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Design of Heat Exchanger - NTU Method

In counter fl...

Question

In counter flow heat exchanger if the heat capacity ratio is one, and NTU is 3, the effectiveness of the heat exchanger is

A

0.25

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B

1

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C

0.5

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D

0.75

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Solution

The correct option is D 0.75
$C = \frac{C_{m i n}}{C_{m a x}} = 1 (g i v e n); N T U = 3$

When C= 1 , effectiveness of counter flow heat exchanger will be become

$ϵ = \frac{N T U}{1 + N T U} = \frac{3}{1 + 3} = \frac{3}{4} = 0.75$

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