Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
[tex]\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W[/tex]
[tex]\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})[/tex]
[tex]h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}[/tex]
[tex]2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}][/tex]
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
