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

  1. D = 30 cm = 0.30 m

    n = 1200 rpm / (60 s/m) = 20 rps

    CQ = Q / (n D3) = 0.25 / [ 20 × (0.3)3 ] = 0.463

    From the given figure, with CQ = 0.463, find CH = 1.24, and CP = 0.72

    ΔH = CH D2 n2 / g = 1.24 (0.3)2 (20)2 / 9.81 = 4.55 m

    ρ = 1 KN ⋅ s2/m4

    P = CP ρ D5 n3 = 0.72 (1) (0.3)5 (20)3       [KN ⋅ m/s]

    P = 14 KN ⋅ m/s = 14 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

    80 / N2 = 95 / (2133.5)2

    N = 2133.5 (80/95)1/2 = 1958 RPM

    (CQ) N = 1958 = (CQ) N = 2133.5

    [Q / (nD3)] N = 1958 = [Q / (nD3)] N = 2133.5

    QN = 1958 / QN = 2133.5 = 1958 / 2133.5 = 0.917

    QN = 1958 = 0.917 QN = 2133.5 = 0.917 × 0.214 = 0.196 m3/s

    n = 1958 / 60 = 32.63 rps

    D = 0.371 m

    CQ = Q / (nD3) = 0.196 / [ 32.63 × (0.371)3 ] = 0.118

    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 (32.63)3 [KN ⋅ m/ s]

    P = 166 KN ⋅ m/ s = 166 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 = 0.35; KE = 1

    hp = (230 - 200) + [ Q2/(2g ((π/4) D2)2) ] [ 0.015 (1000/0.4) + 0.5 + 0.35 + 1 ]

    hp = 30 + [ Q2/(2 (9.81) (π/4)2 (0.4)4) ] (39.35)

    hp = 30 + [ Q2/(0.3098) ] (39.35)

    hp = 30 + 127.02 Q2

    Pump performance curve:   H = 55 - 500 Q 2

    At H = hp:   Q = 0.2 m3/s

    hp = 30 + 127.02 (0.2)2 = 35.08 m.