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Ha (ft) = 80 - 0.3373 Q2where the flow Q is given in 1000s of gallons per minute (1O00s gpm). The difference in reservoir elevations is 57 feet The head loss in the...

Question

Ha (ft) = 80 - 0.3373 Q2where the flow Q is given in 1000s of gallons per minute (1O00s gpm). The difference in reservoir elevations is 57 feet The head loss in the system is given as: hl (ft) = 0.30* Q? where Q is given in 1000s of gallons per minute (1000s gpm): For this system;, the pump operating point is closest toSelect ]The efficiency of the pump can be given with the following equation: Efficiency (%) 100 (-0.0009*Q3+ 0.0081*Q? + 0.0920*Q with Q given in 1000's of gpm. Use the pump

Ha (ft) = 80 - 0.3373 Q2 where the flow Q is given in 1000s of gallons per minute (1O00s gpm). The difference in reservoir elevations is 57 feet The head loss in the system is given as: hl (ft) = 0.30* Q? where Q is given in 1000s of gallons per minute (1000s gpm): For this system;, the pump operating point is closest to Select ] The efficiency of the pump can be given with the following equation: Efficiency (%) 100 (-0.0009*Q3+ 0.0081*Q? + 0.0920*Q with Q given in 1000's of gpm. Use the pump operating point that you estimated, the head added by the pump (Ha in feet) and the efficiency given above; The necessary brake horsepower needed to operate the pump (P in Hp) is closest to: '[Select ] 160 Hp: 135 Hp" B0 Hp 4180 Hp" 145 Hp



Answers

A 70 percent efficient pump delivers water at $20^{\circ} \mathrm{C}$ from one reservoir to another 20 ft higher, as in Fig. P6.102. The piping system consists of $60 \mathrm{ft}$ of galvanized iron 2 -in pipe, a reentrant entrance, two screwed $90^{\circ}$ long-radius elbows, a screwed-open gate valve, and a sharp exit. What is the input power required in horsepower with and without a $6^{\circ}$ well-designed conical expansion added to the exit? The flow rate is $0.4 \mathrm{ft}^{3} / \mathrm{s}$


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