CEE 371

Fall 2009

Homework #3 Water Distribution Pipe Systems

1. For the pipe system shown below (Figure 1), determine the length of a single equivalent pipe that has a diameter of 8 inches. Use the Hazen Williams equation and assume that C_HW = 120 for all pipes. Solve the problem using the following steps:

a. First determine an equivalent pipe (with D=8 in) for pipes #2 and #3 in series. Use a flow of 800 gpm.

b. Second, determine an equivalent pipe for pipe #4 and the parallel equivalent pipe from part (a). Use the head loss resulting from the flow for part (a) as the basis for determining the equivalent pipe length (use D=8 in). What is the flow split between these two parallel pipes? (i.e., for 800 gpm through the part (a) pipe, what is the flow in the parallel pipe, and the total flow)

c. Finally, determine a single equivalent pipe (D = 8 in) for the three pipes in series, pipe #1, the pipe from part (b), and pipe #5.

d. Show that your pipe is hydraulically equivalent by calculating the head loss for this single pipe and comparing it to the sum of the head losses for pipes in the original system.

Figure 1. Pipe System for equivalent pipe problem

Table 1. Pipe Data for Figure 1 Pipe System

	Pipe 1	Pipe 2	Pipe 3	Pipe 4	Pipe 5
Length	500	500	800	1000	700	ft
Diameter	12	6	8	10	12	in

2. Use the Hardy Cross method (and the Hazen Williams equation) to solve for the flows in each pipe of the network shown in Figure 2 and described in Table 2. Also, determine the values of the hydraulic grade line (HGL) and pressure at each node (pipe junction) in the system. Assume that C_HW = 120 for all pipes. The elevation of water in the tank at node A is 250 ft.

Table 2. Pipe & Node Data for Figure 2 Pipe Network

Pipe #	Nodes	Length (ft)	Diam (in)	Node #	Elev. (ft)	External Demand (gpm)
1	A - B	2000	16	A	180	0
2	B - C	800	12	B	50	0
3	B - D	600	10	C	30	400
4	D - E	800	8	D	80	1200
5	C - E	1200	8	E	40	1400
6	C - F	1300	6	F	60	1000
7	E - F	900	8

Figure 2. Pipe Network

Assigned: 30 Sept 09

Due: 7 Oct 09