CIV E 444 - APPLIED HYDRAULICS
SPRING 2016
LAB No. 12 SOLUTION

  1. D = 40 cm = 0.4 m.

    n = 1000 rpm = 1000 / 60 = 16.667 rps

    CH = ΔH / [ (D2n2) / g ] = (3 × 9.81) / [(0.4)2 (16.667)2] = 0.662

    From the given figure, with CH = 0.662, find CQ = 0.65, and with CQ = 0.65, find CP = 0.54

    CQ = Q / (nD3) = 0.65

    Q = 0.65 nD3 = 0.65 × 16.667 × (0.4)  3 = 0.693 m3/s

    CP = P / (ρ D5 n3)

    ρ = 1 KN ⋅ s2/m4

    P = 0.54 ρ D5 n3 = 0.54 (1) (0.4)5 (16.667)3       [KN ⋅ m/s]

    P = 25.6 KN ⋅ m/s = 25.6 KW

  2. From the first figure, at maximum efficiency:   ΔH = 95 m, and Q = 0.214 m3/s

    ( CH ) N = ( CH ) N = 2133.5

    [ ΔH / (D2n2 / g) ] N = [ ΔH / (D2n2 / g) ] N = 2133.5

    [ ΔH / (n2) ] N = [ ΔH / (n2) ] N = 2133.5

    70 / N2 = 95 / (2133.5)2

    N = 2133.5 (70/95)1/2 = 1882 RPM

    (CQ) N = 1882 = (CQ) N = 2133.5

    [Q / (nD3)] N = 1882 = [Q / (nD3)] N = 2133.5

    QN = 1882 / QN = 2133.5 = 1882 / 2133.5 = 0.883

    QN = 1882 = 0.883 QN = 2133.5 = 0.883 × 0.214 = 0.189 m3/s

    n = 1882 / 60 = 31.37 rps

    D = 0.371 m

    CQ = Q / (nD3) = 0.189 / [ 31.37 × (0.371)3 ] = 0.119

    From the second figure:   CP = 0.68

    CP = P / (ρ D5 n3)

    ρ = 1 KN ⋅ s2/m4

    P = 0.68 ρ D5 n3 = 0.68 × 1 × (0.371)5 (31.37)3 [KN ⋅ m/ s]

    P = 147.5 KN ⋅ m/ s = 147.5 KW

  3. hp = (z2 - z1) + [V2/(2g)] [f (L/D) + ∑ KL ]

    hp = (z2 - z1) + [ Q2/(2gA2)]  [ f (L/D) + Ke + Kb + KE ]

    From Table 5-3 (in the text):   Ke = 0.5; Kb = 0.19; KE = 1

    hp = (500 - 450) + [ Q2/(2g ((π/4) D2)2) ] [ 0.016 (2000/0.5) + 0.5 + 0.19 + 1 ]

    hp = 50 + [ Q2/(2 (9.81) (π/4)2 (0.5)4 ] (65.69)

    hp = 50 + [ Q2/(0.7564) ] (65.69)

    hp = 50 + 86.84 Q2

    Pump performance curve:   H = 70 - 700 Q 2

    At H = hp:   Q = 0.159 m3/s

    hp = 50 + 86.84 (0.159) 2 = 52.195 m.