Pressure loss in hoses and connectors for air 2

[Gymnast]

I am trying to improve an air pump to an gymnastics airtrack regarding the pressure drop in the pipes. In general the system consists of a vacuum cleaner motor, some 2" pipes, valve, T's and flexible 2" PVC hoses with Camlock connectors.

I have tried to calculate the pressure loss using these two tools:

http://www.freecalc.com/gasfram.htm

http://www.pneucon.co.kr/Techinform/vmd03.htm

However I get some problems in estimating the equivalent piping material for the hose. The hose is like this seen here:

http://www.burcharth.dk/industry/Product.aspx?ID=45334

Can I expect the manufacturer to give me information regarding surface roughness for flow calculation? What is your estimate?

Furthermore the Camlock connectors for the hose locally reduce the pipe size to about 1.6 inch in diameter and the connector itself is not ideal. Can I in some way estimate the pressure loss from such connectors? Or do someone already hold the right friction factors for such connectors?

Thank you in advance for your help.

 

[Gymnast]

Since yesterday, I got some limited answer on the hose friction issue here:

http://www.eng-tips.com/viewthread.cfm?qid=153047

In my application I get an air velocity of about 25 m/s (82 ft/sec). When I calculate the pressure loss using different calculators and different roughness I get a result within a factor 2, which is acceptable for me.

However the problem regarding the cam lock coupling or hose tail is much worse - here I have a difference of a factor 16 in results - so please help me on that...

If I try to calculate the pressure drop for a similar orifice 2" to 1.6" using this link:

http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm

Then I get a pressure drop of about 2700 Pa or as 13 meters of pipe.

If I try to calculate the problem as two changes of pipe size using these links:

http://www.freecalc.com/gasfram.htm

or

http://www.thermexcel.com/english/ressourc/pdclocal.htm

Then I get a pressure drop of about 155 Pa. as 0.8 meters of pipe.

These two results are so much different. Both ways of calculation may not be the "right way". But I should like an advise on how to move on and perhaps someone can explain the big difference in the two results.

 

[stanlsimon]

The problem with flexible tube or hose is that the corrugations tend to carry over into the inside diameter. Instead of a simple constant velocity you have a series of "semi vena cava" restrictions. Add to that the problem of bending and resultant diameter reduction.

The fundametal relationship for air flow in duct fittings is p=c(v/4005)**2. c is a factor obtained from tables of expansions or contractions. Units are fpm and inch wg.

Are you suffering from insufficient air flow?

 

[Gymnast]

Yes I do suffer from insufficient air flow on a first prototype. On the other hand I don't want to go too high in hose and fitting dimensions for the product.

Some PVC hoses are for protection of burried installations, and they have some large inside spiral slids of about 1-2 mm. I think you shall avoid those hoses. I have tried two other types, which was somewhat smoothed inside:

1) It is a 2" flexible PVC hose normaly used for water suction, and its minimal bending radius is about 5 diameters. It feels smooth inside. This is the hose we use now, and it has a pressure loss of about 2000 Pa on 3 meters and it seems about the double of the calculated loss for a pipe.

2) It is a 2" flexible PVC hose normaly used for central vacuum cleaner installation and similar. The minimal bending radius is about one diameter. On one inside side of the hose it feels smooth but on the other it has a ripple of about 0.5 mm each 9 mm. It might be due to bending a long time after production - I don't know. This tube has a pressure drop of about 5000 Pa and consequently the flow dropped about 20%. The pressure loss is about five times or more than the calculated loss for pipes. When the measurements was performed the hose was not bend significantly. This hose is however lighter and easier to handle.

I'am no professional in such measurement - I just try to explain to the best of my knowledge. Perhaps someone can confirm similar results.

Regarding the fundamental relationship for duct fittings, the formula from stanlsimon seems familiar to the formula previously referenced:

http://www.thermexcel.com/english/ressourc/pdclocal.htm

Stanlsimon, do you have a reference to this formula. What do you mean by "inch wg" - inch water gravity? Then this formula comes close to the reference above. It seems strange to me, that viscosity is no issue for these formulaes. I have no good measurements on this for my case the real pressure loss seems higher than can be calculated by these formulaes. Why is the orifice 16 times worse on pressure drop?

 

[Gymnast]

Well, I just discovered an error in the orifice calculator on the Efunda website:

http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm

When I recalculate using this reference (using Crane):

http://sti.srs.gov/fulltext/tr2001361/tr2001361.html,

then I get a drop of 226 Pa and similar to the results from the pipe size change calculations.

- So now I'am happy again...

But I should still like your comments on the high pressure drop of flexible hose.

 

[katmar]

It is difficult to pinpoint any error without seeing all your data, but it looks to me that you have made an error in calculating the 2700 Pa pressure drop in the orifice calculation. Apart from that it all looks reasonable (within your expected accuracy criteria).

I took your air flow of 25 m/s in a 50 mm ID pipe to give a flowrate of 175 m³/h. For air at 25C and atm press with a density of 1.2 kg/m³ and a viscosity of 0.023 cP I get a pressure drop of around 700 to 1000 Pa for a 3 m section of pipe (depending on the roughness and exact ID). This is close to what you have measured.

