Question

Asked Nov 15, 2019

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- A heat exchanger is being investigated as a waste heat recovery device. A heat exchanger is common device for using a hot fluid to heat a cold fluid without the fluids mixing. The following information is known. The cold fluid stream of liquid A enters at 294.2 K and leaves the device at a temperature of 320.91 K. Liquid A flows at a rate of 0.006 Kg/s and has a specific heat of 4180 J/(Kg K). Liquid B enters the device at a temperature of 350.2 K. Liquid B flows at a rate of 0.005 Kg/s and has a specific heat of 3900 J/(Kg K The dead state is at 293.2 K and 1 bar. The device is adiabatic and the pressure losses are neglected in this problem. Both liquids are incompressible materials.

- Using the first law determine the exit temperature of Liquid B.
- Calculate the rate of exergy destroyed in the heat exchanger
- Calculate the second law efficiency for this device. The exergy product is the exergy increase in liquid A and the exergy input is the exergy change in liquid B.
- If the exiting liquid B is not used for any useful purpose, calculate the exergy destroyed associated with it.

Step 1

Since you have posted a question with multiple subparts, we will solve the first three subparts, i.e. 1, 2, and 3 for you out of the 4 asked. To get the remaining subparts 4 solved, please re-post the question with the remaining subparts.

Consider the case for cold fluid.

Here,

Step 2

Consider the case for hot fluid.

Here,

Step 3

1.

Use the heat balanc...

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