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air temp rise at outlet of a blower 2

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vesselguy

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Feb 25, 2002
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It's been a hell've a long time since I did any thermodynamics so I'm asking for some help here.

How do you calculate the outlet temperature from a rotary air blower? I want to confirm what I'm seeing from a test report is for real or not. In a test report, a 2 lobe blower has 30F ambient air discharging from the blower at 14 psi at 220F. That just don't make sense to me why it should be so high temperature. I've hit back on my books and know that this is a constant volume isothermal process. Using ideal gas law, P1/T1 = P2/T2 (V is cancel out due to constant volume) I only get T2 = (29 psia/14.5 psia)30F = 60F.

 
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Since absolute temperature ratios are about proportional to pressure ratios, it seems you're dealing with an air-cooled or water-cooled blower...
 
Lobe type blowers are isentropic units so the temperature rise should be around Tout=Tin(CR)^((k-1)/k). With two compression ratios I get 137F out with air (k=1.4). To get to 220F I need an inlet temp around 98F. On the other hand, if the actual suction pressure at the inlet plenum was 9.2 psia then the compression ratios go up to 3.15 and the outlet temp is 220F--any chance you are losing 5.3 psi across a suction filter or suction piping?

With your Ideal Gas Law nonsense, T2=(30+460)*2-460=520F.

If it was a dynamic blower then the calculation is polytropic and the polytropic exponent is generally a function of equipment and gas composition (experimentally derived). Even with that, you shouldn't get anywhere near 220F with two compression ratios starting at 30F.

David

 
IRStuff, You're right that all thermodynamic work are in absolute temp. However, for ease, a simple proprotionality funtion like I had it makes no difference whether I converted it to Rankins or stay in F. Also, did I do something wrong here because you put on the forum policy link?

Thanks Zdas04 for you input. I'm a stress guy and I haven't done any thermo for over 20 yrs. so pardon my rusty calc.
 

These blowers are rotary type positive displacement machines which have "slip" (backward leakage) flow, therefore the [Δ]T is higher than expected and is generally proportional to the pressure ratio.
 
vesselguy,
It makes a big difference in your 'simple' proportionality whether you work in common units or absolute units.

For example:
T2=T1*P2/P1 = 30*14.5/0 = Undefined
T2=30*29/14.5 = 60F WRONG
T2=(460+30)*29/14.5=980R = 520F

I'd say that the difference between 60F and 520F is a bit more than "no difference".

David
 
Vesselguy,

Just consider that your formulation makes the pressure go to 0 psi at 0°F, which is obviously erroneous.

TTFN

FAQ731-376
 
Vesselguy, You can't assume constant volume here. The volume does change in a compessor.

Assuming constant volume means the air needs to occupy the same volume at double the pressure. Heating the air to 520F will do just that.

The perfect gas law can't be used to solve a compressor calculation because you don't know the final temperature and volume. Tout=Tin(CR)^((k-1)/k) provides the missing information and allows to solve the case.

 
So it is an isentropic compression ...

Once the ideal isentropic compression is calculated, one can use a compression isentropic efficiency to determine the final temperature outlet value and find out the specific power input requirements for such a case ...

 
One manufacturer advices to take the actually net compressed volume as 75% of the theoretical, i/o to take into account the "slip" effect.

If this is correct then the estimation made by zdas04 of an "inlet" temperature of around 98[sup]o[/sup]F confirms a slip of about 24%.
 
Gents,
I want to thank you all for your valuable input. Truly appreciated. Due to your help, I recalculated the discharge air temperature using your tips provided here and I am comfortable that the test data is valid. I now have to get back to my ball game and which is checking the stress analysis of the piping system and equipment for the new operating temperature.

Thank you all for your patience with me on my extremely rusty thermodynamic.
 
One of the factors not mentioned above is that oil is injected with the air to lubricate and to act as a coolant; regardless of the equations mentioned above, the air temperature will be affected by the oil temperature, so whatever value is calculated in the above compression equations, one must include the cooling effect of the oil.

 

Could a leaking internal relief valve, beside the inherent slip, be the reason? That is, assuming temperature and pressure readings are OK.
 
Chicoppee,
Yes Oil-Flooded screw compressors inject oil into the process. And Ducks Fly. And Fish Swim. And any number of other unrelated factoids. None of which have any relevance to a discussion of 2-lobe blowers which are oil-free.

25362,
I've never known one of these little guys (some not so little, but the compression ratios are low) to have an internal relief valve, but maybe I've missed it. There is a bit of slip due to rotor clearance, but it doesn't seem to have much of an impact on discharge temp. I think the confusion is where they're measuring the suction/discharge pressure. Those measurement points are almost never at the inlet/outlet plenum and the losses through the piping, filters, etc. can cost 3-5 psi easily.

David
 

Zdas04,

From decades-old notes of mine on two-lobe air compressors I gathered the following characteristics, among others not relevant to this discussion:

Sizes: Units were available from a few CFM to about 50,000 CFM.

Single-stage machines: They were generally limited to 15 psig discharge pressures with a few models reaching 20 psig.

Slip: It was constant at constant pressures, and at high pressures and low speeds the slip became a considerable percentage of total capacity resulting in low volumetric efficiencies. These days manufacturers submit tables on "slip velocities" (meaning no discharge, only slip.) Volumetric efficiencies for a pressure range of up to 14 psig and ~590 rpm, were in the range 70-82%. The lower values for the higher pressures. The OP didn't tell us about the machine characteristics, therefore one could assume low rotating speeds.

Temperature rise: As a ROT, temperature rise was taken at about 13[sup]o[/sup]F per psi ([Δ]P).

Relief valves: Whether internal or external, their leaks may contribute to the heating of the sucked air.

Kindly comment for my own education.
 
25362,
Your notes seem good. I've seen people put a blower in front of a regular gas compressor to act as the first stage, but I've not seen anyone put multiple blowers in series so single stage seems to be the norm in my experience.

I looked up 2 lobe blowers and found a really good description at They talk about blowers having a sweet spot between 1.1 and 2.0 compression ratios so with atmospheric air a maximum discharge pressure under 15 psig makes sense.

Rules of Thumb tend to have pretty limited applicability and that one is particularly limited. A 5 psi dP across a suction filter can take a normal discharge pressure from 8 degrees F per psi to 13 very quickly (your ROT almost exactly matched the OP's observation by the way).

One thing I find a lot is people often don't have a clue what their local atmospheric pressure is. I live in the Rockies and atmospheric pressure can change 3 psi from one wellsite to the next. The link above points out that discharge pressure is controlled by downstream resistance to flow so a blower that could develop 29 psia at sea level at 130F discharge would see 170F if it was asked to deliver 29 psia at 6,000 ft. 40F higher temp with the "same" suction/discharge conditions could easily be troubling.

A PSV leak only matters to the discharge temp if the outlet of the PSV is piped to the blower suction pretty close to the inlet plenum. In air service, external PSV's are virtually always atmospheric discharge. I've never seen an internal PSV on a blower so I don't know what the normal practice would be.

David
 
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