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

Answer:

a) $\dot Q = 390\,kW$, b) $\dot m = 6.211\,\frac{kg}{s}$

Explanation:

a) The heat transfer to the water is:

$\dot Q = \dot m_{r} \cdot h_{fg}$

$\dot Q = (3\,\frac{kg}{s} )\cdot (130\,\frac{kJ}{kg} )$

$\dot Q = 390\,kW$

b) The mass flowrate of water is:

$\dot m = \frac{\dot Q}{c_{p,w}\cdot (T_{f}-T_{o})}$

$\dot m = \frac{390\,kW}{\left(4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (35^{\textdegree}C-20^{\textdegree}C)}$

$\dot m = 6.211\,\frac{kg}{s}$