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sam_nich
Apr22-06, 12:36 PM
Does this below seem correct?

Considerations:
1) Airflow/Delivery Pressure chracteristic
2) Speed increased/decreased

Firstly, having drawn a graph of mass flow of air vs. pressure, I have noticed that there is a strong negative correlation. Hence, therefore, the greater mass flow of air passing through the compressor each second, the lower discharge pressure is acheived.

Secondly, when considering the speed of the compressor, an increase in speed will increase the mass flow of air per second, which will in turn decrease the delivery pressure?

Thanks!

Astronuc
Apr22-06, 07:51 PM
Here are some compressor performance curves -

http://www.burtoncorblin.com/BCTB202.pdf

Gokul43201
Apr22-06, 11:46 PM
Does this below seem correct?

Considerations:
1) Airflow/Delivery Pressure chracteristic
2) Speed increased/decreased

Firstly, having drawn a graph of mass flow of air vs. pressure, I have noticed that there is a strong negative correlation. Hence, therefore, the greater mass flow of air passing through the compressor each second, the lower discharge pressure is acheived.

Secondly, when considering the speed of the compressor, an increase in speed will increase the mass flow of air per second, which will in turn decrease the delivery pressure?

Thanks!
Both statements are true (assuming a constant suction/inlet pressure).

engware
Apr23-06, 01:00 PM
Hi there:

Check out eFunda.com at http://www.efunda.com where you can find some useful online caculators and information on the subject matter.

Your observations are pretty much correct.

Thanks,

Gordan

sam_nich
Apr26-06, 12:30 PM
Can someone help to explain why when the mass flow is decreased a higher pressure is achieved?

Thanks.

FredGarvin
Apr26-06, 03:47 PM
First, what kind of compressor are you talking about?

Second, when you reference your mass flow vs. pressure, where is the pressure you are referring? Inlet, outlet...???

sam_nich
Apr26-06, 04:11 PM
Sorry...for the lack of detail.

I am referring to a Rotary Vane Compressor and the delivery pressure (is this the exit pressure by the way?).

Thanks.

ChrisHarvey
May6-06, 08:19 AM
Yes this is confusing me too. I would also be very grateful if someone could explain why a lower mass flow rate gives a higher pressure rise (in a centrifugal compressor).

This is the way I see things: Total pressure rise in a centrifugal compressor is achieved by adding energy to the flow, by giving the initially (nearly) axial flow a tangential component. Once the flow has reached the tip blade (at the exit), it has received as much energy as it can, and now has the maximum possible total pressure. If you decrease the flow rate, the total pressure rise is still limited by this blade speed, which hasn't changed, so surely this can't be the physical mechanism at work? I then think about how losses. Could it be that at high flow rates, the air exit velocity achieves a smaller fraction of the blade tip speed (i.e. low slip factor) than at low flow rates, hence giving a lower pressure rise? This still doesn't seem likely though because if I look at exprimental results for low rotational speeds, and varying flow rates, the same trend is observable... that lower mass flow rates give higher pressure rises.

Could someone please tell me what physical mechanism is at work here?

Thank you very much!

ChrisHarvey
May7-06, 10:00 AM
If anyone is interested here is my solution (right or wrong):

For a compressor, the volumetric flow rate is proportional to the impeller rotational speed: Q = kN. For low flow rates at a given engine speed, the gradient of the line, k is small.

For a simplified analysis of a fan, the change in total pressure (Delta P0) = rho * U^2 (where rho is air density and U is blade tip speed).

U is a linear function of N. Therefore Delta P0 = constant x N^2.

Take the initial relationship Q = kN. Therefore N = Q/k.

Substituting into Delta P0 = constant x N^2 gives:

Delta P0 = constant x (Q/k)^2.

Recall from earlier that low flow rates have a low k value. From the above equation, low k values give higher changes in total pressure, which answers mathematically (I think) my question.

I still don't have a physical explanation for this though so I can really understand it.