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Enthalpy

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Question

A thermal power plant operates on a regenerative cycle withh a single open feedwater heater, as shown in the figure. For the state points shown, the specific enthalpies are: $h_{1} = 2800 k J / k g$ and $h_{2} = 200 k J / k g$ . The bleed to the feedwater heater is $20$ % of the boiler steam generation rate. The specific enthalpy at state $3$ is

A

720 k J / k g

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B

2280 k J / k g

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C

1500 k J / k g

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D

3000 k J / k g

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Solution

The correct option is A $720 k J / k g$
Given data: $h_{1} = 2800 k J / k g; h_{2} = 200 k J / k g$

Using energy balance equation for open feedwater heater
$m h_{3} = 0.2 m h_{1} + 0.8 m h_{2}$
or $h_{3} = 0.2 h_{1} + 0.8 h_{2}$
$= 0.2 \times 2800 + 0.8 \times 200$
$= 560 + 160 = 720 k J / k g$

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