Question

) A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to: (a) pump all of the water over the side

Answer 1

Complete Question

A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to:

(a) pump all of the water over the side

(b) pump all of the water out of an outlet 3 m over the side?

Answer:

a

The work required is $W_d = 14.25*10^6J$

b

The work is  $W_{d3} = 3.6 *10^7 J$

Explanation:

From the question we are told that

      The diameter is $d = 16m$

       The height of the circular side is $H =4m$

      The depth of water is  $D = 3.5m$

       The acceleration due to gravity is $g =9.81m/s^2$

        The density of water is  $\rho_w = 1000kg/m^3$

The radius of the swimming pool is

             $r = \frac{16}{2} = 8m$

The work is mathematically given as

             $W_d = Force * distance$

Now force is mathematically given as

            $F = density * area * height \ of \ pool$

               $= \rho * (\pi r^2 ) dx$

Now the work done to pump all of the water over the side

is

          $W_d = \int\limits^D_0 {\rho (\pi r^2 )(H-x)} \, dx$

               $=\int\limits^D_0 {1000*9.81 *(\pi * 8^2) (4- x)} \, dx$

               $= 64000 *9.8 \pi \int\limits^{3.5} _0 {(4-x)} \, dx$

              $= 64000 * 9.8 \pi [4x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.$

              $= 64000 *9.8 \pi [4(3.5) - \frac{3.5^2}{2} ]$

              $=14.25*10^6J$

Now for 3 meters over the side the new height would be

             $H_{new} = H +3$

                       $= 4+3$

                      $=7m$

Now applying the formula for work that is  above

           $W_{d3 }= \int\limits^D_0 {\rho g (\pi r^2)(H-x)} \, dx$

                 $= \int\limits^{3.5}_0 {1000 *9.8 * (\pi * 8^2)(7 - x)} \, dx$

                 $= 64000 * 9.8 \pi [7x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.$

                $= 64000 *9.8 \pi[7(3.5) - \frac{(3.5)^2}{2} ]$

                $=3.6*10^7J$