Power equation vs Joule-Thompson effect vs Ideal gas law

[DDIV]

It all started with question around heat dissipation of shaft power, that I think got it sorted out thanks to this forum.

But another question came up to surface when a somebody starts to talk about the Joule-Thompson effect.

So if we consider following (this is the actual case):
Ducting with a square area of 1 m2
Air velocity = 10 m/s = 10 m3/2
Air temp = 20 degree Celsius, assume density 1.2 kg/m3
In the middle of the ducting there is an obstruction causing 1000 Pa in pressure drop, so we have same mass flow in and out.
Then I mean that due to this there will be an increase of temperature after the obstruction of:
Power loss over obstruction = volumeflow*dP = 10*1000 = 10 kW
The obstruction is well insulated, therefore I say that this power loss will be transformed into a temperature raise:
dT = power loss / (massflow * cp) = 10 / (12*1.005) = 0.83 degree Celsius.
After the obstruction the temperature is 20.83 degree Celsius.
The above example must be suited to classify as an ideal gas (compression factor roughly 0.9999).

This was my thinking until somebody started to talk about Joule-Thomson effect.
I found here that if you go from 6 atm to 1 atm there is temperature drop around 2 degree Celsius due to the Joule-Thompson coefficient for air at 20 degree.
If I take my example above but change the obstruction to cause a pressure drop of 5 atm.
Then the temperature raise should be roughly 414 degree Celsius:
Power loss over obstruction = volumeflow*dP = 10*500 000 = 5000 kW
dT = power loss / (massflow * cp) = 5000 / (12*1.005) = 414 degree Celsius.
But the Joule-Thomson effect says a temperature drop on roughly 2 degree Celsius.
The above example is almost an ideal gas (compression factor roughly 0.998).

Ideal gas law also states that there should be a temperature drop after the obstruction, but maybe this is not applicable due to that its not a closed volume. I’m thinking of a moving boundary (closed volume) that goes from inlet to outlet and apply ideal gas law to it without adjusting for compression factor.

So my question comes down to following, why there is different outcome depending on approach:
Power equation = 20+414 = 434 degree Celsius (big increase of temperature)
Joule-Thompson effect = 20-2 = 18 degree Celsius (drop of temperature)
Ideal gas law = 6/20=1/x --> x= 3.33 degree Celsius (drop of temperature)

 

[Latexman]

A 6 atm to 1 atm obstruction will result in choked flow at roughly 3 atm, which will be accompanied by a shock wave / pressure discontinuity at the choke. You need to include all that into the evaluation too.

IMHO, a 5 atm pressure drop compared to a 0.01 atm pressure drop is not even close to an apples to oranges comparison.

Good Luck,
Latexman

 

[DDIV]

Thank you Latexman.

It's nice to hear some more deep knowledge to the phenomena "inside" the Joule-Thompson effect.

If we look at the first example when comparing the power equation vs the ideal gas law.
In my example I get a temperature increase of 0.83 degree Celsius, but the ideal gas law says there should be a temperature decrease (if you just look at pressure and temperature), also in litterature I often see the text "reduction of pressure means reduction of temperature", but I'm more thinking that the there will be an energy loss in the reduction of static pressure and this reduction must be converted to an temperature increase, am I'm thinking wrong or do you agree with my first example that there should be an increase in temperature after the obstruction (of 1000 Pa)?

 

[mrev23]

I am not sure this is true: "the Ideal gas law states that there should be a temperature drop after the obstruction."

As I recall:
[ul]
[li]Throttling (without heat transfer) is a constant-enthalpy process[/li]
[li]The enthalpy of an ideal gas is a function only of temperature.[/li]
[/ul]

So I wouldn't expect any temperature change (in an ideal gas) due to the pressure drop created by the obstruction in a perfectly insulated duct.

 

[hydtools]

In general, a reduction in energy would be accompanied by a reduction in temperature. Removing energy from a control volume should reduce its temperature as in chilling a substance.

Ted

 

[Benj107]

Removing energy from a control volume should reduce its temperature

Agreed.

But:
[ul]
[li]The obstruction is not removing energy from the control volume because it is not a turbine or other device capable of transmitting shaft work across the boundary of the control volume.[/li]
[li]Heat transfer is not removing energy from the control volume, per the OP: "The obstruction is well insulated."[/li]
[/ul]

So the temperature of an ideal gas should remain constant throughout the section containing the obstruction.

 

[Latexman]

For a gas that is truly an ideal gas, μ[sub]JT[/sub] is always equal to zero. Air is a real gas that is close to an ideal gas. At 20 C, μ[sub]JT[/sub] of air = 0.2 [sup]o[/sup]C/atm. For a 0.01 atm drop in pressure, that’s only a 0.002 [sup]o[/sup]C drop in temperature for just the JT effect. That’s insignificant.

Good Luck,
Latexman

 

[Benj107]

I don't have much to say about the "power equation" scenario other than to observe:
[ul]
[li]One one hand, "Power loss over obstruction = volumeflow*dP = 10*1000 = 10 kW" -- losing power from the airstream should cool it.[/li]
[li]On the other hand, "dT = power loss / (massflow * cp)" -- that is, the "lost" power shows up as heat added to the airstream[/li]
[/ul]

So the power doesn't actually seem to be lost.

The overall temperature change in a real gas is probably less than "0.83 degree Celsius."

If the actual dT hasn't already been provided by Latexman's application of μJT, then how much less is left as an exercise in this "actual case" -- where the input values are amazingly tidy:
[ul]
[li]Square (not rectangular or round) duct area = 1 m2[/li]
[li]Air velocity = 10 m/s[/li]
[li]Air temp = 20 degree Celsius[/li]
[li]Pressure drop from obstruction = 1000 Pa[/li]
[li]Assume density 1.2 kg/m3 -- Why is the upstream pressure not stated? Density is rarely given as a bare assumption (with no mention of pressure) in an "actual case."[/li]
[/ul]