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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
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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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Mhm, this is an interesting related rates problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Is the shape of the tank given by any chance?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok well list your knowns and unkowns

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Data_LG2 i dont get it

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1This definitely requires Bernoulli's principle & continuity equation most likely, interesting...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Since you have to focus on how much is going in and how much is going out from the hole

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I 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

phi
 one year ago
Best ResponseYou've already chosen the best response.1hole diameter of 5cm , 10 m/s velocity in 1 second we have a cylinder of water dw:1438264204509:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.1radius 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
 one year ago
Best ResponseYou've already chosen the best response.1that 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
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1Ah this was much easier than I thought, thanks phi!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got the answer 1.18

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1thnks for the medal @mathmate
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