wind vs relative wind (A Thermdynamics and aerodynamics issue)


by Brendon_agius
Tags: aerodynamics, airfoil, sailing, thermodynamics
Brendon_agius
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#1
Jul14-09, 10:06 PM
P: 8
The below though experiment which i devised, raises some personal concerns about my understanding of thermodynamics and aerodynamics, based on the fact that they seem to predict different outcomes. If anyone out there could offer some insight into what would actually happen, it would be greatly appreciated.

Its recommended that you read up on the physics of a tack maneuver here http://www.physclips.unsw.edu.au/jw/sailing.html before reading

Bob an aerodynamics expert, invites John a thermodynamics expert on to his sailboat. Its a no wind day and they are late to the physics faculty barbecue, which is down stream. The current is moving at quite a pace and bob notices there is in fact enough relative wind to conduct a tack maneuver, which should allow them to get to there destination sooner. When he tells John of this, John dismisses this saying that it would be impossible to gain extra momentum from nowhere. Bob states that the Bernoulli's principle allows the heat content of the air to be utilized by the airfoil to create greater lift force than drag. John replies by stating that such things are prohibited by the second law of thermodynamics, as such an interaction would reduce entropy.

At this moment in time i would have to side with Bob, i however am quite open to that opinion being wrong.

thank you for taking the time to read
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rcgldr
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Jul15-09, 12:01 AM
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I was thinking the boat had to achieve a VMG (Velocity Made Good) upwind greater than the relative wind speed, but it only needs to achive a VMG upwind > 0, so it taching would reduce the time to arrive at the barbecue.

Assume the stream speed is 5 mph, and the sailboat tachs so it's VMG downstream is 1mph faster.

Air frame of reference:
wind speed = 0 mph
stream speed = 5 mph
sailboat speed = 6 mph

Water frame of reference:
wind speed = -10 mph
stream speed = 0 mph
sailboat speed = 1 mph

heat
heat is not a significant issue here. Using the water as a frame of reference, the reduction of the relative wind speed by the sail's upwind force extracts kinetic energy from the downwind air, and adds it to the sailboat, minus the losses. Using the still air as the frame of reference, the upstream force from the air is transmitted through the sailboat as an upstream force to the water, slowing the water down, extracting kinetic energy from the water, and adding it to the sailcraft.

In the case where the air is the frame of reference, even though the sailcraft's component of speed in the direction of stream flow is greater than the stream, it's forward speed combined with it's hull and keel's angle of attack results in a diverted upstream wash relative to the sailboat, which slows the true downstream and extracts energy from it.

If the entire system (water, air, sailboat) is taken into account then momentum and energy are conserved and just exchanged via interaction between them.

The explanations about "relative wind" at most sailboat sites are misleading. There are two main issues:

1. Forces - The apparent wind experienced by a sailcraft can be broken into components parallel and perpendicular to the path of the sailcraft (with respect to the ground, ice, or water). Since the apparent crosswind is perpendicular to the path of the sailcraft, it's independent of the sailcrafts' speed. The maximum speed is related to how large the ratio of apparent headwind / apparent crosswind can be, for a given wind speed and heading. The limiting factors are the lift to drag ratio of the sail and the aerodyamic and ground interface related sources of drag.

2. Extraction of energy - Using the ground (ice or water) as a frame of reference, the sailcraft must extract energy from the true wind by slowing the wind down. In the cases where a sailcrafts component of speed in the direction of the wind is greater than the wind, the sail is diverting the apparent wind with sufficient upwind component relative to the sailcraft that the true wind is slowed down. For example, if the sailcraft's downwind component of speed is 10 mph, then relative to the sailcraft the component of upwind speed wash diverted from the sail might be 11 mph, which would reduce the true wind speed by 1mph. It's this aspect of diversion of apparent wind that explains how the sail craft extract energy from the wind.

In terms of forces, the apparent headwind is an overhead which the sailcraft must overcome, for a given apparent crosswind. In terms of energy extraction, the apparent wind must be diverted upwind enough to slow the true wind and extract energy from it.

