conversion required for kg/hr to mmscfd 2

[opstech]

hello, i am in need of a conversion factor from kg/hr to mmscfd - have no eng. textsbooks on the plant to look up, i need this to see how much our flash gas compressors are putting through our dew point plant - none of our meters are working yet - plant has just been commisioned. Thanks for your help - lowly tech here, nt an engineer !

 

[zdas04]

Sorry, the real answer is that a not-too-bright (but very cunning) beaurecrat in the state thought that he could get more revenue for the state by slightly overstating the atmospheric pressure. The difference ends up paying the state 2% more severance tax on every chunk of gas extracted so they can have "lower" incremental tax rates than the surrounding states.

An atmospheric pressure of 15.025 (103.539 kPa) would happen at 500 ft (152 m) below sea level. Atmospheric pressure in the Death Valley isn't that high.

TDK2,
A "micro" definition of standard conditions is really easy, every gas-sales contract has one that both parties have agreed to (it may not be the same as the next contract, but who cares?). The problem is that when you try to define a global standard, people keep bringing their own personal axes to grind. It is never technical, simply political.

David

 

[25362]

To zdas04, if 1 psi ~ 2.3 ft of water (less for seawater), then an atmospheric pressure difference of 15.025-14.696 = 0.33 psia, can hardly be 1 ft of water, let alone 500 ft. Am I right ?

 

[mbeychok]

23562:

David (aka zdas04) did not mean submerged 500 feet below the surface of the sea ... he meant land that is 500 feet below sea level.

He was trying to tell you that atmospheric pressure decreases with increasing elevation ... i.e., the atmospheric pressure atop a mountain is lower than the atmospheric pressure on the beach at sea level. I am sure that you know that ... so why keep chewing on this discussion thread which I think we have already gnawed down to the bone.

Milton Beychok
(Contact me at www.air-dispersion.com)
.

 

[25362]

Zdas04, sorry for the misinterpretation. Mbeychock, you are right: more than enough is too much.

 

[ddkm]

Let me give this a shot, hope you guys don't mind me reviving this thread:


Converting from kg/hr to mmscfd

What does MMSCFD stand for? Millions of Standard Cubic Feet per Day (measure of gas produced in a reservoir)

From the Ideal Gas Law
PV = nRT = m/MRT
i.e.
m/V = PM/RT

i.e. the density of the gas
rho = PM/RT

At standard conditions (T=15.556ºC and P=1 atm from Milton's post) (Note: I would have used T=0ºC)

rho = 101325 M / [8314 x(15.556+273.15)]
= 0.04221 M
where:

M = molecular weight of the substance in kg/k-mol

Thus to convert from a value of n kg/hr to mmscfd:

Vol flowrate = Mass flowrate / Density = n kg/hr x 1hr/3600s x 1/0.04221 M = n / 151.956 M (in m3/s) = n / 151.956 M x (3.2808)^3 ft3 x 86400/day = 20078.68 n/M (in ft3/day) = 0.02007868 n/M mmscfd (divide by 1million)

Example: For 1000kg/hr O2
M = 32.0 for O2
then
Vol flowrate = 0.02007868 n/M
= 0.02007868 x 1000/32

= 0.6275 mmscfd



---engineering your life---

 

[ddkm]

Sorry, the editing has gone bonkers. Let me try again:

Converting from kg/hr to mmscfd

What does MMSCFD stand for? Millions of Standard Cubic Feet per Day (measure of gas produced in a reservoir)

From the Ideal Gas Law
PV = nRT = m/MRT
i.e.
m/V = PM/RT

i.e. the density of the gas
rho = PM/RT

At standard conditions (T=15.556ºC and P=1 atm from Milton's post) (Note: I would have used T=0ºC)

rho = 101325 M / [8314 x(15.556+273.15)]
= 0.04221 M
where:

M = molecular weight of the substance in kg/k-mol

Thus to convert from a value of n kg/hr to mmscfd:

Vol flowrate = Mass flowrate / Density
= n kg/hr x 1hr/3600s x 1/0.04221 M
= n / 151.956 M (in m3/s)
= n / 151.956 M x (3.2808)^3 ft3 x 86400/day
= 20078.68 n/M (in ft3/day)
= 0.02007868 n/M mmscfd (divide by 1million)

Example: For 1000kg/hr O2
M = 32.0 for O2
then
Vol flowrate = 0.02007868 n/M
= 0.02007868 x 1000/32

= 0.6275 mmscfd



---engineering your life---