CIV E 444 - APPLIED HYDRAULICS
SPRING 2016
HOMEWORK No. 8 SOLUTION

  1. The flow area when the tunnel is flowing full is:

    A = (18 × 9) + 3.1416 × 92 / 2 = 289.2 ft2

    The wetted perimeter is:

    P = 18 + 9 + 9 + (3.1416 × 9) = 64.27 ft

    The hydraulic radius is:

    R = 289.2 / 64.27 = 4.5 ft

    The mean velocity is:

    V = Q/ A = 4000 / 289.2 = 13.8 fps

    The velocity head is: hv = 13.82 /(2 × 32.17 ) = 2.96 ft

    The kinematic viscosity is:

    ν = 1.06 × 10-5 ft2/s

    The Reynolds number is:

    Re = V (4R) / ν = 13.8 (4 × 4.5) / (1.06 × 10-5 ) = 23,433,962

    The relative roughness is:

    ks / (4R) = 0.01/ (4 × 4.5) = 0.00055

    From the Moody diagram, for Re = 23,000,000 and ks = 0.0005:

    f = 0.017

    The head loss in ft/mile is:

    hf = f [L/(4R)] hv = 0.017 × [5280 / (18)] × 2.96 = 14.7 ft/mi

  2. Write the energy equation between the reservoirs:

    The line losses are: (14.7/5280) × 8550 ft = 23.8 ft

    Other losses are: hv ∑ (losses) = 2.96 (0.09 + 0.03 + 0.72 + 0.25) = 3.23 ft

    Total losses are: 23.9 + 3.23 = 27.03 ft

    Head delivered to turbines = 3250 - 1575 - 27 = 1648 ft

    Maximum power delivered to turbines:

    P (HP) = γ Q H = 62.4 × 4000 × 1648 / 550 = 747,892 HP

    P (MW) = P (HP) × 0.746 /1000

    P = 557.9 MW

  3. Three energy balance equations and continuity equation:

    zA = zD + pD/γ + fAD (LAD /DAD ) VAD2 / (2g)

    zB = zD + pD/γ + fBD (LBD /DBD ) VBD2 / (2g)

    zD + pD/γ = fDC (LDC /DDC ) VDC2 / (2g)

    VAD AAD + VBD ABD = VDC ADC

    AAD = 0.3491 ft2

    ABD = 0.5454 ft2

    ADC = 0.7854 ft2

    State continuity equation as follows:   VAD AAD + VBD ABD - VDC ADC = X

    The solution is by trial and error.

    Assume pD/γ and calculate resulting velocities using energy equations; then plug velocities and areas in continuity equation until X = 0.

    First assume VAD = 0.

    Then pD/γ = 90.0 ft.

    Start by assuming a somewhat lessr value:

    pD/γ = 80 ft:   X = -2.98 Then:

    pD/γ = 70 ft:   X = -2.12

    pD/γ = 50 ft:   X = -0.58

    pD/γ = 44 ft:   X = -0.12

    pD/γ = 43 ft:   X = -0.05

    pD/γ = 42 ft:   X = 0.02

    pD/γ = 42.3 ft:   X = 0.00

    The final velocities are:

    VAD = 5.057 fps.

    VBD = 5.118 fps.

    VDC = 5.801 fps.

    QAD = 0.3491 × 5.057 = 1.765cfs.

    QBD = 0.5454 × 5.118 = 2.791 cfs.

    QDC = 0.7854 × 5.801 = 4.556 cfs.

    pD = 42.3 ft × 62.4 lb/ft 3 = 2,640 lb/ft2

  4. The head loss through either pipes 1 or 2 is the same.

    V1/V2 = [(f2 L2 D1) / (f1 L1 D2)] 1/2

    V1/V2 = [(0.03 × 3200 × 1) / (0.02 × 3000 × 1.3333)] 1/2 = 1.0955

    (D1/D2)2 = (1/1.3333)2 = 0.5625

    Q1 = Q / { { 1 / [(V1/V2) (D1/D2)2]} + 1 }

    Q1 = 20 / { [ 1 / (1.0955 × 0.5625)] + 1 } = 7.625 cfs.

    Q2 = Q / { [(V1/V2) (D1/D2)2] + 1}

    Q2 = 20 / [(1.0955 × 0.5625) + 1 ] = 12.375 cfs.

    Q1 + Q2 = 20 cfs = Q       OK!

    V1 = Q1 / A1 = 7.625 / [(π/4) 12] = 9.708 fps.

    V2 = Q2 / A2 = 12.375 / [(π/4) 1.33332] = 8.863 fps.

    hL, 1 = f1 (L1/D1) [V12/(2g)]

    hL, 1 = 0.02 (3000/1) [9.7082/(2 × 32.17)] = 87.888 ft.

    hL, 2 = f2 (L2/D2) [V22/(2g)]

    hL, 2 = 0.03 (3200/1.3333) [8.8632/(2 × 32.17)] = 87.907 ft.

    hL, 1 = hL, 2       OK!