A 2-kg disk is constrained horizontally but is free to move vertically. The disk is struck from below by a vertical jet of water. The speed

Question

A 2-kg disk is constrained horizontally but is free to move vertically. The disk is struck from below by a vertical jet of water. The speed and diameter of the water jet are 10 m/s and 25 mm at the nozzle exit. Obtain a general expression for the speed of the water jet as a function of height, h. Find the height to which the disk will rise and remain stationary

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Lệ Thu 4 years 2021-09-02T00:02:50+00:00 1 Answers 27 views 0

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  1. Nem
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    2021-09-02T00:04:48+00:00
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    Answer:

    h = 4.281 m

    Explanation

    Given Data,

    Weight of disk = 2kg

    Speed of water jet (V) = 10 m/s

    diameter of water jet = 25 mm

    Cal. the velocity of the water jet as function height by applying Bernoulli’s  eqtn of water surface to the jet

    \frac{P}{\rho } +\frac{V^{2}}{2} + gz = Constant

    \frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g

    V=\sqrt{V_{0}^{2}-2gh}

    Relation between  V_{0} & V

    m=\rho V_{0}A_{0}

    \rho VA=\rho V_{0}A_{0}

    VA= V_{0}A_{0}

    Momentum

    F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA

    -mg = w_{1}[-\rho VA] +w_{2}[\rho VA]

    w_{1} = V

    w_{2} = 0

    mg = \rho V^{2}A

    mg = \rho VV_{0} A

    mg = \rho VV_{0} A_{0}

    mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }

    Solving for h

    h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}]

    g is gravitational acc.

    = \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ]

    = \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}]

    = \frac{100-16.0078}{19.62}

    h = 4.281 m

    h of disk on which it remains stationary.

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