Bernouli Equation

[engpes]

I am designing a liquid outlet line on a separator operating at 350 psig. Immidiately after the separator is a level control valve with a Cv of 33.4 (see attached sketch).

I have an elevation rise of 20 ft and 600 ft of 2" piping. I am flowing light weight crude (SG=0.8).

My question is, if I know the pressure upstream of the level control valve and the Cv, can I use the Bernouli equation to calculate the velocity (in turn flow rate) of the fuid downstream of the valve?

Thanks in advance.

 

[zdas04]

Why does everyone blame Bernoulli for everything? There is no place in the equation for a Cv (or any flow resistance at all).

Look in Crane 410 for the nozzle flow equation. That SG is about 35 API which can be pretty fluffy so you are going to get some noise in the data (the nozzle equations are based on incompressible fluids), but you should be OK.


David

 

[zekeman]

Engpes,

Modified Bernoulli applies as follows which is the classical incompressible flow equation in engineering.

V1^2/2g+p1/rho +z1= V2^2/2g+p2/rho+z2+losses

V1 pipe velocity, ft/sec
V2 velocity of fluid rising in tank, usually neglected
z1 height of pipe at 1
z2 height of top tank surface

losses are :
1) (Q/Cv)^2 for control valve
Q is in gpm so you have to convert
Q=V1*60*pipe area/231
pipe area in in^2

2)fL/D*V1^2/2g for pipe length L
f friction factor obtained from Reynolds number

3)contraction loss fluid entering tank
K*v^2/2g
K approximately 1

 

[desertfox]

Hi engpes

I think you might need some information about the flow rate upstream of the valve before you can use Bernoulli.

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

desertfox

 

[engpes]

Desertfox I dont know the flow rate and am trying to determine this.

I think I agree with zekeman in using (Q/C)^2 to determine the valve losses and get flow.

 

[desertfox]

Hi engpes

I think you need to re-read my post.
To obtain the flowrate downstream of the valve, you require the flowrate UPSTREAM of the valve if yu want to use Bernoulli.

How are you going to use (Q/C)^2 unless you know Q?

Q = gallons per minute = flow rate

desertfox

 

[zekeman]

"Q = gallons per minute = flow rate"

Q=V1*60*pipe area/231
pipe area in in^2

from above post.

 

[desertfox]

Hi engpes

So all we need according to Zekeman's post above is V1 but this is part of what you're trying to find I beleive?

desertfox

 

[engpes]

I will look at it closer, but we should be able to substitute v1xA for Q and have 1 equation and 1 unknown, V1?

Being that it is incompressible I should be able to get flow from V1?

Thanks

 

[Cockroach]

I think DesertFox is correct, you need some information on the state of the upstream condition.

In doing orifice meters for example, the pressure drop gives me a measure of stream velocity. The accurate calibration of the orifice thereby leads me to mass flow. Continuity or the lack of mass lost gives me downstream velocity given the size of pipe and schedule.

But the computation of stream velocity is based on Bernoulli and incompressible flow, corrected for nozzle efficiency. I knew the upstream and downstream pressures, the orifice diameter and density of the working medium.

This is what I interpret DesertFox to be saying, which I believe is the correct approach to your computation.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[zekeman]

"I think DesertFox is correct, you need some information on the state of the upstream condition."

Look at the OP diagram. He gives the upstream pressure

p1=350psi

and

v= v1 is constant in the 2"pipe.

p2= atmospheric pressure

v2 is almost zero but by continuity you could get it.

So we are left with the single equation in v1.

Where is the problem?

This is a 2nd year engineering fluids problem for steady flow.

 

[engpes]

I agree.

 

[desertfox]

hi

Okay, so can you post a solution for the flow rate down stream with the information given in the original question.

regards

desertfox

 

[rb1957]

agreed ... what is the upstream total pressure ? (static + dynamic)

 

[zekeman]

"agreed ... what is the upstream total pressure ? (static + dynamic) "

look at my comment (6 Nov 11 11:10)and the OP sketch.

 

[ione]

Calculate a guess flow rate Q’ assuming a guess downstream pressure P2 (after the control valve) and using the following formula:

Q’= Cv*SQRT(deltaP/SG)

Check Q’ with D’Arcy Weisbach equation (include also entrance effect and differential height).

 

[rb1957]

so the total pressure in the system is 20' head ?

 

[zekeman]

No,

P1/rho +v1^2/2g at station 1 since p1 is the static pressure.

would you please spend some time to read the sketch and note that the flow is from left to right and then maybe you might see it.

 

[rb1957]

thx zeke, i did look at the picture and thought the only place in the circuit that we knew what the total pressure was was at the free surface and following that at the base of the reservoir.

IMHO, it doesn't matter that the flow is left to right, bernie equates total pressure at two places on the circuit, allowing for interchange between static and dynamic pressure (and losses).

 

[zekeman]

"IMHO, it doesn't matter that the flow is left to right, bernie equates total pressure at two places on the circuit, allowing for interchange between static and dynamic pressure (and losses)"

If you did, you would wind up solving an equation where V^2 = a negative number. Try it! The reason is nonlinearity.