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.

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

tallinn tallinn · Answer 2 · 2022-02-15T20:55:42+01:00

The answer is T=194.3C

X=1

So the P is=542.5kPa

Laboratory tests

When Through the laboratory tests, then thermodynamic tables were developed, also these are allowed to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc ..)

When through prior knowledge of two other properties are:

First, we need to find the specific volume knowing the temperature and also the quality of state 1

After that v(p=400kPa, x=0.75)=0.3471m^3/kg

Then As a constant volume process is presented, it keeps constant in state 2, so we find the temperature and also quality with p = 600kPa and v = 0.3471m ^ 3 / kg

After that T(P=600kPa, v=0.3471m ^ 3 / kg)=194.3C

Then x(P=600kPa, v=0.3471m ^ 3 / kg)=1

Then for the second part, we find the pressure, with a quality of 1 and also a volume of

v= 0.3471m ^ 3 / kg

Thus, p=542.5kPa

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