Assume you have created an indoor water fountain. You have connected pieces of pipe with different diameters into a path along which the water will flow. You also have inserted a pump into the circuit. A very simple circuit is shown in the figure below.
Running the pump for a while will accelerate the water and start it flowing. The pump creates a pressure gradient. If we look at a volume V of water in a straight section of pipe while the water is accelerating, then the pressure on side 1 of this volume is different than the pressure on side 2. This results in a net force on the volume of water in that section, and the volume of water accelerates.
Once the water is flowing at the chosen speed, the pump has to do much less work. If the pressure were the same on both sides of the volume V, then the net force would be zero, and the volume of water would continue to move with constant velocity. However, there will still be a small pressure gradient due to frictional forces. The pump now only has to do work against frictional forces. In a frictionless environment the pump would be no longer needed to keep the water flowing. Such a frictionless environment can actually be created. While most liquids freeze at near zero absolute temperature, liquid helium becomes a superfluid. It flows without friction.
For simplicity let us assume a frictionless environment and let us assume that the
water flows steadily through the circuit. The water in different sections of the circuit
has different gravitational potential energy per unit volume.
It also must have different kinetic energy per unit volume. In the narrower sections of
the pipe it must flow faster than in the wider sections, since the same amount of water
must flow across each cross sectional area in the same amount of time.
Is the pressure also different in different sections of the
pipe circuit?
Look at a particular volume of water. As it moves, the boundary 1 moves a distance l1 while boundary 2 moves a distance l2. Since water is incompressible, we have
Volume 1 = Volume 2 = V.
Area 1 ´ l1 = Area 2 ´
l2 = V.
Area 1 ´ dl1/dt = Area 2 ´
dl2/dt = dV/dt.
Area 1 ´ v1 = Area 2 ´
v2
This is the equation of continuity.
 | The volume flow rate of water through a horizontal pipe is 2 m3/min.
Determine the speed of flow at a point where the diameter of the pipe is (a) 10 cm, (b) 5 cm.
|
In different sections of a pipe circuit, a volume V of water can have different potential energy and different kinetic energy. Is the pressure also different in different sections of the pipe circuit?
The potential energy of the water changes as it moves. While all the water moves, the change in potential energy is the same as that of a volume V, which has been moved from position 1 to position 2. The potential energy of the water in the rest of the pipe is the same as the potential energy of the water in this section of the pipe before the movement. We have
change in potential energy = (mass of water in V) ´
g ´ (change in height)
= density ´ V ´ g
´ (h2–h1) = rVg(h2–h1).
The kinetic energy of the water also changes. Again we only have to find the change in kinetic energy in the small volume V, as if the water at position 1 had been replaced by the water at position 2. The kinetic energy of the water in the rest of the pipe is the same as the kinetic energy of the water in this section of the pipe before the movement. We have
change in kinetic energy = ½mv22 – ½mv12 = ½rVv22 – ½rVv12.
If the force on the water at position 1 is different than the force on the water at
position 2, then work is done on the water as it moves. The amount of work done is
W = F1l1 – F2l2.
But force = pressure times area, so
W = P1A1l1 – P2A2l2
= P1V – P2V
.
The work must equal the change in energy. We therefore have
P1V - P2V = rVg(h2–h1)
+ ½rVv22 – ½rVv12,
or
P1V + rVgh1 + ½rVv12
= P2V + rVgh2 + ½rVv22.
Dividing by V we have
P1 + rgh1 +
½rv12
= P2 + rgh2 + ½rv22
or
P + rgh + ½rv2 =
constant.
This is Bernoulli’s equation.
What does Bernoulli's equation mean?
|
If a liquid is at rest, then P1 + ρgh1
= P2 + ρgh2, |
|
If a liquid is at rest in a horizontal pipe, P1
= P2, the pressure is the same everywhere. |
|
If a liquid (or a gas which is not being compressed)
is flowing frictionless in a steady state through a horizontal
pipe, then the pressure depends on the speed of the fluid or the gas.
