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A well-insulated shell-and-tube heat exchanger is used to heat water (cp = 4.18 kJ/kg·°C) in the tubes from 20°C to 70°C at a rate of 4.5 kg/s. Heat is supplied by hot oil (cp = 2.30 kJ/kg·°C) that enters the shell side at 170°C at a rate of 10.8 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of oil.

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Answer:

The heat  transfer can be determine by using water as a system:

[tex]Q=(mcT)_{W} \\=4.5kg/s*4.18kJ/kgC*(70-20)=940.5kW\\[/tex]

To determine the final temperature of the oil is determine from the rate of heat transfer:

[tex]T_{2o} =T_{1o} -Q/(mc)_{o} \\=(170-940.3/10.8*2.3)=129^oC[/tex]

The study of matter is called physics.

The correct answer is 129C

According to the law of thermodynamic, the heat transfer from the high temperature to the low temperature.

[tex]Q =(mcT)w\\=4.5*4.18*(70-20) =940.5[/tex]

Hence, the final temperature of the system change is as follows:-

[tex]T_{20} = T_{10} - \frac{Q}{mc}\\ =(170-\frac{940.3}{10.8*23}) =129^oC[/tex]

Hence, the correct answer is 129.

For more information, refer to the link:-

https://brainly.com/question/20293447

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