Honeywell UDC 3300 Product Manual page 115

Figure 4-1 Example of Mass Flow Compensation using Multiplier/Divider Algorithm, continued
Example - Mass Flow Compensation
Determined value of K:
1
2
K =
90
Therefore K = 0.386
Q = (0.386) (650)
SCFM
Summary of Flow Values At Values Conditions
Temp (T )
( R)
Reference
140 F + 460
Conditions
170 F + 460
170 F + 460
110 F + 460
110 F + 460
4/00
T
600
ref
x
=
P (90) (44.7)
ref
DP (in H O) (IN3 + 14.7)
f
(IN2 + 460)
(Calc - Calc )
K
HI LO
Pressure (T )
f
f
(psia)
30 psi + 14.7
50 psi + 14.7
20 psi + 14.7
50 psi + 14.7
20 psi + 14.7
UDC 3300 Process Controller Product Manual
= 0.14914
2
Flow (SFCM)
DP = 45" H O (50%)
f 2
459
539
395
567
415
DP = 90" H O (100%)
f 2
650
763
559
802
587
22050
103