The faster the fluid is flowing, the lower is the pressure at the same
height. |
| A Venturi tube may be used as a fluid flow meter. If the difference
in pressure P1 - P2 = 21 kPa, find the fluid flow rate in m3/s
given that the radius of the outlet tube is 1cm, the radius of the inlet
tube is 2cm, and the fluid is gasoline (r =
700 kg/m3).
|
Pumps
To accelerate the water in a circuit and to overcome frictional forces while maintaining a steady flow, you use a pump. Pumps create pressure gradients. To accelerate a column of water, a pump either increases the pressure on one side of the column, or decrease the pressure on the other side of the column.
| Consider a simple U-shaped pipe filled with water. The water level is the same in each
leg and the pressure on each surface is the atmospheric pressure. How can you make the
water level rise in the right leg and sink in the left leg?
|
In a vertical, water-filled pipe gravity creates a pressure gradient in the water. The water below has to support the weight of the water above. The pressure exerted by 10 m or 33 feet of water overhead is equal to the atmospheric pressure at sea level. From Bernoulli's equation we have P1+rgh1 = P2+rgh2, where P1 is the pressure at height 1 and P2 is the pressure at height 2. If h1 is zero and the pressure on top of the pipe is 1atm, then P1 = 1atm+rgh2, i.e. the pressure at the bottom of the pipe increases with the height of the pipe. One way of maintaining pressure in plumbing is to have tall columns of water connected to the pipes. Many municipalities use a water tower build at a relative high site within their service region to maintain pressure in the water mains.
Suppose a U-shaped piece of pipe is completely submerged in water, filled with water, and then turned upside down under water. As you slowly pull the top of U-shaped piece of pipe out of the water, the water does not run out of the pipe. Why?
Air cannot enter the pipe. As the water starts running out of the pipe, a near vacuum is created in the topmost region of the inverted U. The pressure here drops to near zero. The atmospheric pressure on the surface of the water in the bucket pushes the water into the U-shaped pipe.
If a U-shaped hose or pipe connects a liquid-filled container at a higher altitude to a container at a lower altitude over a barrier, the liquid can be siphoned into the container at lower altitude. Atmospheric pressure helps to push the liquid over the barrier. In the diagram below P1 > P2, and the fluid is siphoned from the left to the right bucket.
Application: The toilet
Video: Toilet Flushing
Fluid in a pipe experiences frictional forces. There is friction with the walls of the pipe, and there is friction within the fluid itself, converting some of its kinetic energy into heat. The frictional forces that try to prevent different layers of fluid from sliding past each other are called viscous forces. Viscosity is a measure of a fluids resistance to relative motion within the fluid. We can measure the viscosity of a fluid by measuring the viscous drag between two plates.
If you measure the force to keep the upper plate moving with constant velocity v0, you find it is proportional to the area of the plate, and to v0/d, where d is the distance between the plate.
F/A = hv0/d
The proportional constant h is called the viscosity. The units of h in SI units are Pa-s.
| Viscosity of air (20 oC): 1.83*10-5 Pa-s |
| Viscosity of water (20 oC): 1.0*10-3 Pa-s |
| Viscosity of honey (20 oC): 1000 Pa-s |
The work done by viscous forces converts ordered energy into thermal energy. For fluid flowing in a long horizontal pipe, the pressure drops along the pipe in the direction of the flow. The faster the fluid is flowing, the larger is the pressure drop.
Under all circumstances where it has been experimentally checked, the velocity of a real fluid goes to zero at the surface of a solid object. A thin layer of fluid next to the walls of the pipe does not move at all. The speed of the fluid increases with distance from the walls of the pipe. If the viscosity of the fluid is low or the pipe has a large diameter, a large central region will flow with uniform velocity. For a high viscosity fluid the transition takes place over a large distance and in a small diameter pipe the velocity may vary throughout the pipe.
If the water is flowing smoothly through the pipe, it is in laminar flow. The velocity at a given point does not change in magnitude and direction. The water is flowing in a steady state. A small volume of fluid follows a streamline, and different streamlines do not cross. For laminar flow of fluids (and gases under certain conditions) Bernoulli's equation tells us that in the regions where the speed is higher the pressure is lower. If the streamlines are squeezed together in a region, the pressure is lower in that region. (In gases Bernoulli's equation can be applied to laminar flow if the flow speed is much smaller than the speed of sound in the gas. In air we can apply it if the flow speed is less than 300 km/h.)
