Question

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a second point in the line, 18.5 m lower than the first, if the pipe diameter at the second point is twice that at the first. Remember that the density of water is 1000 kg/m3. Please give your answer in units of kPa.

Answer 1

Answer:

The  pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

   The speed at point 1  is  $v_1 = 3.57 \ m/s$

   The  gauge pressure at point 1  is  $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

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

Let the  height at point 1 be  $h_1$ then the height at point two will be

      $h_2 = h_1 - 18.5$

Let the  diameter at point 1 be  $d_1$ then the diameter at point two will be

      $d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as  

         $A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$]

   which can represent as

             $A \ \ \alpha \ \ d^2$

=>         $A = c d^2$

where c is a constant

  so      $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=>          $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=>        $A_2 = 4 A_1$

Now from the continuity equation

        $A_1 v_1 = 4 A_1 v_2$

=>     $v_2 = \frac{v_1}{4}$

=>     $v_2 = \frac{3.57}{4}$

       $v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

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

So  

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

=>    $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

        $P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

       $P_2 = 254.01 kPa$