Comparing the results of Probs. 14–43 and 14–47, the volume flow rate increases as expected when one doubles the inner diameter of the pipe. One might expect that the Reynolds number increases as well. Does it? Explain.

Data from Problem 47

Calculate the volume flow rate between the reservoirs of Prob. 14–43 for the case in which the pipe diameter is doubled, all else remaining the same. Discuss.

Data from Problem 14–43

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. P14–43. 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 shutoff head H₀ = 24.4 m of water column, coefficient a = 0.0678 m/Lpm², available pump head H_available is in units of meters of water column, and capacity V̇ is in units of liters per minute (Lpm). Estimate the capacity delivered by the pump.

Data from FIGURE P14–43

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22- 22-21 7.85 m (elevation difference) D = 2.03 cm (pipe diameter) KL. , entrance = 0.50 (pipe entrance) KL, valve KL, elbow KL, exit L Z2-21 P₁V₁=0 V₁ = 0 Reservoir = 17.5 (valve) = 0.92 (each elbow-there are 5) = 1.05 (pipe exit) = 176.5 m (total pipe length) ε = 0.25 mm (pipe roughness) E 21 Valve Pump V₂=0 Reservoir -D