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 t

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.

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Thu Cúc 4 years 2021-08-13T13:12:56+00:00 1 Answers 4 views 0

Answers ( )

  1. minhkhue
    0
    2021-08-13T13:14:52+00:00
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    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

     

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