The rate at which a fluid flows through a hose or a pipe is proportional to
| the pressure difference, |
| 1/viscosity, |
| 1/hose length, |
| (pipe diameter)4. |
These relationships are expressed in terms of an equation known as Poiseuille’s law.
Volume flow rate = pi*(pressure difference)*(pipe diameter)4/[128*(pipe length)*viscosity)
Problem:
The diameter of a pipe is doubled while the pressure difference across the pipe remains the same. By what factor does the volume flow rate of the pipe change?
|
Solution: |
Think about: What happens when your arteries become clogged?
If a fluid in laminar flow flows around an obstacle, it exerts a viscous drag on the obstacle. Frictional forces accelerate the fluid backward (against the direction of flow) and the obstacle forward (in the direction of flow).
While flying in an open cockpit airplane, you feel the air rushing past you. A person on the ground observes you moving through fairly stationary air. An object moving through a stationary fluid or gas is equivalent to a stationary object submerged in a fluid or gas flowing with the same speed in the opposite direction in another reference frame. The picture above can be viewed as a fluid flowing past a stationary sphere in laminar flow in one reference frame, or a sphere moving through the fluid in another reference frame.
Not all flow is laminar. In turbulent flow, water swirls erratically. The velocity at a given point can change in magnitude and direction. The onset of turbulent flow depends on the fluids speed, its viscosity, its density, and the size of the obstacle it encounters. A single number, called the Reynolds number, can be used to predict the onset of turbulent flow. For the flow past a smooth cylinder of diameter D we have
The Reynolds number has no units, the units on the right hand side of the equation all cancel out. It increases with the flow speed and decreases with the viscosity. Turbulence appears when the Reynolds number is about 2300.
Problem:
About how fast can a small fish swim before experiencing turbulent flow around its body?
| Solution: Flow speed = (Reynolds number * viscosity)/(D * density) Assume the length of the fish is ~10 cm. The viscosity of water is 0.001 Pa-s. The density of water is 1000 kg/m3. If the Reynolds number is greater than 2300, the flow is turbulent. So the critical speed is v ~ (2300*0.001 Pa s)/(0.1 m*1000 kg/m3) = 0.023 m/s |
Videos:
| Transition from laminar to turbulent flow |
| Turbulent flow in a pipe |
Under conditions of turbulent flow Bernoulli's equation is not applicable. It was derived by equating the work done by pressure forces to the change in the potential energy and the ordered kinetic energy of the fluid. Under conditions of turbulent flow the fluid gains disordered kinetic energy. More work is done on the fluid and a greater pressure difference is needed to move the fluid at the same rate.
Only recently have scientist been able to gain a better understanding of the patterns observed in turbulent flow under different circumstances. The study of chaos is providing new insights into many related phenomena exhibiting turbulence, such as global weather patterns, the atmosphere of Jupiter, etc.
When a fluid or gas in laminar flow streams past an object, it exerts a viscous drag on the object. When the objects shape changes the direction of flow, the object is acted on by lift as well as drag forces.
In the above picture, the airplane wing redirects the air flow. The air flowing over the airplane wing moves with greater speed than the air flowing underneath. The pressure above the wing is lower that the pressure below the wing. This results in a net upward force on the wing.
Airflow past ordinary size objects is only laminar at low speed. As the airspeed increases, a boundary layer forms. This boundary layer fills the region behind the object with a turbulent wake.
