Fluids+in+Motion

= Fluids in Motion = toc

Although fluids are commonly associated with only liquids, both liquids and gases share the properties of a fluid. A fluid, in essence, is anything that has the ability to flow.

Flow
The flow of a fluid in motion can be classified in one of two different ways.



__Streamline Flow__
When a fluid's particles all move along the same smooth path, the flow is said to be ** streamline **, or ** laminar **. In streamline flow, each path that the particles follow is called a //streamline//. No two streamlines can cross each other in standard flow. This effect can be seen in wind tunnels. Notice that in pictures (a) and (c), none of the streamlines cross one of the others. Another property of streamlines, is that the streamline will always be in the same direction as the velocity of the fluid.

__Turbulent Flow__
When a fluid surpasses a certain velocity, the flow becomes irregular, or **turbulent**. This also occurs in situations in which the fluid can experience abrupt changes in it's velocity. The motions of a fluid in turbulent flow can become irregular and will not always be in the same direction as the velocity. Such motions are called //eddy currents//. These can be seen in picture (b) where the arrows all begin to curl into themselves.

__Viscosity__

 * Viscosity** is a measure of the degree of internal friction in a fluid. This internal friction is the result of the resistance between adjacent layers of the fluid as they move relative to each other. Viscosity is commonly viewed by the thickness of the fluid, for example maple syrup has a much higher viscosity than water. The SI unit for viscosity is Pascals multiplied by seconds (Pa⋅s), however it is commonly measured in centipoise (cp). Below is a list of many different fluids and their viscosities. Notice the words 'solids' is in quotes because although these fluids are very similar to solids, they are actually classified as fluids because they can flow, just really slowly.



Ideal Fluids
When studying fluids in motion, it is common to only look at the case of an **ideal fluid**, because they have traits that remain constant. According to the textbook (in bold), there are four identifying factors for an ideal fluid.
 * 1) **The fluid is nonviscous**, or there is no internal friction in the fluid. This would mean that viscosity is 0, but in most problems a standard liquid is used under ideal conditions.
 * 2) **The fluid is incompressible**. This means that the density cannot be changed, and this applies to almost any fluid.
 * 3) **The fluid motion is steady**. This means that velocity, density, and pressure must not change over time.
 * 4) **The fluid moves without turbulence**. This requires that there is no motion from any elements in the fluid in any direction other than the direction of velocity. One could also say that this means that the flow must be streamline.

Equation of Continuity
As a fluid moves through a pipe that changes in size, there is a relationship between the cross-sectional area (the area that the fluid moves through) and the velocity of the fluid. This relationship is described in the **equation of continuity** to be:
 * ρ₁A₁v₁ = ρ₂A₂v₂ **

However, when we look at it as an ideal fluid, the densities cannot change, and must remain constant. That is to say, ρ₁ = ρ₂. Therefore the new equation is:


 * A₁v₁ = A₂v₂ **

From this equation, we can say that as the area of the pipe increases, the velocity of the fluid will decrease, and as the area decreases, velocity increases.

Bernoulli's Equation

 * Bernoulli's equation** is a direct result of the law of conservation of energy, only it is being applied to fluids. Bernoulli's equation is another equation dependent upon an ideal fluid, and states that:


 * P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂ **

Or can also be stated as:


 * P + ½ρv² + ρgy = constant **

Notice that some parts of this equation look familiar. That is because ½ρv² and ρgy are the fluid versions of the equations for kinetic and potential energy, ½mv² and mgh. So the equation states that the sum of the pressure, kinetic energy per unit volume and potential energy per unit volume has the same value through the whole streamline.

When combined with the equation of continuity, this equation unveils another relationship. We know that as area decreases, velocity increases. In Bernoulli's equation there is nothing that relates to area, but there is something that relates to velocity. Therefore we see that as velocity increases, kinetic energy per unit volume also increases. Since the sum of pressure, kinetic energy per unit volume, and potential energy per unit volume must remain constant, if the pipe is horizontal and potential energy remains constant, then the pressure varies with velocity. Therefore we come to the conclusion that **fluids with a greater velocity exert less pressure than fluids with a smaller velocity**. As velocity increases, pressure decreases, and as velocity decreases, pressure increases.

