Saturated steam at 100°C is condensing on shell side of a shell and tube heat exchanger. The cooling water enters the tube at 30°C and leaves at 70°C. Calculate arithmetic mean temperature difference in counter flow arrangement.

Question

Question

This question was previously asked in

MPPSC AE Mechanical 2021 Official Paper

Answer 1

Concept:

For the phase change heat exchanger, the LMTD of parallel flow and counterflow will be the same. So taking counter-flow heat exchanger:

F3 Madhuri Engineering 09.05.2022 D8

\(AMTD~=~\frac{θ_1~+~θ_2}{2}\)

where, \(θ_1~=~T_{h_1}~-~T_{c_2}\\θ_2~=~T_{h_2}~-~T_{c_1}\)

Calculation:

Given:

\(T_{h_1}~=~T_{h_2}\) = 100°C, \(T_{c_1}\) = 30°C, \(T_{c_2}\) = 70°C

\(θ_1~=~T_{h_1}~-~T_{c_2}\\θ_2~=~T_{h_2}~-~T_{c_1}\)

\(θ_1~=~100~-~70\) = 30°C, θ₂ = 100 - 30 = 70°C

\(AMTD~=~\frac{θ_1~+~θ_2}{2}\)

⇒ \(AMTD~=~\frac{70~+~30}{2}~\) = 50°C