solve Situation:Water is pumped from one tank to another. Find:r/d= 1, D₁ = 10 in, L₁ = 50 ft.f2 = 0.020, 21 = 10 ft.=D₂ 10 in, L2 = 950 ft, f2 = 0.020.Z₂ = 20 ft, T = 60°F.r/d = 1L = 50 ftD = 10 inf = 0.020T= 60°FElevation 15 ftElevation = 10 fth, ft806040200Discharge (ft³/s).Properties:From Table 10.5: Ke = 0.03; K = 0.35; KE = 1.0.Elevation = 20 ftL = 950 ftD = 10 inf= 0.020Pump curve1000 2000 3000 4000 5000Q, gpm

f = 0.020 L1 = 50 ft L2 = 950 ftD = 10 in or 0.833 ft

Situation: Water is pumped from one tank to another. Find: r/d=1, D₁ = 10 in, L₁= 50 ft. f2 = 0.020, 2₁ = 10 ft. D₂ = 10 in, L₂ = 950 ft, f2 = 0.020. 2₂ - 20 ft, T = 60°F. r/d = 1 L = 50 ft D = 10 in f= 0.020 T = 50°F Elevation 15 ft Elevation = 10 ft h, ft 80 60 40 20 O Discharge (ft³/s). Properties: From Table 10.5: Ke = 0.03; K = 0.35; KE = 1.0. Elevation = 20 ft L = 950 ft D = 10 in f= 0.020 Pump curve 1000 2000 3000 4000 5000 Q. gpm

Situation: Water is pumped from one tank to another. Find: r/d=1, D₁ = 10 in, L₁= 50 ft. f2 = 0.020, 2₁ = 10 ft. D₂ = 10 in, L₂ = 950 ft, f2 = 0.020. 2₂ - 20 ft, T = 60°F. r/d = 1 L = 50 ft D = 10 in f= 0.020 T = 50°F Elevation 15 ft Elevation = 10 ft h, ft 80 60 40 20 O Discharge (ft³/s). Properties: From Table 10.5: Ke = 0.03; K = 0.35; KE = 1.0. Elevation = 20 ft L = 950 ft D = 10 in f= 0.020 Pump curve 1000 2000 3000 4000 5000 Q. gpm

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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