The rate of heat transfer in the heat exchanger is 484 kW
The exit temperature of water is 99.2°C
<u>Explanation:</u>
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