**Fluids and Bernoulli’s Principle**

**General information
about the variables involved:**

P **=**pressure = F Cosq/Area
Units: N/m^{2} = Pa

r
= Density = mass/volume
Units: Kg/m^{3}

F_{at bottom of column of water }= height x area x
density _{of water} x g

P _{at bottom of column of water} = density _{of
water} x g x height
(notice, Area canceled)

F_{input}/A_{input} = F_{output}/A_{output
} or
F_{out} = (F_{i}/A_{i}) A_{o}

__Bottom line:__ You
want a small area of input and a large area of output.
That will give you a large force output for a relatively small force
input.

However, this increase of force output comes at the expense
of the distance over which you must push in the piston.
For example, you will have to push the piston into the hydraulic lift ten
times further than it comes out in order for you to produce 10 times the force
output: F_{out}
= (x_{in}/x_{out}) F_{in}
or F_{o}X_{o} = F_{i}X_{i}

**Buoyant Force:**

F_{buoyant} = r_{fluid}
x g x volume
(volume = height x area)

F_{buoyant} = (r_{fluid}/r_{object})
x weight_{object}
where weight = mg

Therefore, if the r_{object}
> r_{fluid/air}
then F_{buoyant }is small or insignificant.

The net force on any object must take into account the buoyant force. Usually, however, we can ignore the buoyant force because objects are much denser than air.

F_{net} =
F_{g} – F_{b}
where Fb tends toward zero usually.

**Specific Gravity:
**

Specific gravity has no units. It is the ratio between the density of the object in question and water. Since water has a density of 1 g/mL, the specific gravity of any object is equal to its own density (just sans units).

Specific gravity =
r_{object}/r_{water}

NOTE: If you find the specific gravity of an object NOT IN WATER, but in say, oil, then its specific gravity will obviously NOT be equal to its density.

**Bernoulli Principle:
**

Q = flow rate = Area x Velocity = Volume/time

Work of a fluid = Dm/r_{liquid}
( Pressure_{1} – Pressure _{2}) = DE
= KE + U = 1/2 mv^{2} + mgh

**Equation of continutity:** A_{1}Velocity_{1} = A_{2}
Velocity_{2}

Therefore, as area of a pipe increases, the velocity will
decrease. **Flow
rate will remain constant for an incompressible fluid flowing through a pipe of
any diameter. Example:
Water flowing into the hose at some rate Q.
You put your finger on the other end of the hose to decrease the area.
Water will shoot out at a high rate of speed.
This is because the water rate, Q, must be maintained by the same mass of
water entering the pipe as leaving it. This
is accomplished by decreasing its escape area (with your finger) and so the
liquid speeds up to get more mass through in a shorter amount of time.**

**Bernoulli’s equation:** P_{1} + 1/2rv_{1}^{2}
+ rgh_{1}
= P_{2} + 1/2rv_{2}^{2}
+ rgh_{2}

Notice how Bernoulli’s equation is born out of the work of a fluid
equation above which involves KE and U and Pressure.
P means pressure, and v means velocity.
r
is the density of the fluid in question and g is an old friend!
h is the drop in height

**Things to note:** If
there is no **drop in height** (eg. The pipe is horizontal) then h_{1}
= h_{2} and you can drop rgh_{1}
and rgh_{2}
from the equation. Realize this is
the potential energy portion of the equation.

If
there is no change in the **area** of the pipe, then v_{1} = v_{2}
and you can drop the KE portion of the equation.
1/2rv_{1}^{2}
and 1/2rv_{2}^{2}

**Bottom
line:** Bernoulli’s equation
takes into account both the potential and kinetic energies involved in moving
fluids and how that affects pressure. If
there is no height drop, then there is no potential.
If there is no area change, then there is no kinetic energy change.

If the just the area decreases, such that the velocity increases, the pressure will Decrease. (height stays same)

If the just the height drops, Pressure will increase (area must remain constant)

**
Example problem #75 on p. 330**
is an excellent example to look at before the exam:

1. No two
holes have the same pressure. BUT…..Each
hole maintains its own __constant__ pressure because the water is being
replaced in the top at the same rate it is escaping out of the holes.

2. The bottom hole has the most pressure on it. The top has the least.

3. The distance from the top of the container to each hole will give you the pressure of that hole.

4. Since the pressure is constant out of a particular hole, you can use the following equation to determine the velocity of the water shooting out of a particular hole

V_{water}
= _{
}
where h is the
distance the hole is from the top.

5. Once you know the velocity of the water out of a hole, you can use Newton’s laws to determine the downrange distance of that water

remember
y = 1/2gt^{2} so you can solve for t by re-arranging:
t = _{
}

You
can find range in the horizontal direction by x = V_{x}t.
Simply substitute for t and you can determine range x by

X = V_{x
}