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Flow rate calculation using pressure difference and time

by scurveydog
Tags: calculation, difference, flow, pressure, rate, time
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scurveydog
#1
Aug4-10, 03:19 PM
P: 2
I am a second year engineering student out on my first co-op work term. My supervisor's first task for me was to calculate the flow rate of one of his parts that he will use on a system that he is designing. I hooked up the part to a compressed nitrogen tank initially at 6,000psi. I cracked open a valve for 20 seconds and allowed the nitrogen to bleed out of the tank without any restriction except for what was flowing through the part. After 20 seconds, I closed the valve and read the pressure to be 2600psi. I tried to calculate the flow rate using PV=znrt but I got an extremely high flow rate of 28,000L/min. I don't think that this flow rate is correct. Am I approaching the problem the right way with using the ideal gas law?? any suggestions??? Thanks in advance.

This is what I know (my supervisor says that this is all the info that I need to calculate the flow rate):
Initial pressure: 6000psi
Final pressure: 2600psi
Time for pressure drop: 20 seconds
Nitrogen tank volume: 42liters
Test was done at 25 degrees C
inside diameter of part: 3/8"
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abc007
#2
Aug4-10, 05:34 PM
P: 30
What do you mean by ''inside diameter of part: 3/8" '
diameter of pipe or orifice or what??
scurveydog
#3
Aug5-10, 09:27 AM
P: 2
I apologize for the confusion. The part is a cylinder valve with a pressure relief device attached. The inlet and outlet of the part has a 3/8 inch inside diameter (The nitrogen would flow through this part as if it were a tube, so you could assume that it is a 3/8 inch pipe or tube). I hope this helps. Please let me know if you need clarification on anything else.

Q_Goest
#4
Aug5-10, 11:34 AM
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Flow rate calculation using pressure difference and time

This isn't a simple question. As nitrogen escapes, the gas remaining in the vessel cools. The first law reduces to dU = Qin-Hout. Here, the drop in internal energy is calculated by removing the enthalpy of the gas leaving (which cools the remaining gas) then adding the heat transfer from the vessel walls back to the gas remaining in the vessel. This has to be integrated throughout the timeline because the temperature drops significantly.

The way I do this is to do an interative calculation using a spreadsheet that links to a fluids properties database. NIST has a fluids properties database you can use called REFPROP but I use a proprietary one.

Attached is the output trace of pressure and temperature assuming the iniitial conditions you specified and final pressure. I tossed a number into the heat transfer just to take into account some rough estimate of heat transfer, but it isn't very significant in the end because the time is relatively short (20 seconds).

Regarding the 3/8" orifice, that isn't a very good estimate for a cylinder valve. The opening at one or the other end may be that large, but typical cylinder valves have relatively small openings at the seat. Also, the program I use calculates flow out using a valve flow coefficient, Cv. Note that flow is choked throughout the test you performed. That Cv came out to 0.18, which is fairly typical for a cylinder valve. Note also you can obtain the Cv from the manufacturer and it should match this value fairly closely.
Topher925
#5
Aug5-10, 11:34 AM
Topher925's Avatar
P: 1,672
Your supervisor is correct. Try starting with the integral form of the conservation of mass equation and assuming no heat transfer. Then later plug in the ideal gas equation.

The assumption of no heat transfer is arguable. But since the process is only lasting over a time period of 20 seconds it can probably be justified.
BishopUser
#6
Aug5-10, 11:41 AM
P: 165
I can take a stab, but I can't garantee that this is right. You be the judge.

Density of nitrogen at 6000 psi = .3759 kg/L
Density of nitrogen at 2600 psi = .1986 kg/L

Initial mass of nitrogen = .3759kg/L * 42L = 15.79 kg
Final mass of nitrogen = .1986kg/L * 42L = 8.3412 kg

Total mass that was lost = 15.79kg - 8.3412 kg = 7.4488 kg

AVERAGE mass flow rate = 7.4488kg / 20 seconds = .372 kg/s

Average pressure of nitrogen = (6000 psi + 2600 psi) / 2 = 4300 psi
Density nitrogen at 4300 psi = .2998 kg/L

AVERAGE volume flow rate = (.372 kg/s) / (0.2998 kg/L) = 1.24 L/s

I think the flow rate would be highly a function of time. You would expect the flow to be much quicker at a higher pressure right? Maybe the average is only necessary here.

I don't think you need the diameter of the part if you are only interested in volume flow rate. The diameter will be important if you wish to find the velocity of the nitrogen though.

