Pascal Principle
A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. For example, in the diagram below, when you apply an extra pressure to the piston on the left, that extra pressure will be transmitted to the fluid on the right, which will push the piston on the right up. This is exactly how hydraulic system works.
In the diagram below, F1/A1 = F2/A2
Static Pressure and Depth
Put fluid into a container such as a cylinder,
V = Ah
where V is fluid volume, A is cross-sectional area of the cylinder, and h is height of the cylinder. Pressure on the bottom of the container P
P = F/A = pgh
Where F is weight of the liquid in the container, p is liquid density, g is gravity.
Let's see why:
F = mg, where m is mass of liquid, and
m = pV, so
F = pVg = pAhg (since V = Ah)
So
P = F/A = pAhg/A = phg = pgh
Note that this equation can also be derived from the Bernoulli's Equation.
Also note that that pressures of the fluid at different depths are different does not go against Pascal’s principle. Pascal’s principle is about the change in pressure, and a change in pressure will be transmitted uniformly to everywhere. It does not say pressures of the fluid are uniform everywhere. The change in pressure will be added to pressures at different depths.
Archimedes' Principle
Buoyant force: An upward force exerted by a fluid that opposes the weight of an immersed object.
Archimedes principle: The buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid the body displaces.
It should also be noted that a submerged object displaces a volume of fluid equal to its own volume.
For example: an object with an weight of 5N in the air is put in the water and weighs 4.5N. Then we know that the difference in weight, 0.5N, will be the weight of the water displaced.
Rate of Flow
When fuild is flowing in one direction in a pipe, the rate of flow (Q) is defined to be the volume of fluid flowing out of the pipe each second:
Q = vtA/t = vA, where v is velocity of the flow, and A is pipe's cross-secctional area.
Equation of Continuity
If the flow is laminar, then the equation of continuity states that the rate of flow Q will be constant, ie, the same in different places of the pipe. Therefore, as the cross-sectional area decreases, the velocity of the flow must increase:
v1A1 = v2A2
Bernoulli's Equation
In the diagram below, we are looking at the flow at 2 different levels h1 and h2. The pressure, veolocity, cross-sectional area (of the pipe) of the flow at the 2 different levels are P1 & P2, v1 & v2, and A1 & A2. Suppose p is the fluid density.
Note in the diagram, p1 and p2 are not the same, as they they are at different levels. But when you add an extra pressure to p1, that extra pressure will be added to p2 as well.
The Buernoulli's equation is given as:
P1 + pgh1 + 1/2 pv12 = P2 + pgh2 + 1/2 pv22
Let's prove it:
We know work done ΔW = FΔd. In the case of fluid flow, we have F1 at level 1 = P1A1, F2 at level 2 = P2A2 The distance Δd1 at level 1 = v1t and distance Δd2 at level 2 = v2t
Therefore ΔW = P1A1v1t - P2A2v2t
Since Avt is basically fluid volume, Avt = V, which is equal to m/p, v=m/p, where m is fluid mass and p is fluid density, we have
1. ΔW = P1A1v1t - P2A2v2t = P1m/p - P2m/p
When fluid is moved from level 1 to level 2, the change in potential energy
2. ΔPE = mgh2 - mgh1
And the change in kinetic energy
3. ΔKE = 1/2 mv22 - 1/2 mv12
Adding up 2. nad 3. above will give us the same ΔW as in 1.
4. P1m/p - P2m/p = ΔPE + ΔKE = mgh2 - mgh1 + 1/2 mv22 - 1/2 mv12
Therefore we have:
5. P1 + pgh1 + 1/2 pv12 = P2 + pgh2 + 1/2 pv22
Fluid at rest
Applying the Buernoulli's equation by eliminating the items with velocity v (since v = 0), we have:
ΔP = P2 - P1 = pgΔh = pg(h1-h2)
This means the difference in pressure is just proportional to the difference in fuild levels.
Fluid escaping through a small orifice
When an orifice is opened, the PE = pgh at orifice will all be converted to KE = 1/2 pv2. So we have
1/2 pv2 = pgh, eliminating p, we have
1/2 v2 = gh
As fluid flows out, h will become smaller, and therefore v will become slower too.
At orifice, flow rate R = vA, from which you can derive the relationship between flow rate R and height h.
Fluid moving horizontally
Applying the Buernoulli's equation, h1 = h2, we can eliminate the pgh terms:
P1 + 1/2 pv12 = P2 + 1/2 pv22
P1 - P2 = (1/2) p (v22 - v12)
As velocity of moving fluid increases, its static pressure descreases.