__Sample #1__
Water ideally flows through a pipe of radius 6 centimeters at a rate of 5 meters per second. The pipe then narrows to a radius of 2 centimeters. What is the new velocity of the water?

__ Sample #2 __
A pump forces water at a constant flow rate out through a pipe whose cross-sectional area gradually decreases. At the exit point, the cross-sectional area, A₂, is 1/3 the cross-sectional area at the entrance, A₁. If the difference between the entrance and exit heights is equal to 60cm, and the flow speed just after the the water leaves the pump is 1m/s, what is the gauge pressure (or pressure minus that due to the atmosphere) at the entrance point?



__Sample #3__
A tank of water releases its water through a small hole near the bottom of a tank. If the difference between the top of the water and the hole is 80cm, what is the velocity of the water as it escapes the tank?



__Solution #1__
First convert everything to basic metric units.

(6cm)((1m)/(100cm)) = .06m (2cm)((1m)/(100cm)) = .02m

Now find the cross-sectional area in both sections of the pipe.

A = π(r)² π(.06)² ≈ 0.011m² π(.02)² ≈ 0.001m²

Next, use the equation of continuity with the areas you have obtained and the given velocity.

A₁v₁ = A₂v₂ (0.011m²)(5m/s) = (.001m²)v₂ 0.055m³/s = (.001m²)v₂ 55m/s = v₂

__Solution #2__
Once again, begin by changing everything to basic metric units.

(60cm)((1m)/(100cm)) = 0.6m

Now, because the area decreases to 1/3 it's size, the velocity must increase to 3 times it's initial velocity, due to the continuity equation. Since the initial velocity is equal to 1m/s, three times that means that the final velocity is 3m/s. If we use the entrance point as the reference point for height, y₁ = 0 and y₂ = 0.6m. We also know that the density of water is 1000 kg/m³. From Bernoulli's equation we now obtain:

P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂ P₁ - P₂ = ½ρv₂² + ρgy₂ - ½ρv₁² - ρgy₁

P₁ - P₂ = ½(1000kg/m³)(3m/s)² + (1000kg/m³)(10m/s²)(0.6m) - ½(1000kg/m³)(1m/s)² - (1000kg/m³)(10m/s²)(0m) P₁ - P₂ = 4500kg/(m⋅s²) + 6000kg/(m⋅s²) - 500kg/(m⋅s²) - 0kg/(m⋅s²) P₁ - P₂ = 10000 N/m²

The question asks for the pressure at the entrance without the atmospheric pressure. Well, we know that at the exit point all of the pressure is caused by the atmosphere, because that is the only external source of pressure. Therefore, P₂ = P atm and P₁ - P₂ is the same as P₁ - P atm which is representative of gauge pressure, or pressure minus the atmospheric pressure. Thus we come to the conclusion that:

P = 10000Pa

__Solution #3__
First step, as always, is to convert to basic metric units.

(80cm)((1m)/(100cm)) = 0.8m

Now, let's try and find out as much as we can about each part of Bernoulli's Law.

The difference of height between the two points of interest is 0.8m, so if we set y₁ = 0, then y₂ = 0.8m. Next we notice that the only external source of pressure on both points of interest is the atmosphere, therefore P₁ = P atm and also that P₂ = P atm. Therefore, P₁ = P₂, and they will both cancel out of the equation. Next we look at the difference in area between the hole, and the top of the tank. Because the cross-sectional area of the water is so much greater than that of the hole, or A₂ >> A₁, We can say that the difference in velocity is so great, or v₂ << v₁ , that v₂ is basically 0, which takes that out of the equation as well. We end up with:

ρgy₂ = ρgy₁ + ½ρv₁²

The densities will cancel out as well, so solving for v₁, we get:

gy₂ = gy₁ + ½v₁² gy₂ - gy₁ = ½v₁² 2g(y₂ - y₁) = v₁² √(2gh) = v₁

This is similar to the same equation that can be derived from setting kinetic energy equal to potential energy and is actually called **Torricelli's Theorem**. Now by plugging in the numbers we have, we can solve for v₁ :

√(2(10m/s²)(0.8m)) = v₁ √(16m²/s²) = v₁  4m/s = v₁