## 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

1. anonymous

2. anonymous

@ganeshie8

3. anonymous

@Data_LG2

4. anonymous

@mathmate

5. anonymous

@rajat97

6. Astrophysics

Mhm, this is an interesting related rates problem

7. Astrophysics

Is the shape of the tank given by any chance?

8. anonymous

no

9. Astrophysics

Ok well list your knowns and unkowns

10. anonymous

@Data_LG2 i dont get it

11. Astrophysics

This definitely requires Bernoulli's principle & continuity equation most likely, interesting...

12. Astrophysics

Since you have to focus on how much is going in and how much is going out from the hole

13. 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

14. anonymous

15. phi

hole diameter of 5cm , 10 m/s velocity in 1 second we have a cylinder of water |dw:1438264204509:dw|

16. 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

17. phi

radius of the cylinder is 5cm/2 = 2.5 cm = 0.025 m and volume is $V= \pi r^2 h = \pi (0.025)^2 \cdot 10$

18. phi

that is in m^3/sec multiply by 60 to get the volume that flows out in 1 minute $V = \pi \cdot (0.025)^2 \cdot 600$

19. 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

20. Astrophysics

Ah this was much easier than I thought, thanks phi!

21. anonymous

22. 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 $$\Large =V_{pipe} *( \frac{pipe~diam}{tank diam})^2$$ $$\large =10~m/s \times ( \frac{0.05 m}{1.18 m})^2 \times 1000~mm/m \times 60~s/min$$ $$\large =1077.3~mm/min$$

23. 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

24. rajat97

thnks for the medal @mathmate