Water at 400 kPa with a quality of 75% has its pressure raised 200 kPa (to 600 kPa) in a constant volume process. What is the new state (X, T) ? At what pressure does the water go into the single phase region.

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