Physics 125 Dierker Lecture 24

Today's lecture: FLUID DYNAMICS

Last lecture we discussed the static properties of fluids. Today we will discuss the properties of moving fluids. Fluid dynamics involves the properties of moving fluids. There are two main ideas behind our understanding of the behavior of moving fluids.

1. Conservation of mass. This will lead us to the Continuity Equation.

2. Conservation of energy. This will lead us to Bernoulli's equation.

Continuity Equation (or, "What goes in, must come out!"):

Consider fluid flowing into a pipe whose entrace has an area A_{1} and whose exit has an area A_{2}.

The streamlines in this picture show the direction that the fluid flows in.

What is the speed of the fluid flowing into the pipe, v_{1}, and the speed of the fluid flowing out of the pipe, v_{2}? Consider how much fluid goes into and out of the pipe in a given amount of time t. Since mass is conserved (we don't lose any or make any additional mass inside of the tube), what goes in in the time t must come out in that time.

[mass flow / unit time]_{in} = [mass flow / unit time]_{out}

How much mass, m_{1}, goes in in the time t?

m_{1} = V_{1} (this is from the equation for the density, = mass/volume)

m_{1} = (A_{1}x_{1}) (from looking at the first disk's volume, denoted V_{l})

m_{1} = A_{1}(v_{1}t) (from velocity = displacement / time)

m_{1}/t = A_{1}v_{1} (mass per unit time entering the pipe)

This says that if the velocity is high, then more mass per unit time enters. Likewise, if the area through which the fluid passes is large, then more mass per unit time enters too. Think about it, it makes sense.

Similarly, the mass per unit time leaving the pipe is given by

m_{2}/t = A_{2}v_{2} (mass per unit time leaving the pipe)

Since we must have m_{1}/t = m_{2}/t, it must be that

A_{1}v_{1} = A_{2}v_{2}

The density cancel out here because we are dealing with the same fluid at both ends of the pipe. Therefore we are left with:

A_{1}v_{1} = A_{2}v_{2}

This is known as the Continuity equation. For example, if I place my thumb over the end of a hose, i.e., reduce the area, which has water coming out of it, my experience tells me that the water will come out more quickly. That is just what the continuity equation predicts.

Example 11.9 from the text - Blood flow through a heart.

The heart pumps blood out of the aorta into the arterial network. The area of the aorta and velocity of the blood coming out of it, and the area of a single artery are

A_{ao} = 2.5 x 10^{-2} m^{2}

v_{ao} = 0.33 m/s

A_{art} = 1.25 x 10^{-3} m^{2}

What is the velocity of the blood flowing through each artery?

We will apply the continuity equation here. We will also assume that all the arteries have the same cross-sectional area. Then, since there are 32 arteries, we have

A_{ao} v_{ao} = 32 A_{art} v_{art}

So,

v_{art} = A_{ao} v_{ao} / (32 A_{art})

v_{art} = (2.5 x 10^{-2} m^{2}) (0.33 m/s) /[(32) (1.25 x 10^{-3} m^{2})]

= 0.21 m/s

This tells us that as the blood enters each artery, the velocity of the blood flow is slower as compared to the blood flow in the aorta.

*Bernoulli's equation*: Energy conservation for fluids.

Consider a fluid flowing into a pipe again, as above, but now let the two front and back of the pipe be at different heights, y_{1} and y_{2}, respectively. Also, in general, the pressure in the fluid at the entrance and the exit, P_{1} and P_{2}, respectively, might differ. Bernoulli's equation is a statement of energy conservation for the fluid flowing in this pipe:

P_{1} + 1/2 v_{1}^{2} + g y_{1} = P_{2} + 1/2 v_{2}^{2} + g y_{2}

Do any of these parts of the equation look familiar?

P (pressure) = the work done by the pressure per unit volume

1/2 v^{2} = the kinetic energy per unit volume

gy = the potential energy per unit volume

Note that it is more convenient to express the various contributions to the energy of moving fluids as energy densities, (i.e., energy per unit volume) rather than total energies. See the text for a derivation of this equation. Is there an analog of friction for fluids? Yes, it is called viscous drag and can lead to energy loss. But we will ignore this in our discussions, as it is very complicated.

Example: Blood pressure

We know that pressure in the aorta is different than the pressure in the foot. Why? Let's examine this using Bernoulli's equation. The pertinent facts are

_{blood} = 1.1 x 10^{3} kg/m^{3}

A_{ao} = 2.5 x 10^{-2} m^{2}

v_{ao} = 0.33 m/s

P_{ao} = 1.3 x 10^{4} Pa

v_{foot} = 0.21 m/s

What is the blood pressure in the arteries in the foot, given that your feet are about 1.5 meters below your heart when you are standing up?

