Answer:
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Explanation:
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Answer:
T=194.3C
X=1
P=542.5kPa
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties
First we find the specific volume knowing the temperature and quality of state 1
v(p=400kPa, x=0.75)=0.3471m^3/kg
As a constant volume process is presented, it keeps constant in state 2, so we find the temperature and quality with p = 600kPa and v = 0.3471m ^ 3 / kg
T(P=600kPa, v=0.3471m ^ 3 / kg)=194.3C
x(P=600kPa, v=0.3471m ^ 3 / kg)=1
for the second part we find the pressure, with a quality of 1 and a volume of
v= 0.3471m ^ 3 / kg
p=542.5kPa
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Answer:
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