Buoyancy

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Buoyancy

Take the block of water inside the bucket on the left hand side. This block of water is being upheld by the pressure in the liquid underneath the block of water, while being pressed down by the force of gravity. Since this block of water is stationary, that means that the force of gravity equals the force of buoyancy. So at this point:

\(F_{gravity}=F_{buoyant}=m_f g\)

Now, regardless of the object that replaces this block of water, the pressure underneath the block doesn’t change. So the buoyant force doesn’t change, as long as the new object replaces the volume of the block of water. So regardless of the object, the buoyant force is:

\(F_{buoy}=m_f g=\rho_f V_f g\)

Now the volume of the block of water is the volume of water that is displaced by the submerged object. The object replaced that volume of water. So now we have a way of measuring the buoyant force. We simply find the amount of water that is displaced when an object is submerged, found by the increase in volume of the container. Now if the density of the object is less than the density of the water the object will float.

The buoyant force here is equal to the gravitational force acting on the object, so the object floats. If the object is pushed down further the buoyant force overcomes the gravitational force and pushes the object to the surface.

If an object is more dense than the fluid then the object will sink to the bottom of the container.

Question

If we have a container full of water (\(\rho= 1000 \frac{kg}{m^3}\)) and a submerged object with a mass of 200 grams, which takes up 0.1 litres of volume, how much does the object weigh?

The weight is given by the net force divided by 10. The net force is: \(F_{net}=F_{Gravity}-F_{Buoyant}\)

The force of gravity is simply calculated using: \(F_{Gravity}=m g\)

The buoyant force we know is calculated using: \(F_{Buoyant}= \rho_f V_f g\)

The volume needs to be in units of \(m^3\), \(V=0.1 L = 0.00001 m^3\)

Bringing it all together gives us:

\(F_{net}=mg-\rho_{f}V_{f}g=0.2\times10-1000\times0.00001\times10=1.9 N\)

And then the apparent weight of the object is:

\(Weight = m_{apparent}=\frac{1.9}{10} kg\)

The block has an apparent weight of 190 grams.

Non-viscous Flow

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Non-viscous Flow

Equation of Continuity

Let us take the closed pipe that we can see above. Closed means that no mass can enter or leave the pipe between surface 1 and surface 2. Since no mass can enter or leave the pipe we have conservation of mass. So the amount of water that flows through surface 1 per second must equal the amount of water that flows through surface 2.

\(\frac{mass_1}{sec}=\frac{mass_2}{sec}\)

Now we know from earlier that the density of a fluid is equal to the mass per volume:

\(\frac{\rho_1 V_1}{sec}=\frac{\rho_2 V_2}{sec}\)

If the pipe retains its cross section long enough then the volume swept out per second is simple the area times the length of the pipe:

\(\frac{\rho_1 A_1 l_1}{sec}=\frac{\rho_2 A_2 l_2}{sec}\)

We now recognize that length per second is the same as distance per second which of course is velocity. so:

\(\rho_1 A_1 v_1 = \rho_2 A_2 v_2\)

Now for an incompressible fluid the density will not change between the two surfaces, so: \(\rho_1=\rho_2=\rho\) And the density term on both sides cancels. This gives us the equation of continuity of flow as:

\(A_1 v_1=A_2 v_2\)

Bernoulli’s Equation

The fluid must also obey the conservation of energy. Let us take the following pipe:

Firstly, we will say this is a closed system, i.e. no heat will enter or escape the fluid. Secondly, there is no viscosity in the fluid, that is we are ignoring the internal forces of the liquid. So then the total energy of the liquid must remain constant at all locations. The total energy of the fluid at point 1 is the same as the total energy at point 2:

\(E_1=constant=E_2\)

The total energy is comprised of 2 parts, Kinetic energy and Potential energy. The Kinetic Energy is given by: \(KE=\frac{1}{2} m v^2\) For a fluid we like to think in terms of density so: \(KE=\frac{1}{2} \rho V v^2\)

The Gravitational Potential Energy is given by: \(GPE=m g h\) Again we want to use density: \(GPE=\rho V g h\)

In fact we actually have 2 forms of Potential energy, the first given by gravity, and the second due to the pressure in the liquid. Remember that pressure changes within the liquid, and pressure is force per area. We would then expect that the pressure is also a form of energy. This turns out to be very simply: \(PPE=P V\)

Putting all these terms together we have: \(TE=\frac{1}{2} \rho V v^2 + \rho V g h + P V = constant\)

Now recognize that the volume component in all these is the same, so we can separate it out and bring it across to the other side. For an incompressible liquid, the volume of the bit of water that we are following does not change, so volume remains constant. We are then left with the famous Bernoulli Principle:

\(\frac{1}{2} \rho v^2 + \rho g h + P = constant\)

Question

Using the pipe above can you extract Pascal’s Principle from continuity of flow and Bernoulli’s Principle?

