Bernoulli's Equation - Efflux question

AI Thread Summary
The discussion revolves around calculating the speed of efflux and the volume discharged per unit time from a hole in a water tank. The speed of efflux is correctly determined to be 16.6 m/s using Bernoulli's equation. The area of the hole, calculated as π(0.003 m)², is essential for finding the volume discharged per unit time. A participant initially miscalculated the volume due to confusion in unit conversion, but the correct volume discharged per second is ultimately clarified to be 4.69 x 10^-4 m³. Accurate unit conversion is emphasized as a critical aspect of the calculations.
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A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air.

A) What is the speed of efflux?

sqrt(2*g*h) = 16.6 m/s

B) What is the volume discharged per unit time.

This is the equation i believe, dV/dt = A*v

I don't know exactly how to get A..

any Hints?
 
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Are you serious? hole is a circle with radius 3 mm. It's area is pi(r2)=
9 pi square mm.
 
I used (0.009 m)*pi * 16.6 m/s = 0.469 for the volume discharged per unit time. what am I doing wrong?
 
Imagine a "one second" cylinder of water coming out the hole: it will be a cylinder 16.6 m long and with base area the area of the hole.

But the area of a circle is pi r2, not pi r! The area of the whole is
(0.009 m)2*pi and so the volume of water discharged in one second is
(0.009)2*pi*16.6
 
First off

@hallsofivy: dude that's so retarded. he already squared it it's (3mm)^2 which becomes 9mm squared.

Final answer: your answer is in fact right you're only off by a degree of 10^-3. such that the final answer should be 4.69*10^-4. I think this happened when you were converting
mm^2 to m^2. [unit conversion's a ***** isn't it :cool:]
 
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