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    Hydrodynamics: Equation of continuity, rate of flow of fluid

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    Q: A meter long hose of 14cm diameter is connected to a second hose which is 4 meters long and 7cm in diameter. At the open end of the second hose water flows out at a rate of 8 cm^3/sec. What is the ratio of the speed of the water flowing in the second hose to the speed of the water flowing in the first hose?

    © BrainMass Inc. brainmass.com December 24, 2021, 5:41 pm ad1c9bdddf
    https://brainmass.com/physics/hydrodynamics/hydrodynamics-equation-continuity-rate-flow-fluid-63268

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    Q: A meter long hose of 14cm diameter is connected to a second hose which is 4 meters long and 7cm in diameter. At the open end of the second hose water flows out at a rate of 8 cm^3/sec. What is the ratio of the speed of the water flowing in the second hose to the speed of the water flowing in the first hose?

    1) (v2/v1) = 4
    2) (v2/v1) = 1/3
    3) (v2/v1) = 1
    4) (v2/v1) = 1/2
    5) (v2/v1) = 1/4
    6) cannot be determined from the information given.
    7) (v2/v1) = 8
    8) (v2/v1) = 1/8
    9) (v2/v1) = 3
    10) (v2/v1) = 2

    Answer:

    This is the problem about the flow of a non-viscous incompressible (ideal) fluid through a pipe of varying cross-section. As the fluid is incompressible the rate of volume flowing from any cross-section will remains same. Consider a tube of varying cress-section. At points P and Q let the cress-section areas A1 and A2. If the velocities at these points are v1 and v2 respectively then the volume flowing in time dt from these cross-section are A1*v1*dt and A2*v2*dt respectively. (the column PQ shifts to P'Q') As the fluid is incompressible, the two volumes will be same and we get

    dQ/dt = A1*v1*dt = A2*v2*dt

    Where dQ/dt is volume flow rate in the tube, which is same everywhere in the tube.

    Or A1*v1 = A2*v2

    Or at different points of a pipe of varying cross-section product A*v is constant.
    This equation is called equation of continuity.
    This is why by reducing opening the speed of water from hose increases.

    Now to our problem

    Area of cross-section of first hose A1 = r12 =  (14/2)2 = 49 cm2, and
    Area of cross-section of second hose A2 = r22 =  (7/2)2 = 49/4 cm2.

    Hence according to equation of continuity

    v2/v1 = A1/A2 = 4

    The correct answer is 1) (v2/v1) = 4
    A vector can be written in components form.

    Consider a vector with magnitude A and in the direction making an angle  with the x-axis. This vector can be resolved (broken away) in two components, which are mutually perpendicular. We are doing this because a vector has no projection (effect) in the direction perpendicular to it, so the two components will have no effect in others direction and we can add of subtract them separately in the two directions easily.

    The component of vector in x direction is having magnitude A cos and to make it vector in x direction we multiply it by unit vector in x direction i Cap or . Similarly the component in y direction is A cos(90 -) = A sin  and to make vector multiplied by unit vector in y direction . Hence vector can be written as the sum of two vectors
    A cos  and A sin  and hence we write

    = A cos  + A sin 

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:41 pm ad1c9bdddf>
    https://brainmass.com/physics/hydrodynamics/hydrodynamics-equation-continuity-rate-flow-fluid-63268

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