The figure above shows the airflow past a cylinder as the airspeed and therefore the Reynolds number increase. Remember:
| The first two pictures show laminar flow at low speed. The air directly
before and behind the cylinder comes to a stop. The pressure is highest
here, but the net force on the cylinder due to pressure differences is
approximately zero. There is no pressure drag
due to pressure differences between the front and back side of the
cylinder. The cylinder experiences only viscous drag due to air friction. |
| In the fourth picture a turbulent wake has formed. The air behind the
cylinder no longer slows down and the pressure no longer rises behind the
cylinder. Due to the high pressure in front of the cylinder it now
experiences a large pressure drag. This
happens for Reynolds numbers of approximately 2000 to 100000. The pressure
drag is much larger than the viscous drag. It can decrease the forward
component of velocity of an object moving through a fluid or gas very
rapidly. A thrown object can seem to stop in midair and drop straight to
the ground. You can easily observe this by throwing an air-filled balloon. |
| As the airspeed increases and the Reynolds number becomes larger than 100000, the turbulent region works itself forward as shown in the fifth picture. We have what is called a turbulent boundary layer. The flow lines now separate from the cylinder and follow the turbulent boundary layer, as shown in the fifth picture. We have something similar to laminar flow around an object of a different shape. The pressure behind the object rises again and the pressure drag is drastically reduced. |
The condition of the surface determines when the Reynolds number becomes critical and a turbulent boundary layer forms. The rougher the surface, the lower the speed at which the Reynolds number becomes critical. The surface of many balls used in a variety of sports is intentionally roughed up. Golf balls have dimples and grooves, tennis balls have hair, etc. This decreases the speed at which the Reynolds number becomes critical, and a turbulent boundary layer forms even at moderately high speeds. In this way pressure drag can be largely eliminated and only the viscous drag acts on the ball.
 |
| The shape of an object can redirect airflow, thus
producing lift. Spinning symmetrical objects can also produce lift.
Even for laminar flow, a thin layer of
air next to the object does not move with respect to the object. A
thin layer of air next to a spinning ball spins with the ball. As
the distance from the ball increases, the speed of the air changes,
so that the airflow around the ball exhibits the patterns shown in
the figure on the right.
If the ball is spinning clockwise as seen in the diagram, the air flows faster over the top of the ball. The pressure on the top is lower than the pressure on the bottom of the ball, and there is a net lift force on the ball towards the top of the diagram. This force is called the Magnus Force. Note: The direction of the lift force depends on the direction of the ball's spin. The lift force does not have to point upwards. Animation: Magnus Effect |
 |
||
| Turbulent regions can also form. The wake behind
the ball can be deflected because the spinning surface pulls air
with it. This again can result in a net lift force. This force is
called the wake deflection force.
|
 |
||
| We can also just use
Newton's third law to explain the lift force on the ball. The spinning ball interacts with the air molecules. The total momentum of interacting objects is conserved. If the air molecules are deflected in on direction, then the ball must be deflected in the other direction. |
Spinning baseball in a water tunnel |
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| Problem:
Suppose you are practicing baseball and throw the ball with a spin so that it moves forward and curves to the right. Now when the actual game takes place you are pitching into the wind. Please explain how you could alter your throw so as to make the ball take the same path.
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Demonstration:
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| An airplane's weight always acts in the downward direction. By
interacting with the surrounding air, the airplane's wings push some of the air
moving relative to them downward. The reaction force (Newton's third law)
pushes the wings and therefore the airplane in the upward direction.
|
|
| The interaction between the air and a wing is complicated. The air is a viscous fluid flowing with high speed around the wing. The Reynolds number is certainly much larger than 2300 and a turbulent boundary later forms. A wing shaped like an airfoil can prevent the turbulent boundary layer from separating over a large range of air speeds. The remaining air then flows past this boundary layer in approximately laminar flow. |
|
| To generate lift with a symmetric airfoil, the
airfoil must be turned, so that its symmetry axis makes an angle
with the direction of the airflow. This angle is called the
angle of attack of the wing with respect to the free-stream
flow. The airfoil now deflects the airflow downward and the reaction force pushes the airfoil upward. The air also flows faster over the upper surface than over the lower surface. The pressure therefore is lower near the top than near the bottom of the wing. Again we conclude that the wing is pushed upward. |
 |
| Lift can also be generated by giving the airfoil an
asymmetric shape. Again , the airflow is deflected and the air
flows faster over the upper surface than over the lower surface.
The wing is pushed up. Link: Shape effect on lift |
|
| The relative velocity of airplane and air is the most important factor in producing lift. The lift will increase proportional to the square of the velocity. Increasing the angle of attack increases the lift, but also increases the drag. When the airspeed falls below a critical speed, (i.e. the Reynolds number falls below approximately 100000), a turbulent wake begins to detach. The wing looses all lift, and pressure drag increases dramatically. The airplane stalls, and without intervention by the pilot falls nearly straight to the ground. |
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| Link: Flow over airfoils | |
Link:
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Bernoulli
Ball (This is an experiment you can try at home.) |
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