How to assign thermodynamic properties after splitting two phase into gas and liquid

[Chef]

Chillerman, Mad is right and the whole question is just about how to get the properties of the split streams. If I read the previous posts correctly it seems it should be possible to simply move the 2 streams to their respective lines on the bubble but does that meet the no work and no heat input criteria is met. So I propose this:- (which is way too complicated for a Sunday morning I should add)

As no work or heat has been applied to either of the separated streams we can write the state function as

dS=dQ-dW

and the process as dh=dQ-dW+pdv+vdp

As dQ the heat input is zero and dW the work done is zero the dS must be zero

So x*S4-S3+S5-x*S5 = 0

And as both p the pressure and v the volume are constant dh can be set to zero so there is no gain or loss of enthalpy

So m*(h4-h3)-m*(1-x)*(h4-h5) = 0

Where h3 is the enthalpy at a quality of x
And h4 is the enthalpy at the gas line
And h5 is the enthalpy at the liquid line
Also S3, S4 and S5 are the entropy at the same locations.

This boils down to the fact if these state functions are equal to zero then we can indeed split the gas out and position directly on the gas line whilst the liquid must then be plotted on the liquid line with the caveat that both streams must have their respective masses carried with them and new processes can then be started.

But I am still mystified why the gas can have its entropy changed without work or heat input?

Chef

[chillerman2006]

Chillerman, Mad is right and the whole question is just about how to get the properties of the split streams. If I read the previous posts correctly it seems it should be possible to simply move the 2 streams to their respective lines on the bubble but does that meet the no work and no heat input criteria is met. So I propose this:- (which is way too complicated for a Sunday morning I should add)

As no work or heat has been applied to either of the separated streams we can write the state function as

dS=dQ-dW

and the process as dh=dQ-dW+pdv+vdp

As dQ the heat input is zero and dW the work done is zero the dS must be zero

So x*S4-S3+S5-x*S5 = 0

And as both p the pressure and v the volume are constant dh can be set to zero so there is no gain or loss of enthalpy

So m*(h4-h3)-m*(1-x)*(h4-h5) = 0

Where h3 is the enthalpy at a quality of x
And h4 is the enthalpy at the gas line
And h5 is the enthalpy at the liquid line
Also S3, S4 and S5 are the entropy at the same locations.

This boils down to the fact if these state functions are equal to zero then we can indeed split the gas out and position directly on the gas line whilst the liquid must then be plotted on the liquid line with the caveat that both streams must have their respective masses carried with them and new processes can then be started.

But I am still mystified why the gas can have its entropy changed without work or heat input?

Chef

Afternoon Chef

Thanks for such a detailed answer to work with,

(This is going to take me all day to get my head around as I have to reference all the letter symbols first, before i can attempt to work out what you mean.)

Morning MF

But I will, you's have me hooked now.

side tracking slightly away from the original reason for posting as i always look at the final acheivement

the research that you's are doing is for the design of 'A'

the problem i have is where they (liquid/vapour) end up and thus system balanceing

the component that does the system work has a limited ability

with this in mind

the more work it does removing the vapour from 'A'

the less work it can do in the usual way

ideally the liquid will be subcooled before it leaves 'A' moving it away from the bell

but it appears to me that this will reduce overall system ability to cool

and there is a fine line of how much vapour can be removed from 'A' without this reduction

I know you's must have already considered this so,

Just asking, can enough vapour be removed from the liquid to increase cooling without loosing cooling ability in the conventional way ?