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Because of the great difference in elevation (658 to 922 feet), it is necessary to divide the distribution system into two zones so that the maximum pressure in pipes and at fixtures will not be excessive. In this problem, all water is supplied the distribution system at elevation 922 feet. A suitable dividing point would be at elevation 790 feet. All dwellings above this point would have water pressure directly from the reservoir, and all below would be served through a pressure-reducing valve to provide not less than 15 lb/in.2 at the highest fixture or more than 60 lb/in.2 at the lowest fixture. If two-thirds of the dwellings are in the upper zone and one-third is in the lower zone, it can be assumed that the peak or maximum hourly demand rate of flow will be similarly divided (Figure 2.20).

Assume the total maximum hourly or peak demand rate of flow for an average daily water consumption of 30,000 gpd to be 210 gpm. Therefore, 70 + 140 gpm can be taken to flow to the upper zone and 70 gpm to the lower zone. If a 3-inch pipe is used for the upper zone and water is uniformly drawn off, the head loss at a flow of 210 gpm would be about 0.33 x 20 feet per 100 feet of pipe. And if 2-1/2-inch pipe is used for the lower zone and water is uniformly drawn off in its length, the head loss at a flow of 70 gpm would be about 0.33 x 6.2 feet per 100 feet of pipe. If the pipe in either zone is connected to form a loop, thereby eliminating dead ends, the frictional head loss would be further reduced to one-fourth of that with a dead end for the portion forming a loop. Check all head losses.

In all of these considerations, actual pump and motor efficiencies obtained from and guaranteed by the manufacturer should be used whenever possible. Their recommendations and installation detail drawings to meet definite requirements should be requested and followed if it is desired to fix performance responsibility.

In another instance, assume that all water is pumped from a deep well through a pressure tank to a distribution system. See Figure 2.21. The lowest pumping water level in the well is at elevation 160, the pump and tank are at elevation 200, and the highest dwelling is at elevation 350. Find the size pump, motor and pressure storage tank, operating pressures, required well yield, and size mains to supply a development consisting of 100 two-bedroom dwellings using an average of 30,000 gpd.

Use a deep-well turbine pump. The total pumping head will consist of the sum of the total lift plus the friction loss in the well drop pipe and connection to the pressure tank plus the friction loss through the pump and pipe fittings plus the maximum pressure maintained in the pressure tank. The maximum pressure in the tank is equal to the friction loss in the distribution system plus the static head caused by the difference in elevation between the pump and the highest plumbing fixture plus the friction loss in the house water system, including meter if provided, plus the residual head required at the highest fixture.

With the average water consumption at 30,000 gpd, the maximum hourly or peak demand was found to be 210 gpm. The recommended pump capacity is taken as 125 percent of the maximum hourly rate, which would be 262 gpm. This assumes that the well can yield 262 gpm, which frequently is not the case. Under such circumstances, the volume of the storage tank can be increased two or three times, and the size of the pump correspondingly decreased to one-half or one-third the original size to come within the well yield. Another alternative would be to pump water out of the well, at a rate equal to the safe average yield of the well, into a large ground-level storage tank from which water can be pumped through a pressure tank at a higher rate to meet maximum water demands. This would involve double pumping and, hence, increased cost. Another arrangement, where possible, would be to pump out of the well directly into the distribution system, which is connected to an elevated storage tank. Although it may not be

FIGURE 2.21 A water system flow diagram with booster station.

economical to use a pressure-tank water system, it would be of interest to see just what this would mean.

The total pumping head would be

Lift from elevation 160 - 200 ft = 40 ft 40 ft

Figure 2.21 shows a distribution system that forms a rectangle 1,000 x 1,500 feet with a 2,000-foot dead-end line serving one-third of the dwellings taking off at a point diagonally opposite the feed main. The head loss in a line connected at both ends is approximately one-fourth that in a dead-end line. The head loss in one-half the rectangular loop, from which water is uniformly drawn off, is one-third the loss in a line without drawoffs. Therefore, the total head loss in a 3-inch pipeline with a flow of about 210 gpm is equal to

Renewable Energy 101

Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

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