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5.2 Fluid Transport
A family of straight lines of equal slope is obtained, therefore, when s is plotted against Q on log–log paper for specified
diameters d. Two points define each line. Pairs of coordinates for a 12 in. pipe, for example, are as follows:
(c) Q = 100,000 gpd, s = 0.028 ft∕1,000 ft.
(d) Q = 1,000,000 gpd, s = 2.05 ft∕1,000 ft.
0.63 0.54
2. Written in logarithmic form, Eq. (5.34), v = 0.115Cd
,isasfollows:
s
log v = 1.0607 + 0.63 log d + 0.54 log s
If the diameter d is eliminated from the logarithmic transforms of Eqs. (5.36) and (5.34), then
(a) log Q = 0.180 + 4.17 log v – 1.71 log s.
(b) log s = 0.105 + 2.43 log v – 0.585 log Q.
3. A family of straight lines of equal slope is obtained when s is plotted against Q on log–log paper for specified velocities v.
Two points define each line. Pairs of coordinates for a velocity of 1 ft/s, for example, are as follows:
(a) Q = 100,000 gpd, s = 1.5ft∕1,000 ft.
(b) Q = 10,000,000 gpd, s = 0.10 ft∕1,000 ft.
EXAMPLE 5.38 CIRCULAR CONDUIT EQUIVALENCE TO A HORSESHOE CONDUIT
2
2
A tunnel having a horseshoe shape (Fig. 5.18) has a cross-sectional area of 27.9 ft (2.59 m ) and a hydraulic radius of 1.36 ft (0.41
m). Find the diameter, hydraulic radius, and area of the hydraulically equivalent circular conduit.
0.5D
D

D d
D

Figure 5.18 Horseshoe conduit section.

Solution 1 (US Customary System):
r
D = 1.53 a 0.38 0.24 (5.38)
D = 1.53 × (27.9) 0.38 × (1.36) 0.24 = 5.85 ft (1.78 m).
Hydraulic radius r = D∕4 = 1.46 ft (0.45 m).
2 2 2
Area a = D ∕4 = 26.7ft (2.48 m ).
Solution 2 (SI System):
r
D = 1.53 a 0.38 0.24 .
D = 1.53 (2.59) 0.38 (0.41) 0.24 = 1.78 m (5.85 ft).

Hydraulic radius r = D∕4 = 0.45 m (1.46 ft).
2 2 2
Area a = D ∕4 = 2.48 m (26.7ft ).
Note that neither the cross-sectional area nor the hydraulic radius of this equivalent circular conduit is the same as that of the
horseshoe section proper. Equation (5.38) is applicable to both the US customary system and the SI system.