For sailcraft β (beta) is the angle defined as the angle between the apparent wind and a sailcrafts true heading. It can also be defined as

β = tan-1(apparent_crosswind / apparent_headwind)


Earlier posts about sailcraft, mostly about iceboats, but they include some of the math:

http://www.physicsforums.com/showthread.php?t=283813

http://www.physicsforums.com/showthread.php?t=273787


Bernoulli
(xxChrisxx was right, I should just keep a copy of my standard response to Bernoulli principle):

In most (real world) cases, the interaction between a solid moving through a fluid or gas violates Bernoulli principle. Bernoulli principle applies to the interaction between the surrounding fluid or gas and the fluid or gas that interacts directly with the solid.

But at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (by the propeller) violates an assumption used to derive the equation.

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html
Brendon_agius
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#3
Jul15-09, 01:44 AM
P: 8
Thank you for your response. I however am not getting a grip on your momentum transfer explanation. If one considers the reference frame of the shore, one cannot extract macroscopic momentum from the still air, as it has no macroscopic momentum. If a sailboat can conduct a tack maneuver under this circumstance, the energy initially has come from the water current to allow the tack, but once the the sailboat exceeds the speed of this water during the tack, it will only serve to sap energy from the sailboat due to drag as the water flows relatively in the opposite direction.

If a tack can be preformed from under such circumstances energy will have to move from the still air, into the sailboat and then into the water. Such an event to me seems to defy the second law of thermodynamics.

rcgldr
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Jul15-09, 02:01 AM
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wind vs relative wind (A Thermdynamics and aerodynamics issue)


Quote Quote by Brendon_agius View Post
If one considers the reference frame of the shore, one cannot extract macroscopic momentum from the still air.
In this case it's extracting momentum from the water, by slowing down the rate of downstream flow. The air provides a resistant force to the downstream movement of the sailcraft which the sailcraft utilizes to exert an upstream force against the water, slowing the water down.

I explained that here:

"In the case where the air (or the shore) is the frame of reference, even though the sailcraft's component of speed in the direction of stream flow is greater than the stream, it's forward speed combined with it's hull and keel's angle of attack results in a diverted upstream wash relative to the sailboat, which slows the true downstream and extracts energy from it."

The sailcraft makes use of the differential in speed between air and water. It has to extract more energy from one of them than it adds to the other in order to have a net inflow of energy that it can use to propel itself. The amount of net inflow of energy will be the same regardless of the frame of refence used (air, shore, or water).

Instead of a sail, the boat could use a windmill to drive a prop and achieve faster than downstream travel while traveling directly downstream. The effeciency doesn't have to be that high. Say the windmill boat moves 1mph faster than a 10mph stream. Relative to the windmill boat, the apparent headwind is 11 mph, while the apparent headstream is 1 mph. Say the effective gearing is 10 to 1 between windmill and prop. Ignoring losses, the speed at the prop is 1.1mph and the torque at the prop is 10 times the torque from the windmill. From a work standpoint, the power input is the force times speed of the headwind through the windmill, and the power output is the higher force but much lower speed of the prop thorugh the water. The output power can be much less than the input power because the headwind speed is so much higher than the relative stream speed. Despite losses in the system, the energy ouput requirement is much less than energy input available, so the system can be lossy and still work.

The Brennan torpedo is a similar example, using wires that were retracted to drive propellers that drove the torpedo forwards:

http://en.wikipedia.org/wiki/Brennan_torpedo
Brendon_agius
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#5
Jul15-09, 06:59 PM
P: 8
Thank you once again for your reply. My main point of concern is highlighted by the title "wind vs relative wind". Under standard wind, the sail acts as an airfoil when conducting a tack maneuver that achieves greater sidewards pull than drag force. What it seems like you are stating is that relative wind does not allow an airfoil to achieve greater sidewards pull than drag force, as you state no energy is taken from the air.

