water in the tank is 2.5 ft. Water enters the tank through a top port of diameter 6 in. at a velocity of 24 ft/s. The water leaves the tank

Question

water in the tank is 2.5 ft. Water enters the tank through a top port of diameter 6 in. at a velocity of 24 ft/s. The water leaves the tank through two exit ports each with a diameter of 6 in. If the scale shows a reading of 585 lbf, calculate the weight of the tank when it is empty

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Hải Đăng 6 months 2021-08-03T14:23:53+00:00 1 Answers 3 views 0

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    2021-08-03T14:24:53+00:00

    Answer:

    The weight of the tank when it is empty is 501.38 lbf

    Explanation:

    The velocity of outlet is equal to:

    A_{i} v_{i}=2A_{o} v_{o}  \\\frac{\pi d_{i}^{2} v_{i}  }{4} =\frac{2\pi d_{o}^{2} v_{o}  }{4}\\v_{o} =\frac{d_{i}^{2}  v_{i} }{2d_{o}^{2}  }

    Where

    vi = 24 ft/s

    di = 6 in = 0.5 ft

    do = 6 in = 0.5 ft

    v_{o} =\frac{0.5^{2}*24 }{2*0.5^{2} } =12ft/s

    The total weight of the water excluding the tank weight is equal to:

    W=pv_{i}^{2}  +pghA_{tank} -pv_{o}^{2} =p(v_{i}+gh(\frac{\pi d_{tank}^{2}  }{4} -v_{o}^{2} ))

    p = 62.43 lb/ft³

    dtank = 20 in = 1.67 ft

    Replacing:

    W=62.43(24^{2} +(32.17*2.5*(\frac{\pi 1.67^{2} }{4} )-12^{2} )=37929.5 lb=83.62lbf

    The weight of the tank is:

    Wtank = 585 – 83.62 = 501.38 lbf

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