Edit: the above posters seem more knowledgable. I wonder how good of an estimation this calculation gives though.
abc007
#7
Aug5-10, 02:43 PM
P: 30
- Rate of mass = Rate of accumulation of mass

- Q[tex]\rho[/tex] = d/dt (V[tex]\rho[/tex]) ** Q is volumetric flow rate & V is vessil vloume
= V d[tex]\rho[/tex]/dt + [tex]\rho[/tex] dV/dt ** [tex]\rho[/tex] is density
= V [d[tex]\rho[/tex]]/[dt]

** since Volume of vessel is constant with time ==> dV/dt = 0

** since [tex]\rho[/tex] = M p/RT ** M is molecular weight
so ,

- Q [M p]/[R T] = V [M]/[RT]

dp/dt = -Q p/V
rearange equation then integrate

∫ dp/p = - Q/V ∫ dt

==> Ln (p/p0) = -Qt/V
rearrange
V Ln (p0/p) / t = Q

know substitute numbers

42 Ln(6000/2600) / 0.333 min = 105.47 L/min
volumetric flow rate Q = 105.47 / 1000 = 0.10547 m3/min

i hope that if you get the ansewer , you will share it with use.
Calculost
#8
Dec27-10, 10:34 AM
P: 5
Quote Quote by abc007 View Post
- Rate of mass = Rate of accumulation of mass

- Q[tex]\rho[/tex] = d/dt (V[tex]\rho[/tex]) ** Q is volumetric flow rate & V is vessil vloume
= V d[tex]\rho[/tex]/dt + [tex]\rho[/tex] dV/dt ** [tex]\rho[/tex] is density
= V [d[tex]\rho[/tex]]/[dt]

** since Volume of vessel is constant with time ==> dV/dt = 0

** since [tex]\rho[/tex] = M p/RT ** M is molecular weight
so ,

- Q [M p]/[R T] = V [M]/[RT]

dp/dt = -Q p/V
rearange equation then integrate

∫ dp/p = - Q/V ∫ dt

==> Ln (p/p0) = -Qt/V
rearrange
V Ln (p0/p) / t = Q

know substitute numbers

42 Ln(6000/2600) / 0.333 min = 105.47 L/min
volumetric flow rate Q = 105.47 / 1000 = 0.10547 m3/min

i hope that if you get the ansewer , you will share it with use.

Thanks ABC007,

I used the same equations, but to plot the flow rate of a tank pressurizing over time (ln(P/Po) instead). The difference in my problem was that the mass flow rate into the tank was known. I simply added this initial flow rate to the rate from your equations. Otherwise the flow approches zero as the pressure approaches the target. At the instant the tank pressurizes to the target a back pressure regulator opens maintaining constant pressure. I'd like to say I'm 100% sure of this approach but I'm not. What do you think?

Ken
abc007
#9
Dec27-10, 02:00 PM
P: 30
actually i already forget how i solved it but anyway
downlod the following file and see the Example 2.1. i hope it will help
http://faculty.kfupm.edu.sa/CHE/zaid...dule2_lec4.doc
coucher630
#10
Dec27-10, 02:36 PM
P: 4
http://www.air-dispersion.com/feature2.html

At this link, you should use the Rasouli-Williams model. The equation they give is set up for iterating to give you a time-dependent response. (Pay careful attentions to units here...if you run into trouble you can PM me, and I'll forward you a derivation for the equation that should clarify any discrepancies based on units).

Basically, the thing to realize here is that at your pressure levels, you have have choked flow. Choked flow occurs when the ratio of absolute upstream pressure to absolute downstream pressure exceeds ((k+1)/2)(k/(k-1)). At this point, your exit velocity is sonic and will remain constant until the pressure ratio drops below that critical pressure ratio. However, conservation of mass says the flow rate does not stop increasing just because the velocity is no longer increasing. That's where compressibility effects come into play. So essentially, your dealing with changes in the nitrogen's density at the outlet of the tank, and ideal gas law won't take that into account by itself.

Look into choked flow more for your own understanding, but the link above should be everything you need to solve this problem....well that and MATLAB.

Good luck!
Calculost
#11
Dec27-10, 03:05 PM
P: 5
Quote Quote by abc007 View Post
actually i already forget how i solved it but anyway
downlod the following file and see the Example 2.1. i hope it will help
http://faculty.kfupm.edu.sa/CHE/zaid...dule2_lec4.doc
Thanks ABC007

-ken


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