Closed const Volume system.

[KesTheHammer]

This is my first post to the list.

The system is a 10 m3 container with 1 m3 water and 9m3 steam at 10 MPa.

It is a closed(no mass flows), constant volume system.

If heat is added to this system (say add 1% of current enthalpy), the pressure and the temperature should rise.

My question is how is this rise in P and T related and how is it calculated?



Kestell Laurie
South Africa

 

[pmover]

very suspicious in that this may be a homework problem . . .

initially have enthalpy h.

add 1% enthalpy to a closed system - i.e. a constant volume

follow constant volume line to enthalpy at h+1%

determine p & t.

-pmover

 

[Montemayor]

Kestell:

I have to ask this: is this an academic or student problem?
I'll assume it isn't until you reply.

You haven't given us all the basic data, but I'll assume you have equilibrium established within the container - in other words, the water is SATURATED (as well as the steam above it) - and there is perfect insulation on the system (no heat leaks). Without going into the mundane math details, I suspect the energy and mass balance will reveal that you have reiterative equations that require you do a trial-and-error exercise to resolve the ultimate, equilibrium condtion at the elevated energy condition.

 

[KesTheHammer]

Thanks for very quick replies.

It is not a student problem. I am trying to better understand what happens in a boiler.

pmover, I also thought it could/should be a homework problem, but could not find an example like that in my textbook. The solution you give exactly what i did. The problem just comes in with that last step.

"determine p and t".

Montemayor, you are entirely correct in your assumptions. I want to understand the simple scenario entirely before moving on.

The method I used with the trial and error is:
I used the heat of evaporation (h_fg) to determine how much water turned to steam. Then I have a known quality (q) of the mixture.

Now I tune the pressure up until the volume equation matches.

V = q*V_g+(1-q)*V_f
Then however the enthalpy equation is off.
h= q*h_g+(1-q)*h_f

Now at this point I know the reason why my enthalpy is off is because my q is not correct (h_fg != constant).

This is how I tried solving it. I was sort of successful but not satisfied with my result. I wondered if there is something simple that I was missing, so I came here.

Kestell


Kestell Laurie
South Africa

 

[Montemayor]

Kestell:

Your logic -or your ability to communicate it- is impressive. Your analysis is also correct. The basic input is that the saturated water's enthalpy is increased. This leads to a an increase in steam generation. However, THIS IS NOT A STEADY STATE PROCESS. It is a closed, batch process. This is what makes it typically hard and tedious to calculate because the system will seek its own equilibrium in accordance with satisfying all the other parameters: the water and steam remain saturated, the increase in vapor inventory increases the pressure (closed system) while the enthalpy remains at its increased value. I believe I just described about 3 or 4 simultaneous equations that have to be resolved. A steady state process is far easier: what goes in comes out. There is no accumulation.

That's why you have the reiterative situation - just as you've described it. The important thing is not the correct, numerical answer. The important item is that you have an accurate and insightful concept of what exactly is happening in a boiler if the the steam outlet is suddenly blocked while the fuel continues to be burned in the furnace.

 

[sailoday28]

With a closed rigid container and the total mass fixed, the specific volume is fixed.

change in enthaly=heat added +INTEGRAL of vdp or
delta h=Q+v(P-Pi)
Where h=enthalpy, P=press, v=spec vol
h-hi =Q +v(P-Pi)
Assuming hi, Q and Pi are known
hf+xhfg -vP= constant

v=vf+xvfg

hf+ (v-vf)/vfg +vP =constant

Above assumes final condition is two phase. With v known

vary Pressure, P, which yields hf, vf,vfg until solution is obtained.
Regards

 

[sailoday28]

correction to my last post.
The last equation hf+ (v-vf)/vfg +vP =constant has a typo

It should be corrected to read

hf+ (v-vf)hfg/vfg - vP =constant

Regards

 

[KesTheHammer]

In Sailoday28's reply he notes that:
h-hi =Q +v(P-Pi)

or rearranged:

Q = delta_h - v(delta_P)

This shows that the heat added (Q) is less than the enthalpy gain.

I don't think my calculation took that into account.

Thanks again for your help. Everyone that made a contribution.

Kestell Laurie
South Africa