fore to obtain the following dimensionless parameters: Tr 2 = pw(gD^3 )1/2/^a, and Tr 3 =
Pa(SEP)^112 I^a = (gD^3 )l/2/va> where va = kinematic viscosity of air.
HYDRAULIC SIMILARITYAND
CONSTRUCTION OF MODELS
A dam discharges 36,000 ft^3 /s (1,019,236.7 L/s) of water, and a hydraulic pump occurs on
the apron. The power loss resulting from this jump is to be determined by constructing a
geometrically similar model having a scale of 1:12. (a) Determine the required discharge
in the model, (b) Determine the power loss on the dam if the power loss on the model is
found to be 0.18 hp (0.134 kW).
Calculation Procedure:
- Determine the value of Qm
Two systems are termed similar if their corresponding variables have a constant ratio. A
hydraulic model and its prototype must possess three forms of similarity: geometric, or
similarity of shape; kinematic, or similarity of motion; and dynamic, or similarity of
forces.
In the present instance, the ratio associated with the geometric similarity is given, i.e.,
the ratio of a linear dimension in the model to the corresponding linear dimension in the
prototype. Let rg denote this ratio, and let subscripts m andp refer to the model and proto-
type, respectively.
Apply Eq. Ua to evaluate Qm. Or Q = C 1 Z?/*^15 , where C 1 is a constant. Then QJQP =
(bmlhp)(bmlhp)^. But bmlbp = hmlhp = rg; therefore, QJQp = ^= (Vi 2 )^2 -^5 = 1/499; Qm =
36,000/499 = 72 ftVs (2038.5 L/s). - Evaluate the power loss on the dam
Apply Eq. 16 to evaluate the power loss on the dam. Thus, hp = C 2 Qh 7 where C 2 is a con-
stant. Then hP//hpw = (QJQ1n)(H^h1n). But Qp/Qm = (l/rg)^2 -^5 , and hjhm = l/rg; therefore,
hp/ip™ = (l/^)^3 5 = 12 3/= 5990. Hence, hpp = 5990(0.18) = 1078 hp (803.86 kW).
P ART 2
PUMP OPERATING MODES, AFFINITY
LAWS. SPEED, AND HEAD
SERIESPUMPlNSTALLATlONANALYSlS
A new plant requires a system pumping capability of 45 gal/min (2.84 L/s) at a 26-ft
(7.9-m) head. The pump characteristic curves for the tentatively selected floor-mounted
units are shown in Fig. 15; one operating pump and one standby pump, each 0.75 hp (0.56
kW) are being considered. Can energy be conserved, and how much, with some other
pumping arrangement?