Engineering, 26.02.2020 18:04 bri1710

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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Answer from: gabriellax995

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:

Specific heat capacity of oil = 2.20 kJ/kg°C.Specific heat capacity of water = 4.18 kJ/kg°C.Initial temperature = 40°C.Final temperature = 150°C.Rate for oil = 2 kg/sTemperature in = 22°CRate for water = 1.5 kg/s

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

$\Delta E = E_{in} -E_{out} =0$

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

$Q_o = m_oc_o\theta\\\\Q_o= 2 \times 2.2 \times (150-40)\\\\Q_o = 4.4 \times 110$

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:

$Q_o = Q_w\\\\Q_o = m_wc_w(T_{out} - T_{in})\\\\T_{out}=\frac{Q\; +\;T_{in} }{m_wc_w}$

Substituting the given parameters into the formula, we have;

$T_{out}=\frac{484\; +\;22 }{1.5 \times 4.18}\\\\T_{out}=\frac{506}{6.27}$

Exit temperature = 80.71°C

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A thin-walled double-pipe counter-flow heat exchangeris used to cool oil (cp=2.20 kJ/kg·°C) from 150...

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