# Bernoulli's Equation & Fluid Dynamics LAB

1. Dec 6, 2009

### Danneskjöld

Before I describe the problem, I would like to apologize for my lack of comprehension
of this subject. It is the first semester of Physics, but the end, and has become quite
complicated.

1. The problem statement, all variables and given/known data

There is a 2.0 L soda bottle used, with a hole a certain height (hhole) from
the bottom (and a small mini-hose to assist in the motion of the exiting of the fluid from the
bottle). The bottle is aligned with the opening of a sink, and a hose is attached to the top
so as to fill the bottle to a certain height and remain at that height constantly (h1).
Then the distance the water travels out the hole is measured and recorded as a value
of x1.
Subsequently, this is done one more time with the water level at a new higher height (h2) and a new value of x2.

Okay, hopefully that makes sense.

Next, we are given two equations that are used to derive some other new equations
meant to discover the velocity (I'm uncertain if I should include the messy method of
proving these equations weren't pulled out of my ear, or just cut to the chase).

THEN... we are to solve for velocity.
Then we are asked to comment on our results, giving probable reasons for difference
in values obtained from equations 1 & 2.

2. Relevant equations

Now for the equations.

Equation 1:

Point 1_____Point 2_____Point 3
___P0___= P0 + $$\rho$$gh1 = P0+($$\rho$$vH2O2)/2

OR

$$\rho$$vH2O2 = $$\rho$$gh1 ; vH2O = $$\sqrt{(2gh[sub1])}$$
___2

Equation 1: vH2O = $$\sqrt{(2gh[sub1])}$$

Equation 2:

h2 = ½ gt2 OR t = $$\sqrt{[2h(sub2)]/g}$$ ; X = vH2Ot ; OR t = x/vH2O = $$\sqrt{(2gh[sub1])}$$

THEN

vH2O = X*$$\sqrt{g/(2h(sub2))}$$ ; vH2O = 2.21 * X/$$\sqrt{h(sub2)}$$

Equation 2: vH2O = 2.21 * X/$$\sqrt{h(sub2)}$$

3. The attempt at a solution

hhole = h2 = 6.9 cm
h1 = 8.4 cm
x1 = 8.6 cm
v1H2O=________?

hhole = h2 = 6.9 cm
h1 = 11.1 cm
x2 = 9.8 cm
v2H2O = _______?

Using Equation 1:

v1H2O = $$\sqrt{2gh(sub1)}$$ = $$\sqrt{(2)(9.8)(0.084m)]}$$ = 1.6464 m/s

v2H2O = $$\sqrt{2gh(sub2)}$$ = $$\sqrt{(2)(9.8)(0.11m)]}$$ = 2.156 m/s

Using Equation 2:

v1H2O = (2.21) (X/$$\sqrt{h(sub2)}$$ = [(2.21)(0.086m)]/$$\sqrt{0.084m}$$ = 0.655769301 m/s

v2H2O = (2.21) (X/$$\sqrt{h(sub2)}$$ = [(2.21)(0.098m)/$$\sqrt{0.11m}$$ = 0.65302 m/s

So to discuss these differences and probably reasons, I came to the conclusion
that the 1st equation is lacking the value of X. I purport that because we normally
consider volume (not cross-sectional area) when we are dealing with pressure,
and also because the X value is not taken into consideration within the first equation,
that perhaps it is wrong for those reasons.

However, the truth is, I don't even know what I am talking about, and am simply
left in a complete state of perplexity. Any and all help is appreciated.

P.S.: I have even asked my class mate and my father for help, both of whom
are equally confused.
P.P.S.: My roommate mentioned something along the lines of the potential for
the two Bernoulli equations in question to be from two different relatives from
the Bernoulli family (the roommate has read many a Mathematician's biography,
which is how they know about the relations)... I believe this to be nonsense. ;)

2. Dec 7, 2009

### Andrew Mason

What does Bernouilli's equation predict about the value of X as a function of H? That is the question you have to answer.

Since the hole is at the same height above the bottom of the bottle and bottom of the sink, the distance of fall of the exiting water is constant for all H (height of the water in the bottle above the hole). This means that the time of fall is the same. So X is proportional to the horizontal speed of the exiting water: $$X \propto v_{H_2O}$$

What does Bernouilli say the relationship is between H and v? So what is the relationship that you should find between H and X? Does your data support that relationship?

AM