Would a wind turbine on a moving car generate energy efficiently?

[Xtensity]

I was talking on another forum about an idea someone proposed. They said if you put a miniature wind turbine as a Hood Ornament on a car, as the car drives, it would produce extra usable energy.


I said this would not work as if the wind is not blowing at the correct angle and velocity then the wind turbine on the car would actually slow it down to a minute degree. Meaning, the friction between the air and the wind turbine would slow down the car(to a very small degree) and the car would have to speed up to compensate for the loss of kinetic energy. The energy used to speed up the car to make up for the loss would be greater than the energy produced by the turbine meaning the energy consumption is being increased even more.

Who is correct and can you explain why? I've already tried to explain that the energy has to come from somewhere to move the wheel, and if it isn't coming from the non-moving air then it must be coming from the car.

 

[Danger]

You have correctly explained it yourself, in useful terminology. Although elaboration is possible, it certainly isn't necessary. You're right; they're wrong. (Feel free to quote the Laws of Thermodynamics if you feel so obliged.)

 

[Xtensity]

Yes but I need someone to explain it in clearer terms. I have tried to explain it to the person (same as I did above) and they didn't understand me. I'm going to be linking them to this topic so if anyone else has a clearer explanation, feel free.

 

[dbell5]

Tanstaafl!

 

[OmCheeto]

Yes but I need someone to explain it in clearer terms. I have tried to explain it to the person (same as I did above) and they didn't understand me. I'm going to be linking them to this topic so if anyone else has a clearer explanation, feel free.

You could build a model.
Take the motor-generator I built this winter for example.

On the left is the generator. Driving the generator via a fan belt is the motor.
When the fan belt was removed, the motor consumed 200 watts.
When the fan belt was installed, the motor consumed 300 watts.
This was just drag from the idling generator and belt.
If the generator was then hooked to a 40 watt load, the power consumed by the motor would probably go up another 100 watts.
The same thing is happening in your wind driven generators.
For every watt they generate, your vehicles primary engine is going to consume more than that watt. (Unless of course you turn off the car while driving or going down a hill and turn off the car, which would be unwise as power steering and power brakes would fail and you would probably die trying this experiment)

I was trying to explain the concept of regenerative braking to someone once, and they could not get it into their head that a generator under electrical load exerts a torque. Unless that torque is countered, the generator will quickly stop rotating.

The torque required to keep your friends little hood ornament fans rotating under load is called wind drag. That wind drag has to be countered by your vehicles primary engine.

 

[brainstorm]

Any energy you got from the generator could more efficiently be subtracted from the wind drag of the vehicle by increasing its aerodynamics. I.e. I think removing the hood ornament altogether would generate convert more energy into forward motion than it would deliver as electricity. If you want more electricity, why not just increase the output of your alternator and take the energy directly from the engine instead of first converting it into forward momentum and then getting it back from the wind drag?

 

[cjl]

Yes but I need someone to explain it in clearer terms. I have tried to explain it to the person (same as I did above) and they didn't understand me. I'm going to be linking them to this topic so if anyone else has a clearer explanation, feel free.

Wind turbines have a rather phenomenal amount of drag, and the amount of drag produced by the turbine will always outweigh the amount of energy you can get out of it.

 

[ErikD]

Just quote the http://en.wikipedia.org/wiki/Conservation_of_energy" . When the car is standing still the turbine isn't moving. When the car is moving the turbine will move. So the energy for the turbine's movement comes from the car's movement. And because of the law of conservation of energy either the car will slow down a bit or will use more gas. If we would live in a frictionless and dragless world and you had a 100% efficient generator attached to the turbine you would still only get back the same amount of energy the car lost (in speed or gas). Now we don't live in such a world, so a portion of that energy will for example be lost to heat due to friction with the air and friction inside the turbine. So the turbine could never generate extra energy.

 

[Lsos]

One the most fundamental laws of physics states that this won't work. You can't make energy, you can just convert it from one form to another...and every time you do so, you WILL lose some. What your friend is proposing is a perpetual motion machine: something that creates more energy than it consumes. It doesn't work and it will never work. If he doesn't understand this then it's going to be very difficult to explain.

The main reason a car uses fuel in the first place is because of air resistance. If it wasn't there, you could just accelerate to speed and practically coast most of the way. A wind turbine on a car will increase air resistance, requiring more fuel. What's more, it will ALWAYS increase the air resistance by such an amount, that the car will require to burn more fuel than the amount of energy the turbine puts back in. Always. No matter what elaborate scheme you come up with...you can use turbines, magnets, slopes, gears, levers, thermocouples, you can split atoms and enlist all the forces of nature, and you will ALWAYS lose.

 

[Danger]

There is one practical application—a turbine mounted on a stationary vehicle in the presence of a significant breeze. You could, for instance, stick one on the roof of a parked RV to power your TV or computer. In that case, there would be no parasitic losses to the vehicle.

 

[Bob S]

The total wind power incident on a stationary horizontal axis wind turbine (HAWT) is P = ½ρAv³ (Newton-meters/sec), where ρ is the air density, A is the frontal area of the HAWT, and v is the wind velocity. The maximum theoretical efficiency of converting this wind power to turbine power is 16/27 = 59% (called the Betz limit). because the air needs sufficient residual energy (velocity) to move away from behind the turbine after passing through the blades. See

http://www.wind-power-program.com/betz.htm

Actual HAWT maximum efficiencies are ~40 to 45%. So putting a HAWT on the hood of a car will produce more drag force (F = ½ρC_dAv², C_d is drag coefficient) than it produces in electrical power. See

http://en.wikipedia.org/wiki/Drag_(physics )

Bob S

 

[brainstorm]

What if the wind turbines were placed in such a way as to reduce drag by redirecting the flow of air, such as at the back of a vehicle. If the turbines shifted the air flow so that it reduced the low pressure behind the vehicle, couldn't the vehicle theoretically move forward more easily while the turbine motion also generated electricity?

