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<7. Operation>
7-4
IM 01C25K01-01E
 Reynolds Number Calculation
Re = 354 (3)
W
D
.
µ
Where, Re: Reynolds number at normal ow rate
W: Weight ow at normal ow rate (kg/h)
(Note)
D: Orice bore (mm)
µ: Viscosity (mPa·s)
Note: Determination of W
· For liquid, W=Qf·ρf
· For gas, W=Qn·ρn
 Differential Pressure Correction using
Reynolds Number
∆P =
2
.
∆P
0
1
Kaf/Ka
Where, ∆P: Corrected differential pressure
∆P
0
: Differential pressure obtained from
procedure (b)
Kaf/ka: Correction factor obtained from
Figure 7.5.3
For details concerning determination of
differential pressure correction using Reynolds
number, pressure loss, etc., refer to TI
01C20K00-01E.
7.5.2 Example of Calculation
Fluid: N
2
gas (Nitrogen gas)
Flow range: 0 to 25 Nm
3
/h
(ow rate at 0°C, 1 atm)
Normal ow rate: 18 Nm
3
/h
Specic density: 1.251 kg/Nm
3
(specic density at 0°C, 1 atm)
Temperature: 30°C
Pressure: 100 kPa
Viscosity: 0.018 mPa·s
From Equation (2), air equivalent volumetric ow
Qo is:
Q
o
= 0.5356 × 25 1.251 ×
= 18.38 Nm
3
/h = 306.3 Nl/min
273.15 + 30
101.325 + 100
A differential pressure range of 0 to 2400 mmH
2
O is
obtained from Figure 7.5.2 applying an orice bore
of 6.350 mm (where, Zf/Zn=1 is assumed).
From Equation (3), Reynolds number at normal ow
rate Re is:
Re = 354 × = 6.97 × 10
4
18 × 1.251
6.35 × 0.018
Since the correction factor (1.00) is constant at
this Reynolds number, no differential pressure
correction is required. Consequently, the differential
pressure range is determined as 0 to 2400 mmH
2
O.
F0705.ai
0.05
0.508 0.864 1.511 2.527 4.039 6.350
0.01 0.05 0.1 0.5 1 5 10 40
200
100
50
10
5
1
200
100
50
10
5
1
0.1 0.5 1 5 10 40
Equivalent water flow(liter/min at 0°C, 1 atm)
Differential
Pressure
(kPa)
Figure 7.5.1 Relationship between Equivalent Water Flow and Differential Pressure