Air compressor - choked flow

[49078]

Hello all,

I have been trying to find the temperature of air after exiting a nozzle from an air compressor. The air in the tank is at 125psig and 25deg C and is exiting to a slight vacuum ~ 13psia. Nozzle exit area is 1/4". Flow rate from the tank is ~0.5cfm.

B/c the back pressure is so low the flow will be choked in the nozzle (and the M# = 1)

I have searched engtips and found many references to,

http://www.air-dispersion.com/feature2.html

which is a great site and gives this formula

(T2 / T1) = (P2 / P1)^((k – 1) / k)

Using this information I get a temp ~-120deg C. I went through some old fluid books but cannot find a similar approach. I am concerned with this sol'n b/c the books approach the problem by finding a stagnation temperature, which requires a M# at the entry to the nozzle, and then finding the exit temperature.

If anyone can help me out that would be great, fluid dynamics is not my area and I'm worried I have overlooked some conditions.

The application is for a heat exchanger where we use compressed air to clean the tubes.

 

[zdas04]

For a reality check, compressing gas raises it's temp, blowing it down to vacuum lowers the temperature. To go from 125 psig to 13 psia is a lot more of a Joule-Thompson cooling than an Adiabiatic heating. I don't have the Joule-Thompson equations here, but that is where to look.

By the way the mixed units were entertaining.



David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[iainuts]

Hi 49078,
The equation you provided is for a polytropic expansion (or compression). If k = ratio of specific heats, then its also an isentropic expansion, ie: no heat transfer, completely reversible such that there is no change in entropy.

Flow through a nozzle is not isentropic, its isenthalpic (change in enthalpy is zero). Yes, it's adiabatic too (ie: no heat transfer) but its not reversible. If you put a control volume around the orifice, you can reduce to Hin = Hout because there's no work, no heat and no stored internal energy.

From that, assuming 125 psig in, 77 F (25 C) and 13 psia out, the temperature out for air is 73 F. Not much cooling affect, but that's to be expected under these conditions.

Note also that if the area of the nozzle is 0.25 in2 (diameter = 0.565"), then the flow isn't going to be 0.5 SCFM (I'm assuming you're using standard units, not something else) the flow is around 500 SCFM, depending on discharge coefficient for your case. For a well rounded nozzle, it may be as high as 600 SCFM, but regardless, it's a lot more than 0.5 CFM no matter what units you are using.

 

[sailoday28]

I believe you are using the 13 psia as a choked back pressure.

For a perfect gas, constant specific heats and Adiabatic flow.

Ao^2=A^2 +(gamma-1)/2 * U^2 (1)

Where A = sound speed
gamma = Cp/Cv
U velocity

subscript o refers to stagnation conditions.

For choked flow (Ao/A)^2 = 1 + (gamma-1)/2

or To/T = 1 + (gamma -1 )/2
With
To= 25+273.15 T at throat is easily found.

Calculation of conditions downstream of the throat requires another equation. I am pressed for time and will give more input later.
Regards

 

[49078]

zdas04, I will look into JT effect. I also appreciate the reality check and I am glad the units entertained you.

Iainuts, fluids is not my thing but I have been doing some reading on to get some sort of background on this problem. I thought that since no shock can occur (since not a CD nozzle) that this can be considered isentropic?? In regards to area and flow this has been concerning me but was not sure if I had to look at this. I was using the number from our supplier regarding the air delivery and it was quoted at 5 cfm (for 10 nozzles).

As Sailoday28 mentioned, I was using the 13psia as a chocked back pressure. What I do not understand is why the temperature in the tank is the stagnation temperature. That is the approach I initially took to find the temperature. I came to Engtips after to check and went to the webpage (shown in the original post) and found the different formula which results in a much different answer.

 

[49078]

I should really stress that I only have minimal fluid background plus I have not really used much of it since school over 3 years ago.

I do appreciate everyone's help.

 

[TD2K]

An old rule of thumb is for every 100 psi dP you will get about 7F temperature drop. I'd expect about a 10F/6C drop for your case.

 

[sailoday28]

Stagnation temperature.
Basically the tank fluid velocity is negligible compared to all other flow velocities.
A simple approach is to consider adiabatic flow
An energy balance for perf gas, const specific heats will yield
Ao^2= A^2 + (gamma-1)/2 *U^2

Where the sound speed A and U, the velociy are at any location. In the tank, U is approx =0
which yields (Ao/A)^2 = To/T.
Therefore----Local tank temp is therefore stagnation temp.

