A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value residence uses 800 g/day/house and has a fire flow requirement of 1000 gpm. If the water system has a daily peaking factor = 1.64 and an hourly peaking factor = 2.5, determine (a) the average daily demand of this subdivision and (b) the design-demand used to design the distribution system.

Respuesta :

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

[tex]\text{Number of house in 1100 acres} = 4\times1100[/tex]

[tex]\text{Number of house in 1100 acres} = 4400[/tex]

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

[tex]\text{average daily demand}=house\times\text{Per house water demand}[/tex]

[tex]\text{average daily demand}=4400\times800\ gallon/day[/tex]

[tex]\text{average daily demand}=3520000\ gallon/day[/tex]

[tex]\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min[/tex]

[tex]\text{average daily demand}=2444.44\ gallon/min[/tex]

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

[tex]\text{design demand}=(Q_{max})daily\times\text{fire flow}[/tex]

[tex]\text{design demand}=1.64\times2444.4\times1000[/tex]

[tex]\text{design demand}=4008816\ gallon/m[/tex]

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.