Question

A water tower is being used to provide water to a subdivision. To ensure that it provides sufficient water pressure for the homes, fire hydrants, etc., it is built on top of a hill that, at an elevation 69.40 ft higher than the subdivision, hhill. The water in the tank is 55.70 ft deep, hwater, and the tower upon which it sits is 70.40 ft tall (htower), as shown in the figure below. The 12.200 in (inner diameter) pipe that conveys the water to the subdivision can be considered to have an effective length of 2.3600 mi, for the purposes of calculating friction losses. Friction loss can be assumed to vary with pipe velocity, v, by the following equation: ℱ = 0.01 L v2 / D What is the absolute pressure for the water delivered to the subdivision if the flow rate is to be 91,600 gph gallons per hour? (You can assume that the tank is open to the atmosphere and is kept completely full with water pumped into it from a nearby spring.)

Answer 1

Theory: we will calculate first velocity inside the pipe and then we will use bernouli's equation to calculate the pressure drop

Q = 91600 gph* (m3/s/gph) = 91600* 0.000001052 = 0.09634 m3/s

inner dia of pipe = 12.2 inch =12.2*0.0254 m = 0.31 m (since 1 inch = 0.0254 m)

area of cross secion = PiD²/4 = 3.14*0.31*0.31/4 = 0.0754 m2

velocity through pipe = Q/A = 0.09634/0.0754 = 1.277 m/s

Friction loss F = 0.01Lv2/D where L is the effective length of pipe = 2.36 mi = 2.36*1609.3 m = 3798.1 m

F = 0.01LV²/D = 0.01*3798.1*1.277²/0.31 = 199.8

Total height from top of tank to subdivision = 69.4+55.7+70.4 = 195.5 ft = 195.5*0.3048 = 59.59 m (1 ft = 0.3048 m)

Applying bernoulis equation

P1/Rho*g + V₁²/2g +Z1 = P2/Rho*g + V₂²/2g +Z2 + F/2g

ssince V1 = V2 (same volumetric flow rate), Z1-Z2 = 59.59 m & F = 199.8

P1 = Patm = 101325 pa

101325/(1000*9.8) + (59.59) = P2/(1000*9.8) +199.9/(2*9.8)

P2 = 585357 Pa = 5.77 atm