More later
Regards

 

[iainuts]

Hi 49078,
It sounds like you're using a nozzle which isn't a converging/divirging (CD) nozzle to restrict air flow into a large vessel of some kind, perhaps your heat exchanger. A CD nozzle allows energy to be converted to kinetic energy by increasing the velocity of the gas. In a CD nozzle, the flow can be isentropic, and the flow may get very cold, but note that this 'cold' gas which has been isentropically expanded is moving at supersonic velocity. If the nozzles in your apparatus result in exceedingly high velocities (many times the speed of sound) then it might be you're using a nozzle that results in an isentropic expansion, but if not, if the nozzles are more typical and result in some air blowing around inside your receiving tank, the gas energy hasn't been converted to kinetic energy so the result is an isenthalpic expansion. In this case, the enthalpy remains constant (isenthalpic flow) and temperature drop for air at the conditions you suggest is fairly slight. Blowing air directly from a low pressure (~100 psi) source across your hand will show this to be the case.

I looked over the web page you mentioned, but I'm not sure from reading it what is being isentropically expanded. For most cases of a gas escaping from a high pressure source to atmosphere (ie: through a leak) the process is isenthalpic, not isentropic. Hope that helps.

 

[49078]

My pressure ratio is (13/139) ~= 0.1 which is <0.528 so am I correct that this flow is choked?

We are using compressed air, so I use a K value of 1.4.

Using the formula given by Sailoday on his 19 Jun 06 post, T = ~-25deg C.

Using the formula given on the webpage I refer to, and using p1/p2 = 0.53 (where the flow would become choked) I get ~-25degC.

However, this is off by TD2K's rule of thumb and Iainuts 18 Jun 06 post.

Another comment about the application, the compressed air is released only in quick blasts, lasting less than a second. These are similar to blowers used to clean boilers.

I appreciate all the input

 

[iainuts]

The flow will be choked as you say, when the pressure ratio p1/p2=.53 This is true regardless of what type of nozzle you have.

Why are you concerned about the temperature? Are you worried that the air temperature may cool the boilers and create a problem such as a material/temperature issue? If so, the temperature the boiler will see is the temperature determined from an isenthalpic process. If your question is "what temperature is the gas at the throat of the nozzle right where the shock wave is", then I think you could see some significant cooling at that point due to the conversion of energy from pressure to high velocity. But that temperature suddenly rises as it exists the nozzle and velocity drops back to essentially zero. That's not a reversible process unless you have a diffuser that continues to increase the velocity in the downstream portion to supersonic conditions. So although the process of converting the 125 psi to velocity in the throat may (if the nozzle is properly designed) be isentropic, the expansion is not unless the flow maintains supersonic velocity.

 

[49078]

Iainuts,

Thanks for your quick response.

And I am concerned about problems caused by this temperature. I guess the temperature I found to be -25C(-13F) would be the temp right at the throat of the nozzle.

This is not a CD nozzle so I will look into isenthalpic processes now and not isentropic as I had been before. Is this how you got 73F in your 18 Jun 06 post? Is this the temperature once the air has expanded after exiting the nozzle or where is this point physically?

Do you know of any good online references? I will look for more information later today as well.

 

[49078]

Let me rephrase my second paragraph.

The reason I am looking at this, is a corrosion problem we have seen in a recent project at this blasting point. The -25C temp, would this be seen right at the exit plane of the nozzle prior to the velocity decreasing?

 

[sailoday28]

For adiabatic flow, perfect gas constant specific heat ratio
To stagnation temperature.
Po stagnation temp upstream of choked area.
Gamma = Cp/Cv

G1=2/(gamma-1) G2=(gamma+1)/2 G3=G1*G2

alpha th = choked area
alpha = area downstream station of known back pressure, p, where temperature, T, is to be determined.

B= [(alpha * p)/(alpha th * Po)]^2 * G2^G3 * G1

X=T/To X^2 +BX -B = 0

T/To = [ -B + sqrt( B^2+4B)]/2

For air with gamma = 1.4
B= 14.93*[(alpha * p)/(alpha th * Po)]^2

For calc purposes let [{alpha * p)/(alpha th * Po)]= 1.5
B=33.59 T/To= .972
To= 298K T=289.6K

Note for this case P/Po approx 13/140
Use correct ration of downstream area to throat area.

Regards

 

[iainuts]

"Is this how you got 73F in your 18 Jun 06 post?"
Yes, I assumed an isenthalpic process.

"Is this the temperature once the air has expanded after exiting the nozzle or where is this point physically?"
It's the temperature of the air once it has slowed back down below sonic velocity and expanded to 13 psia.

"Do you know of any good online references?"
This one in Wikipedia is valid:
Link:

http://en.wikipedia.org/wiki/Joule-Thomson_effect

"In physics, the Joule-Thomson effect, or Joule-Kelvin effect, is a process in which the temperature of a real gas is either decreased or increased by letting the gas expand freely at constant enthalpy (which means that no heat is transferred to or from the gas, and no external work is extracted)."

