Refer to Table 7.8 below, and note how much water can be conserved by installing

Question

Refer to Table 7.8 below, and note how much water can be conserved by installing efficient fixtures and leak detection (units in this table are gallons per day per capita or gallons per day per person). Assume the water entering the house is from a drinking water treatment plant, and the water exiting the house flows to a wastewater treatment plant. Ignore any other uses of water by the household. Using the information about electricity consumption to produce water and treat wastewater, calculate the reduction in two criteria air pollutants, SO2 and NO, that results from water conservation in for 10,000 people. Assume the following emissions factors for electricity: 1.4 lbs SO2/MWh and 1.0 lbs NO/MWh.

Table 7.8

Table/7.8 Water Usage in U.S. Households: Typical and Efficient Alternatives Typical water usage in a U.S, house ar greater than when water-efficient fixtures are installed and households pay attention to leak detection. Percentage based on total use. ouse old Water Usage with Water-Efficient Fixtures and Leak Detection, gpdc 1% total use) | Typical Water Usage, gpdc (% total use) | Activity Showers Clothes washing Dishwashing Toilets Baths Leaks Faucets Other domestic uses Total 11.6 (16.8%) 15.0 (21.7%) 1.0 (1.4%) 18.5 (26.7%) 1.2 (1.7%) 9.5 (13.7%) 10.9 (15.7%) 1.6 (2.2%) 69.3 gpdc 8.8 (19.5%) 10.0 (22.1%) 0.7 (1.5%) 8.2 (18.0%) 1.2 (2.7%) 4.0 (8.8%) 10.8 (23.9%) 1.6(3.4%) 45.3 gpdc | | 4

Explanation / Answer

For the question given above amount of water in gallons per person per day that could be saved by employing efficient system is 69.3- 45.3 = 24 gpdc

Assuming 0.04 kWh per gallon of electricity consumption required to produce drinking water and treat waste water already taking into consideration the assumption that 90 percent of water utilised is converted to waste water

For a population of 10,000 people total amount of water saved = 24* 10000 = 240000 gallons

Amount of electricity consumption reduced = 240000*0.04 = 9600 kWh or 9.6 MWh

Thus reduction in SO2 emission = 1.4*9.6 =13.44 lbs per day

Reduction in NO emission= 1*9.6 = 9.6 lbs per day

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