Respuesta :
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]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
