Cold water Cp 418 kJkgmC is used to cool hot oil in a well

Cold water (C_p = 4.18 kJ/kgm-C) is used to cool hot oil in a well insulated industrial heat exchanger (C_p = 2.0 kJ/Kg_m-C) from 150C to 40C. The water flow rate is 1.75 kg_m/sec and it enters the heal exchanger at 20C. The oil flow rate is 2.0 kg_m/sec. Calculate: the rate of heat transfer in the heat exchanger; and kW the exit temperature of the water (C). degree C

Solution

Hello there.

In this question the inlet and exit temperature of the hot oil is given

ie. T ( inlet oil ) = 150 ⁰ C &  T ( exit oil ) = 40 ⁰ C

while inlet water temperature is given = 20 ⁰ C

C_{p oil} = 2 KJ/kgK

C_{p water} = 4.18 KJ/kgK

mass flow rate water ( m_w) = 1.75 kg / s

mass flow rate oil (m_o) = 2 kg / s

So , Solving

(a) The rate of heat transfer in the heat exchanger

= m_o * C_{p oil *} ( T_{inlet oil} - T_{exit oil} ) =2*2*(150 - 40) = 440 KJ/s= 440 KW ( ANS)

(b) The exit temperature of water

Since the heat released by oil in the heat exchanger = the heat taken by water in the heat exchanger

Therefore, equating both

m_o * C_{p oil *} ( T_{inlet oil} - T_{exit oil} ) = m_w * C_{p water *} ( T_{exit water} - T_{inlet water} )

2*2*(150 -40) = 1.75 * 4.18 * ( T_{exit water} - 20 )

=> T_{exit water} = 80.15 ⁰ C (ANS)

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