Energy Loss from Regulator
Energy Loss from Regulator
(OP)
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?
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?





RE: Energy Loss from Regulator
hope this helps . . .
good luck!
-pmover
RE: Energy Loss from Regulator
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?
RE: Energy Loss from Regulator
- 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.
RE: Energy Loss from Regulator
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?
RE: Energy Loss from Regulator
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.
RE: Energy Loss from Regulator
a star is given . . .
-pmover
RE: Energy Loss from Regulator
RE: Energy Loss from Regulator
RE: Energy Loss from Regulator
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.