assignment6_solutions.pdf - CE 356 Elements of Hydraulic Engineering Fall 2019 Assignment 6 Pumps Solutions Problem 1 Write the equation for N P

assignment6_solutions.pdf - CE 356 Elements of Hydraulic...

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CE 356 Elements of Hydraulic Engineering Fall 2019 Assignment 6: Pumps Solutions Problem 1 Write the equation for NPSH available : Z S = P 0 - P V γ - NPSH available - h L At 20 C , the vapor pressure of water is p v = 0 . 339 h lb 2 in 2 i . To avoid cavitation NPSH available NPSH required . Substitute in known values and unit conversion terms: h L = 14 . 7 lb 2 in 2 - 0 . 339 lb 2 in 2 144 in 2 ft 2 62 . 3 lb ft 3 - 10 ft - 5 ft h L = 18 . 2 ft Write the equation for energy losses in the system, between the lower reservoir and the pump: h L = V 2 2 g f 1 L 1 D + K S + 4 K b From Table 3.1, for cast-iron pipes the roughness height (e) is 0.00085ft. Thus the relative roughness value is: e 1 D = 0 . 00085 ft 1 ft = 0 . 00085 With Q = 10 cfs and the pipe diameter equal to 1 ft , the water velocity is: V = Q A = 4 · 10 ft 3 s π (1 ft ) 2 = 12 . 7 ft s From Table 1.3, at 20 C the kinematic viscosity of water is 1 . 003 × 10 - 6 m 2 /s , which equals 1 . 080 × 10 - 5 ft 2 /s . Thus the Reynolds’s number is: Re = DV ν = (1 ft ) ( 12 . 7 ft s ) 1 . 08 × 10 - 5 ft 2 s = 1 . 1 × 10 6 From the Moody diagram, the friction factor is then f 1 = 0 . 019. From page 96 in the book, K B is approximately 0.19. Substituting known values into the equation for enery loss yields: 18 . 2 ft = ( 12 . 7 ft s ) 2 2 ( 32 . 2 ft s 2 ) h 0 . 019 L 1 1 ft + 2 . 5 + 4 × 0 . 19 i 6-1
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Assignment 6 6-2 Solving for the length L 1 yields: L 1 = 211 ft Therefore if the distance between the supply reservoir and the pump is at or below 211 ft, cavitation within the pump will be avoided. Problem 2 1. To generate a system head equation, start with the basic energy equation for the sys- tem: H P = H B - H A + h L h L = V 2 2 g ( f L D + K ) = ( 4 Q πD 2 ) 2 2 g ( f L D + K ) Use the Swamee-Jain approximation to determine the friction factor as a function of the unknown pipe diameter: f = 0 . 25 log 2 e D 3 . 7 + 5 . 74 Re 0 . 9 Finally use the equation for Reynolds number ( Re ) written in terms of the flow: Re = DV ν = DQ νA = 4 DQ νπD 2 = 4 Q νπD Combine these equations into a single equation relating pump head ( H p ) and flow (Q): H P = H B - H A + ( 4 Q πD 2 ) 2 2 g f L D + K From Table 3.1, the roughness height (e) for commercial steel is 0.045 mm. Kinematic
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