(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!
(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!
(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)
(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!