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Vacuum Process Pipe Sizing

kimjh0 (Mechanical) 
24 Aug 06 12:40 
I would like to know how to size VACUUM process pipes.
I was told by a facility engr friend to the following;
1. determine the total flow rate (scfm) 2. determine the size of the pump accordingly 3. convert scfm to acfm 4. look at the pressure loss chart (crane chart out of Flow of Fluids > given at 100 psig) to determine psig loss per 100 ft.
He said he's done it this way for over 15 years and has worked well.
Now, my question.
The chart is for 100 psig of air. The chart reads as follows;
"to determine the pressure drop for inlet or average pressures others than 100 psi and at temperatures other than 60F, multiply the values given in the table by the ratio:" (100+14.7/ P+14.7)(460+t / 520) "Pressure drop is inversely proportional to the absolute pressure."
For 25" Hg = 12.27 psi. (2.43 psi absolute) This means I have to multiply the pressure loss number by 47.2 ? For example, @175 scfm, I would see 1.58 psig drop per 100 ft. If I multiply 1.58 by 47.2, I won't have any negative pressure left. My friend said 1.58 psig (3.2 "Hg) is the pressure loss per 100 ft. I don't need to multiply anything. The number sounds right.
However, it doesn't make sense. My friend's explanation is that the density of air doesn't really come into play when calculating the pressure loss. So, we can use the scfm number and use the chart valve (@100 psig) I disagree. Why would you correct for positive air system, but not for negative air system.
Can anyone shed some light for me? I tried to dig the web for a while, but I could not find any info. Thank you.


katmar (Chemical) 
24 Aug 06 14:40 
Your factor of 47.2 is approximately correct, but subject to some conditions. I only have a metric copy of Crane 410, so I cannot comment on your actual numbers but I presume you are using Appendix B14/15? I cannot find it stated in Crane, but I would not use their ratio method if the predicted pressure drop was more than about 20% of the upstream absolute pressure. This is why your predicted pressure drop of 74.6 psi (=1.58 x 47.2) under vacuum is impossible. You do not state the pipe size for which you calculated the 1.58 psi pressure drop, but it seems to correspond to an ID of about 1.3" for 100 psi air. It is impossible to get 175 scfm air through 100 ft of this pipe if the supply pressure is only 2.43 psia. To achieve this you would have an inlet velocity of about 1920 ft/s, and an outlet velocity much higher. This is supersonic and not possible. A more reasonable pipe would be 4" Sched 40. With 100 psig air your pressure drop would be 0.0074 psi per 100 ft. With 2.43 psia air the pressure drop would be 0.41 psi per 100 ft. (Assuming isothermal flow) This is a ratio of 55.4, which is reasonably close to your value of 47.2. With a 6" pipe the pressure drop as a fraction of the supply pressure would be less than in the 4" pipe and therefore Crane's pressure ratio method would be more applicable. In a 6" pipe the ratio (by my calcs) would be 48.6, i.e. getting closer to the "ideal" 47.2. The Crane manual has been around almost forever, and it was designed for engineers who did not have computers on their desks. There is no longer any excuse to use shortcut methods like Appendix B14/15. The correct equation to use is 16 (or 17) and this can be very easily put into a spreadsheet. The era of shortcut methods and nomograms is over. Katmar Software Engineering & Risk Analysis Software http://katmarsoftware.com 

kimjh0 (Mechanical) 
24 Aug 06 16:55 
Hi, Katmar. Yes, pipe ID 1.38" (1.25" sch 40). And yes Appendix B14/15 is what I was looking at. I have 1981, 20th edition. Amazingly, the page numbers are still the same. I think there is a mistake in your statement; The velocity I get is 280 ft/sec. Speed of sound at 68F is 1126 ft/sec. So, it is far less. And the velocity would be the same at the inlet as well as outlet since the pipe diameter is the same. Cross sectional area of 1.38" => .0104 ft^2. Q=VA, so V = Q/A => 175 cfm / .0104 ft^2 / 60 = 280 ft/sec. Why do you apply the multiplier only when the drop is less than 20% of inlet? I think you either apply all the time or not. Physics is the same. As you stated, if it comes out to be too much, you have to upsize the diameter. 175 cfm is Actual CFM, not SCFM. You sized it for SCFM, right? Under vacuum @ 25"Hg, the multiplier to convert SCFM to ACFM is 6. 30"Hg / (25"Hg + 30"Hg). It's high compared to compresssed air. @100 psig, the pressure multiplier is 7.8. 1 1/4" sch 40 may be bit small, but workable over few hundred feet. I do know for sure because I have seen an installation like that. Actually, if you look at the pump connection for 195 ACFM unit, it's only 2". http://www.powerexinc.com/dsp_specsheet.asp?product_subcategory_id=14I am just trying to make sense out of all this. Anyone else who has done this before? Thanks in advance. 

