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# Gas flow rate Test separator

## Gas flow rate Test separator

(OP)
Hello,

On a test separator the gas flow rate is 1,187 kg/hr i m trying to convert this gas flow rate to scf/d

The gas specific gravity of the gas is 0.98

As per my calculation the gas flow rate will be around 2.2 MMSCFD as per below calculation

1,187/ 0.98 = 1,212 kg/hr

1,212 * 22,42 - 2,712 m3/hr
2,712*35 = 94,920 scf/hr
94,920 * 24 = 2.2 MMSFCD

However this gas flow rate seems to be overestimated ....

Is the calculation above correct?

### RE: Gas flow rate Test separator

What's the 22.42?

Also in you calculation you've used 2.242.

What's gas specific gravity? Specific to what base gas?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Gas flow rate Test separator

(OP)
Hi,

The gas specific gravity is the specific gravity of the process gas as compared to air.

I have used on the second equation the factor 22.42 to convert from kg/hr to m3/hr

### RE: Gas flow rate Test separator

Density air at STP is 1.293 kg/ m3.[EDIT. This is at 0C. It's 1.225 at 15C/ 60F]

So divide 1212 by 1.293 to get m3/hr.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Gas flow rate Test separator

(OP)
I have redone the calculation taking into consideration your comments and the gas rate I got now is around 0.78 MMSCFD which make more sense

Thanks a bunch !

### RE: Gas flow rate Test separator

Gas mol. wt = 0.98*29 = 28.42 kg/kg.mole
Gas weight rate/day = 1187x24 = 28488 kg/day
Gas rate, kgmoles/day = 28488/28.42 = 1002 kgmoles/day
Gas rate, sm3/day = 1002 x 23.645 = 23692 sm3/day ; [1.0kgmole of an ideal gas = 23.645 sm3 at 1atm,298degK]
Gas rate, scfd = 23692 x 35.3 = 836388scfd = 0.84mmscfd

### RE: Gas flow rate Test separator

Now you just need to determine which is your standard conditions of reference in your region, or company.

https://www.engineeringtoolbox.com/stp-standard-nt...

OPECand a majority of the natural gas industry in North America have adopted 60 °F and 14.73 psia as their standard reference conditions for expressing natural gas volumes and flow rates (rather than the 60 °F and 14.696 psia commonly used previously).
https://www.chemeurope.com/en/encyclopedia/Standar...

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

### RE: Gas flow rate Test separator

I get same as georeverghese at 0.84 mmscfd as follows:

Using ideal gas equation: PQ = mRT at 14.7 psia and 60 F, 1187 kg/hr = 2617 lb/hr

14.7(144)(Q) = 2617(1545/((.98)(28.96)))(520)

Q = 34997 scfh = 839929 scfd = 0.84 mmscfd

### RE: Gas flow rate Test separator

My air density was at 0C. At Standard conditions (15C, 101.125 kPa) it's 1.225.

So this makes it 0.84 Mmscfd

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Gas flow rate Test separator

Hi,
Define the standard condition (P,T) first then it's a piece of cake using the perfect gas law (PV=nRT).
Calculate the density of air at STP then the density of the gas at STP, knowing the specific gravity of the gas thus the volumetric flow rate at STP.

STP (101.325 Kpa, 288.15 K); Ro gas 1.202 kg/m3, Qv = 987.50 m3/h (15C, 101.325 Kpa)
Qv= 0.8369 MMSCFD
Pierre

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