Handbook of Civil Engineering Calculations

2. Compute the velocity at each section
   Applying Eq. 8 gives F 1 = 10/0.785 = 12.7 ft/s (387.10 cm/s); F 2 = 10/0.349'= 28.7 ft/s
   (874.78 cm/s).


3. Compute p 2 by applying Eq. 9
   Thus, (p 2 -P 1 )Iw = (F^2 - Fi)/(2g) + Z 1 -Z 2 - hF = (12.7^2 - 28.7^2 )/64.4 + 140 - 106 - 9 =
   14.7 ft (448.06 cm); p 2 = 14.7(62.4)/144 + 18 = 24.4 lb/in^2 (168.24 kPa).

FLOW THROUGH A VENTURI METER

A venturi meter of 3-in (76.2-mm) throat diameter is inserted in a 6-in (152.4-mm) diam-
eter pipe conveying fuel oil having a specific gravity of 0.94. The pressure at the throat is
10 lb/in^2 (68.95 kPa), and that at an upstream section 6 in (152.4 mm) higher than the
throat is 14.2 lb/in^2 (97.91 kPa). If the discharge coefficient of the meter is 0.97, compute
the flow rate in fallons per minute (liters per second).

Calculation Procedure:

1. Record the given data, assigning the subscript 1
   to the upstream section and 2 to the throat
   The loss of head between two sections can be taken into account by introducing a dis-
   charge coefficient C. This coefficient represents the ratio between the actual discharge Q
   and the discharge Q 1 that would occur in the absence of any losses. Then Q = CQ 1 , or (F^2
   -Ff)/(2g) = CV
   Record the given data: D 1 = 6 in (152.4 mm); P 1 = 14.2 lb/in^2 (97.91 kPa); Z 1 = 6 in
   (152.4 mm); D 2 = 3 in (76.2 mm);p 2 = 10 lb/in^2 (68.95 kPa); Z 2 = O; C = 0.97.


2. Express V 1 in terms of V 2 and develop velocity
   and flow relations
   Thus,

T 2g/zv lo.5

r

'-

c

[r^\ <•**

Also

f 2ghv lo.5

fi-^brew] ^)

If F 1 is negligible, these relations reduce to

V 2 = C(2ghv)™ (ISd)

and

Q = CA 2 (Wy)Q-* (ISb)

3. Compute hv by applying Eq. 9
   Thus, hv= (P 1 -P 2 )Iw + Z 1 -Z 2 = 4.2(144)/[0.94(62.4)] + 0.5 = 10.8 ft (3.29 m).