Question

In a counterflow heat exchange water is heated by flue gas. Heat capacity rate of water and gas are 5.86 kJ/kg and 3.52 kJ/kg respectively. The water enters at ${40}^{\circ }C$ and leaves at ${70}^{\circ }C$. If gas enters at ${110}^{\circ }C$, heat exchanger's LMTD is

• A. 333.05 K
• B. 301.85 K
• C. 28.85 K
• D. 244.15 K

Answer 1

The correct option is C 28.85 K



$C_c(T_{\infty }-T_0)=C_h(T_{h_i}-T_{h_0})$

$5.86\times [70-40]=3.52(110-T_{h_0})$

$T_{h_0}\simeq {60}^{\circ }C$

${\theta }_i={40}^{\circ }C$, ${\theta }_0={20}^{\circ }C$

LMTD $=\frac{40-20}{\text{ln}\left(\frac{40}{20}\right)}={28.85}^{\circ }C$

There will be no change in value of LMTD in Kelvin because it is temperature difference