Handbook of Civil Engineering Calculations

DETERMINATION OF FLOW IN A PIPE

Two reservoirs are connected by a 7000-ft (2133.6-m) fairly smooth cast-iron pipe 10 in

(254.0 mm) in diameter. The difference in elevation of the water surfaces is 90 ft

(27.4 m). Compute the discharge to the lower reservoir.

Calculation Procedure:

1. Determine the fluid velocity and flow rate
   Since the secondary items are negligible, the entire head loss of 90 ft (27.4 m) results
   from friction. Using Eq. 216 and solving for F, we have 90 = 0.38(7)F
   L86
   /0.833
   L25
   ;
   V= 5.87 ft/s (178.918 cm/s). Then Q=VA = 5.87(0.545) = 3.20 ft
   3
   /s (90.599 L/s).


2. Alternatively, assume a value of f and compute V
   Referring to Fig. 8, select a value for/ Then compute V by applying Eq. 20. Next, com-
   pare the value of/corresponding to this result with the assumed value of/ If the two val-
   ues differ appreciably, assume a new value of/and repeat the computation. Continue this
   process until the assumed and actual values of/agree closely.

PIPE-SIZE SELECTION BY

THE MANNING FORMULA

A cast-iron pipe is to convey water at 3.3 ft^3 /s (93.430 L/s) on a grade of 0.001. Applying
the Manning formula with n- 0.013, determine the required size of pipe.

Calculation Procedure:

1. Compute the pipe diameter
   The Manning formula, which is suitable for both open and closed conduits, is

v__ MS(H^ (22)

n

where n = roughness coefficient; R = hydraulic radius = ratio of cross-sectional area of
pipe to the wetted perimeter of the pipe; s = hydraulic gradient = dH/dL. If the flow is uni-
form, i.e., the area and therefore the velocity are constant along the stream, then the loss
of head equals the drop in elevation, and the grade of the conduit is s.
For a circular pipe flowing full, Eq. 22 becomes

/2.1590«\3/8

D

=(-^-)

(22a)

Substituting numerical values gives D = (2.159 x 3.3 x 0.013/0.001

1/2

)

3/8

= 1.50 ft (45.72

cm). Therefore, use an 18-in (457.2-mm) diameter pipe.