The idea that relative wind would differ from actual wind, is something I'm finding almost impossible to comprehend.
rcgldr
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#6
Jul15-09, 10:21 PM
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Quote Quote by Brendon_agius View Post
"wind vs relative wind". The idea that relative wind would differ from actual wind, is something I'm finding almost impossible to comprehend.
Velocity is alway relative to some frame of reference. For a sailcraft the 3 most common frame of references would be the air, the shore, or the water.

For a more general thought about "relative", note that at the equator of the earth, the surface of the earth is rotating eastward at about 1040 mph. Earth orbits the Sun at about 67,000 mph, and the Sun orbits the Milky Way at about 486,000 mph. So what is the speed of the "wind" in this example?

Under standard wind, the sail acts as an airfoil when conducting a tack maneuver that achieves greater sidewards pull than drag force.
Depending on the frame of reference, the sail is moving through the air, or the air is moving past the sail, or both the air and the sail are moving with respect to some ground based reference.

What it seems like you are stating is that relative wind does not allow an airfoil to achieve greater sidewards pull than drag force, as you state no energy is taken from the air.
Using the shore or the still air as a frame of reference, I only stated that the energy added to the air must be less than the energy extracted from the water, in order for there to be a net energy input source. Even though kinetic energy is relative to a frame of reference, the calculated net energy input should be the same regardless of which reference frame is use to make the energy calculations.

I didn't mention this before, but what an airfoil can't do is to convert an apparent headwind into thrust. Only the apparent crosswind component of the apparent wind can be converted into thrust. I covered this in my "forces" issue. From a "force" perspective, the apparent headwind component is an overhead.
Brendon_agius
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#7
Jul16-09, 01:26 AM
P: 8
If its not much trouble, it would be greatly appreciated if you could offer some perspective on this particular thought experiment. If it requires more clarity please ask

A low friction puck has placed on top of it an airfoil designed so that*side-wards*pull on the device is greater than drag(device resembles a wind surfer). This device is accelerated on a track and then exits on to a level, low friction plain (no wind present) where it is no longer has thrust applied to it. One may predict that such a device upon entering the the plain would change its direction as it is pulled to one side and*gain momentum*due to the fact that the*side-wards*pull on the device is greater than drag.

This gain in momentum would defy the second law of thermodynamics, as the only energy source available is the heat content of the air.
If such a gain in momentum is in fact impossible, it would mean that an airfoil could not produce grater lift force than drag.
thank you once again
rcgldr
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Jul16-09, 02:52 AM
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Quote Quote by Brendon_agius View Post
A low friction puck with an airfoil on top, frictionless plane.
The instant the puck is sliding on the plane it is losing kinetic energy, transfer its kinetic energy to the air. Assuming some type of control system that maintains the angle of attack, the path of the puck is a decreasing radius spiral. The greater the lift, the tighter the radius. The lower the drag, the lower the deceleration in speed and radius of the spiral. The momentum changes direction, but decreases in value along with the speed depending on the drag.

While the puck is being accelerated inwards (lift) and backwards (drag), the air affected by the puck is being accelerated outwards and forwards, the Newton third law pair of forces. This interaction is how the kinetic energy is transferred from the puck to the air.
Brendon_agius
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#9
Jul16-09, 05:58 AM
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So an airfoil cannot achieve greater lift force than drag force
rcgldr
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Jul16-09, 05:19 PM
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Quote Quote by Brendon_agius View Post
So an airfoil cannot achieve greater lift force than drag force
I never stated that. In the case of high end gliders, such as a Nimbus 4T:

http://en.wikipedia.org/wiki/Schempp-Hirth_Nimbus-4

the glide ratio of the entire glider is 60:1, which corresponds to a lift to drag ratio of 4:1.

If the Nimbus weight was 1500lbs, and acheived 60:1 glide ratio at 60.0083327547 mph, the decent rate would be 1 mph, the forward speed would be 60mph, and the power consumed would be only 4hp (4 hp = 1500 lbs x 1 mph / 375 conversion factor). The power input is 4 hp from gravity, and the glider in turn transfers the energy gained from gravity to the air at the rate of 4 hp.

