Volumetric flow calculation

[Sa-Ro]

Hi..

Can we calculate volumetric flow for given data:

Case 1:

Orifice diameter, mm = 5

Inlet pressure, Bar = 6

Pressure drop, Bar = 1

Medium: Compressed air

Case 2:

Orifice diameter, mm = 5

Inlet pressure, Bar = 6

Pressure drop, Bar = 0

Medium: Compressed air

Thank you.

 

[LittleInch]

Yes.

This is choked flow. The flow will be the same as the inlet poseur is the same. Look up choked flow.

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[Sa-Ro]

Can you please provide the formula?

Thank you.

 

[LittleInch]

https://www.eng-tips.com/faqs.cfm?fid=1201

Also many online calculators

When you say pressure drop, do you really mean downstream pressure on the other side of the orifice?

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[Sa-Ro]

Not down stream pressure.

Say upstream pressure is 6 bar and downstream pressure is 5 bar for case 1

 

[Sa-Ro]

For case 2 upstream pressure is 6 bar. Downstream pressure is not controlled. It is fully opened.

So I am not sure how to tell the downstream pressure. Whether same 6 bar or zero bar

 

[Sa-Ro]

We are practically measuring volumetric flow as per ISO 6358-2

Discharge method.

Upstream pressure is controlled by a regulator.

Downstream pressure is controlled by a ball valve.

Upstream Pressure gauge is mounted between regular and test specimen.

Downstream Pressure gauge is mounted between test specimen and ball valve.

We are maintaining 5 bar at this downstream pressure gauge while maintaining 6 bar at upstream pressure gauge.

 

[hydtools]

The flow is not choked until p(downstream)/p(upstream) reaches 0.528.

Ted

 

[LittleInch]

Sa-Ro,

Ok, Case 1 is for a 1 bar pressure drop from 6 barg to 5 barg. This e.g. gives you 5.46 l/sec

https://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm#calc

It is dependant on the flow discharge co efficient

This with a smaller Cd give about 10% less.

https://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html

If the d/stream pressure is less than about 2 barg, then you'll have choked flow.

Throttling with a ball valve is poor practice. They are not good control valves. You should use a globe or plug valve.


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Also: If you get a response it's polite to respond to it.

 

[hydtools]

Some reading material.

Ted

 

[Sa-Ro]

The flow is not choked until p(downstream)/p(upstream) reaches 0.528.

Our practical results:

Upstream pipe dia: 7 mm

Orifice dia (Test specimen - plain bore): 5 mm

Discharge tank capacity: 1000 liters.

Method: Charged the tank to 12.5 barg.

Since 1 bar = 1000 liters, 12.5 barg = 12, 500 liters of free air.

Set the upstream pressure to 6 barg.

Control the downstream pressure to 5 barg by ball valve.

Noted the time taken to drop pressure from 11.5 barg to 10.5 barg (1000 liters)

Volumetric flow, LPM = 1000 liters / time in minutes.

Volumetric flow rate: 1200 LPM (20 LPS) (Inlet:6 barg, outlet: 5 barg, Pressure drop: 1 barg)

Critical pressure ratio (b): 0.65

Choked flow: 1800 LPM at 3.9 barg

gives you 5.46 l/sec

practical result: 20 LPS

https://www.engineeringtoolbox.com/orifice-nozzle-...

This formula gives results of 4.5 LPS.

I will check with globe or plug valve.

One more doubt:

Downstream pressure showing as 5 barg when upstream pressure showing 6 barg.

The pressure passing thru the controlling ball valve is 5 barg or 1 barg?

Thank you

 

[LittleInch]

The but I'm not getting here is

"Set the upstream pressure to 6 barg"
How?
When the tank pressure is 11.5 barg??

Also that inlet pipe diameter makes a big difference. It's very small.[pre][/pre]

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Also: If you get a response it's polite to respond to it.

 

[hydtools]

Are you confusing absolute pressure with gage pressure? 1bar=1000liters. 12.5barg=12500 liters.

Ted

 

[Sa-Ro]

@LittleInch

Even though the tank pressure is 11.5 barg, we are controlling the input pressure to test specimen by a regulator.

Test circuit:

Air source ---> Tank ---> Pressure gauge ---> Filter ---> Regulator ---> Pressure gauge (P1) ---> Test specimen ---> pressure gauge (P2) ---> Ball valve.

Inlet diameter (before test specimen): 7 mm

test specimen diameter: 5 mm

@Hydtools

If the volume of tank is 1000 liters (diameter * length), it can contain 1000 liters of free air,
i.e., 0 bar gauge pressure = 1 bar atmospheric = 2 bar absolute pressure.

Then tank pressure is 12.5 barg, it has 12.5 * 1000 liter = 12,500 liters of free air.

Thank you

 

[hydtools]

@Sa-Ro: bara=barg+bar(atm)

Ted

 

[LittleInch]

Also I think those calculations are giving you actual volume, whereas you've essentially converted the flow to standard volumes.

Also when you put in the actual density of the air entering the orifice at 6 barg (about 8.4 kg/m3), this gives you a different figure.

If you work on mass then I'm getting values of 0.016 to 0.018 kg/sec.

When you convert that to litres per second at atmospheric pressure you get about 15 l/sec.

So all the equations need to recognise the pressure that you're working at and avoid accidentally converting to standard conditions when everything else is in actual.

Discharge co efficient has a big impact as well.


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[Sa-Ro]

bara=barg+bar(atm)

I am clear with this concept.

Can you please explain with numbers.

Say, my gauge pressure is zero barg. Standard atmospheric pressure is 1 bar.

Hence, absolute pressure = 0+1 = 1 bar. right?

(Earlier I have mentioned 2 bar as absolute)

orifice at 6 barg (about 8.4 kg/m3)

Can you explain the calculation steps?

My calculation steps as below:

q = cd A2 [ 2 (p1 - p2) / ρ (1 - (A2 / A1)2) ]1/2

[URL unfurl="true"]https://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html[/url]

Cd=0.7
D1 = 7 mm, A1 = (Pi/4)*(D1*D1) = 0.000038465 Sq.m
D2 = 5 mm, A2 = (Pi/4)*(D2*D2) = 0.000019625 Sq.m
P1 = 6 barg = 5,88,600 N/Sq.m
P2 = 5 barg = 4,90,500 N/Sq.m
Rho = 1.22 Kg/Cu.m

Q, LPS = (0.7*0.000019625)*(2(588600-490500)/1.22*(1-(0.000019625/0.000038465)^2))60.5 = 6.4

Refer the attached MS excel file for flow calculation.

Thank you.

 

[LittleInch]

I'm fairly sure that the density used is the actual density at the upstream point which at 6 barg is appro 8.4 kg/m3

Also that the flow rate is the volumetric flow at 6 barg.

You need to amend it to be the volumetric flow at standard conditions, then compare it to your drop in pressure in the tank.

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[hydtools]

Standard atmosphere pressure is 1.013 bars. Bara=barg+1.013

Ted

 

[LittleInch]

At sea level

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