Three water reservoirs A, B, and C are connected by circular pipes as shown. The

ID: 1714203 • Letter: T

Question

Three water reservoirs A, B, and C are connected by circular pipes as shown. The primary water line

leaving reservoir A is a 24 inch diameter smooth steel pipe. At point P, this line divides into two 18 inch

diameter smooth steel pipes discharging into reservoirs B and C. It is known that the elevation at point

P is about 700 feet. Find the required elevation of reservoir C. Ignore minor losses.

Res A, El 1010 ft Res B, El = 500 ft L2 = 5,000 ft D2 = 18 in P L1 = 10,000 ft D1 = 24 in Res C, EI =? L3 = 4,000 ft D3 = 18 in

In the given question we have been given steel pipes and let friction factor be f

Also neglecting all the minor losses we have frictional losses and head loss between two points be equated to frictional loss then

Hl = f l v^2 / 2gD = 8f l Q^2 / ^2g D^5 ---------equation 1

Using equation 1 between points A and P where Hl = 310 ft ,we get Qa = 3.465/ sqrt (f)

Using equation 1 between points P and B where Hl = 700 - 500 =200 ft we get Qb = 1.917/ sqrt (f)

Now maintaining continuity at point P we get Qa= Qb + Qc

Thus Qc = 1.548/ sqrt( f)

But using equation 1 between points P and C we get Hl = 104.268 ft

Thus elevation of C = 700- 104.268 = 595.732 ft

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