Re: [foxboro] Typical Orifice Flowmeter Compensation Calculations

Pablo,

Thank you for the feedback.  This is much along the lines of what Fox
placed in the calc blocks, and is a "non-rigorous" (no compressibility,
etc) approach that is good enough for control.

Some questions:

1) I assume that if your "FI" has already been transformed with the sqrt at
the transmitter, your equation would look a bit different?  Specifically

Qcomp = Qref Sqrt[( FI/FImax) ((PI+1.013)/(Pref + 1.013)) *
((Tref+273.15)/(TI+273.15))]

would become

Qcomp = Qref ( FI/FImax) Sqrt[((PI+1.013)/(Pref + 1.013)) *
((Tref+273.15)/(TI+273.15))]

2) Assuming Qref is standard conditions, and the flow rate at those
conditions would be considered "standard" (e.g.  Std-M^3/hr), when
compensated at other conditions, is the flow rate still in "standard
units"?

3) (and this if everyone) Why can't one simply use PV = nRT to compensate a
flow?  Thus  since

n*R is constant

(P*V)/(z*T) = (Pb*Vb)/(zb*Tb)
where Tb, Pb, zb = density, temp, pressure, and compressiblity at base
conditions

solving for Vb
Vb = V * (zb*Tb* P)/(z*T*Pb)
and Qv = V/t
thus
Qb = Qv  * (zb*Tb* P)/(z*T*Pb)
Qv = Qb * (z*T*Pb)/(zb*Tb* P)






BTW, here is the derivation of the equation Pablo is using:

Given:
(A) Qm = K * sqrt[ h * r] = mass flow of a fluid through an orifice (r =
density of fluid flowing, h = dp)

(B) Qv = Qm/rho = K * sqrt[h / r] = volumetric flow at "r" conditions.

(C) P*V = z*n*R*T

Then,

since n = mass/mol-wt
and V = mass/r
and n/V =  r/(mol-wt)
then
r = (mol-wt * P)/(z*R*T)

mol-wt/R = z*r*P/T = constant
rb, Tb, Pb, zb = density, temp, pressure, and compressiblity at base
conditions
thus
z*r*P/T = zb*rb*Pb/Tb
r = rb * [(zb*Pb*T) / (z*Tb*P)]

substituting into (B)

Qv = K * sqrt[h / {rb * [(zb*Pb*T) / (z*Tb*P)]} ]
Qv = K * sqrt[h / rb] * sqrt[ (z*Tb*P) / (zb*Pb*T)]
since K * sqrt[h / rb] = Qb (base volumetric flow)
Qv = Qb * sqrt[ (z*Tb*P) / (zb*Pb*T)]



Hook 'Em,
Kirk

Kirk Carver, PE
ExxonMobil Development Company
Facilities -  Instrumentation and Controls
12450 Greenspoint Drive
GP6-315
Houston, TX 77060
(281) 654-4881 (Greenspoint Office)
(713) 350-1158 (Worley Parsons)
(713) 962-2549 (mobile)
kirk.d.carver@xxxxxxxxxxxxxx (best chance of contacting me)


                                                                           
             "Pablo Lioi"                                                  
             <plioi@hotmail                                                
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                                                                   Subject 
                                      Re: [foxboro] Typical Orifice        
             01/23/06 04:12           Flowmeter Compensation Calculations  
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Kirk,
Hope this helps:

CALCA BLOCK

RI01  FIxxxx.PNT
RI02  PIxxxx.PNT
RI03  TIxxxx.PNT

M01 1.013      (Base Pressure, Bar)
M02 273.15    (Base Temp, Kelvin)
M03 298.15    (Reference Temp, in Kelvin)
M04 18.103    (Reference Pressure, Bara)
M05 350        ( max differential pressure, mBar)
M06 6.39       (Flow @ max dif pressure, Ref. Pressure and Ref. Temp)

ADD RI02 M01
ADD RI03 M02
DIV
MUL M03
DIV M04
MUL RI01
DIV M05
SQRT
MUL M06
OUT RO01
END


The formula is

Qcomp = Qref Sqrt[( FI/FImax) ((PI+1.013)/(Pref + 1.013)) *
((Tref+273.15)/(TI+273.15))]

Pref+1.013 is stored in M04
Tref+273.15 is stored in M03
FImax is stored in M05
Qref is stored in M06

Determine Qref with a flow calculation program, using FI = max differential

pressure, P = work pressure T = work temperature

Pref is the work pressure and Tref the work temperature

Note in the above formula that if (FI=FImax, PI = Pref and TI = Tref) then
Qcomp = Qref, so the formula compensates for the deviations from the
reference pressure and temperature.

You can put this formula in Excel and make a table of Qcomp vs. P and Qcomp

vs. T and compare results against your flow calculation program to see how
it works

I am using SI units. If you need otherwise, just change the values in
M01-M06
As you see, the transmitter is not calculating the square root of the diff
pressure.

If you have square root at the transmitter then the MUL RI01 line has to be

moved after the SQRT, leaving the rest unchanged (you still need the SQRT
for the FI range, pressure and temp).

Good luck
Pablo Lioi
System Engineer
TOTAL AUSTRAL
ARGENTINA

>From: kirk.d.carver@xxxxxxxxxxxxxx
>Reply-To: foxboro@xxxxxxxxxxxxx
>To: foxboro@xxxxxxxxxxxxx
>Subject: [foxboro] Typical Orifice Flowmeter Compensation Calculations
>Date: Sun, 22 Jan 2006 16:32:17 -0600
>
>I am comparing some different configurations on our project and wanted to
>get a quick survey of how others perform flowmeter compensation in I/A.
>
>Assuming two flavors of 4-20mA flowmeter signal:
>a) in H2O (or psi) differential
>b) volumetric flow (square root in the transmitter)
>How do you compensate the flow for pressure and temperature?
>
>Calc block steps would be appreciated.
>
>
>Hook 'Em,
>Kirk
>
>Kirk Carver, PE
>ExxonMobil Development Company
>Facilities -  Instrumentation and Controls
>12450 Greenspoint Drive
>GP6-315
>Houston, TX 77060
>(281) 654-4881 (Greenspoint Office)
>(713) 350-1158 (Worley Parsons)
>(713) 962-2549 (mobile)
>kirk.d.carver@xxxxxxxxxxxxxx (best chance of contacting me)
>
>
>
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