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12 November, 23:55

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.

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  1. 13 November, 02:35
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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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