Suppose that the two reservoirs in Prob. 14–39E are 1000 ft farther apart horizontally, but at the same elevations. All the constants and parameters are identical to those of Prob. 14–39E except that the total pipe length is 562 ft instead of 124 ft. Calculate the volume flow rate for this case and compare with the result of Prob. 14–39E. Discuss.

Data from Problem 14–39E

A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure, as sketched in Fig. P14–39E. The dimensions and minor loss coefficients are provided in the figure. The pump’s performance is approximated by the expression H_available = H₀ – aV̇², where the shutoff head H₀ = 125 ft of water column, coefficient a = 2.50 ft/gpm², available pump head H_available is in units of feet of water column, and capacity V̇ is in units of gallons per minute (gpm). Estimate the capacity delivered by the pump.

FIGURE P14–39E

Transcribed Image Text:

22-21 = 22.0 ft (elevation difference) D 1.20 in (pipe diameter) KL, entrance = 0.50 (pipe entrance) = 2.0 (valve 1) = 6.8 (valve 2) KL, valve 1 KL, valve 2 KL. elbow = 0.34 (each elbow-there are 3) KL, exit = 1.05 (pipe exit) L=124 ft (total pipe length) 8 = 0.0011 in (pipe roughness) V₁=0 Reservoir 22- 22-21 Valve 1 Z1 <-D Pump Reservoir V/₂=0 Valve 2