Question

A closed, rigid, 0.45 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the pressure is p2 = 4 bar. Determine the initial quality, x1, and the heat transfer, in kJ.

Answer 1

Answer:

initial quality = 0.3690

heat transfer = 979.63 kJ/kg

Explanation:

Given data:

volume of tank 0.45^3

weight of water 12 kg

Initial pressure 20 bar

final pressure 4 bar

Specific volume $v = \frac {0.45}{12} = 0.0375 m^3/kg$

At Pressure = 20 bar, from saturated water table

$v_f = 0.01177 m^/kg$

$v_g = 0.099587 m^3/kg$

$x = \frac{v -v_f}{v_g -v_f} = \frac{0.0375 - 0.001177}{0.099587 - 0.001177}$

inital quality is x =0.3690

Heat transfer is calculated as

$u_1 = h_f + x(h_g - h_f) = v_f + x( h_{fg})$

from saturated water table, for pressure 20 bar ,

$h_f = 908.79 kJ/kg, h_{fg} = 1890.7 kJ/kg$

     =908.79 + 0.0357(1890.7)

      = 979.63 kJ/kg