Volumetric Flow Rate from Pressurized Header 1

[Diggergrad]

I have a 3" nitrogen header maintained at 150 psig, and I am trying to figure out what the volumetric flow rate would be through a 1" I.D. 100 ft branch exiting to atmosphere (12 psia). There will be a ball valve isolating it from the atmosphere. So in other words, I am wondering if there is a way to calculate flow rate based on the pressure difference of 150 psig though a 1" line that runs 100 ft when I open the valve completely?

 

[zdas04]

Yes there is.

Ok, that was too easy. It is a multi-step process that is significantly more involved than your problem statement would indicate. First you need to determine the sonic velocity out the ball valve. Then assume (incorrectly) that the 150 psig exists at the ball valve and determine a mass flow rate out the ball valve. Then calculate the pressure drop at that flow rate down the 100 ft of 1-inch. It is a big number, subtract it from the header pressure and recalculate the mass flow rate out the ball valve (sonic velocity stays the same, but mass flow rate is a function of upstream pressure). Now calculate the pressure drop down the 1-inch at the new mass flow rate. That gives you a new pressure at the ball valve so you go through it again. Eventually you get to the point that change from one step to the next is acceptably small and that is your mass flow rate. Divide it by density at standard conditions to get a volume flow rate at standard conditions or actual density to get flow rate at actual conditions.

One thing to keep in mind as you're doing these iterations is that you need to make sure that at each step the relationship between upstream pressure and downstream pressure allows choked flow (look in the FAQ's for a good write up on what those conditions are). Usually, if you are going to fall out of choked flow it is on step 2 and you might be able to pretend that you didn't as long as the choked-flow assumption works on step 3.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

"It is always a poor idea to ask your Bridge Club for medical advice or a collection of geek engineers for legal advice"

 

[Diggergrad]

Thanks for the response. I am a young engineer and this is a good learning process for me.

Would you be able to extrapolate on this subject a bit more for me. I understand that sonic velocity is when the fluid velocity is equal to Mach 1 or the speed of sound. Would the speed of sound be considered through the medium of air or would it be through nitrogen? Also, would the sonic velocity be an absolute value or would it be a different value through different valves?

Could you also give the formula or formulas to determine mass flow rate from the upstream pressure?

Thanks again.

 

[zdas04]

Take a look at faq378-1201. If you still have questions ask them.

David

 

[Diggergrad]

Thanks for the link.. I checked that out earlier today and am still a bit fuzzy on a few of the concepts. To confirm that the flow is going to be choked, I would use the upstream pressure as 162 psia (150 psig) and use the downstream pressure as 12 psia (atm). If i did that, I would have a ratio of 13.5 indicating that it is well above the choked criteria. Then I would go on to calculate the mass flow rate using the given equation.

Now when I look at the equation, is the discharge coefficient always going to be 0.72 or is it case specific? Also what would I use for the orifice hole area.. would it be the opening area of the ball valve or just the internal pipe area for the discharge?

To further confuse myself, if I were able to use all of the given information to determine a mass flow rate, would I need to continue through the iterative process as you described in your previous post?

Thanks in advance for your help!

 

[Latexman]

Your little problem is not for the faint of heart. Definitely a two martini problem if calculated by hand.

One problem you got is it doesn't fall into one well defined model. Is it an orifice or is it a pipe? Well, it's both. If it was a pure orifice, no pipe, it'd choke at dP/P1 ~ 0.5. If it was pure pipe, no reduced port ball valve (orifice), it'd choke at dP/P1 ~ 0.85.

So, which is it closest to? Look at the resistance of the pipe alone and the ball valve alone. Which one offers the most resistance? That's the one that will rule. And the larger the difference, the closer to that model your real world problem will approach. I'd put my money on it being more pipe-ular than orifice-like.

I'd calculate the "equivalent length" of the valve and the brach tee and add them to the length of pipe and solve it like a pipe flow problem.

Let us know how it turns out.

Good luck,
Latexman

 

[Diggergrad]

Thanks Latexman,

Thanks for your input to this problem! After doing some further searching on this type of problem around google, I came across some college slides that had similar information. As you indicated, my confusion was knowing whether it was an orifice or a pipe and how they each impacted the system. At the time I was not sure if I could look at one or the other or if I had to include both. Like you recommended, I used equivalent lengths for the entrance, exit, and valve. The assumptions I made were as follows:
-Isothermal Flow
-Fully developed turbulent flow so I could get a friction factor without a Reynolds number on the Moody diagram because I didn't have a velocity.

The steps I took were as follows:
1) Determine friction factor
2) Determine sum of K values using equivalent lenghts
3) Determine sonic pressure drop by use of (P1-P2)/(P1), the sum of the K values, and the use a provided graph that would give you the ~0.5 to ~0.85 that you indicated to confirm that the flow was indeed choked.
4) I then used the following equation:

M flow = A * [(p1*P1*144in2/ft2*gc)/(Ksum)]^1/2

where:

M flow = mass flow (lbm/s)
A = pipe area (ft^2)
p1 = density at actual conditions (lbm/ft^3)
P1 = Inlet Pressure (lbf/ft^2)
gc = 32.17 ft-lb/lbf-s^2
Ksum = Sum of pipe length and fittings equivalent lengths multiplied by f and divided by D.

5) I then took my answer in lbm/sec multiplied it by 3600 sec/hour and divided it by the density at actual conditions to get my answer in ACFM.

I ended up getting 3532 ACFH which seems reasonable for that pressure.

However, in my calculations I had to guess at what I thought the equivalent lengths for the valve would be and also what the absolute roughness would be for braided steel hose which I took to be 1.5x 10-4 ft, so the calculations may be off a little but I figure I am within 20%.

Thanks again for your input!

 

[Latexman]

Is this a smooth wall hose or corrugated hose (e/d ~ 0.187 for 1")? It makes a big difference.

Is the ACFH for inlet conditions (150 psig) or exit conditions? Or, what was your lbm/sec?

Good luck,
Latexman

 

[Diggergrad]

Latexman,

It appears as though the hose I have is going to be smooth. The ACFH was for the inlet condition of 150 psig, and I came up with a value of 0.7987 lbm/sec.

Now, if I were to take a 1" branch 100 ft off of the 3" header, and then split that 1" branch into two 3/4" branches each spanning 100 ft, would I use the same methodology as the former problem? With this new problem, I am wondering if I should approach the system using the analogy of an electric circuit in parallel to determine an overall k value; in other words, use the equation 1/Ktot = 1/K1 + 1/K2 for the 3/4" branches. I would then solve for Ktot, and use it in the mass flow equation that I used before. Is this the correct way to approach the problem? The other portion of this scenario that confuses me, is that before I had a single pipe area that I plugged into the equation, whereas, now I have the area of the 1" and the area of the two 3/4" pipes. I'm not sure how to use those appropriately now.

If you could give an insight into this problem, it would be appreciated.

Thanks,
Diggergrad

 

[Latexman]

0.8 lb/sec looks good.

It's solved just like an electric circuit where you convert two resistors in parallel to one in series. Ktot applies to one 3/4" hose (the resistor) with the total flow (the current) at the pressure drop (voltage drop).

Good luck,
Latexman