Question

In a parallel-flow heat exchanger, the NTU is calculated to be 2.5. The lowest possible effectiveness for this heat exchanger is

• A. 91.79 %
• B. 49.66 %
• C. 99.32 %
• D. 10%

Answer 1

The correct option is B 49.66 %

Effectiveness of parallel flow heat exchanger
$ϵ=\frac{1+e^{(1+C)NTU}}{1+C}$
When C = 0, $ϵ=\text{maximum}$
When C = 1, $ϵ=\text{minimum}$
$ϵ=\frac{1+e^{-\left[\right(1+1)\times 2.5]}}{1+1}=49.66\text{\%}$