(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!