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Fluids

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.