"As a gas expands, the average distance between molecules grows. Because of intermolecular attractive forces, expansion causes an increase in the potential energy of the gas. If no external work is extracted in the process ("free expansion") and no heat is transferred, the total energy of the gas remains the same because of the conservation of energy. The increase in potential energy thus means a decrease in kinetic energy and therefore in temperature."

"The reason I am looking at this, is a corrosion problem we have seen in a recent project at this blasting point. The -25C temp, would this be seen right at the exit plane of the nozzle prior to the velocity decreasing?"
If the nozzle you have is 100% efficient at converting the pressure to velocity, then the temperature of the gas as it goes into the shock wave will be roughly -25 C. Once it hits that point, and assuming it slows back down after the shock, the temperature will quickly rise. The velocity would have to continue to increase at supersonic velocity in order for the gas temperature to continue to decrease. You might be able to set up a test using a thermocouple placed in the flow stream. I doubt it would show a temperature as low as -25 C, and the temperature would probably be very much higher than that any significant distance downstream of the shock wave.

 

[sailoday28]

iainuts (Mechanical)
The joule-thomson coefficient is a partial derivative which describes the change in temperature with pressure at constant enthalpy.
Flow through a horizontal nozzle, with no heat transfer is not isenthalpic--- because velocity and therefore kinetic energy change.
Hin +KEin = constant

49078 (Materials)Please refer to my previous post

For air with gamma = 1.4
B= 14.93*[(alpha * p)/(alpha th * Po)]^2
This derivation is based upon isentropic conditions upto and including the throat or minimum area of the nozzle. To correct for nozzle losses, use

B= 14.93*[(alpha * p)/(Cd*alpha th * Po)]^2
where Cd is the coefficient of discharge for the nozzle.

With a converging nozzle, alpha would be the downstream pipe area. And yes, with the back pressure, 13psia, you have specified, there will be a shock.

Does the nozzle exit to a pipe or a large container, or room that is at the negative pressure?

Regards

Regards.

 

[iainuts]

"Flow through a horizontal nozzle, with no heat transfer is not isenthalpic--- because velocity and therefore kinetic energy change.
Hin +KEin = constant "
In the case of a CD nozzle, I could agree, which is why I suggested that up to the shock you might find isentropic conditions (energy is being converted to KE and you might actually get a temperature as low as -25C right at the nozzle discharge). However, from the description provided for this particular nozzle, it sounds as if no effort is being made to maintain supersonic velocity downstream of the shock. There is no diffuser and this is not a CD nozzle. Velocity must drop after the choke point. Once velocity drops completely and pressure is at 13 psia, dKE ~ 0 If this is the case, dH=0.

I think the question is, "What temperature could the material of the boiler see where it is impinged by the flow stream?" It seems there is still considerable velocity, so it may not be exactly isenthalpic, but some where in between isenthalpic and isentropic. But if there is no diffuser, the velocity must drop before it impinges on the boiler, so I can't see it being less than -25 C, in fact it is probably considerably warmer.

 

[sailoday28]

"Another comment about the application, the compressed air is released only in quick blasts, lasting less than a second. These are similar to blowers used to clean boilers."
I completely missed the above statement. AND more information is needed since this is an unsteady flow problem.

Is the length of piping upstream of the nozzle significantly long compared to the length of the nozzle?
If so- then the nozzle MIGHT be analyzed as a quasi-steady flow device and the upstream piping analyzed thru transient analysis.

Regards

Does the reservoir or upstream pressure, remain constant during the short blasts?

More later.

Regards

 

[49078]

I have not really had a chance to read through the latest posts thoroughly but I have noticed a couple questions taht I can answer right away.

There is no diffuser. We make no attempt to keep the velocity up. The nozzle exits into a tube(pipe) into the heat exchanger which is slightly below atmospheric.

I imagine the upstream pressure drops slightly after a blast but I am not aware of how much. We do have a large resevoir and I would think the pressure would be maintained close to 100psi. The length of tube leading to the nozzle is significantly long, I am not sure what the actual length on this particular project however.

There is a lot of information in these posts and I will read them asap. I appreciate the help.

 

[49078]

Alright, well i went through the very helpful posts.

I have gone through them all now, and started applying my numbers. Referring to Sailoday's post on June 20 and June 21 I have gone through and got T of ~22degC. This assumes a cd of 0.72 (saw this on another post regarding nozzles, not sure if this acceptable) and a choked area for 1/4"D and a pipe diameter of 2.5".

This seems to be reasonable as most people do not seem to expect significant cooling. This makes me wonder though if I were to have a 3" pipe if the air would actually be warmer?

I appreciate everyone's contributions.