For a 1.6" orifice I get a permanent pressure drop of 450 Pa. Be careful in using the flowmeter formula for orifices because it will give you the pressure drop between the tapping points and not the permanent pressure drop. There is always some pressure recovery after the tapping point.

I would think that your approach of calculating the connectors as two changes in diameter would be better than taking it as a sharp orifice plate, but doing this I get a higher value than you did. But it depends on the actual air flow you used and you have not specified that.

If you give the actual ID of your pipe and the air flowrate you have used I can do a closer comparison with my calcs.

BTW, "inch wg" is inch water gauge. This is the height of the column of water if you measure the pressure with a manometer.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

Gymnast - I see you found the error I suspected while I was busy replying to you!

It is hard to reconcile your measured pressure drops in the pipe with the calculated values. Even by taking very high roughnesses I cnnot get the values you measured. How accurately have you been able to measure such low pressure drops (I assume using an inclined manometer?). What is the actual ID of the pipe?

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Gymnast]

I measured the pressure using a Bourdon Gauge Monometer for the range 0-16000 Pa, and it is attached to a small silicon hose, that was inserted into a small hole in the larger pipe. The system beeing measured actually also consisted of a swing Check Valve some fittings and the hose of 3 meters to the free.

I don't have the exact ID (It will be hard to get through the supply chain) on the two hoses measured, but I shall try to include two photos of the hoses to this message. The black one is the one with the unexpected large pressure loss. This large pressure drop cannot be explained by the normal pipe formula with a large roughness. So I believe when stanlsimon explained something about "a series of "semi vena cava" restrictions".

 

[katmar]

If you have got a check valve and other fittings included in the 2000 Pa the pressure drop is not out of the ordinary at all. The 5000 Pa is on the high side. Are there any other differences (ID, fittings, kinks etc) between the two pipes?

Stanlsimon could be right. If you take each turn of the ribbing as an orifice with a bore of 49 mm in a pipe with an ID of 50 mm you get a theoretical pressure drop of 3 Pa for each one. If you have 110 turns per metre of length that gives close to 1000 Pa over 3 metres. If we add in some interference effects it could make sense.



Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Gymnast]

Sorry Katmar, I believed that ID was Identification and I see now that you ment Inner Diameter.

The inner diameter of the black hose is 49 mm, and it is 51 mm on the grey one. But there is no other difference in the two set-ups.

 

[vzeos]

Gymnast,

The attached file deals with Flow in Corrugated Hose and may be of some help to you. It was excerpted from Fluid Power Design Handbook by Frank Yeaple (1995)(

http://cgi.ebay.com/Fluid-Power-Des...720655QQcmdZViewItem?IMSfp=TL0803230848a4294)

 

[Gymnast]

Vzeos, thank you for this reference.

Frank Yeaple names other references suggesting 3-4 times higher pressure drop with a corrugated hose than with an "ordinary" pipe. This rule should be OK, if you have turbulent flow in both cases. With the formulaes of Frank Yeaple I get a pressure drop for my black hose case of about 3100 Pa. I measured 5000 Pa - but i have no laboratory conditions nor good measuring practice. Anyway these equations gives much closer results to my measurements that just assuming a 0.5 mm high roughness inside of a normal pipe.

When you read the specifications in the manufacturers catalog on my two hoses, they are both having a smooth inside. When you look on a picture of them you may suggest that both of them are corrugated. And my "subjective" feeling with a finger inside the hoses would say that the grey is smooth and the black i corrugated. Before choosing a hose type, I suggest you get a sample piece and feel with your finger inside (360 degrees). I you have bending feel on the bending, which may have changed the inner surface.

 

[vzeos]

Gymnast,

From your pictures, it looks like the tube reinforcement is a helical rib. This would induce a swirl effect to the fluid flow. You might try adjusting the coefficient of S so that the calculated pressure drop matches your measured pressure drop more closely.

 

[Gymnast]

The pictured hoses has helical rib, and I don't know of non-helical types of flexible hoses for a long time. I believe the formulaes developed already incorporate the helical nature. S is the distance between each rib, and in this case it is about 8,5 mm. According to Frank Yeaple this is like an orifice each 8,5 mm with a diameter change of 3.7 mm.

Frank states, that you should not use higher changes of diameter in the calculation eventhough the internal rib hight provides a higher diameter change. On the other hand if the internal rib hight is lower that 3.7 mm Frank states nothing. In my case I assume the diameter change to be below 1 mm and only in one side, and it still has a very high pressure drop. Perhaps the shape of the internal ribs are imortant, but it is just speculation.

 

[vzeos]

Gymnast,

Both helical and non-helical flexible hoses are in use today. Your belief that the formulas developed already incorporate the helical nature of hoses may not be correct. Look at figure 9.3, left image. Note that the corrugation lobes are diametrically opposed. There is no helical lead. The picture resembles an accordion rather than a screw thread. This is the kind of corrugated hose that Frank is talking about.

You say ‘Frank states that you should not use higher changes of diameter in the calculation even though the internal rib height provides a higher diameter change’. This means that the fluid contained in the lobe beyond distance k is stagnant and does not participate in the bulk flow. This is similar to boundary layer theory but on a macro level.

