Can anyone help me on determining the flow rate of argon? 1

[haydenle]

I am an intern at a steel mill. I very new to this, so bear with me. I currently have a project to size a flowmeter. I know the pressure upstream is around 200psig, this is coming from and Airliquide vessel outside the plant. Down stream we have a pressure regulator and it is regulated down to 80 psig. The headlosses for the system will take forever to determine. The temperature leaving the vessel is -20 degrees F and I am not sure what the temp is at the regulator. Please any help would be greatly appreciated.

 

[insult2injury]

oh, forgot to mention...

"I believe it is a gas supply. It is supplied from a 6050 gallon vessel, which then flow through a vaporizer."

If is flows through a vaporizor, then the tank must contain a liquid.

 

[haydenle]

insult

Yes it does contain liquid in the vessel. Are you saying that I could have a two phase mixture on my hands. If pressure and temperature are small and the height doesn't change then, along a stream line it is constant. I still dont know what the velocity upstream or downstream. All I have is:

Outside (at the vessel)
Pressure after the vaporizer: 180 psig
Flow rate at the regulator: 24000 scfm(assumed from 48000/2)
Temp leaving vaporizer: -20 degrees below ambient


Inside (near the purging system)
Pressure regulated down to: 80 psig
Temp( Assuming Ambient): 70 F

Vessel Vaporizer Inside
_____ _ _ _
[ ] || || || (+100ft of piping)
[ ] || || || (elevation change of 15ft)
[ ] ||_||_|| (numerous elbows)
[___ ]---| |---0-----/ /-------0------= exit

P=180psig P=80psig
T=70F T=70F
Q=24000scfh Q=?

 

[BigInch]

by the time it gets in the pipe, it should be all vapor.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[insult2injury]

After the vaporizer it should be all vapor. I was addressing your application of bournoulli's equation which you were applying from the tank and setting one of the velocities to 0.

Try this:

Point 1:
P_1 = 180 psig T_1 = 70 F z_1 = 15 ft

Point 2:
P_2 = 80 psig T_2 = -20 F z_2 = 0 ft

For conservation of mass, m_dot_1 = m_dot_2; therefore, rho_1*A_1*v_1 = rho_2*A_2*v_2

Assuming A_1 = A_2, then v_1 = v_2*(rho_2/rho_1)

Plug everything into bournoulli:

P_1 + rho_1*{[v_2*(rho_2/rho_1)]^2}/2 + rho_1*g*z_1 = P_2 + rho_2*(v_2^2}/2

Look up or calculate the values for the densities and solve for v_2. Just be careful with units.

 

[BigInch]

Finally, the map to the end of the rainbow.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[insult2injury]

 

[haydenle]

insult

Thank you very much. Please excuse my ignorance and lack of experience.

 

[BigInch]

Cheers. We like this. Better than working a real job.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[haydenle]

One last question, the velocity to calculate the flowrate, does this mean its acfm or scfm? I am assuming acfm.

 

[BigInch]

OK last answer.

Everything is always acfm, acfm acfm, unless you're writing up a report or selling the stuff. Then its acfm converted to scfm at the sales contract custody transfer conditions.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[haydenle]

thanks

never assume anything?

 

[insult2injury]

your bill is in the mail!

 

[haydenle]

insult

you are an insult to injury

 

[haydenle]

correction :you add insult to injury

 

[haydenle]

I am getting 1646 ft/s. Is this number to high?

 

[insult2injury]

That number seems a bit larger than expected. I would have expected something much less. Double check your math and units.

 

[kenvlach]

Too high. 1646 ft/sec = 1122 miles/hr (Mach 2). Hard to believe 80 psig producing that flow through any significant length piping. The problem isn't entirely defined, but easy to solve if 24000 scf/hr is correct. Simplified calculation using ideal gas law in form V' = (T'/T) x (P/P') x V:

Convert 24000 scf/hr (scf @ 60^oF (288.7 K, 14.73 psia) to cf/hr at at 70^oF (294.3 K) and 80 psig (94.7 psia):

= (294.3/288.7) x (14.73/94.73) x 24000
= 3804 cf/hr at 70^oF (294.3 K) and 80 psig (94.7 psia).
= 1.0567 cf/sec.

A 1" Sch 40 pipe has inside cross-section [π](1.049)²/4 = 0.864 sq. inches = 0.006002 sq. ft.
Hence, linear flow rate
= 1.0567(cf/sec)/0.006002 sq. ft
= 176 ft/sec

 

[haydenle]

Here is how I set up the problem:

P1=(180+14.7)psia
T1=(70+460)R
Z=0

P2=(80+14.7)
T2=(70+460)R
Z2=15 ft


I got the density for the different pressures. Then, like insult said, used convervation of mass to be able to solve for V2.

P1+rho1*(rho2/rho1*V2)^2/2=P2+rho2*V2^2/2+rho2*g*Z2

I also used mathcad to check my calculations. I went through the calcualation several times. I dont see anything wrong, but I could be wrong.

 

[kenvlach]

Excuse me, but it seems you are treating the 80 psig regulator as a pressure measured on a piping system.* Unless the upstream portion isn't adequately supplying the regulator to maintain the 80 psig output, it doesn't matter. Could be a 12" diameter pipe at 2000 psi upstream -- doesn't matter.
What is downstream of the 80 psig regulator??

*OK if 80 psig was measured with the regulator valve fully open, with max usage downstream.

 

[insult2injury]

WayneLee,

Is the pipe the same dimension at the 180 and 80 psi? If not, the bournoulli simplification won't be right. Also, I just realized that I omitted the head loss term in that equation so don't forget about that. Even without that term though, I don't think it should produce such a high velocity.