10 replies to this topic

[juuichi]

Hi everyone

I have been working on an Excel spreadsheet on the piping design for compressed air. I am required to calculate the pressure at every point in the different segments of the system, such as the pipes, fittings etc. On the portion of the piping along the riser, I calculated that there is a rise in pressure as the air flows down the riser, due to the decrease in elevation (which makes sense as the higher the elevation, the lower the pressure). However I also realised that the air would be flowing from a lower pressure point (at the higher point in the pipe) to a higher pressure point (at the lower point in the pipe), and this had me becoming very frustrated for some time already as I cannot figure out that a flow can occur from low pressure to high pressure, even as I read articles and forums online.

I would like anyone who is sure to enlighten me. It may be that I have got some of the concepts wrong, but I am awaiting your corrections. Please help and thanks in advance.

[sheiko]

Hi,
Usually elevation pressure drop (or rise) is negligible for gases (because of very low density).
But to be sure you should rigourously apply the generalized Bernoulli equation (including frictionnal pressure drop in the riser) between top and bottom.
You could also consider sharing your Excel spreadsheet to receive more specific advice.

Edited by sheiko, 04 January 2013 - 02:19 PM.

[juuichi]

Hi,

The density for the compressed air is about 10kg/m3 (at 8barg and 30degC), and the drop in height along the riser piping is about 6m. Calculating the pressure gain from the drop in elevation using density*height*gravitational acceleration = 10*6*9.81 = 589Pa. Which I agree is small relative to the compressed air pressure.

Then I calculated the pressure drop due to friction from the flow, for a flow speed of about 9.7m/s , the fanning friction factor is about the magnitude of 0.004 (which is quite typical), for pipe diameter 108mm, using deltaP=2*f*density*pipe length*speed^2/diameter = 2*0.004*10*6*(9.7^2)/0.108 = 418Pa. This gives me the pressure drop from frictional force.

Hence the net pressure drop = (-589P)Pa + 418Pa = -171Pa. The negative pressure drop means that pressure increases as the air flow from the top of the riser pipe to the bottom of the riser pipe. However like what I stated earlier, what puzzled me was that the flow of the air was from lower pressure to higher pressure, which is very counter-intuitive.

I would like to know if it is absolutely possible for the flow to occur from low pressure to high pressure? The pressure of the gas that I am calcuating here refers to the total pressure of the gas contributed by the static pressure, kinetic pressure and the gas head pressure right?? Please help as I am really puzzled, and I would definitely appreciate it if the experts can help explain.

[demank]

Dear Juuichi (eleven?)
when the air is in not flowing condition, you are right, that at the lower point have pressure high a little bit (171 Pa = 0.00171 bar) than the high point.
but in case of the air is in flowing condition, it is impossible lower point have higher pressure which refer to the air flow is from higher to lower point.

Then I want to know, in case of the air in flowing condition, how much pressure at the end of user (instrument device? maybe 5 - 7 bar, or to atmospheric pressure?)

Edited by demank, 05 January 2013 - 06:13 AM.

[sheiko]

For flow of a compressible fluid (gas or vapour) in a pipe the Darcy-Weisbach equation (using Darcy/Moody or Fanning friction factors) is not applicable because both density and velocity change as the fluid flows down the pipe.

[Bobby Strain]

And the rain falls down from the sky, too.

Bobby

[Steve Hall]

Juuichi,

I think you are missing a key concept.

First of all, your pressure drop due to friction calculation is OK. You can use the incompressible flow correlation if the total pressure drop is less than 10% of the pressure. In your case, the pressure drop is approximately 0.5 kPa with a pressure of 900 kPa (absolute). The density change is too small to make a significant difference.

The static head is 0.6 kPa as you stated. Therefore, the pressure at the bottom of the drop will be 0.1 kPa higher than at the top if measured with pressure gages at each point. This is only true during flowing conditions of course, since that's the only time you get the frictional pressure drop.

What you're missing is that you can only achieve the flow when there is an overall driving force. Where is this pipe leading to and what is the pressure at its terminus? The bottom of the 6 m drop is clearly not the terminal point since there is flow through it. Continue your thought experiment and assume the pipe turns upward and proceeds 6 m. The pressure at the top of this leg will be reduced another 0.5 kPa due to friction, and also be reduced 0.6 kPa due to the static head. Thus, the static head portion of your question cancels out.

Let me change the problem description to liquid water. It's easier for me to envision. Think of an open reservoir (tank) containing water at a constant height. Say the level in the tank is 1 m. Now, connect a 100 DN pipe to the bottom and extend it 5 m down below the tank where it discharges to atmosphere. In this case, the pressure that is equivalent to your 8 Bar of air is 6 m of water, or about 60 kPa. As water flows through the pipe it experiences a pressure drop due to friction. If you neglect the inlet and outlet transitions to the 5 m pipe, the flow rate equilibrates to the flow that results in a pressure drop of 60 kPa, or about 100 liters/s.

Hope this clears up your confusion.

Steve

[henryleung]

As you stated

"Then I calculated the pressure drop due to friction from the flow, for a flow speed of about 9.7m/s"
The fact that you have a flow speed of 9.7m/s tells there is an external driving force other than the static head.
If the flow is due to the static head only, you can calculate what the flow rate should be and I am sure it is not 9.7m/s but something lower. Which means you won't have the friction loss as you calculated.

The problem statement stated the flow to be 9.7m/s, so your friction loss calculation shows a bigger number than the static head. But that's because of what is not stated in the question: There is an external driving force giving the 9.7m/s flow

Edited by henryleung, 05 January 2013 - 11:46 PM.

[Ajay S. Satpute]

Dear juuichi,

Try using PIPENET or HYSYS for solving this problem, not because of it is complicated, but because it would help you appreciate the importance of back pressure in your problem.

Regards.

Ajay

[juuichi]

Thank you Steve Hall. I think you have unlocked this puzzle to me. Your answer cleared up my doubts. Thank you.

[narendrasony]

There is nothing wrong. Flow can take place from lower to higher pressure.
BUT flow can not take place from lower to higher Energy level or the Total head (which is the sum of dynamic, elevation and pressure head in Bernoulli's equation). Elevation for higher point is more than the lower point.


Regards
Narendra Kumar