_{3} + w_{p} = 190 + 1.789 = 191.789 k J / k g$

$Hence Rankine efficiency = \frac{(h_{1} - h_{2}) - w_{p}}{(h_{1} - h_{4})}$

$= - \frac{(3084.2 - 2420) - 1.789}{(3084.2 - 191.789)} = 0.229$

$∴ efficiency ratio = \frac{indicated thermal efficiency}{Rankine efficiency} = \frac{0.2140}{0.229} = 0.9344$

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