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anonymous

  • one year ago

water being drained through a hole at the bottom of a tank at a velocity of 10m/s. for a tank diameter of 1.18m and hole diameter of 5cm, find the rate change of the water level in the tank in m/min

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  1. anonymous
    • one year ago
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    help please :(((

  2. anonymous
    • one year ago
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    @ganeshie8

  3. anonymous
    • one year ago
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    @Data_LG2

  4. anonymous
    • one year ago
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    @mathmate

  5. anonymous
    • one year ago
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    @rajat97

  6. Astrophysics
    • one year ago
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    Mhm, this is an interesting related rates problem

  7. Astrophysics
    • one year ago
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    Is the shape of the tank given by any chance?

  8. anonymous
    • one year ago
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    no

  9. Astrophysics
    • one year ago
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    Ok well list your knowns and unkowns

  10. anonymous
    • one year ago
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    @Data_LG2 i dont get it

  11. Astrophysics
    • one year ago
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    This definitely requires Bernoulli's principle & continuity equation most likely, interesting...

  12. Astrophysics
    • one year ago
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    Since you have to focus on how much is going in and how much is going out from the hole

  13. Astrophysics
    • one year ago
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    I might be overthinking it though, I'm sort of tired as well, I think @phi would be better than me at this, as I sort of am thinking of it as being an easy related rate problem, but it could also require some fluid dynamics, I'm not entirely sure right now :P

  14. anonymous
    • one year ago
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    help me please :(

  15. phi
    • one year ago
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    hole diameter of 5cm , 10 m/s velocity in 1 second we have a cylinder of water |dw:1438264204509:dw|

  16. phi
    • one year ago
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    we should find the volume of that cylinder. because they want: find the rate change of the water level in the tank in m/min we probably should work in meters

  17. phi
    • one year ago
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    radius of the cylinder is 5cm/2 = 2.5 cm = 0.025 m and volume is \[ V= \pi r^2 h = \pi (0.025)^2 \cdot 10\]

  18. phi
    • one year ago
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    that is in m^3/sec multiply by 60 to get the volume that flows out in 1 minute \[ V = \pi \cdot (0.025)^2 \cdot 600 \]

  19. phi
    • one year ago
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    the equivalent volume in the tank is also a cylinder \[ V = \pi r^2 \Delta h\\ \pi (0.59)^2 \Delta h \] equate the volumes to get \[ \pi (0.59)^2 \Delta h= \pi \cdot (0.025)^2 \cdot 600 \] and solve for change in height

  20. Astrophysics
    • one year ago
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    Ah this was much easier than I thought, thanks phi!

  21. anonymous
    • one year ago
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    i got the answer 1.18

  22. mathmate
    • one year ago
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    You may want to check your answer. In engineering and physics, the units (dimensions) are critical. 1.18 could be in m/min, mm/s, etc. In fact, since both the input pipe and the tank are circular, the water rises at a speed inversely proportional to the square of the ratio of diameters, i.e. Rise in the tank \(\Large =V_{pipe} *( \frac{pipe~diam}{tank diam})^2\) \(\large =10~m/s \times ( \frac{0.05 m}{1.18 m})^2 \times 1000~mm/m \times 60~s/min\) \(\large =1077.3~mm/min\)

  23. rajat97
    • one year ago
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    i think we can use equation of continuity \[A _{1}v_1 = A_2v_2\] where A1 is the cross section area of the tank, v1 is the speed of the water at the top and A2 is the cross section area of the hole and v2 is the speed of the water from the hole. This equation is derived by assuming that water is incompressible and volume out = decrease of volume inside. v1 is the speed of the topmost layer of the water i.e. the rate of change of height of the water. If you are interested in the derivation, i can post it. My answer may not be clear so feel free to ask anything. Well i got the answer that of @mathmate 's i.e. 1077.3mm/min

  24. rajat97
    • one year ago
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    thnks for the medal @mathmate

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