Question

cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small hole has occurred in the side at a point 16.5 m below the water level and that water is flowing out at the volume flow rate of 2.90 10-3 m3/min. Determine the following. (a) the speed in m/s at which water is ejected from the hole

Answer 1

Answer:

a

The velocity is  $v =17.98 \ m/s$

b

The diameter is  $d = 0.00184m$

Explanation:

The diagram of the set up is shown on the first uploaded image  

From the question we are told that

    The height of the water tank is $h = 20.0 \ m$

    The position of the hole $p_h = 16.5m$ below  water level

     The  rate  of water flow $\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s$

  According to Bernoulli's theorem position of the hole

              $\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

Where  u is  the initial speed the water through the hole = 0 m/s

              $P_o$ is the atmospheric pressure

            $\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

                   $v = \sqrt{2gh}$

Substituting value

           $v = \sqrt{2 * 9.8 * 16.5 }$

              $v =17.98 \ m/s$

The Volumetric flow rate is mathematically represented as

          $\r V = A * v$

     Making A the subject

              $A = \frac{\r V}{v}$

 substituting value  

             $A = \frac{0.048 *10^{-3}}{17.98}$

                 $= 2.66*10^{-6}m^2$

Area is mathematically represented as

        $A = \frac{\pi d^2}{4}$

  making d the subject

         $d = \sqrt{\frac{4*A}{\pi} }$

  Substituting values

        $d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }$

          $d = 0.00184m$