A water pipe tapers down from an initial radius of R1 023 m

A water pipe tapers down from an initial radius of R₁ = 0.23 m to a final radius of R₂ = 0.08 m. The water flows at a velocity v₁ = 0.84 m/s in the larger section of pipe.

1)

What is the volume flow rate of the water?

2)

What is the velocity of the water in the smaller section?

3)

Using this water supply, how long would it take to fill up a swimming pool with a volume of V = 159 m³? (give your answer in minutes)

4)The water pressure in the center of the larger section of the pipe is P₁ = 273510 Pa. Assume the density of water is 10³ kg/m³.

What is the pressure in the center of the smaller section of the pipe?

5)

If the pipe was turned vertical and the volume flow rate in the larger section is kept the same, which answers would change? (select all that apply)

1) the speed of water in the smaller section

2) the volume flow rate in the smaller section

3) the pressure in the smaller section

Solution

a) Volume flow rate is,

Av = ( )(0.23)^2 * 0.84 = 0.1396 m^3 / sec

b)the velocity is calculated as follows:

         A₁V₁= A₂V₂

      0.1396 = V₂( )( 0.08)²

                 V₂ = 6.94 m/s

c) time is calculated as follows:

    t = 159 / 0.1396 = 1138.968 s = 18.98 min

d) use bernoulli\'s equation and solve pressure:

     273510 + 0.5( 1000)(0.84)^2 = P₂+ 0.5(1000)(6.94)^2

                                               P2 = 249781 Pa