Honeywell UDC 3300 manual Where

Figure 4-1 Example of Mass Flow Compensation using Multiplier/Divider Algorithm

Example - Mass Flow Compensation

A gas flow rate of 650 SCFM develops a differential pressure of 90" H 2O across an orifice plate at reference conditions of 30 psig and 140F. Compensate this gas flow for temperature and pressure variations.

Flow = K DPf x Pf x Tref

Tf Pref

Apply Multiplier/Divider Algorithm:

Where:

f = flowing conditions

ref = reference conditions (in absolute units)

PV = K		(Input A x Ratio A+ BiasA ) x (Input C x Ratio C + Bias C )
				(Input B x Ratio B + Bias B )				X (Calc HI – Calc LO )
Assign inputs using Engineering units:
Let:
Input A = DPf = IN1 (in H 2O)
Input B = Tf = IN2 + Bias2 = IN2 F + 460 ( R)
Input C = Pf= IN3 + Bias3 = IN3psig + 14.7(psia)
Tref		= 140 F + 460 = 600 R
Pref		= 30 psig + 14.7 = 44.7 psia
Calc Hi		= 650.0		Flow in SFCM at Reference Conditions
Calc Lo = 0.0
	K = to be determined next
Note: If temperature and pressure signals are already ranged in absolute units,
no Bias is required for inputs B and C.
PV = Q	SCFM		=	DPf x (IN3 + 14.7)	x	K2	x	(650.0 - 0.0)
				(IN2 + 460)

Note: When IN2 and IN3 are at the reference conditions of 600 R (140 F) and 44.7psia (30 psig) respectively and DPf = 90" H2O, the equation must calculate 650 SCFM. To accomplish this, divide the DP value by "90" to normalize the equation.

Q

SCFM

=

DP f

x

(IN3 + 14.7)

x

Tref

x

650

90

(IN2 + 460)

Pref

Rearranging terms:

Q SCFM =

DPf

x (IN3 + 14.7)

x

1

x

Tref

x 650

Example continued

(IN2 + 460)

90

Pref

on next page

Variable

Constant = K

22049

2

102 UDC 3300 Process Controller Product Manual 4/00