Answer:
initial quality = 0.3690
heat transfer = 979.63 kJ/kg
Step-by-step 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](https://img.qammunity.org/2020/formulas/engineering/college/53ex8rk54gie6czulpvlb4odv4l8xotst6.png)
At Pressure = 20 bar, from saturated water table
![v_f = 0.01177 m^/kg](https://img.qammunity.org/2020/formulas/engineering/college/mk3moqg8arb1vzh8sfaw2u9ilx2wsrsoi8.png)
![v_g = 0.099587 m^3/kg](https://img.qammunity.org/2020/formulas/engineering/college/yw3snlj82qjigaqraxnga3n7kmv91nu0vh.png)
![x = (v -v_f)/(v_g -v_f) = (0.0375 - 0.001177)/(0.099587 - 0.001177)](https://img.qammunity.org/2020/formulas/engineering/college/y26cuyhi8wx0ek2a5ernzkax0xubrru9tl.png)
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))](https://img.qammunity.org/2020/formulas/engineering/college/dzmfhzplqsuhh7390xosmdug48pf84xyqr.png)
from saturated water table, for pressure 20 bar ,
![h_f = 908.79 kJ/kg, h_(fg) = 1890.7 kJ/kg](https://img.qammunity.org/2020/formulas/engineering/college/46l9et8rersr7vf0mpq52943ikgywzipcs.png)
=908.79 + 0.0357(1890.7)
= 979.63 kJ/kg