P_{ao} + 1/2 v_{ao}^{2} + g y_{ao} = P_{foot} + 1/2 v_{foot}^{2} + g y_{foot}

P_{foot} = P_{ao} + 1/2 (v_{ao}^{2} - v_{foot}^{2}) + g (y_{ao} - y_{foot})

P_{foot} = 1.3 x 10^{4} Pa + 1/2 (1.1 x 10^{3} kg/m^{3}) [(0.33 m/s)^{2} - (0.21 m/s)^{2}]

+ (1.1 x 10^{3} kg/m ) (10 m/s^{2}) (1.5 m)

P_{foot} = 1.3 x 10^{4} Nt/m^{2} + 36 Nt/m^{2} + 1.65 x 10^{4} Nt/m^{2}

P_{foot} = (Pressure/volume) + (KE/volume) + (PE/volume)

P_{foot} = 3 x 10^{4} Nt/m^{2} = 30 kPa

There are several points worth noting about this example.

1) KE/volume << Pressure/volume and PE/volume. That is, the KE makes a negligible contribution to the pressure difference between the aorta and the foot.

2) P_{foot} > P_{aorta} mostly due to the weight of the column of blood above your foot which pushes down and increases the blood pressure in your feet.

3) If we were to calculate the blood pressure in your brain, we would find it less than that of the aorta since it is higher up. The phenomena of fainting results because of a lack of blood going to the head. Fainting makes people fall down (the body goes from a vertical to a horizontal position; standing to lying). This horizontal position allows the low pressure of blood in the head to increase to that of the heart. Because the head and heart are at the same height when in a horizontal position, blood flows easier into the head region. Fainting is a clever response regulator of low blood pressure in the head. (Of course, lying down also increases the blood supply to your feet, but that doesn't affect most peoples brains!)

*VENTURI EFFECT*

Why do airplanes fly? Airplanes fly because of something called the Venturi Effect. To understand the Venturi effect, lets reconsider the problem of fluid flowing into the large end of a pipe and out of the small end of the same pipe. As we have seen, the velocity and pressure of the fluid at the inlet and outlet are not necessarily the same.

The Continuity equation gives us the relationship

A_{1} v_{1} = A_{2} v_{2}

Let's take the inlet and outlet of the pipe to be at the same height, y, so that the gravitational potential energy density for the fluid flowing into and out of the pipe is the same. Bernoulli's equation then gives us

P_{1} + 1/2 v_{1}^{2} + gy = P_{2}+1/2 v_{2}^{2} + gy_{}

The pressure drop in the fluid as it flows through the pipe is thus

P_{1} - P_{2} = 1/2 [v_{2}^{2} - v_{1}^{2}]

We can use the Continuity Equation to substitute for v_{1}, giving

P_{1} - P_{2} = 1/2 [v_{2}^{2} - v_{2}^{2} ( A_{2}/A_{1})^{2}]

= 1/2 v_{2}^{2}[1 - ( A_{2}/A_{1})^{2}]

Since in our example, A_{2} < A_{1}, then, A_{2}/A_{1} < 1 and [1 - ( A_{2}/A_{1})^{2}] > 0. Thus, P_{1} - P_{2} > 0, or P_{1} > P_{2}. We conclude that the pressure at the large end of the pipe is greater than the pressure at the small end of the pipe. The reason is that the fluid must flow more quickly at the small end of the pipe in order for the same amount of mass per unit time to pass out of the pipe as passes into the pipe. Thus, a greater fraction of the total energy density of the fluid is taken up by its kinetic energy density at the outlet than at the inlet, leaving less energy density left over in the form of pressure at the outlet than at the inlet.

Although we arrived at this conclusion by considering the specific case of a horizontal pipe with different areas on its inlet and outlet, our conclusion is a general one. That is, the pressure will generally decrease as a fluid flows more quickly.

Now, finally onto the airplane question. Why do planes fly?

Let's examine the air flow around an airplane wing. Here is a picture of an airplane wing's cross-section.

Imagine that the plane is moving to the left. The streamlines, which show the lines of flow of the fluid (in this case the air) around the wing are close together above the wing. This means the fluid is flowing more quickly there (it has to flow more quickly on top because it has further to go due to the shape of the wing) and this causes low pressure. The streamlines are further apart under the wing and this implies a high pressure. The high pressure pushes upward on the wing, causing the plane to rise. This is called LIFT.

What if an airplane wing took a symmetrical shape? The streamlines above and below the wing would be the same, therefore creating no difference in pressure. This situation will not give the plane any lift.

Demonstration: Funnel and ball.

Here we have a funnel attached to an air hose. Air is blowing out of the funnel. When a ball is placed into the funnel, what happens? We would expect the ball to be blown out of the funnel, however this is not what happens! Instead the ball stays stuck inside the funnel. Let's examine the streamlines. The difference in pressure pushes the ball into the funnel. This is exactly opposite of the example with the airplane and lift. Notice where the high and low pressures are and notice the direction of the force.

Lecture 24 - 5 DEC 1996