Page 461 - Chemical process engineering design and economics

440 Chapter 8

p D v 49.42 lb M/ft 3 2.067 in 6.692 ft/s
Re = —— = ——————————— ———— ————— = 2.52xl0 5
u. 2.265X10" 4 Ibw/ft-s 12 in/ft 1

70 gal/min 1
3
Q = ————— ————— = 0.1560 ft /s (4.42 mVs)
60 s/min 7.48 gal/ft 3

Select 50 in (127 cm) of water as the pressure drop across the orifice. The
pressure drop in force per unit area is related to the pressure drop in terms of a
liquid height by

g 32.17 ft/s 2 lb M 50 in
Pi - Pa = — Pw Az = —————————— 62.4 —— ———
2
gc 32.17 lb Mft/s -lb F ft 3 12 in/ft
Pi - Pa = 260.0 Ibp/ft 2 (12.45 kPa)

After substituting the values of Q, D, P! - p 2, p, and g c into the first equation
above, we obtain

( 2 (32. 17) (260.0) V 2
2
0.1 56 = C D Oi/4) (2.0667/12) (3 2 I ———————————— I
4
I 0.792 (62.4) (1 - P )J
Considine [16] gives equations for the orifice coefficient, CD, for several
ways of measuring pressure drop across the orifice. For corner pressure taps,
shown in Figure 8.2.1, the equation is

8
2
''
'' + 91.71p /Re
' -
C D = 0.5959 + 0.0312 p ' 2 1 1 - 0.184 p 8 00 22 5 5 a 7 5
Solving these two equations simultaneously for C D and p using Polymath
[27], the orifice coefficient, C = 0.6035 and P = 0.6690. Thus, the orifice diame-
D
ter,
DO = 0.6690 (2.067) = 1.383 in (3.51 cm).

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