CIV E 444 - APPLIED HYDRAULICS
FALL 2009
HOMEWORK No. 6 SOLUTION


  1. Use P (KW) = 8 Q (m3/s) ht (m)

    P = 8 × 1.5 × 17 = 204 KW.

  2. D = 6 in = 0.5 ft.

    Calculate area:   A = π (D2/4) = 3.1416 × 0.5 2 / 4 = 0.19635 ft2

    Calculate velocity:   V = Q / A = 0.7 / 0.19635 = 3.56 fps.

    Calculate Reynolds number:   R = V D/ ν = 3.56 × 0.5 / (1.0 × 10-3) = 1780.

    Since R < 3000, the flow is laminar. The Hagen-Poiseuille equation applies.

    hL = 128 ν L Q / (g π D4)

    hL / L = 128 × 1.0 × 10  -3 × 0.7 / (32.2 × 3.1416 × 0.5 4 )

    hL / L = 0.01417 ≅ 0.0142

  3. D = 0.3 m

    Q = 60 L/s = 0.06 m3/s

    V = Q / [(π/4) D2] = 0.06 / (0.7854 × 0.3 2 ) = 0.849 m/s

    For T = 15o, ν = 1.14 × 10-6 m2/s

    Reynolds number:   R = VD / ν = 0.849 × 0.3 / (1.14 × 10-6) = 223,421

    From Fig. 5-5:   ks / D = 0.0009

    From Fig. 5-4:   f = 0.0195

    hf = f (2000/D) V2/(2g) = 0.0195 × (2000/0.3) [(0.849) 2/(2 × 9.81)]

    hf = 4.77 m.

  4. ν = 1.0 × 10-6 m2/s

    D = 0.2 m.

    From Fig. 5-5: ks / D = 0.00065

    (D3/2/ν) (2ghf /L)1/2 = [0.23/2 /(1.0 × 10 -6 )] [(2) (9.81) (0.01)]1/2 = 39,618

    From Fig. 5-4: f = 0.019

    V = [ (hf/L) 2g D / f ]1/2 = [(0.01) (2) (9.81) (0.2) / 0.0195 ]1/2 = 1.44 m/s

    Q = VA = V (π/4) D2 = 1.44 (3.1416/4) (0.2)2 = 0.045 m3/s.

    Q = 45 L/s.