Wet steam expansion 1

[andyross333]

I believe this should be a straightforward calculation but it has been a long time since I've done this stuff and am having trouble wrapping my head around it.

I have one tank at equilibrium that is partially filled with 240C water at 33 bar (state 1). A valve is opened to a second equivalent sized tank that is empty. I need to calculate the final temperature and pressure once the new equilibrium is reached (state 2). The tanks can be considered adiabatic.

Thanks.

 

[ione]

You do have conditions (pressure and temperature) which identify state 1. From steam tables you can get the specific volume and, knowing the volume of tank 1, you can calculate the mass of your fluid. Now after opening the valve when the equilibrium is reached, that mass will occupy a new volume (volume of tank 1 + volume of tank 2). You can easily calculate the new specific volume and find conditions of state 2 from the tables.

 

[zekeman]

Not so simple.
You have to write 2 equations
1 conservation of weight
2 conservation of energy

1 weight of steam(1) + weight of water(1)= weight of steam(2) + weight of water(2)

2 internal energy Of steam(1)+internal energy of water(1)=internal energy Of steam(2)+internal energy of water(2)

Best way I have found is to get v(T), u(T) from steam tables along saturation line so the 2 equations have 2 variables, T and water(2)

Might do it iteratively,by first assuming the steam pressure drops according to standard gas laws. Then find the incremental effect of the evaporated water.

 

[sailoday28]

What is initially in the empty tank and at what state conditions?
Where is the outlet on the source tank? Is it below or above liquid level?
Regards

 

[zekeman]

As an additional hint,if the water in the tank is not much greater than the weight of the steam, and use the first iteration, to get the new temperature.
From this you could write the amount of water flashed to steam would be
(T2-T1)/hg*W
For example, if the expansion tank were equal to the original tank, then 1st iteration would change the temperature from 464F to 400F and
(464-400)W=1115delta W
deltaW=64/1115= .057

 

[zekeman]

correction
deltaW=64/1115= .057
should be
deltaW=64/1115= .057W

 

[zekeman]

The reason that the amount of water is important and must be included is that if you consider a huge amount, then, any change of volume would not drop the water temperature that much so the system would essentially remain at the initial state and enough water would flash off to fill the new volume.

 

[zekeman]

If I may,let me show an example

Consider a volume of 1 lb of steam at saturated conditions 33 bar (465 F)
, you initial state. Let the volume of the expansion be V2=V1 the same as the volume,V1 (V1=0.947 from steam tables) of steam in the initial state. Let there be 2 lb saturated water

First assume gas laws for 1st iteration. Then, the new sat temperature is
400 F.
The amount of water flashed would be
447*W=373(W-delW)+1115delW=373*W+delW*(ufg)=373W-delW*742
delW=(447-373)*W/742
For W=1
I get
delW=0.097 lb
Now we have 1.097 lb steam which would have a specific volumne of
2V1/1.097=(2*.947)/1.097=1.586=vg
using steam tables, the new sat point is T=415 The new delW would be
delW=(447-391)/726W=0.077lb
vg=2V1/1.077=1.615
new T=413 F; p=287psi= 19.5 bar
Another iteration would get only slightly closer.

 

[rmw]

First off, did you go back to school? This sounds like a question right out of a first semester thermo textbook. What you didn't tell us was whether the valve was between the vapor space or well below the level of the water. And in the case of the latter if the vapor spaces were connected. And at equilibrium with what? Is the first tank at equilibrium with water vapor only or with a mix of air and ater vapor, and is the second tank under a vacuum, or atmospheric pressure full of air.

So many possibilities...

rmw

PS: but reading back through before posting, I guess Sailoday already ask all this.

 

[racookpe1978]

Yes, those were my questions as well:

Did the first tank, once opened, bleed saturated steam to the empty tank, or did it bleed water?

If it bled steam (or water), when was the second tank vented? Or are we to assume that the second tank - and now the first when they equalize - needs to accomodate that unvented air? (which will change the final pressure)

When the steam (or water) hit the second tank, what/how much (how large are the tanks with what wall thicknesses) was needed to raise the temperature of the second tank walls - we are told it (the systems ?) is adiabatic, so the text (probably) wants us to ignore the energy needed to heat the second tank (which must come from the energy in the sat steam in the first tank), but you can't really ignore it. I've bled too many steam pipes and turbines and drains for too long during cold plant startups to ignore that much water mass & energy & heat ... 8<)

 

[ione]

I agree one should know the steam dryness in its initial state, but I’m wondering weather we can consider the enthalpy of the initial state equal to that of the final state. If the valve connecting the two tanks acts as a throttling valve, then this should be plausible (adiabatic process with no works nor any sensible increase of kinetic energy). The process is not properly isenthalpic, as enthalpy won’t remain constant through the whole process, but initial and final enthalpy can be considered equal with a very good approximation.

 

[sailoday28]

A simple energy balance will yield Integral [hdm]= mu -(mu)i (1)
where h is stagnation entalpy of souce tank of exiting fluid, dm is decrease in mass of source tank
mu is total internal enery of receiver with i, the initil conditions in receiver.
If steam is vented from source tank, I beleive including quality will be difficult.

As a start
If saturated steam is being vented from the source tank
get steam table data for enthalpy vs pressure, say from 33 bars to 28 bars. If the variation in entalpy is small then approximate the above integral in (1)
Also if the receiver tank starts out with low pressure, again use steam tables and Daltons Law to approximate right hand side.
Yes--there are a lot of ifs.

Regards