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.

a. The rate of heat transfer in the heat exchanger is equal to 484 kW.

b. The exit temperature of water is equal to 80.71°C.

Given the following data:

a. To determine the rate of heat transfer in the heat exchanger, we would apply the law of conservation of energy:

Thus, rate of heat transfer from oil in the heat exchanger is given by the formula:

Heat transfer = 484 kW

b. To determine the exit temperature of water:

Heat lost by oil is equal to the heat gained by the water according to the law of conservation of energy:

Substituting the given parameters into the formula, we have;

Exit temperature = 80.71°C

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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

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explanation: