Question

An adiabatic heat exchanger receives 3 kg/s of R12 as a saturated vapor at 960 kPa and condenses it to a saturated liquid at this pressure. The cooling water at 100 kPa enters the heat exchanger at 20 C and leaves at 35 C. Determine the heat transfer to the water in kW and the water flowrate in kg/s.

Answer 1

a)
`\dot Q = 390\,kW`, b)
`\dot m = 6.211\,(kg)/(s)`

Step-by-step explanation:

a) The heat transfer to the water is:

`\dot Q = \dot m_(r) \cdot h_(fg)`

`\dot Q = (3\,(kg)/(s) )\cdot (130\,(kJ)/(kg) )`

`\dot Q = 390\,kW`

b) The mass flowrate of water is:

`\dot m = (\dot Q)/(c_(p,w)\cdot (T_(f)-T_(o)))`

`\dot m = (390\,kW)/(\left(4.186\,(kJ)/(kg\cdot ^(\textdegree)C) \right)\cdot (35^(\textdegree)C-20^(\textdegree)C))`

`\dot m = 6.211\,(kg)/(s)`