CIV E 444 - APPLIED HYDRAULICS
SPRING 2016
LAB No. 9 SOLUTION

  1. Use orifice equation:   Q = K Ao (2g Δh)1/2

    Assume d, find K from Fig. 5-21, and calculate Q. Repeat for another d.

    First trial: d = 2 ft

    d / D = 2 / 4 = 0.5

    Δh = Δp/γ = (2000 lbs/ft2)/(62.4 lbs/ft3) = 32.05 ft

    (2 g Δh)1/2 = (2 × 32.17 × 32.05) 1/2 = 45.41

    For T= 20oC:   ν = 1.27 × 10-5 ft2/s

    (2 g Δh)1/2 d/ν = 45.41 × 2 / (1.27 × 10-5) = 7.15 × 106

    From Fig. 5-21:   K = 0.63

    Q = 0.63 (π/4) (2)2 × 45.41 = 90 cfs       [Discharge is too low; orifice too small]

    Second trial:   d = 2.5; d / D = 0.625.

    Δh = Δp/γ = (2000 lbs/ft2)/(62.4 lbs/ft3) = 32.05 ft

    (2 g Δh)1/2 = (2 × 32.17 × 32.05) 1/2 = 45.41

    For T= 20oC:   ν = 1.27 × 10-5 ft2/s

    (2 g Δh)1/2 d/ν = 45.41 × 2.5 / (1.27 × 10-5) = 8.94 × 106

    From Fig. 5-21:   K = 0.66

    Q = 0.66 (π/4) (2.5)2 × 45.41 =147 cfs       [Discharge is too low; orifice too small]

    Third trial:   d = 2.1; d / D = 0.525.

    Δh = Δp/γ = (2000 lbs/ft2)/(62.4 lbs/ft3) = 32.05 ft

    (2 g Δh)1/2 = (2 × 32.17 × 32.05) 1/2 = 45.41

    For T= 20oC:   ν = 1.27 × 10-5 ft2/s

    (2 g Δh)1/2 d/ν = 45.41 × 2.1 / (1.27 × 10-5) = 7.51 × 106

    From Fig. 5-21:   K = 0.63

    Q = 0.63 (π/4) (2.1)2 × 45.41 = 99 cfs       [Discharge is ok now; orifice diameter ok]

  2. Red / K = (2 g Δh) 1/2 (d/ν)

    Red / K = (2 × 9.81 × 10) [(0.25 / (1.0 × 10-6)]

    Red / K = = 3.5 × 106

    From Fig. 5-21: K = 0.61

    Q = K Ao (2 g Δh) 1/2

    Q = 0.61 (π/4) (0.25)2 (2 × 9.81 × 10) 1/2

    Q = 0.419 m3/s.

  3. Red = 4 Q / ( π d ν)

    Red = 4 × 0.6 / (3.1416 × 0.25 × 1.0 × 10-6)

    Red = 3.05 × 106

    K = 0.61

    Δh = [Q / (K Ao) ]2 / (2 g)

    Δh = [0.6 / (0.61 × (π/4) 0.252 ]2 / (2 × 9.81)

    Δh = 20.46 m

  4. ks / D = 0.004 ft / 1 ft = 0.004

    Assume very high Reynolds number. Estimate f from Moody diagram:   f = 0.028.

    Velocity head is lost in the free jet:   KE = 1

    Write the energy equation between the reservoir water surface elevation and [the elevation of the] free jet.

    100 - 64 = [f (L/D) + Ke + Kb + KE] [V2/(2g)]

    36 = [ 0.028 × (100/1) + 0.12 + 0.35 + 1 ] [V2/(2 × 32.17)]

    36 = 4.27 (V2/64.34)

    V = 23.29 fps.

    V2/(2g) = (23.29)2/(2 × 32.17) = 8.43 ft.

    The velocity head is 8.43 ft.

    A = (π/4) D2 = (3.1416/4) (1)2 = 0.7854 ft2

    Q = 23.29 × 0.7854 = 18.29 cfs.

    For T = 60oF :   ν = 1.217 × 10-5 ft2/s

    Calculate Reynolds number:   Re = VD/ν = 23.29 × 1 / (1.217 × 10-5) = 1.9 × 106

    From Fig. 5-4:   f = 0.028.       [Assumed value is OK].

    The point of maximum pressure pmax is right above (upstream) of the 90o bend.

    Write the energy equation between a point immediately upstream of the bend (max) and a point (2) at the exit.

    The velocity heads are the same, so they cancel.

    (pmax/γ) + zmax = (p2/γ) + z2 + Σ hL

    p2/γ = 0

    (pmax/γ) + 44 = 0 + 64 + [f (L/D) + Kb ][V2/(2g)]

    (pmax/γ) + 44 = 0 + 64 + [0.028 (28/1) + 0.35](8.43)

    pmax/γ = 20 + 9.56 = 29.56

    pmax = γ (29.56) = 1845 lb/ft2

    Or, alternatively, write the energy equation between the reservoir water surface elevation (1) and a point immediately upstream of the bend (max)

    The velocity heads are the same, so they cancel.

    (p1/γ) + z1 = (pmax/γ) + zmax + Σ hL

    p1/γ = 0

    0 + 100 = (pmax/γ) + 44 + [ f (L/D) + Ke + 1] [V2/(2g)]

    56 = (pmax/γ) + [ 0.028 (72/1) + 0.12 + 1] [8.43]

    56 = (pmax/γ) + 26.43

    pmax/γ = 29.57 ft.

    pmax = 1845 lb/ft2

    Immediately upstream of the pipe entrance, the pressure head is 5 ft and the velocity head is 0.

    Immediately downstream of the pipe entrance, the velocity head is 8.43 ft, and the drop in pressure head is 8.43(1 + 0.12)

    Therefore, the point of minimum pressure is immediately downstream of the pipe entrance.

    pmin = 62.4 × [5 - 8.43(1 + 0.12)] = -277 lb/ft2.