GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

Question : 59 of 65

Marks: +1, -0

Hot and cold fluids enter a parallel flow double tube heat exchanger at

100^{∘} {C}

and

15^{∘} {C}

, respectively. The heat capacity rates of hot and cold fluids are

C_{h}=2000 {W} \/ {K}

and

{Cc}=1200

{W} \/ {K}

, respectively. If the outlet temperature of the cold fluid is

45^{∘} {C}

, the log mean temperature difference (LMTD) of the heat exchanger is ___________K (round off to two decimal places).

Solution:

Given data

T_{h_{i}}=100^{∘} {C}, T_{c_{i}}=15^{∘} {C}_{'} T_{c_{e}}=45^{∘} {C}

C_{h}=2000 {W} {/} K, C_{c}=1200 W {/} K

Using the energy balance equation for the heat exchanger

C_{h}(T_{h_{i}}-T_{h_{e}})=C_{c}(T_{c_{e}}-T_{c_{i}})

⇒ 2000(100-T_{h e})=1200(45-15)

⇒ T_{h_{e}}=82^{∘} {C}

∆ T_{i}=T_{h_{i}}-T_{c_{i}}=100-15=85^{∘} {C}

∆ T_{e}=T_{h e}-T_{c_{e}}=82-45=37^{∘} {C}

L M T D=\;{∆ T_{i}-∆ T_{e}}/{\ln \;{∆ T_{i}}/{∆ T_{e}}}=\;{85-37}/{\ln \;{85}/{37}}=57.71^{∘} {C}

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