What I stated was that an airfoil can't produce thrust from an apparent headwind. Continuing with the glider example, imagine there's a an updraft of 1 mph. The glider can then maintain level flight at 60 mph forward. The updraft of 1 mph is the apparent crosswind, which the wings divert into thrust sufficient to propel the glider forward at 60mph. At this speed the thrust from the wings is equally opposed by the drag, which is related to the apparent headwind, so the glider ceases to accelerate once at 60 mph. The wings cannot generate any thrust from the apparent headwind of 60mph. If the updraft ceases, and the glider maintains horizontal flight, it slows down (until it can no longer maintain horizontal flight) because the wings can't generate thrust from the apparent headwind, only from an apparent crosswind (the updraft in this case).

The text at some sailcraft web sites make the implication that the faster the glider moves forward, the more thrust (forwards force) the wings can produce, which is false. In the case of a sailcraft, the total force will increase, but not the component of this force in the direction of travel (thrust). Getting back to the glider, the amount of thrust the wings can produce is related to the apparent crosswind (in this example the updraft) only; the apparent headwind is an overhead, and actually reduces the thrust produced as forward speed increases due to drag. What these web sites don't make clear is that the role that the apparent headwind plays is that it and the apparent crosswind are diverted with sufficient "upwind" component against the "true wind" to slow down the true wind (in this case the updraft) and extract energy from it.
Brendon_agius
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#11
Jul16-09, 10:28 PM
P: 8
I now understand that I have over generalized my original thought experiment. I am sorry as I have difficulty understanding the discourse, its been a while since I've studied physics which did not cover much aerodynamics. The second thought experiment is my original, with the first devised in an effort to make the problem more understandable and less pretentious by applying it to something somewhat everyday. When I devised the first presented thought experiment, I had envisioned a tack occurring at 45 degrees to the wind direction. I now understand that a lift/drag ratio exceeding 1 is not necessary for a tack to occur, as one can conduct a tack at angles closer to the perpendicular of wind direction.
I do however still believe that the ability for an airfoil to achieve lift/drag ratios exceeding 1 such as 4 you say is exhibited by the Nimbus defy the second law of thermodynamics. The only reason that I had the water moving instead of there being actual wind, was to aid understanding as there is no real difference based on the principle of relativity, to me it seems that anything that can perform some kind of tack where by it has moved a greater distance forward than perpendicular has absorbed some energy from the heat content of the air.
If you consider the device described in the second thought experiment, has the same lift drag ratio as the Nimbus, it would experience as soon as it enters the plane, 4 times the acceleration to one side than deceleration at its front and hence increase in momentum at that particular moment.
rcgldr
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Jul16-09, 11:29 PM
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Quote Quote by Brendon_agius View Post
experience 4 times the acceleration to one side than deceleration at its front and hence increase in momentum at that particular moment.
By definition, lift is perpendicular to the direction of travel, so it doesn't change the overall speed of an object. Thrust and drag are in the direction of travel and these change the overall speed of an object. The amount of drag determines the rate of deceleration of the object. The amount of lift affects the radius of the path, radius = mass x speed2 / lift. If the drag was zero, the object would simply travel in a circle, never gaining or losing speed or momentum.

Going back to the 1500 lb Nimbus 4T in a 1 mph updraft:

Using a ground based frame of reference.

apparent headwind = 60 mph
apparent crosswind = 1 mph
thrust generated by the wings due to diversion of apparent crosswind = 24.9930575 lbs.
component of drag of entire glider opposing thrust due to deceleration of apparent headwind = -24.9930575 lbs
vertical force generated by the wings due to diversion of the apparent wind = 1500 lbs
gravity force opposing lift = -1500 lbs

The forces cancel, so the glider does not accelerate.

