Question

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its original diameter, what is (a) the speed and (b) the pressure of the water?

Answer 1

Answer:

a

  $v_2 = 5.6 \ m/s$

b

   $P_2 = 80600 \ Pa$

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

      The speed of the water  is $v_1 = 1.4 \ m/s$

       The original area of the pipe is  $A_1 = \pi \frac{d^2 }{4}$

       The  new area of the pipe is  $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

         

Generally the continuity equation is mathematically represented as

       $A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity  

So

        $\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=>     $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=>    $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=>    $1.4 = 0.25 * v_2$

=>     $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             $P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value  $\rho = 1000 \ kg /m^3$

             $P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=>          $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=>          $P_2 = 80600 \ Pa$