anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
help please :(((
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@Data_LG2

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anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@rajat97
Astrophysics
  • Astrophysics
Mhm, this is an interesting related rates problem
Astrophysics
  • Astrophysics
Is the shape of the tank given by any chance?
anonymous
  • anonymous
no
Astrophysics
  • Astrophysics
Ok well list your knowns and unkowns
anonymous
  • anonymous
@Data_LG2 i dont get it
Astrophysics
  • Astrophysics
This definitely requires Bernoulli's principle & continuity equation most likely, interesting...
Astrophysics
  • Astrophysics
Since you have to focus on how much is going in and how much is going out from the hole
Astrophysics
  • Astrophysics
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
anonymous
  • anonymous
help me please :(
phi
  • phi
hole diameter of 5cm , 10 m/s velocity in 1 second we have a cylinder of water |dw:1438264204509:dw|
phi
  • phi
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
phi
  • phi
radius of the cylinder is 5cm/2 = 2.5 cm = 0.025 m and volume is \[ V= \pi r^2 h = \pi (0.025)^2 \cdot 10\]
phi
  • phi
that is in m^3/sec multiply by 60 to get the volume that flows out in 1 minute \[ V = \pi \cdot (0.025)^2 \cdot 600 \]
phi
  • phi
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
Astrophysics
  • Astrophysics
Ah this was much easier than I thought, thanks phi!
anonymous
  • anonymous
i got the answer 1.18
mathmate
  • mathmate
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 \(\Large =V_{pipe} *( \frac{pipe~diam}{tank diam})^2\) \(\large =10~m/s \times ( \frac{0.05 m}{1.18 m})^2 \times 1000~mm/m \times 60~s/min\) \(\large =1077.3~mm/min\)
rajat97
  • rajat97
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
rajat97
  • rajat97
thnks for the medal @mathmate

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