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

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: