Thermodynamics-T& P & v relationship

[yecko]

Homework Statement

Homework Equations

The Attempt at a Solution

Why is the temperature of the first state is 133.52^oC?

If at 300kPa & 133.52^oC, from the ‘Saturated water—Pressure table’, it would give vf=0.001073m3/kg, far different from the specific volume as stated in the question which is 0.5m3/kg. Why don’t we need to consider the specific volume v1 in this question?

If mass, temperature and pressure are all stated the same, is it reasonable that the specific volume is much larger than the theoretical one?

Thank you very much!

 

[Chestermiller]

What is the specific volume of saturated water vapor at 300 kPa?

 

[yecko]

vf:0.001073 vg:0.60582
do you mean it is a mixture of liquid and vapor?
but why is the temperature just equal to the sat. temp. @300kPa, regardless the specific vol.?

 

[Chestermiller]

vf:0.001073 vg:0.60582

I asked the question to provide a hint for solving your problem. Apparently, you didn't get the hint. Think about it.

 

[yecko]

do you mean it is a mixture of liquid and vapor?
but why is the temperature just equal to the sat. temp. @300kPa, regardless the specific vol.?

 

[Chestermiller]

do you mean it is a mixture of liquid and vapor?
but why is the temperature just equal to the sat. temp. @300kPa, regardless the specific vol.?

Yes. That's what I mean. The 0.5 is the weighted average of the vapor and liquid specific volumes.

 

[yecko]

but why is the temperature just equal to the sat. temp. @300kPa, regardless the specific vol.?
as for v=0.5, isn't that should increase the temperature first before all molecules achieving sat. temp., then have latent heat change in state 2 for the liquid part change to gas state?
what i mean is that isn't the state 1 temp should be lower than 133.5^oC, then it achieve 133.5^oC only in state 2?
thanks

 

[Chestermiller]

but why is the temperature just equal to the sat. temp. @300kPa, regardless the specific vol.?
as for v=0.5, isn't that should increase the temperature first before all molecules achieving sat. temp., then have latent heat change in state 2 for the liquid part change to gas state?
what i mean is that isn't the state 1 temp should be lower than 133.5^oC, then it achieve 133.5^oC only in state 2?
thanks

If the liquid is saturated, at equilibrium, the vapor must be saturated too. In this system, the average specific volume of the mixture of liquid and vapor is somewhere between that of saturated liquid and saturated vapor. This means that part of the water is saturated liquid and part of the water is saturated vapor.

In going from state 1 to state 2, all the remaining liquid evaporates, and the temperature and pressure remain constant.