Pascal's Principle

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Density

Density is defined as the amount of mass per unit volume:

\(Density=\frac{Mass}{Volume}\)

The units for Volume are \(\frac{kg}{m^3}\). Density is represented by the Greek letter \(\rho\) (rho) The air density is approximately \(1.2\frac{kg}{m^3}\). The Average density of muscle in the human body is \(\approx1200\frac{kg}{m^3}\)

Force, Pressure and Density

We can state the force of an object due to gravity in terms of density as being:

\(F=\rho V g\)

So if we know the density of an object we can calculate the force it exerts. Now if we know the area this force is acting on we can also calculate the pressure of this object as:

\(P=\frac{\rho V g}{A}\)

Now if we take a cylindrical or rectangular shape for the object then the volume of the object is given by the area times it’s height (h). so, \(P=\rho g h\)

Pascal’s principle

We can use the expression of pressure in terms of density to calculate the pressure inside a liquid at any point. We assume here a stationary liquid with no sideway forces.

If we draw a surface in the liquid the pressure exerted down through that surface is \(\rho gh\). On top of that comes the pressure exerted on the top surface of the liquid (usually atmospheric pressure). So the pressure at any height in the liquid is given by the relationship:

\(P(h)=P_{atmosphere} + \rho g h\)

This relationship is known as Pascal’s principle.

Since the liquid is at rest, the force at a point in the liquid downward needs to be countered with an equal force upwards. You can show that at any point in a liquid at rest the force in all directions is the same.

Worked Example

17. If the difference in height of the liquid is h=10cm, the external pressure is 100Pa, and the liquid is water with density 1000 \(\frac{km}{m^3}\), what is the unknown pressure? Calculate using Pascal’s Principle.

18. Since the liquid is at rest the pressure at the same height is the same. So if we take our baseline at surface A, the pressure is the same as the pressure on the other side of the bend at the same height. This pressure we know to calculate using Pascal’s principle:

\(P_{unknown}=P_b=P_{surface} + \rho g h\)

Using the values \(\rho=1000 \frac{kg}{m^3}\), \(h=0.1 m\) and \(P_{surface}=1000Pa\) we can calculate the unknown pressure to be:

\(P_{unknown}= 1000 + 1000\times0.1\times10 = 2000Pa\)

Pressure

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Pressure is defined as force per unit area and represented with an SI unit, the pascal.

\(p=\frac{F}{A}\)

Where \(p\) is the pressure, and \(F\) is the magnitude of force applied normal to the area \(A\). This means pressure is a scalar quantity.

For comparison \(1 Pa\) is approximatelly the pressure exerted by a stamp with a fly on top of it. The systolic blood pressure of the human body is \(\approx16 000 Pa\)

But what exactly does pressure mean? And why is it useful? In the figure below 4 different surfaces are drawn each with the same pressure.

We can see that as long as we keep the ratio of force and area the same the pressure remains constant. What’s interesting is we can keep shrinking them both until we essentially have a scalar value at a point. It’s often beneficial to think of pressure this way.

You may wonder why pressure is a scalar, why it has no concept of direction. Direction is certainly important when considering forces (imagine if gravity wasn’t pulling you down towards the earth).

So why isn’t it important when we talk about pressure? The reason is that the force is understood to act normal to the surface of the object on which the pressure acts.

Example 1: Ball in a Bath Tub

Imagine a tub of liquid (say water). Now just suppose that at every point in the tub the pressure is \(10\) Pa. What would happen if we put say a tennis ball right in the middle. In other words what would be the net force on the ball?

The figure above shows a small patch of area on the surface of the ball lets call it \(A_p\). Now we know the pressure is \(10\) pa, the force must be \(10\times A_p\) and its direction is normal to the surface. This means the force points straight towards the center of the ball.

So what is the total force that the liquid exerts on the ball? If it’s not immediately obvious think about it?

Its zero right? This is because for every small patch that causes a force toward the ball’s center there is always one on the other side of the ball that directly cancels it. If we include gravity the net force is \(mg\) downward.

So the ball will sink! But wait wouldn’t the tennis ball float? What assumption above is incorrect?

