Question

•• Unfiltered olive oil must flow at a minimum speed of 3.0 m/s to prevent settling of debris in a pipe. The oil leaves a pump at a pressure of 88 kPa through a pipe of radius 9.5 mm. It then enters a horizontal pipe at atmospheric pressure. Ignore the effects of viscosity. (a) What is the speed of the oil as it leaves the pump if it flows at 3.0 m/s in the horizontal pipe? (b) What is the radius of the horizontal pipe?

Answer 1

Answer:

a

The velocity at which the oil leaves the pump is $v_1 = 5.15 m/s$

b

The radius of the horizontal pipe  $r_2 = 0.01244 m$

Explanation:

From the question we are told that

    The rate of flow of unfiltered olive oil is  $v = 3.0 m/s$

    The pressure on the pump is $P_1 = 88 kPa = 88 *10^{3} \ Pa$

     The radius of the pipe is  $r = 9.5 = \frac{9.5}{1000} = 0.0095 \ m$

     

Generally we can define this motion with Bernoulli's Equation as

       $P_1 + \frac{1}{2} \rho v_1 ^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2 + \rho g h_2$

          $P_2$  is the  pressure inside the pipe which is the atmospheric pressure which has a value of $P_2 = 1.01 *10^{5} \ Pa$

           $\rho$ is the density of olive oil which has a value of $\rho = 980 kg/m^3$

           $g$ is the acceleration due to gravity

           h the height of the pipe since from the question we are not told that it is place on anything hence the height is the diameter of the pipe

           $v_1 \ and \ v_2$ are the speed at which it leaves the pump and the speed at which it flows in the pipe respectively

Making $v_2$  the subject of the equation

              $v_1 = \sqrt{[\frac{2}{\rho} (P_2 -P_1 - \frac{1}{2} v^2_2)]}$

Substituting values

              $v_1 = \sqrt{[\frac{2}{980} (1.01 *10^5 -88 *10^{3} - \frac{1}{2}(3.0)^2)]}$

              $v_1 = 5.15 m/s$

Using continuity equation to define the motion of this fluid we have

              $A_1 v_1 = A_2 v_2$

Where  $A_1$ is the area of the pump which is circular so mathematically it is

        $A_1 = \pi r^2_1$

         $A_2$ is the area of the horizontal pipe  which is circular so mathematically it is

            $A_2 = \pi r^2_2$

So the continuity equation becomes

             $\pi r^2 _1 v_1 = \pi r^2 _2 v_2$

Making $r_2$ subject of formula

            $r_2 = \sqrt{\frac{r_1^2 v_1}{v_2} }$

Substituting values

           $r_2 = \sqrt{\frac{ 0.0095^2 5.15 }{3.0} }$

           $r_2 = 0.01244 m$