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A water tower that is 90 ft high provides water to a residential subdivision. The water main from the tower to the subdivision is 6 inches, schedule 40, and 3 miles long.

If each house uses a maximum of 50 gallons per hour (at peak demand) and the pressure in the water main is not to be less than 30 psig at any point, how many homes can be served by the water main?

Answer :

Answer:

number of houses = 3751.243

Explanation:

given data

tower high H = 90 ft

pipe length L = 3 mile

pipe dia d = 6 in

solution

we consider here loss is neglected by dia 6 in pipe

so we apply here bernaulis equation from top to bottom height 90 ft

[tex]\frac{P1}{\rho g} + \frac{V1^2}{2 g} + Z1 = \frac{P2}{\rho g} + \frac{V2^2}{2 g} + Z2[/tex] ..........................1

here P1 is = o gauge pressure

and P2 = 30 Psi = 206.843 × [tex]10^{3}[/tex] Pa

and Z1 = 27.432 m

and Z2 = 0 and V1 = 0

so from equation 1

0+0+27.432 = [tex]\frac{206.843*10^3}{1000*9.81}[/tex] × [tex]\frac{V2^2}{2*9.81}[/tex]

solve we get

V = 11.16 m/s

V = 36.6 ft/s

and

flow will be here

flow Q = AV ............2

Q = [tex]\frac{\pi}{4} (0.15)^2[/tex] × 11.16

Q = 0.19723 m³/s

Q = 187562.157 gal/hr

we have given house use maximum = 50 gal/hr

so total home served = [tex]\frac{total flow}{need 1 home}[/tex]

number of houses = [tex]\frac{187562.157}{50}[/tex]

so number of houses = 3751.243

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