- Compute Q by applying Eq. 14b
Thus, (A 2 IA 1 )^2 = (D 2 ID 1 )^4 =^1 Xi 6 ; A 2 = 0.0491 ft^2 (0.00456 m^2 ); and Q = 0.97(0.049)[64.4
x 10.87(1 -^1 Xw)]^0 -^5 = 1.30 ft^3 /s or, by using the conversion factor of 1 fWs = 449 gal/min
(28.32 L/s), the flow rate is 1.30(449) = 584 gal/min (36.84 L/s).
FLOW THROUGH AN ORIFICECompute the discharge through a 3-in (76.2-mm) diameter square-edged orifice if the wa-
ter on the upstream side stands 4 ft 8 in (1.422 m) above the center of the orifice.
Calculation Procedure:
- Determine the discharge coefficient
For simplicity, the flow through a square-edged orifice discharging to the atmosphere is
generally computed by equating the area of the stream to the area of the opening and then
setting the discharge coefficient C = 0.60 to allow for contraction of the issuing stream.
(The area of the issuing stream is about 0.62 times that of the opening.) - Compute the flow rate
Since the velocity of approach is negligible, use Eq. I5b. Or, Q = 0.60(0.0491)(64.4 x
4.67)^0 -^5 = 0.511 fWs (14.4675 L/s).
FLOW THROUGH THE SUCTION PIPE
OFA DRAINAGE PUMPWater is being evacuated from a sump through the suction pipe shown in Fig. 7. The en-
trance-end diameter of the pipe is 3 ft (91.44 cm); the exit-end diameter, 1.75 ft (53.34
cm). The exit pressure is 12.9 in (32.77 cm) of mercury vacuum. The head loss at the en-
try is one-fifteenth of the velocity head at that point, and the head loss in the pipe due to
friction is one-tenth of the velocity head at the exit. Compute the discharge flow rate.
Calculation Procedure:
- Convert the pressure head to feet of water
The discharge may be found by comparing the conditions at an upstream point 1, where
the velocity is negligible, with the conditions at point 3 (Fig. 7). Select the elevation of
point 1 as the datum.
Converting the pressure head at point 3 to
feet of water and using the specific gravity
of mercury as 13.6, we have p 3 /w =
-(12.9/12)13.6 = - 14.6 ft (-4.45 m). - Express the velocity head at 2
in terms of that at 3
By the equation of continuity, V 2 = A 3 V 3 IA 2 =
(1.75/3)^2 V 3 = 0.34F 3. FIGURE 7