FLOW IN PIPES AND CHANNELS 21
IfR/ u^2 is taken as 0.002,Rewill vary from (7. 05 ð 105 )to(4. 07 ð 105 ). From
Fig. 3.7 this corresponds to a range ofe/dof between 0.004 and 0.005 or a roughness of
between 0.92 and 1.15 mm, which is too high for a commercial pipe.
Ifeis taken as 0.05 mm,e/dD 0 .0002, and, for Reynolds numbers near 10^6 ,R/ u^2 D
0 .00175. SubstitutingR/ u^2 D 0 .00175 and integrating gives a time of 398 s for the level
to fall from 3 m to 1 m. IfR/ u^2 D 0 .00175,Revaries from (7. 5 ð 105 )to(4. 35 ð 105 ),
and from Fig. 3.7,e/dD 0 .00015, which is near enough to the assumed value. Thus the
time for the level to fall is approximately 400 s.
PROBLEM 3.4
Two storage tanksAandBcontaining a petroleum product discharge through pipes each
0.3 m in diameter and 1.5 km long to a junction atD.FromDthe product is carried by
a 0.5 m diameter pipe to a third storage tankC, 0.8 km away. The surface of the liquid
inAis initially 10 m above that inCand the liquid level inBis 7 m higher than that in
A. Calculate the initial rate of discharge of the liquid if the pipes are of mild steel. The
density of the petroleum product is 870 kg/m^3 and the viscosity is 0.7 mNs/m^2.
Solution
See Volume 1, Example 3.4
PROBLEM 3.5
Find the drop in pressure due to friction in a glazed porcelain pipe 300 m long and
150 mm diameter when water is flowing at the rate of 0.05 m^3 /s.
Solution
For a glazed porcelain pipe,eD 0 .0015 mm,e/dD 0. 0015 / 150 D 0 .00001.
Cross-sectional area of pipeD/ 4 0. 15 ^2 D 0 .0176 m^2.
Velocity of water in pipe,uD 0. 05 / 0. 0176 D 2 .83 m/s.
Reynolds numberD ud/ D 1000 ð 2. 83 ð 0. 15 / 10 ^3 D 4. 25 ð 105
From Fig. 3.7,R/ u^2 D 0 .0017.
The pressure drop is given by equation 3.18:PfD 4 R/ u^2 l/d u^2
or: 4 ð 0. 0017 300 / 0. 15 1000 ð 2. 832 D 108 ,900 N/m^2 or 1 MN/m^2
PROBLEM 3.6
Two tanks, the bottoms of which are at the same level, are connected with one another
by a horizontal pipe 75 mm diameter and 300 m long. The pipe is bell-mouthed at each