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4/00 UDC 3300 Process Controller Product Manual 103
Figure 4-1 Example of Mass Flow Compensation using Multiplier/Divider Algorithm, continued
Example - Mass Flow Compensation
Determined value of K:
K
2
=x
1
90
T
ref
P
ref
==
0.14914
600
(90) (44.7)
Therefore K = 0.386
SCFM
Q =
(0.386) (650)
(Calc - Calc )
HI LO
K
DP
f
(in H O) (IN3 + 14.7)
(IN2 + 460)
2
140 F + 460
170 F + 460
170 F + 460
110 F + 460
110 F + 460
30 psi + 14.7
50 psi + 14.7
20 psi + 14.7
50 psi + 14.7
20 psi + 14.7
459
539
395
567
415
Flow (SFCM)
DP = 45" H O (50%)
f2
650
763
559
802
587
DP = 90" H O (100%)
f2
Temp (T )
( R)
f
Summary of Flow Values At Values Conditions
Pressure (T )
(psia)
f
Reference
Conditions
22050