CE444-APPLIED HYDRAULICS
FALL 2009
SECOND MIDTERM
NOVEMBER 16, 2009, 0730-0845

Name: ______________________ Red ID __________________ Grade: _________

Instructions: Closed book, closed notes. Use engineering paper. When you are finished, staple your work in sequence (1 to 4), and return this sheet with your work.

  1. (25%) Calculate the discharge Q (cfs) for an asphalted cast-iron pipe of diameter D = 10 in and head loss per unit of length hf/L = 0.008. The water temperature is 65oF.

    D = 10/12 ft = 0.8333 ft.

    By interpolation in Table A-5:  ν = 1.14 × 10-5 ft2/s

    From Figure 5-5:  relative roughness ks/D = 0.0005

    Re f1/2 = (D3/2/ν) (2ghf /L)1/2 =  [0.83333/2 /(1.14 X 10-5) ]   [(2) (32.17) (0.008)]1/2 = 47,872

    From Fig. 5-4:   f = 0.018

    hf = f (L/D) V2/(2g)

    V = [ (hf/L) 2gD / f ]1/2= [(0.008) (2) (32.17) (0.8333) / 0.018]1/2 = 4.88 ft/s

    Q = VA = 4.88 [(3.1416/4) (0.8333)2] = 2.66 cfs.

    Check:

    Re = VD/ν = 4.88 × 0.8333 / (1.14 × 10-5) = 356,711

    From Fig. 5-4, for ks/D = 0.0005 and Re = 356,711:   f = 0.018

    hf = f (L/D) V2/(2g)

    hf/L = (f/D) V2/(2g) = (0.018/0.8333) [4.882/(2 × 32.17)] = 0.008    OK!

  2. (25%) Using an explicit equation, calculate the pipe diameter D (m) for the following conditions: Q = 0.075 m3/s, head loss per unit of length hf/L = = 0.008, pipe roughness ks = 0.00012 m, and water temperature T = 18oC.

    B* = g (hf/L) = 9.81 (0.008) = 0.07848

    ν = 1.06 × 10-6 m2/s

    D = 0.66 [(ks1.25 Q 9.5 / B* 4.75) + (ν Q 9.4 / B* 5.2) ] 0.04

    D = 0.66 [(0.00012 1.25) (0.075 9.5) / (0.07848 4.75) + (0.00000106) (0.075 9.4) / (0.07848 5.2) ] 0.04

    D = 0.26 M.       OK!

  3. (25%) Determine the diameter of a pipe to conduct a Q = 2.5 m3/s between two reservoirs A and B. Reservoir A is at elevation 520 m, and reservoir B at elevation 390 m. The distance between the reservoirs is 4200 m. Neglect all minor losses. Assume f = 0.015 and T = 20oC.

    hf = f (L/D) V2/(2g)

    hf = 520 - 390 = 130 m.

    L = 4200 m.

    f = 0.015

    V = Q/A

    V2 = Q2/A2

    hf = f (L/D) Q2/ [A2(2g)]

    A = (π/4) D2

    A2 = (π/4)2 D4

    hf = f (L/D) Q2/ [(π/4)2 D4(2g)]

    D5 = f Q2/ [(π/4)2 (2g) (hf/L) ]

    D5 = 0.015 × (2.5)2/ [(0.7854)2 (2 × 9.81) (130/4200) ] = 0.25

    D = 0.76 m

    A = (π/4) D2 = 0.454 m2

    V = Q/A = 2.5/0.454 = 5.5 m/s

    Re = VD/ν = 5.5 × 0.76 / (1.0 × 10-6) = 4,180,000 (f constant)

  4. (25%) What is the change in head (of water) for a 0.7-m diameter orifice in a 1-m diameter pipe carrying a discharge of 2 m3/s of water? Assume T = 20oC.

    Red = 4Q/(πdν) = (4 × 2)/(3.1416 × 0.7 × 1.0 × 10-6) = 3,637,818

    From Fig 5-21, for d/D = 0.7:   K = 0.7.

    Q = K Ao (2gΔh)1/2

    Δh = [Q/(KAo)]2 / (2g)

    Δh = [2/(0.7 × (π/4) 0.72)]2 /(2 × 9.81)

    Δh = 2.81 m.

    Proof:

    Red/K = (2g Δh)1/2 (d/ν)

    Red/K = (2 × 9.81 × 2.81)1/2 [0.7/(1.0 × 10-6)]

    Red/K = 5,197,574

    From Fig 5-21, for d/D = 0.7:   K = 0.7.

    Q = K Ao (2gΔh)1/2

    Q = 0.7 × (π/4) × 0.72 [2 × 9.81 × 2.81]1/2

    Q = 2.0 m3      OK!