Figure 3-1 Mass Flow Example - Honeywell UDC3500 Product Manual

Configuration
Example - Mass Flow Compensation
A gas flow rate of 650 SCFM develops a differential pressure of 90" H O across an orifice plate
at reference conditions of 30 psig and 140 F. Compensate this gas flow for temperature and
pressure variations.
DP
Flow = K
Apply Multiplier/Divider Algorithm:
(Input A x Ratio A + Bias A) x (Input C x Ratio C + Bias C)
PV = K
Assign inputs using Engineering units:
Let:
Input A = DP = IN1 (in H O)
Input B = T = IN2 + Bias2 = IN2 F + 460 ( R)
Input C = P = IN3 + Bias3 = IN3psig + 14.7(psia)
T = 140 F + 460 = 600 R
ref
P = 30 psig + 14.7 = 44.7 psia
ref
Calc = 650.0
Hi
Calc = 0.0
Lo
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
Note: When IN2 and IN3 are at the reference conditions of 600 R (140 F) and 44.7psia (30
psig) respectively and DP = 90" H O, the equation must calculate 650 SCFM. To accomplish
this, divide the DP value by "90" to normalize the equation.
Q
SCFM
Rearranging terms:
Q
SCFM
80

Figure 3-1 Mass Flow Example

x P T
ref
f f
x
T P
f
ref
(Input B x Ratio B + Bias B)
f 2
f
f
Flow in SFCM at Reference Conditions
DP x (IN3 + 14.7)
=
f
(IN2 + 460)
f
2
= DP (IN3 + 14.7)
f
x
90 (IN2 + 460)
(IN3 + 14.7)
=
DP x
f
(IN2 + 460)
Variable
UDC3500 Universal Digital Controller Product Manual
Where:
f = flowing conditions
ref = reference conditions (in absolute units)
2
x
K x
(650.0 - 0.0)
T
ref
x x 650
P
ref
1 T
ref
x x x 650
90 P
ref
2
Constant = K
2
X (Calc – Calc )
HI
LO
Example continued
on next page
22049
10/05