katmar (Chemical) 
25 Aug 06 2:04 
The reason I based all my calcs on 175 scfm and not on acfm was because you gave that information in your first post. Also, the 1.58 psi/100ft is reasonable if the flow is 175 scfm but it is not possible to get 175 acfm of air at 100 psig down a 1.38" ID pipe. You can disregard all my calculated numbers from the previous post, including the velocity, because of this wrong basis. GIGO. When a gas flows through a pipe the pressure decreases from the inlet to the outlet, i.e. this is the pressure drop. As the pressure on a gas decreases its density decreases and its volume increases. As you say, the pipe diameter is constant so as the volume of the flowing gas increases towards the outlet its velocity has to increase to maintain a constant mass flow. This increase in velocity down the pipe length is the basic difference between calculating a pressure drop for a gas and a liquid. With the liquid the pressure drop is all due to friction (neglecting static head differences). With a gas you have friction and acceleration effects. As a rough rule of thumb, if the pressure drop is less than 20% of the inlet pressure the acceleration effects are minor and you can use the ratio method proposed by Crane. For higher flowrates and pressure drops the acceleration effects are proportionally higher and because they are nonlinear you cannot use ratios any more. The intended meaning in my earlier post was that you should not use ratios when the pressure drop is over 20% of the inlet pressure. I did not mean that you should not apply the ratio method in those cases. I meant that the ratio method is not applicable and you should not use it. You cannot get 175 acfm through several hundred feet of 1.38" ID pipe at a supply pressure of 2.43 psia. You will probably need a 2 or 2 1/2 inch pipe. Katmar Software Engineering & Risk Analysis Software http://katmarsoftware.com 

kimjh0 (Mechanical) 
25 Aug 06 8:32 
Thanks, Katmar.
The length I am looking at is only 300 ft. Well, it's really confusing. I noted "scfm" initally because my friend told me that's the right way. acfm number is higher in vacuum. 6 times @ 25"Hg. acfm is 1/7.8 of scfm number @ 100 psig of air.
If I use the ratio method as the crane book says (with the multiplier), I get a huge pressure drop. So, it does not work.
If I use the chart as my friend described, I seemed to get a number that's close. Does this mean I should intepret acfm number in vacuum as scfm and ignore the chart numbers @100 psig? The chart numbers are then at 1 atmosphere?
That was my initial question. That's all. 

katmar (Chemical) 
25 Aug 06 9:38 
If your chart is of a similar format to the metric version I have, there should be columns down the left hand side showing both acfm and scfm. But I still maintain that you should ignore the chart. It is difficult to extrapolate from 100 psig all the way to vacuum. Use the correct formula from Crane (or any other fluids text) and put it in a spreadsheet or purchase some existing software. I would also recommend very strongly that you do all your calculations on a mass flow basis rather than on a volumetric basis. Over the years I have seen so many misunderstandings over the scfm vs acfm debate. And there is also the debate over what standard conditions are. A pound of fluid remains a pound of fluid irrespective of the pressure and temperature. Katmar Software Engineering & Risk Analysis Software http://katmarsoftware.com 

Agusta (Civil/Environmental) 
27 Nov 06 10:04 
This may be a late post but I have found a site that has a description of these calculations using different equations. Read the description of the equation, I believe it addresses your discussion. This website charges if you want to actualy use their equation, but it is simple enough to create yourself. http://www.lmnoeng.com/Flow/weymouth.htmA good source for engineering information is engineeringtoolbox.com 



