# Fluids(Bernoulli and Continuity)

1. Sep 16, 2011

### knowNothing23

A large keg of height H and cross-sectional area A1 is filled with root
beer. The top is open to the atmosphere. There is a spigot opening of area A2,
which is much smaller than A1, at the bottom of the keg.
(a) Show that when the height of the root beer is h, the speed of the root beer leaving the spigot is
approximately (2 gh)^(1/2) **square root of 2gh** .
(b) Show that if A2 << A1, the rate of change of the height h
of the root beer is given by dh /dt = −( A2 /A1 )(2gh )^(1/2) .

(c) Find h as a function of time if h = H at t = 0.
(d) Find the total time needed to drain the keg if
H = 2.00 m, A1 = 0.800 m2, and A2 = 1.00 × 10–4 A1. Assume laminar nonviscous
flow.

I have problems with the answers of (C) and (D).

My attempt at (C):
I just had to integrate (B) and solve for "h", but my answer is different than the one from the book. Please, help.
h(t)= ( -(A2/A1)t(2g)^(1/2) + H^1/2 ) ^2, where A2,A1 are the areas, t is time, and (2g)^1/2 is the square root of 2g.

My solution for (D) follows from (C), but since my solution for C is presumably wrong, then my solution for D is wrong.

2. Sep 16, 2011

### pabloenigma

Why do you integrate B?
I mean you should not assume that A2<<A1 for C. Solve for the general case

3. Sep 22, 2011

### knowNothing23

A2<<A1 is not an assumption. It's a fact in the problem.