Pressure relief question (thermo)

[Smed]

suppose I have a pressurized container filled halfway with water and the rest filled with a noncondensible gas. In one case, I would open a valve at the top and relieve only gas, and in the other case I would open a valve at the bottom and relieve only liquid. If I relieve an equal amount of energy in both of these cases, will the pressure drop by the same amount? What I'm wondering is whether the phase has any impact on pressure or if it's entirely dependent on internal energy.

Thanks.

 

[Mech_Engineer]

The pressure will relieve until the tank's pressure equalizes with the surrounding air. Whether the valve is on top of the tank or on the bottom is immaterial, at least to the simplest approximation.

 

[Smed]

I'm more interested in not relieving the tank's pressure all the way until atmospheric. Suppose this is a 10 ft3 tank at 1000 psia and I relieve 1000 Btu worth of air in case 1 and 1000 Btu worth of water in case 2. Will the tank in both cases end up at the same pressure?

 

[Mech_Engineer]

I'm more interested in not relieving the tank's pressure all the way until atmospheric. Suppose this is a 10 ft3 tank at 1000 psia and I relieve 1000 Btu worth of air in case 1 and 1000 Btu worth of water in case 2. Will the tank in both cases end up at the same pressure?

What do you mean when you say "release 1000 btu"? Do you mean release stored energy for the tank system, thermal energy, molecular energy?

 

[Smed]

I mean releasing stored internal energy by opening the valves.

 

[maddog1964]

suppose I have a pressurized container filled halfway with water and the rest filled with a noncondensible gas. In one case, I would open a valve at the top and relieve only gas, and in the other case I would open a valve at the bottom and relieve only liquid. If I relieve an equal amount of energy in both of these cases, will the pressure drop by the same amount? What I'm wondering is whether the phase has any impact on pressure or if it's entirely dependent on internal energy.

Thanks.

when you say noncondensible gas would that be the same as non-compresable, if so to create the pressure other than amosphereic would it only be the water or fluid that is compressed to create the pressure? or does it work different?

 

[maddog1964]

I mean releasing stored internal energy by opening the valves.

looks like mech enger is offline, work with valves and pressure.. but not the science jargan... but if i understand your thought (and please correct me) the gas is Not compressable so would not afford any energy, so the stored energy would be from the fluid? would that be correct?

 

[Smed]

looks like mech enger is offline, work with valves and pressure.. but not the science jargan... but if i understand your thought (and please correct me) the gas is Not compressable so would not afford any energy, so the stored energy would be from the fluid? would that be correct?

I don't mean that the gas is incompressible, only that it's non-condensable. In other words, it won't change to liquid by interacting with the liquid water. So in this case, it could be Nitrogen. I'm really just trying to make it clear that the top half of the tank is gas, and the bottom half is liquid.

 

[Mech_Engineer]

looks like mech enger is offline, work with valves and pressure.. but not the science jargan... but if i understand your thought (and please correct me) the gas is Not compressable so would not afford any energy, so the stored energy would be from the fluid? would that be correct?

There is far more stored energy in a compressed gas system than a pressurized fluid system at the same pressure. Thing is, releasing fluid from your "combined" system reduces the pressure in the system, meaning the stored energy in the gas goes down anyway. So in a nutshell, what you're asking isn't possible because releasing fluid affects the gas' stored energy anyway.

 

[surrelative]

the energy content of the system is given by the enthalpy i.e H=U+PV, it can be considered as sum of enthalpies of the two phases of the closed system i.e
Hgas=Ugas+P*(Vgas)
Hwater=Uwater + P*(Vwater)
(consider mechanical equm)
Now 'Ugas' depends on 'Cgas'(specific heat) and Uwater depends on Cwater
and Cgas is not equal to Cwater
and density of gas and water are different

Now if on relieving the pressure in the two cases, he same net enthalpy drop of the system occurs, it will still result in different pressures as the changes in internal energies and volumes will be different due to unequal specific heat and densities of the two fluids.

try to deduce results using properties of known fluids.
The pressure relieving process for the gas will be isochoric resulting in cooling of the system and for relieving water it will be a simple quasi-static adiabatic expansion of the gas(similar to an expanding piston), the internal energy of water won't change appreciably.

 

[xandro101]

Good answer, surrelative. Others have been out of point.

 

[xandro101]

But the pressure relieving process for the gas is not isochoric as the exit of water at the bottom provides extra volume for gas to occupy.

 

[surrelative]

the process is isochoric only in case 1 when only the tap to relieve gas has been opened. if you open the taps for gas and water simultaneously, then it would be a polytropic process.

 

[xandro101]

That's true. I agree. Thanks.