Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. The velocity

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. The velocity profile is uniform at sections 1 and 2. The correct statement about the static pressure p2 relative to p1 and the velocity V2 relative to V1 along the streamline down the center of the pipe is

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Lệ Thu 5 years 2021-08-30T12:07:15+00:00 1 Answers 38 views 0

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  1. trungdung22
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    2021-08-30T12:08:51+00:00
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    Answer:

    v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

    Explanation:

    As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

    Rate of mass flow can be written as,

    m = \rho A v

    where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

    Now, according to the problem, as the density of the fluid does not change, we can write

    && m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

    where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

    As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

    Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

    So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

    P_{2} > P_{1}.

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