Question

A thin-walled double-pipe counter-flow heat exchangeris used to cool oil (cp=2.20 kJ/kg·°C) from 150 to 40°Cat a rate of 2 kg/s by water (cp=4.18 kJ/kg·°C) that enters at22°C at a rate of 1.5 kg/s. Determine the rate of heat transferin the heat exchanger and the exit temperature of water.

Answer 1

The rate of heat transfer in the heat exchanger is 484 kW

The exit temperature of water is 99.2°C

Explanation:

Given -

Specific heat of oil, Coil = 2.2 KJ/Kg°C

Specific heat of water, Cwater = 4.18 KJ/Kg°C

ΔEsystem = Ein - Eout

Where E is the energy

Here, change in kinetic and potential energy of the fluid stream are negligible.

Therefore,

ΔEsystem = Ein - Eout = 0

Ein = Eout

mh₁ = Qout + mh₂ (since Δke ≅ Δpe ≅ 0)

Qout = mCp (T1 - T2)

The rate of heat transfer from oil -

Q = [m X Coil (Tin - Tout)]

Q = (2 kg/s) (2.2 KJ/Kg°C) (150°C - 40°C)

Q = 484 kW

Heat lost by oil is gained by water.

Therefore, the outlet temperature of the water is

Q = [m X Cwater (Tout - Tin)]

Tout = Tin + Q/ m X Cwater

Tout = 22°C + 484 kJ/s/ (1.5 kg/s) (4.18 kJ/kg°C)

Tout = 99.2°C