Example 2: A pressure change

For now lets just make a small modification to the bath tub example. What would happen if the pressure on the left of the ball was greater than the pressure on the right? For some help look at the picture below

The balls surface hasn’t changed, so all the forces still all point to the centre of the ball. But what has changed is that forces on the left hand side are all larger than the forces on the right. If we added up all the forces we would find that the total pressure force points to the right. The ball would start moving that way.

So coming back to the sinking tennis ball problem from the last example and in light of what we have just discovered, what is one way to make the tennis ball float? What can we say about how the pressure changes in a tub with real liquid?

Here we can see that if we know what the pressure is at every point in a fluid we can predict what forces will act on a submerged object. Also we can see that a change in pressure also known as a pressure gradient or a pressure differential can cause motion. For example a blood pressure differential is what causes blood to flow through our veins.

Pressures of Solids, Liquids and Gasses

The pressures that solids, liquids and gasses exert on their surroundings are all different.

Without other external forces acting on them, solids exert a pressure on the surface thay are resting on equivalent to the weight of the object acting on the area of material that touches the surface.

A liquid not only exerts a pressure downwards but also sidewards against the sides of the container that holds the liquid.

A gas exerts a pressure against all sides of it’s container (assuming a closed container), created by molecules bouncing around on the inside.

Viscous Flow

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Viscosity

Imagine that the fluid is made up of an infinite number of layers of fluid, each of which moves with their own velocity.Each of these layers of fluid experiences a kind of friction with the layers above and below it. For normal friction we have the friction coefficient \(\mu\). For fluids we have viscosity \(\eta\). Viscosity exists because of the interaction between the fluid molecules.

Viscosity is a property of the fluid. Fluids with a low viscosity flow easily, while fluids with a high viscosity require a large driving force to make the fluid flow. Viscosity has units \(\frac{N s}{m^2}\) which is also equal to \(Pa\cdot s\).

Fluids have the property that at a solid/liquid boundary the fluid velocity has to be zero. For low viscosity fluids the fluid reaches it’s maximum velocity at a small distance from the boundary. A high viscosity fluid requires a larger distance from the boundary to increase velocity.

To understand why you need to understand that the velocity has to be continuous across the cross section of the container. Now if instead of a solid, let us think of the boundary of being an infinitely-high viscosity fluid, that simply does not move. Then for the fluid to have a continuous velocity the fluid needs to approuch zero velocity at the boundary, or else there is a discontinuity of veloctity at the boundary.

Poiseuille’s Law

Consider an artery with blood of viscosity \(\eta\) flowing through it With Volumetric flow rate (velocity) H. The pressure difference between each end of the artery is going to have to be larger if the artery is longer. For a smaller artery the pressure needs to be higher to reach the same velocity (at the centre of the artery) since the fluid drops to zero at the boundary. This actually turns out to be a \(\frac{1}{r^4}\) correlation. Obviously the higher the viscosity the greater the pressure difference needs to be. Then there is also a constant factor which is \(\frac{8}{\pi}\).

Combining these terms we have the result known as Poiseuille’s law:

\(H=\frac{\Delta P \pi r^4}{8 \eta l}\)

Turbulence

Our approximation of the fluid existing as an infinite number of sheets us called laminar flow. However there comes a point where the fluid is not laminar anymore but becomes turbyllent. The layers of fluid mix and eddies, sections where the velocity is in a different direction than the direction of net fluid flow, occur.

The velocity at which turbulence starts is dependent on the viscosity and density of the fluid, as well as the shape and size of the pipe it is flowing through. We can define a factor that takes into account all of these properties, which is called the Reynolds Number:

\(Re=\frac{\rho v L}{\eta}\)

Where L is the characteristic length of the pipe, dependent on the shape of the pipe. For a circular cross section we will use the diameter of the pipe.

For a fluid in a closed pipe the flow is laminor for Reynolds number below 2000. Turbulence occurs for Reynolds number above 3000, with a semi-turbulent section between.

Question

The Aorta is the main artery that runs from the heart to the gut, most of the arteries in the body branch from the Aorta. Let’s say the Aorta is approximatelly 2 cm in diameter and extends for about 70 cm. If the density of blood is \(1060 \frac{kg}{m^3}\), and the viscosity of blood is \(4\times10^{-3}Pa\cdot s\), find:

1. the pressure required to achieve a Volumetric flow rate of \(5\times10^{-6} \frac{m^3}{s}\).
2. the velocity at which the blood flow becomes turbulent.