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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 in. sch 40, 3 miles long. If each house uses a maximum of 50 gal/hr (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?

1 Answer

2 votes

Answer:

number of houses = 3751.243

Step-by-step 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


(P1)/(\rho g) + (V1^2)/(2 g) + Z1 = (P2)/(\rho g) + (V2^2)/(2 g) + Z2 ..........................1

here P1 is = o gauge pressure

and P2 = 30 Psi = 206.843 ×
10^(3) Pa

and Z1 = 27.432 m

and Z2 = 0 and V1 = 0

so from equation 1

0+0+27.432 =
(206.843*10^3)/(1000*9.81) ×
(V2^2)/(2*9.81)

solve we get

V = 11.16 m/s

V = 36.6 ft/s

and

flow will be here

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

Q =
(\pi)/(4) (0.15)^2 × 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 =
(total flow)/(need 1 home)

number of houses =
(187562.157)/(50)

so number of houses = 3751.243

User Lizarisk
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