What I was trying to point out to you is that the coefficient 0.438 (ie 2*0.219) is not written in stone. This coefficient was determined by a linear regression of experimental data. Look at figure 9.3, right image. This is a plot of the radial expansion, k, verses pitch for a number of tests. By the way, S is the pitch, which is defined as the distance from an arbitrary point on one lobe to the corresponding point on the adjacent lobe. The plot shows the equation k=0.219 * S, where 0.219 is the slope of the regression line. Now, assume that your hose has a pressure drop that is larger than predicted because your radial expansion does not fall on the regression line, you have the ability to modify the equation to fit a particular case (or type of hose) so long as you adhere to the equation’s limitations.

 

[Gymnast]

Hi Vzeos

Thank you for your help in this matter. And how you may change k in relation to S for a particular hose type.

I should like to pay attention on more to my first problem regading the eFunda calculation on this site:

http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm

I have had a discussion with the eFunda staff regarding this issue via e-mail. They were not convinced by my arguments and of cause it makes me uncomfortable. I have tried these two examples:

Their formulaes are included in their website. Do you think that these calculations seems right? What should convince them otherwise?

 

[vzeos]

Gymnast,

I don’t see any problem regarding the eFunda calculation. I think the problem is that you are trying to compare apples and oranges.

As katmar correctly pointed out above, the orifice meter programs do not calculate the permanent pressure loss. Let me explain. When a fluid is flowing in a pipe at a constant rate, its flow rate is the cross sectional area multiplied by the velocity. As the fluid approaches the orifice, the fluid streamlines converge toward the orifice bore. Since the area of the orifice bore is smaller than area of the pipe, the fluid velocity must increase to maintain the constant flow rate. Since the fluid velocity increased, the kinetic energy of the fluid also increased. And since energy is conserved, the pressure at the orifice plate drops considerably corresponding to the increased velocity. This pressure drop, however, is temporary. Once the fluid resumes flowing in the straight pipe it will slow down, loose its kinetic energy and recover most of the pressure lost through the orifice. Some of the pressure loss due to frictional effects will be permanent.

Measurement engineers are able to accurately correlate the pressure drop across an orifice to the fluid flow rate through the orifice. In contrast, piping engineers are mostly interested in the permanent pressure loss in the piping system.

Why don’t you post a cross-sectional diagram of your Camlock connector? Maybe we can help you choose a calculation technique.

 

[Gymnast]

vzeos, thank you for clarifying this issue regarding the orifice. Yes, I have been comparing apples and oranges. However, according to this reference:

http://sti.srs.gov/fulltext/tr2001361/tr2001361.html

For the 50mm/40mm case the permanent pressure drop should be about 38% of the near orifice drop. With this reference I get at pressure drop of 970 Pa with the same flow.

However we may leave this orifice issue as suggested by vzeos and try the real case. I have made a drawing below:

First consider an air flow of 50 l/s making a mean air velocity of 25.7 m/s at 50 mm diameter and 38.2 m/s at the 41 mm diameter. When the air is released into the airtrack, I believe that the kinetic energy is somehow lossed - am I right?

How much pipe is needed for the pressure to recover from the small diameter of 41 mm?

 

[vzeos]

Gymnast,
The document you reference attempts to calculate the permanent pressure drop due to an orifice plate. The diagram of the Camlock connector you posted does not indicate any element that resembles an orifice plate. From left to right, the connector has a sudden contraction (50mm to 41mm), then a sudden expansion (41mm to 46mm) and another sudden expansion (46mm to 50mm) except this expansion is complicated by a conical cavity which will have a noticeable effect on the pressure drop.

It is possible to calculate the resistance coefficient for sudden expansions and sudden contractions. The conical cavity is more difficult. The connector is further complicated since the flow elements are close together. The usual rule of thumb is that disturbances due to piping elements require about 10 to 20 pipe diameters of straight pipe to dissipate. The connector has close coupled elements that will make calculation unreliable.

The best way to establish the pressure drop due the connector would be by experimentation. If you establish the pressure drop in, say, 4 meters of plain pipe and then establish the pressure drop in 4 meters of pipe with the connector installed in the middle of the run, then the difference in the pressure drops is the pressure drop attributable to the connector.

The attached document may be of some help to you.

http://www.engr.iupui.edu/me/courses/hydraulicresistance.pdf

 

[Gymnast]

Vzeos

Thank you for this explanation and the link.

I guess, that when you only have 3-4 diameters of pipe after the low inner diameter of 41 mm, a large portion of the kinetic energy is lost. As you say - normally you need 10 to 20 diameters for the pressure to recover. At 49 l/s in the 41 mm diameter, no recovery equals 826 Pa. If the diameter is 50 mm this loss would be 374 Pa. And then you get the other elements with some loss.

For the next design attempt a flow of 90 l/s and estimate to be able to manage that with 2 1/2" fittings, connectors and and smooth hose and with a total pressure loss of about 3000 Pa. The inner diameter of the 2 1/2" connector is 57 mm.