Flowmeter equation

[zesaint]

Hello engineers!

I ve got an easy question for u, but this one is drivin me nuts...
So, i ve just bought a new flowmeter at Dwyer and the propose an equation to correct nonstandard operating conditions:
Q2=Q1x(sqrt(P1*T2/P2*T1)) (1)
EX:
a float sitting at 60 grad on the flowmeter scale have a exit pressure of 5psig, and exit temperature ate 85F, so the actual flowrate will be :68.8SCFH, with a flowrate calibrated with P1=14.7 psi and T1=530R...

BUT WHY THE DONT USE THE PERFECT GAS LAW: PV=NRT, i find the same formula, without the square root... Where is this square root in (1) coming from??

Thank you so much for an answer

 

[mbeychok]

zesaint:

I am not sure that I understand what you are asking. If you are asking about how to convert an ideal gas flow rate from one set of temperature and pressure to another set of temperature and pressure , then using the square root in your formula is incorrect. Wherever you found that formula, it was in error.

I would also point out that you are using 530 degrees R (70 degrees Fahrenheit) and 1 atmosphere as your standard cubic foot. In the USA, 1 atmosphere and 60 degrees Fahrenheit are used much more often to define the standard cubic foot than are 1 atmosphere and 70 degrees Fahrenheit.





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

 

[abeltio]

see Flow of Fluids, technical paper 410, published by Crane.

for compressible flow through and orifice the square root is there... the equations are all derived from:
Q = CA[√](2g[Δ]p)

for compressible flow the flow is proportional to the square root of P/T, so to convert from flow 1 to flow 2 the equation given by the supplier seems correct.
cheers

saludos.
a.

 

[25362]

[•] Standard conditions for fan and the gas metering industries are dry air, 70[sup]o[/sup]F and 29.92 in. Hg.

[•] Abeltio is right. Rotameters, as well as orifices, work on pressure drops.
It is the pressure drop (substantially constant) through the variable annular constriction around the float that keeps it floating, and one gets a reading on the calibrated tube scale.

 

[zesaint]

abeltio:

are u sure we can consider here that air at my condition , around 68.8SCFH, is compressible?
thank you everybody for the replys, i am going to check rigt away in crane!

 

[25362]

Zesaint, a square root relation between flowrate and pressure drop applies for flow measurements of incompressible as well as compressible fluids (remember Bernoulli's formula ?).

There are multiplying correction factors to account for Re numbers and geometrical dimensions.
Compressible fluids include an additional correction factor

[ε] = f (p[sub]2[/sub]/p[sub]1[/sub], Cp/Cv, m)​

where:

p[sub]2[/sub]/p[sub]1[/sub] is pressure ratio
Cp/Cv is the specific heat ratio
m is the squared ratio of the orifice to the pipe diameter; for orifices, m <0.55

The elevation of the float in rotameters depends on the flow velocity, the density and the viscosity of the fluid.

Thus, they require individual calibration. Standard rotameters -for liquids and gases- are given for connection to pipes with diameters from 3 to 150 mm. Air flows at atmospheric pressures and room temperature range from 0.005 to 17,000 m[sup]3[/sup]/h.

 

[mbeychok]

It seems to me that a rotameter or other measuring device must have been designed to convert the pressure drop of the air to cubic feet per hour at some temperature and pressure. Thus, the reading of 60 cubic feet per hour that zesaint got from his device is valid only for that design temperature and pressure. To convert that 60 cubic feet per hour to another temperature and pressure, he does not need the square root.

My interpretation of zesaint's original posting was that he was not asking us how to reset the design factor for converting the rotameter's pressure drop to cubic feet an hour ... rather he was asking us simply how to convert a gas flow rate from the design set of conditions to another set of conditions (for which the square root is not required).

Perhaps, zesaint should clarify just exactly what he was asking.

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

 

[zesaint]

mbeychok :

zesaint will clarify his question :

i've bought a new rotamater. I know that a rotamer is good for one precise condition: the one he has been calibrated for.
I not using mine for the standard conditions proposed by the manufacturer, so i've just aplied the perfect law gas to convert my copnditions regarding the conditions of the "standardistion". The last post by mbeychok represents very well my question.
Unfortunatly i dont have the crane handbook, and user :25362 i didnt know that we have to use a factor for Re :
m is the squared ratio of the orifice to the pipe diameter; for orifices, m <0.55, do u have the reference for that?
thank you again everyone

 

[abeltio]

zesaint:
the original post was asking about the validity of the equation supplied by the manufacturer.
i think that has been established - the equation seems correct.

if you are trying to apply the equation for perfect gases as an academic exercise... sure, no problem.

if the question is to confirm the flow and you still have doubts about the formula... perhaps your best bet is to consult with the manufacturer.

cheers


saludos.
a.

 

[25362]

The correction factors I mentioned are used on orifices, venturi meters, nozzle meters, pitot tubes, ie, instruments that measure [&Delta;]P[sub]f[/sub].

Rotameters measure cross flow areas with [&Delta;]P[sub]f[/sub] kept constant.

I recommend Perry VI chapter 5-20 as a good source.

As for the question in hand I think there is a difference between correcting a false meter signal and converting a false reading to standard conditions.

 

[mbeychok]

25362:

As for the question in hand I think there is a difference between correcting a false meter signal and converting a false reading to standard conditions.

The above quote from your last response is exactly the point I have been trying to make ... only you said it more succinctly. Thank you!

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

 

[25362]

Mbeychok, you're welcome.

 

[Hephaistos]

Hello!
I have the same problem as zesaint.

And I did not find the answer of zesaints question:

Where is this square root in (1) coming from??

As zesaint, I also find this Formular (based on perfect gas law), but without any square root!

Has this somehing to do with compressible/ non compressible states?

Thanks!
Heph

 

[mbeychok]

Hephaistos:

If you read all of the responses in this thread carefully, I think you will find that:

(1) If you want to re-calibrate or reset the design factor of the flowmeter so that its readings correspond to a new set of temperature and pressure, then the equation with the square root should be used.

(2) If you simply want convert the reading obtained from the existing design set of temperature and pressure to another set of temperature and pressure, then you use the universal gas law.

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

 

[Hephaistos]

@mbeychok:
Thanks for your answer.

Indeed, I want to re-calibrate corresponding to a new set of pressure and temperature.

And I am already using the equation with the square root.

But: Do have any source, where the equation might come from?

I tried to derive this equation from the perfect gas law, but it didn´t work. I thought, you might know, where it´s coming from.

Hephaistos

 

[UmeshMathur]

All:

I believe the best reference to settle all flow metering questions of the type we have seen in this discussion is Richard W. Miller: "Flow Meter Engineering Handbook", (3rd edition, McGraw-Hill, 1996). Specifically, variable area flow meters are dealt with extensively on pp. 14.40-14.52.

When discussing questions regarding the fundamentals, I still use the venerable McCabe, Smith, and Harriott: "Unit Operations of Chemical Engineering", (5th edition, McGraw-Hill, 1993), as I used the first edition in undergraduate school and still think it is one of the most lucid textbooks around.