Bernoulli's Equation - Efflux question

[sb_4000]

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?

 

[HallsofIvy]

Are you serious? hole is a circle with radius 3 mm. It's area is pi(r²)=
9 pi square mm.

 

[MAPgirl23]

I used (0.009 m)*pi * 16.6 m/s = 0.469 for the volume discharged per unit time. what am I doing wrong?

 

[HallsofIvy]

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

 

[fearthebob]

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 ]