Question

A decorative fountain was built so that water will rise to a hieght of 8 feet above the exit of the pipe. the pipe is 3/4 diameter galvanized schedule 40 steel pipe. The resistance coefficiant of each elbow is 1.5. the height, is h1 is 4 inches above of the centerline of the pipe. Determine the required pressure guage to achieve the result.

Answer 1

Step-by-step explanation:

given,

height of rise of water = 8 ft

1 ft = 0.3048

8 ft = 8 × 0.3048 = 2.44 m

pipe diameter = 3/4 of galvanized schedule 40 steel pipe

resistance coefficient = 1.5

h_1 = 4 inch

1 inch = 0.0254 m

4 inch = 4 ×0.0254 = 0.1016

velocity of water

`v = √(2gh)`

`v = √(2* 9.8 * 2.44)`

v= 6.92 m/s

loss of head due to bend

`h_L=K(v^2)/(2g)`

`h_L=1.5(6.92^2)/(2* 9.81)`

`h_L = 3.66 m`

applying Bernoulli's equation

`\Delta P = (1)/(2)\rho v^2+\rho g h_1 + h_L`

`\Delta P = (1)/(2)1000 * 6.92^2+1000* 9.81 * 0.1016 + 3.66`

`\Delta P = 24943\ Pa`

`\Delta P = 24.9\ KPa`