CE444-APPLIED HYDRAULICS
SPRING 2016
SECOND MIDTERM
APRIL 8, 2016, 1500-1550

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%) 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)

  3. (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!

  4. (25%)

    1. What are the two main reasons for the 17th Stree Canal levell failure during Hurrocane Katrina?

      (1) Overestimating the in-situ soil strength and (2) using a design factor of safety (FS = 1.3) which was too low.

    2. According to the ASCE Hurricane Katrina External Review Panel, what institutional deficiencies led to the failure of the levee system?

      (a) Piecemeal approach to project, (b) lack of external review, (c) failure to evaluate and take into account risks associated with budget limitations and congressional authority.

    3. What are the units of kinematic viscosity?

      L2/T

    4. What is the main condition leading to cavitation? When is this condition produced?

      (a) Increase in velocity leading to reduction in pressure below atmospheric, (b) changes in pressure from low to high as a result of changes in velocity.

    5. What forces are considered in the static analysis ((stability) of a pipe bend?

      (1) Force due to changes in momentum, (2) force due to changes in pressure, (3) the gravitational force, and (4) the static force required to keep the bend in place.