Using an air based frame of reference:

air speed = 60.0083327547 mph
horizontal speed = 60 mph
vertical speed = 1 mph
glide slope = 60:1
glide angle = 0.95484 degrees down from horizontal = tan-1(1/60)
lift = 1499.79171 lbs (1499.583449 lbs vertical, 24.9930575 lbs horizontal)
drag = 24.99653 lbs (0.416551 lbs vertical, -24.9930575 lbs horizontal.
lift + drag = 1500 lbs vertical, 0 lbs horizontal.
gravity = -1500 lbs
lift + drag + gravity = 0 lbs vertical, 0 lbs horizontal

The forces cancel, so the glider does not accelerate.

power output to air = drag x speed / 375 = 24.99653 * 60.0083327547 / 375 = 4 hp
power input from gravity = weight x vertical_speed / 375 = 1500 * 1 / 375 = 4 hp

Back to your question, assume there is no gravity and zero drag on the glider, and that it's speed is 60 mph = 88 fps. Since there is zero drag, the glider maintains it's 88 fps speed. Since there is 1500 lbs of lift, the glider experiences 1 g (32.174 ft / sec2) of centripetal acceleration, flying in a circle with radius 240.691 feet. If there is drag, then the glider's speed decreases over time regardless of it's path.
Brendon_agius
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#13
Jul18-09, 04:05 AM
P: 8
I always considered that the behavior of the device in my second presented thought experiment, would be somewhat like that of a projectile launched perpendicular, though I am now skeptical that this is the case.

If you consider the x axis (Parallel with track) and y axis, initially upon exiting onto the plane, its velocity may be described by these formulas.

(velocity in x axis) = (Track exit speed) - ( - Deceleration due to drag)(Elapsed time after exiting track)

(velocity in y axis) = 4( - Deceleration due to drag)

(Magnitude of velocity)^2 = (velocity in y axis)^2 + (velocity in x axis)^2
= 17( - Deceleration due to drag)^2( Elapsed time after exiting track)^2 + (Track exit speed)^2 - 2( - Deceleration due to drag)(Elapsed time after exiting track)

From this formula one can determine that the net speed of the device increases over time and hence has gained energy.

This formula wont be appropriate for long based on the fact that the relative wind direction will change and lower the lift/drag ratio.
rcgldr
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Jul18-09, 07:46 AM
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Even without drag in the direction of travel, once the puck is no longer headed in the direction of the x axis, the lift from the sail slows down the component of speed in the x direction while it games speed in the y direction, such that x2 + y2 = radius of circular path.

The sail can't accelerate the puck in the y direction without also slowing down it's x direction component.

Without any drag factor a good analogy of the sail would be the tires of a turning car. The cars overall speed doesn't change, but it's direction does.

The lift to drag ratio is relative to the current path of the puck. It's not relative to the x-axis except for the instantaneous moment the puck path is in the direction of the x axis.
Brendon_agius
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#15
Jul18-09, 05:51 PM
P: 8
I guess I have always considered that the angle of attack would not change and so as a consequence the sideward pull would not be perpendicular to device velocity beyond the initial instant.
If this is not the case I can understand how it would not gain any energy from the force applied, like an object in orbit. It's just that I always considered that it would behave more like an object falling towards earth that has a low horizontal velocity, that although it slowing down in the x axis the y axis is speeding it up at a greater magnitude.
rcgldr
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#16
Jul18-09, 06:23 PM
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The best a sail can do is to generate a force perpendicular to the direction of the apparent wind. If there is any drag, then the force will include a component in the direction of the apparent wind ("backwards"). A sail cannot generate a component of force forwards relative to the apparent wind, only a powered interface such as a propeller could accomplish this.

Once the puck has a +y component of direction, then the lift force from the sail will have a -x component of direction.

If the angle of attack of the sail was adjusted so that it remained constant with respect to the x axis, then lift approaches zero as the direction of the puck lines up with the sail. The angle of attack with respect to the apparent wind is decreasing over time as the puck changes direction. As mentioned above, at best the sail can only generate a force perpendicular to the direction of the apparent wind, which in this case is the negative of the direction of the puck. So with even with zero drag, the same rule applies: once the puck has a +y component of direction, then the lift force from the sail will have a -x component of direction. The only difference in this case is that the lift force is decreasing over time. The path of the puck would be similar to a hyperbola, and if there is zero drag, then the puck's speed would remain constant.


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