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

Answer:

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

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 2.44}$

v= 6.92 m/s

loss of head due to bend

$h_L=K\dfrac{v^2}{2g}$

$h_L=1.5\dfrac{6.92^2}{2\times 9.81}$

$h_L = 3.66 m$

applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho v^2+\rho g h_1 + h_L$

$\Delta P = \dfrac{1}{2}1000 \times 6.92^2+1000\times 9.81 \times 0.1016 + 3.66$

$\Delta P = 24943\ Pa$

$\Delta P = 24.9\ KPa$