C = 100 and the flow rate were reduced to 2000 gal/min (126.2 L/s)? Explain how the
Hardy Cross method is applied to the water-supply piping system in Fig. 11.
Calculation Procedure:
- Compute the domestic water flow rate in the system
Use an average annual domestic water consumption of 150 gal/day (0.0066 L/s) per capi-
ta. Hence, domestic water consumption = (150 gal per capita per day)( 100,000 persons) =
15,000,000 gal/day (657.1 L/s). To this domestic flow, the flow required for fire protec-
tion must be added to determine the total flow required.
- Compute the required flow rate for fire protection
Use the relation Qf= 1020(P)
05
[1 - 0.01(P)
05
], where Qf= fire flow, gal/min; P = popu-
lation in thousands. Substituting gives Qf= 1020(10O)
05
[1 - 0.01(10O)
05
] = 9180, say
9200 gal/min (580.3 L/s).
- Apply a load factor to the domestic consumption
To provide for unusual water demands, many design engineers apply a 200 to 250 percent
load factor to the average hourly consumption that is determined from the average annual
consumption. Thus, the average daily total consumption determined in step 1 is based on
an average annual daily demand. Convert the average daily total consumption in step 1 to
an average hourly consumption by dividing by 24 h or 15,000,000/24 = 625,000 gal/h
(657.1 L/s). Next, apply a 200 percent load factor. Or, design hourly demand =
2.00(625,000) = 1,250,000 gal/h (1314.1 L/s), or 1,250,000/60 min/h = 20,850, say
20,900 gal/min (1318.6 L/s).
- Compute the total water flow required
The total water flow required = domestic flow, gal/min + fire flow, gal/min = 20,900 +
9200 = 30,100 gal/min (1899.0 L/s). If this system were required to supply water to one
or more industrial plants in addition to the domestic and fire flows, the quantity needed by
the industrial plants would be added to the total flow computed above.
- Select the flow rate for each pipe
The flow rate is not known for either pipe in Fig. 9a. Assume that the shorter pipe a has a
flow rate Qa of 12,100 gal/min (763.3 L/s), and the longer pipe b a flow rate Qb of 18,000
gal/min (1135.6 L/s). Thus, Qa + Qb = Qt= 12,100 + 18,000 = 30,100 gal/min (1899.0
L/s), where Q = flow, gal/min, in the pipe identified by the subscript a or b; Qt = total
flow in the system, gal/min.
- Select the sizes of the pipes in the system
Since neither pipe size is known, some assumptions must be made about the system. First,
assume that a friction-head loss of 10 ft of water per 1000 ft (3.0 m per 304.8 m) of pipe is
suitable for this system. This is a typical allowable friction-head loss for water-supply
systems.
Second, assume that the pipe is sized by using the Hazen-Williams equation with the
coefficient C = 100. Most water-supply systems are designed with this equation and this
value of C.
Enter Fig. 10 with the assumed friction-head loss of 10 ft/ 1000 ft (3.0 m/304.8 m)
of pipe on the right-hand scale, and project through the assumed Hazen-Williams
coefficient C= 100. Extend this straight line until it intersects the pivot axis. Next, en-
ter Fig. 10 on the left-hand scale at the flow rate in pipe a, 12,100 gal/min (763.3 L/s),
and project to the previously found intersection on the pivot axis. At the intersection
with the pipe-diameter scale, read the required pipe size as 27-in (686-mm) diameter.
