Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.78 m/s through a pipe 5.0 cm in diameter.

Question

Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.78 m/s through a pipe 5.0 cm in diameter. The pipe tapers down to 2.8 cm in diameter by the top floor, 16 m above, where the faucet has been left open.
Calculate the flow velocity and the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.

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Edana Edana 6 months 2021-07-17T09:14:00+00:00 2 Answers 29 views 0

Answers ( )

  1. Amity
    0
    2021-07-17T09:15:11+00:00
    Reply

    Answer:

    The flow velocity is   v_A = 2.4869 \ m/s

    The  gauge pressure is  P_B = 2.249 \ atm

    Explanation:

    The diagram for this question is shown on the first uploaded image

    From the question we are told that

           The gauge pressure P_A = 3.8 \ atm = 3.8 * 101325 = 385035 \ Pa

            The speed of flow is  v_A = 0.78 m/s

            The diameter of the pipe is d = 5.0 cm = \frac{5}{100} = 0.05 \ m

             The diameter at the top floor is d_1 = \frac{2.8}{100} = 0.028 \ m

              The the height from the ground is  h_B = 16 m

    So we are going to make some assumption

     We would assume that the position on the street is A

        and the position on the top floor is B

    So from continuity equation the velocity of the flow in the street is

                 v_A = \frac{V_B * a_B}{a_A}

    Where  

               a_A is the area of the pipe at the base

     So    

                a_A = \frac{\pi d^2}{4}

                a_A = \frac{ 3.142 * (5)^2}{4}  

                a_A =19.64 m^2  

      and  a_B i the area of the pipe at the top floor

              a_B = \frac{\pi d^2}{4}

    substituting values

              a_B = \frac{ 3.142 * (2.8) }{4}

              a_B = 6.16 \ m^2

    So

           v_A = \frac{0.78 * 19.64 }{6.16}

          v_A = \frac{0.78 * 19.64 }{6.16}

           v_A = 2.4869 \ m/s

    Applying Bernoulli’s equation

            P_A + \rho g h_A + \frac{1}{2} \rho v_A ^2 = P_2 + \rho g h_B + \frac{1}{2} \rho v_B^2

    Since the pipe started from the floor h_A = 0m

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

    Substituting values

      385035 + (1000* 9.8 * 0 ) + \frac{1}{2} * 1000 * 0.78^2 = \\

                                                       P_B + 1000 * 9.81 *16 + \frac{1}{2} * 1000 * 2.487^2

    P_B = 225446.6 \ Pa

     Converting back to atm

             P_B = \frac{223446.6}{101325}

            P_B = 2.249 \ atm

         

  2. mocmien2
    0
    2021-07-17T09:15:14+00:00
    Reply

    Answer:

    v_2 = 2.49 m/s  

    P_2  = 2.19 atm

    Explanation:

    By using continuity equation:

    v_2 = (A_1[tex]v_1[/tex]) / A_2 = (d_1 / d_2)²v_1 = (5/2.8)² x 0.78 = 2.49 m/s  

    By using Bernoulli’s Equation:

    P_1 + ρgh_1 + ½ρ (v_1 )²= P_2 + ρgh_2 + ½ρ(v_2 )²              

    [v_1 ² –v_2 ²] /2  = (P_2–P_1 )/ ρ + g(h_2 – h_1)

    (0.78²- 2.49²)/2 = (P_2–P_1 )/ 1000 + (9.8 x 16)  

    -5.6= (P_2–P_1 )/ 1000 + (156.8)

    (P_2–P_1 ) = – 162400 Pa  

    P_2 = P_1 -162400,   P_1 = 3.8atm = 385035 Pa  

    P_2  = 385035-162400 = 222635 Pa ( gauge  pressure)

    P_2  = 222635 Pa=> 2.19 atm

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