a) We use the table of saturated water - temperature . In the quality region the pressure as "p = 3976.2 kPa"
b) To find the mass of the vapor we must determine the quality. We use the next equation-
"U = Uf + x(Ug - Uf)"
Then,
"0.3 m3\/kg = 0.00125 m3\/kg + x( 0.05008 - 0.00125 ) m3\/kg"
"0.3 = 0.00125 + 0.04883x"
"0.04883x = 0.3 - 0.00125"
"0.04883x = 0.29875"
"x = 0.29875 (m3\/kg) \/ 0.04883 (m3\/kg)"
"x = 6.11816"
Using the relationship of-
"X = mg \/ m ,"
we find the vapor mass
"mg = xm = 6.11816 \\times10 kg = 61.1816 kg"
To find the enthalpy of the vapor-
Using the relationships
"dinP = -\\frac{d(AHvap)}{R}\\frac{1}{T}"
"\\Delta \u041dvap = (\\frac{1}{3976.2}) \\times (523?) \\times8.314"
"\\Delta \u041dvap = 571.93\u0408"
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