(d)
Loss of heod A to B
Xh (clockwise) =84(2780 + 2960) = 46,000
Zh (counterclockwise) = 8.4(3580 + 2350) = 49,000
h(tt) = Jo^|§§ 00)'^85 =34 ft (10.4 m)FIGURE 11. Application of the Hardy Cross method to a water distribution system.Related Calculations. Two pipes, two piping systems, or a single pipe and a
system of pipes are said to be equivalent when the losses of head due to friction for equal
rates of flow in the pipes are equal.
To determine the flow rates and friction-head losses in complex waterworks distri-
bution systems, the Hardy Cross method of network analysis is often used. This
method^1 uses trial and error to obtain successively more accurate approximations of the
flow rate through a piping system. To apply the Hardy Cross method: (1) Sketch the
piping system layout as in Fig. 11. (2) Assume a flow quantity, in terms of percentage
of total flow, for each part of the piping system. In assuming a flow quantity note that
(a) the loss of head due to friction between any two points of a closed circuit must be
the same by any path by which the water may flow, and (b) the rate of inflow into any
section of the piping system must equal the outflow. (3) Compute the loss of head due
to friction between two points in each part of the system, based on the assumed flow in
(a) the clockwise direction and (b) the counterclockwise direction. A difference in the
calculated friction-head losses in the two directions indicates an error in the assumed di-
rection of flow. (4) Compute a counterflow correction by dividing the difference in
head, Ah ft, by n(Q)n~l, where w = 1.85 and Q = flow, gal/min. Indicate the direction of
this counterflow in the pipe by an arrow starting at the right side of the smaller value of
h and curving toward the larger value. Fig. 11. (5) Add or subtract the counterflow to or
(^1) O (^5) RoUTkC—General Engineering Handbook, McGraw-Hill.