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Friction of a pipe and the effects on compressed air? |
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| Jan29-13, 04:40 PM | #1 |
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Friction of a pipe and the effects on compressed air?
Let's say a receiving device that requires compressed air is 1.5 miles away from the source and demands 600 cubic feet per minute. Let's say the pipes have a diameter of four inches and already contain air at 90psi.
If friction in the pipes would be ignored, it would require the same amount of energy to supply air regardless of pipe distance correct? SO how would one understand the properties of pipe friction on the compressed air? How would I figure out the energy required to keep pushing the air to meet the demand of 600 cubic feet per minute through 1.5 miles of piping? Any insight would be appreciated, thanks in advance |
| Jan29-13, 04:53 PM | #2 |
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I think what you are asking is, what is the pressure drop in the pipe. To do this problem, you need to know the viscosity of the air at the operating temperature. Do you know the operating temperature?
The first thing to do is to assume that the air is incompressible, so that its density is constant. If the pressure drop you calculate is significant compared to your 90 psi, then you need to get more serious, and to redo the calculation taking into account the compressibility. Use the ideal gas law to calculate the density of the gas. Divide the volumetric throughput rate 600 cfm by the cross sectional area of the pipe to get the velocity of the air in the pipe. Use the velocity, density, viscosity, and pipe diameter to get the Reynolds number. Use the friction factor-Reynolds number relationship to get the friction factor at the wall. Use the calculated wall friction factor, gas density, and gas velocity to get the shear stress at the wall. Use the shear stress at the wall to calculate the pressure drop. |
| Jan29-13, 05:16 PM | #3 |
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Sorry if I'm not understanding correctly I'm pretty new to this stuff
I'm trying to determine total cost to run a compressor based of the energy required to push compressed air to the receiving device plus the cost of the compressed air being used at the receiver. The actual of the compression of the air causes the temperature to heat up a lot so air is first compressed to 60 psi, cooled, compressed to 90 psi, cooled, and sent through a dryer and out through the pipes. (I'm ignoring the dryer because the compressors are being replaced but not the dryers and to simplify the problem slightly) |
| Jan29-13, 05:44 PM | #4 |
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Friction of a pipe and the effects on compressed air? |
| Jan29-13, 07:26 PM | #5 |
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Ok I think I've calculated everything...
Would converting this pressure drop to pounds per square foot then using that in (pounds per square foot * cfpm)/33000 = theoretical horsepower allow me to quantify the energy needed to compensate for this drop? |
| Jan29-13, 07:38 PM | #6 |
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Disregard that last post...
I could convert the pressure drop to volume drop, and using the efficiency of the compressor find out how much energy it takes to generate that much volume correct? |
| Jan29-13, 10:03 PM | #7 |
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What pressure drop did you calculate?
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| Jan29-13, 11:49 PM | #8 |
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I found Reynolds number and relative roughness then used the moody diagram to find the friction factor, but not sure how to get to pressure drop from there. Why is shear stress needed?
(Have velocity, distance, diameter, viscosity, flow rate also) |
| Jan31-13, 09:13 AM | #9 |
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Recognitions:
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The shear stress is not required. You already have a friction factor from the Moody Diagram.
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| Jan31-13, 09:53 AM | #10 |
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Ok found the formula Δp = (v² × f × L × ρ)/2D
With the pressure drop how would I convert to volume drop or energy required to accomidate for that pressure drop? (I tried ideal gas law but that definitely doesn't work) |
| Jan31-13, 12:15 PM | #11 |
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I think you're missing a factor of 4 in the equation. What number of psi do you calculate for the pressure drop?
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| Jan31-13, 12:31 PM | #12 |
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Mentor
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Sorry I missed this thread before, but the discussion is waaaaay too complicated. Engineers don't figure this stuff out from scratch every time they design a system, they use tables and/or specialized calculators or slide rules. Just Google "compressed air pipe friction loss" and pick one.
For the energy, select an air compressor and see how much power it uses at the different pressures. |
| Jan31-13, 02:00 PM | #13 |
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It's good to understand what's going on and what factors come into play in these things, but I'm with Russ on this one. Pressure losses in pipelines is a pretty well established science.
Here's a calculator on Eng. Toolbox |
| Jan31-13, 03:14 PM | #14 |
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Chet |
| Jan31-13, 03:31 PM | #15 |
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Mentor
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Dia was four inches.
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| Jan31-13, 04:40 PM | #16 |
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Yea, Engineering Toolbox is great. And yep, OP said ID was 4 inches, it results in an approximate pressure drop in the viscinity of 5.4 psi. This of course doesn't include the losses from any valves, elbows, tees, flanges, etc.
Compressibility isn't really a huge issue here anyway. The difference in the density of air (at otherwise standard conditions) between 95 psi and 90 psi is only something like .026 lb per cubic foot. It's not going to alter the pressure drop too much, and since these calculations are only good as close approximations anyway, it most likely falls within the margin of error. Edit: The one issue I see is that the the airflow the calculator (indeed, most calculations) uses is the airflow at atmospheric conditions. So it really depends on whether the OP meant 600 ACFM or SCFM. If the OP is expecting 600 ACFM (at the 90 psi condition), then that means that the actual, standard conditions, flow is in the realm of 3600 SCFM, making the pressure drop more significant, probably prohibitively so in a 4" ID pipe. |
| Jan31-13, 10:04 PM | #17 |
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