DISCHARGE OFLOOPING PIPES
A pipe carrying 12.5 ft
3
/s (353.90 L/s) of water branches into three pipes of the following
diameters and lengths; D 1 = 6 in (152.4 mm); L 1 = 1000 ft (304.8 m); D 2 = 8 in (203.2
mm); L 2 = 1300 ft (396.2 m); D 3 = 10 in (254.0 mm); L 3 = 1200 ft (365.8 m). These pipes
rejoin at their downstream ends. Compute the discharge in the three pipes, considering
each as fairly smooth.
Calculation Procedure:
- Express Q as a function of D and L
Since all fluid particles have the same energy at the juncture point, irrespective of the
loops they traversed, the head losses in the three loops are equal. The flow thus divides it-
self in a manner that produces equal values of hF in the loops.
Transforming Eq. 2IZ?,
kD2.ei
Q=Jj^ (24)where Hs a constant.
- Establish the relative values of the discharges; then determine
the actual values
Thus, Q 2 IQ 1 = (8/6)^267 /1.3^0538 = 1.87; Q 3 SQ 1 = (10/6)^267 /1.2^0538 = 3.55. Then Q 1 + Q 2 +
Qi = Si(I + 1.87 + 3.55) = 12.5 ft
3
/s (353.90 L/s). Solving gives Q 1 = 1.95 ft
3
/s (55.209
L/s); Q 2 = 3.64 ft^3 /s (103.056 L/s); Q 3 = 6.91 ft^3 /s (195.637 L/s).
FLUID FLOW IN BRANCHING PIPES
The pipes AM, MB, and MC in Fig. 9 have the diameters and lengths indicated. Compute
the water flow in each pipe if the pipes are considered rough.
Calculation Procedure:
- Write the basic equations governing the discharges
Let subscripts 1,2, and 3 refer to AM 9 MB 9 and MC, respectively. Then hFi + Hp 2 =110;
hFl + Hp 3 = 150, Eq. a; Q 1 = Q 2 + Q 3 , Eq. b. - Transform Eq. 21 c
The transformed equation is
(
hir\0.5l3
~T^ / <^25 )- Assume a trial value for hF1 and find the discharge;
test the result
Use Eqs. a and 25 to find the discharges. Test the results for compliance with Eq. b. If we
assume hFl = 70 ft (21.3 m), then Hp 2 = 40 ft (12.2 m) and Hp 3 = 80 ft (24.4 m); Q 1 =