How Does Water Flow Rate Change Over Time in a Draining Cylinder?

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SUMMARY

The discussion focuses on calculating the flow rate of water draining from a cylinder over time, specifically a cylinder with a height of 5.5 meters and a radius of 1 meter, featuring a hole 0.5 meters from the bottom with a radius of 0.02 meters. The key equations involve the relationship between the height of the water (h(t)), the volume of fluid (V(t)), and the flow rate, which is defined as the product of the cross-sectional area and velocity. Participants seek to derive the rate of change of flow rate (dflowrate/dt) and the rate of change of height (dh/dt) as the water level decreases.

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Homework Statement


Ok, so the height of the cylinder is 5.5 meters. The radius is 1 meters.
There is a hole, .5 meters from the bottom(5 meters from the top), its radius is .02 meters.

The question is how fast does water flow out, relative to time(this because the height of the water keeps dropping, so the flow rate decreases). [/B]

Thank You :)

Homework Equations


Am i doing it right?
If i am, can you please solve it(i can't quiet figure it out)?
Please give the rate of change of the flowrate(dflowrate/dt).
Also, rate of change of height(dh/dt).
[/B]

The Attempt at a Solution


Height = total volume - (integral sign) flowrate dt
flowrate = (cross sectional area * Velocity(
18632c090016d6d87c3f7653a95cf02a.png
) * density

dflowrate/dt = (deritive of the expression above)[/B]
 
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Suppose that h(t) is the height of the fluid level in the cylinder at time t, and R is the radius of the cylinder. What is the volume of fluid in the cylinder V(t) at time t?

Chet
 

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