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Back to basics flow vs density relationship in an orifice 1
- Thread starter bouk
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LittleInch
Petroleum
Pressure?
Maybe it's me but I'm not understanding the question. Can you elaborate a bit more than one line please?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Maybe it's me but I'm not understanding the question. Can you elaborate a bit more than one line please?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #4
Apologies, here's a bit more background. If I have a mass rate of gas (Mol wt. 20) and want to read off a graph with the x-axis being a rate of air (StdL/hr) what the equivalent rate of air is, how would one do this?
I used the ideal gas relationship (PV = nRT) to take the mass rate of gas (kg/hr), divide by the mol weight of air and convert this molar rate to StdL/hr.
My colleague said that flow through an orifice was proportional to the square root of the density (and thus molecular weight) and not directly proportional, and I wondered what (physically) makes this the case and why my logic isn't following?
I used the ideal gas relationship (PV = nRT) to take the mass rate of gas (kg/hr), divide by the mol weight of air and convert this molar rate to StdL/hr.
My colleague said that flow through an orifice was proportional to the square root of the density (and thus molecular weight) and not directly proportional, and I wondered what (physically) makes this the case and why my logic isn't following?
LittleInch
Petroleum
What's on the Y axis?
An orifice as I'm sure you know measures flow by looking at the delta P and other physical attributes of the orifice.
Flow is proportional to the sqr root of the delta P. So given that density is a function of pressure, then the flow could be said to be proportional to the sqr root of the DELTA of the densities.
however you are measuring the pressure so just use the delta P to establish flow rate. Not this will be actual flow at whatever pressure you're flowing at so would then need a further conversion to get to standard conditions
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
An orifice as I'm sure you know measures flow by looking at the delta P and other physical attributes of the orifice.
Flow is proportional to the sqr root of the delta P. So given that density is a function of pressure, then the flow could be said to be proportional to the sqr root of the DELTA of the densities.
however you are measuring the pressure so just use the delta P to establish flow rate. Not this will be actual flow at whatever pressure you're flowing at so would then need a further conversion to get to standard conditions
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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1
- #7
Hi,
The mass flowrate is a function of the square root of the density (upstream the orifice )* coefficients related to the pipe, the orifice *square root of the Pressure difference due to orifice.
Note : you can refer to ISO 5167.
Pierre
The mass flowrate is a function of the square root of the density (upstream the orifice )* coefficients related to the pipe, the orifice *square root of the Pressure difference due to orifice.
Note : you can refer to ISO 5167.
Pierre
- Thread starter
- #8
Thankyou all for the responses. That has given me the useful equations and background information I needed.
My curiosity now stems even further as to how those relationships/equations were derived in the first place, but I guess that's like asking a question why does gravity act the way it does?! I'll just accept that it's forces at play that cause the behaviour of the gas in the way presented in the equations.
PS - No need to reply to my rhetorical question above
My curiosity now stems even further as to how those relationships/equations were derived in the first place, but I guess that's like asking a question why does gravity act the way it does?! I'll just accept that it's forces at play that cause the behaviour of the gas in the way presented in the equations.
PS - No need to reply to my rhetorical question above
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