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Convection

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Question

The steam consumption of a steam power plant is 40 tonnes per shift of 8 hours when developing 300 kW power.
Saturated steam enters the turbine at 16 bar and leaves at 0.2 bar. The properties of steam are given below.

At 16 bar ,

$h_{g} = 2976.2 k J / k g$

$S_{g} = 6.282 k J / k g K$

At 0.2 bar ,

$h_{f} = 180.1 k J / k g$

$s_{f} = 0.612 k J / k g K$

The steam at outlet of turbine has 11% moisture in it. Neglect pump work thermal efficiency of the plant is

A

6.82%

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B

7.72%

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C

8.41%

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D

9.86%

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Solution

The correct option is B 7.72%
$M_{s} = \frac{40 t o n n e s}{8 h o u r s} = 5 t o n e s / h r = 5000 k g / h r$

$h_{1} = h_{g} a t 16 b a r = 2976.2 k J / k g$

$h_{3} = h_{f} a t 0.2 b a r = 180.1 k J / k g$

$Power = 300 k W$

$Thermal efficiency = \frac{Power}{Heat supplied}$

$η_{t h} = \frac{3600 \times P}{m_{s} (h_{1} - h_{f s})}$

$= \frac{3600 \times 300}{5000 (2976.2 - 180.1)} \times 100 % = 7.72 %$

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