Energy Loss from Regulator 2

[triage]

If I have a canister of compressed air, at To and 300 bar, and expanded it through an ideal turbine to 1 bar, (into the atmosphere), like discharging a capacitor, the energy produced would be Eo.

If I placed a 6 bar regulator on to the same canister at To and 300 bar, then fed the output at 6 bar to the same ideal turbine expanding again to 1 bar, how would this Energy output compare to the original Eo, and is there a calculation based on Pressures / Temperatures that can be made directly?

 

[pmover]

look at the air tables . . . which are similar to steam tables (perhaps you remember steam tables). your answers can be determined from the air tables. for the regulator, it is a constant enthalpy process.

hope this helps . . .
good luck!
-pmover

 

[triage]

Thank you pmover,
I do remember steam tables. I will consult air tables, which I do have a book on.

What it seems to indicate, is that if a regulator takes me down in pressure with constant enthalpy, its temperature will drop per Joule Thompson, however its overall ability to do work through isentropic expansion (constant Entropy) is relatively unchanged.

Yes?

 

[iainuts]

My understanding is you want to compare a turbine expanding from To and 300 bar down to 1 bar -
- to a turbine expanding from To and 6 bar down to 1 bar.

As pmover mentioned, the expansion across the regulator is isenthalpic (ie: it is NOT isentropic). There's a temperature drop of roughly 80 F associated with that, but the entropy is not the same.

I calculate that for a 100% efficient expander going from 300 barg to 1 barg with 70 F inlet, power is 119 hp for a flow of 1 lbm/s.

Similarly, for a 100% efficient expander going from 6 barg to 1 barg with 70 F inlet, power is 54 hp for a flow of 1 lbm/s.

If however, you don't heat up the air after expanding from 300 barg to 6 barg through your regulator, temperature exiting the regulator and going into the expander is about -10 F and power drops to 45 hp for a flow of 1 lbm/s.

So going from 300 barg to 6 barg on the inlet of the expander doesn't drop the power linearly, but it certainly curtails the amount of power you can obtain. Also, not heating the air back up after it's gone through the regulator makes another significant (negative) impact on the expander.

 

[triage]

So from your explanation, I lose over half of the available energy, 119 hp to 45 hp.In my scenario, there will be no heat added after the regulator, however in each case, the source container is at 300 bar and To, conditions.I also understand how to calculate the potential work available from the source container, using isentropic expansion of an ideal turbine.What I don't understand is the equation for work output from the same canister at 300 bar, To; through a 6 bar regulator then through a turbine.

If pmover is correct and the regulator expands the air from 300 bar to 6 bar with constant enthalpy, then I see from the air tables recommended, that the isentropic expansion work remains about the same.

iainuts poses that there will be a significant loss in the ability of the compressed air to do work, from 119hp to 45. Can you reference the equation or method used for this please?

 

[iainuts]

Briefly, method used is simply application of first law of thermo. The problem is always with having fluid properties, not the problem of solving the equations.

First law for an expander reduces to:
dU = Hin – Hout + Qin – Qout + Wout

Neglecting heat transfer and noting that no energy is stored within the expander control volume (dU = 0) this results in:

Wout = Hin – Hout

So all you need is Hin and Hout to determine work. The rate of flow gives power.

Power = Wout * mdot

Now all you need are the various states and you’re done. That’s where a good computer program that can access fluid properties comes in. There are a few on the market such as the NIST REFPROP database but I have a proprietary one that I use. Here are the values I have for fluid properties of air:
Condition 1: 300 barg, 70 F
H = -629.85 Btu/mole
S = 34.95 Btu/mole F

Condition 2: 6 barg, -9.4 F
H = -629.85 Btu/mole
S = 42.34 Btu/mole F

Condition 3: 6 barg, 70 F
H = -71.02
S = 43.48

Condition 4: 1 barg, -302 F
H = -3066.16
S = 34.95

Condition 5: 1 barg, -146 F
H = -1559.91
S = 42.34

Note that properties such as these are not absolute, they are relative, so your database may differ significantly in the absolute values listed, but the difference between those values should be the same.

For the case of 300 barg expanding to 1 barg, W = H(1)-H(4) (where number in paren is condition #). Note S(1) = S(4)

For the case of 300 barg regulated down to 6 barg, note H(1) = H(2)

For the case of 6 barg expanding to 1 barg, W = H(2) – H(5). Note S(2) = S(5)

Hope that helps.

 

[pmover]

good job iainuts!

a star is given . . .

-pmover

 

[triage]

I agree pmover, good job iainuts
:-D

 

[dcasto]

In real terms, I don't think you can find a 300 to 1 turbine. A two stage with 17 to 1 whould be possible. And the air had better be dry, the exit temp from a 60 deg F source will be -312 F (at 80% expander eff).

 

[triage]

good point dcasto,
In honesty, the scenario was hypothetical so that I could understand the effect on isentropic work from regulation from a high pressure to a lower pressure with comensurately greater volume.

The contributions gave me plenty to contemplate, although the iainuts solution treats the problem as two steps, whereas the system will be continuous.

In both cases, the the original canister will expand from 300 bar to 1 bar, driving a turbine. In case 1, the turbine will have 300 to 1 expansion with less volume, and in case 2, the turbine will have 6 to 1 expanion with more volume. What I am trying to understand from this scenario is the effect of using a regulator in the system on the resultant work output of the turbine.

Iainuts case treats the system like the entire canister was expanded through a regulator to 6 bar, with a 79.4 degree temp drop to a total new volume, then that volume, at 79.4 degrees, is expanded through a turbine to -146 degrees.

In reality, the canister is going to be droping in temperature during the entire course, so the very first stream of air out of the regulator will be -9.4 degrees, but as the canister air cools from expansion, so will the air from the regulator.