2 example of calculation, Example of calculation -4 – Yokogawa EJA115 User Manual

Page 34

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IM 01C22K01-01E

7-4

7. OPERATION

᭿ Equivalent Air Flow Calculation

Q

o

= 0.5356 . Qn

ρn . .

(2)

273.15 + t

101.325+ p

Zf

Zn

Where, Qo: Air equivalent volumetric flow at 0

°C,

1 atm (Nm

3

/h)

Qn: Volumetric gas flow at 0

°C, 1 atm (Nm

3

/h)

␳n: Specific gas density at 0°C, 1 atm (kg/Nm

3

)

Zn: Compression factor of gas at 0

°C, 1 atm

Zf: Compression factor of gas at operations

conditions (t

°C, p kPa)

(b) Obtain a differential pressure from the above

equivalent water or air flow using the nomograph
shown in Figure 7.5.1 or 7.5.2. In this procedure,
multiply Qw or Qo by 1000/60 to convert the flow
unit into liter/min.

(c) Select an orifice bore, taking into considerations

pressure loss, etc.

(d) As necessary, calculate Reynolds number at normal

flow rate and correct the differential pressure
obtained from the procedure (b).

᭿ Reynolds Number Calculation

Re = 354

(3)

W

D.

Where, Re: Reynolds number at normal flow rate

W: Weight flow at normal flow rate (kg/h)

(Note)

D:

Orifice bore (mm)

␮: Viscosity (mPa·s)

Note: Determination of W

· For liquid, W=Qf·

␳f

· For gas, W=Qn·

␳n

᭿ Differential Pressure Correction using

Reynolds Number

⌬P =

2

.

⌬P

0

1

Kaf/Ka

Where,

⌬P: Corrected differential pressure
⌬P

0

: Differential pressure obtained from proce-

dure (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.

F0713.EPS

7.5.2 Example of Calculation

F0714.EPS

Fluid: N

2

gas (Nitrogen gas)

Flow range:

0 to 25 Nm

3

/h (flow rate at 0°C, 1 atm)

Normal flow rate: 18 Nm

3

/h

Specific density: 1.251 kg/Nm

3

(specific density at 0°C,

1 atm)

Temperature: 30°C
Pressure: 100

kPa

Viscosity: 0.018

mPa·s

From Equation (2), air equivalent volumetric flow Qo
is:

Q

o

= 0.5356

Ч 25 1.251 Ч

= 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 orifice bore of
6.350 mm (where, Zf/Zn=1 is assumed).

From Equation (3), Reynolds number at normal flow
rate Re is:

Re = 354

Ч

= 6.97

Ч 10

4

18

Ч 1.251

6.35

Ч 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.

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