Fluids and Bernoulli’s Principle
about the variables involved:
P =pressure = F Cosq/Area
Units: N/m2 = Pa
= Density = mass/volume
Fat 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)
Finput/Ainput = Foutput/Aoutput or Fout = (Fi/Ai) Ao
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
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: Fout = (xin/xout) Fin or FoXo = FiXi
Fbuoyant = rfluid
x g x volume
(volume = height x area)
Fbuoyant = (rfluid/robject)
where weight = mg
Therefore, if the robject
then Fbuoyant 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.
Fnet = Fg – Fb where Fb tends toward zero usually.
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 =
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.
Q = flow rate = Area x Velocity = Volume/time
Work of a fluid = Dm/rliquid
( Pressure1 – Pressure 2) = DE
= KE + U = 1/2 mv2 + mgh
Equation of continutity: A1Velocity1 = A2 Velocity2
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: P1 + 1/2rv12
= P2 + 1/2rv22
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.
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 h1 = h2 and you can drop rgh1 and rgh2 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 v1 = v2 and you can drop the KE portion of the equation. 1/2rv12 and 1/2rv22
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
Vwater = 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
y = 1/2gt2 so you can solve for t by re-arranging:
can find range in the horizontal direction by x = Vxt.
Simply substitute for t and you can determine range x by
X = Vx