§ 87. Branched Pipe connecting Reservoirs at Different Levels.—Let A, B, C (fig. 98) be three reservoirs connected by the arrangement of pipes shown,—l1, d1, Q1, v1; l2, d2, Q2, v2; h3, d3, Q3, v3 being the length, diameter, discharge and velocity in the three portions of the main pipe. Suppose the dimensions and positions of the pipes known and the discharges required.
If a pressure column is introduced at X, the water will rise to a height XR, measuring the pressure at X, and aR, Rb, Rc will be the lines of virtual slope. If the free surface level at R is above b, the reservoir A supplies B and C, and if R is below b, A and B supply C. Consequently there are three cases:—
| I. | R above b; Q1 = Q2 + Q3. |
| II. | R level with b; Q1 = Q3; Q2 = 0 |
| III. | R below b; Q1 + Q2 = Q3. |
To determine which case has to be dealt with in the given conditions, suppose the pipe from X to B closed by a sluice. Then there is a simple main, and the height of free surface h′ at X can be determined. For this condition
ha − h′ = ζ (v12/2g) (4l1/d1) = 32ζQ′2l1 / gπ2d15;
h′ − hc = ζ (v32/2g) (4l3/d3) = 32ζQ′2l3 / gπ2d35;
where Q′ is the common discharge of the two portions of the pipe. Hence
(ha − h′) / (h′ − hc) = l1d35 / l3d15,
from which h′ is easily obtained. If then h′ is greater than hb, opening the sluice between X and B will allow flow towards B, and the case in hand is case I. If h′ is less than hb, opening the sluice will allow flow from B, and the case is case III. If h′ = hb, the case is case II., and is already completely solved.