 

[russ_watters]

What if the wind turbines were placed in such a way as to reduce drag by redirecting the flow of air, such as at the back of a vehicle. If the turbines shifted the air flow so that it reduced the low pressure behind the vehicle, couldn't the vehicle theoretically move forward more easily while the turbine motion also generated electricity?

What you are describing is a fan, not a turbine. A turbine impedes airflow.

 

[Bob S]

What if the wind turbines were placed in such a way as to reduce drag by redirecting the flow of air, such as at the back of a vehicle. If the turbines shifted the air flow so that it reduced the low pressure behind the vehicle, couldn't the vehicle theoretically move forward more easily while the turbine motion also generated electricity?

This is a very interesting point. The power loss due to air drag by a vehicle is P = ½ρAC_vv³, where C_v is the coefficient of drag, and A is the frontal area of the vehicle. The object of streamlining a vehicle is to reduce C_v. If a wind turbine could be placed in a location such that the reduction in the vehicle air drag power loss was >= the output power of the turbine, then you could theoretically have turbine "efficiency" over 100%. Any comments?

Bob S

 

[RonL]

This is a very interesting point. The power loss due to air drag by a vehicle is P = ½ρAC_vv³, where C_v is the coefficient of drag, and A is the frontal area of the vehicle. The object of streamlining a vehicle is to reduce C_v. If a wind turbine could be placed in a location such that the reduction in the vehicle air drag power loss was >= the output power of the turbine, then you could theoretically have turbine "efficiency" over 100%. Any comments?

Bob S

Are the moderators ahead of the send action ?
Just had a post disappear as I started to hit send. Anyhow I'll just say, warmer air flowing through a turbine of some sort, converting energy and pushing cold air into the draft area behind the vehicle will reduce drag to a point that the turbine should power the vehicle and have a surplus energy.

I'm sure I could have said it better, but that does not happen often.

Ron

 

[Xtensity]

This is a very interesting point. The power loss due to air drag by a vehicle is P = ½ρAC_vv³, where C_v is the coefficient of drag, and A is the frontal area of the vehicle. The object of streamlining a vehicle is to reduce C_v. If a wind turbine could be placed in a location such that the reduction in the vehicle air drag power loss was >= the output power of the turbine, then you could theoretically have turbine "efficiency" over 100%. Any comments?

Bob S

The situation we're talking about involves 0 air movement. Meaning all of the energy MUST be coming from the cars kinetic energy as it moves forward. Even if drag and friction on the turbine were minimized, the turbine still could never produce 100% of the energy that it takes to spin it to generate a select amount of energy. If it somehow manages to generate more than 100% of the energy which it took to spin it, then you just made a perpetual motion machine.

 

[russ_watters]

Are the moderators ahead of the send action ?
Just had a post disappear as I started to hit send.

No...

Anyhow I'll just say, warmer air flowing through a turbine of some sort, converting energy and pushing cold air into the draft area behind the vehicle will reduce drag to a point that the turbine should power the vehicle and have a surplus energy.

I'm sure I could have said it better, but that does not happen often.

Well it happens all the time in a jet engine, but a car driving down the highway is not a jet engine. A turbine by itself doesn't do any pushing and even in a jet engine is removing energy from the exhaust that would otherwise be propelling the plane.

 

[brainstorm]

The situation we're talking about involves 0 air movement. Meaning all of the energy MUST be coming from the cars kinetic energy as it moves forward. Even if drag and friction on the turbine were minimized, the turbine still could never produce 100% of the energy that it takes to spin it to generate a select amount of energy. If it somehow manages to generate more than 100% of the energy which it took to spin it, then you just made a perpetual motion machine.

Perpetual motion machine is an exaggerated claim. If the vacuum forming behind the vehicle is increasing the drag, then it is already taking unnecessary energy from the engine. All this turbine would be doing is getting that energy back from the drag and reducing the drag at the same time. Probably it would be more efficient to just reduce the drag without the turbine, though, I'm guessing.

 

[RonL]

There is more than one thing going on in respect to the car and the hood ornament windmill.
Wind blowing straight at a stationary car at the speed of 50mph will produce the same exact drag friction as if the car is moving at 50mph. Drag is calculated using the square factor.
Now consider a windmill, it is stationary and if a wind of 50mph blows straight at it, the energy of the wind is calculated using a cube factor.

In my mind, energy in the air motion affects, at the expense of drag on the car shell, can be two different values. Proper ducting and turbine design can extract this energy to a high level of efficiency.

My understanding is that a volume of warm air flowing at some velocity and performing work on a turbine of flow through design, will discharge that same volume at a colder temperature.
Will energy required to move a volume of air, not be proportional and related to it's temperature ? if so then that amount of work (heat) has to be transferred into some action. I would propose on an electric car a Regen system.

Ron

 

[Lsos]

I'm not sure what the power formula for a turbine spinning in air is. But, I'm going to go ahead and guess that it's similar to the formula for how much power a car needs to push that turbine through the air. In fact, the formulas are probably damn near identical.

There's probably an air pressure, an area, and a V^3 in there.

That said, I just don't see how any air pressure/ air deflection/ strategic placement trick would result in the power extracted by the turbine being greater than the power used to get it to spin.

 

[brainstorm]

Try this simple intuitive thought experiment:

1) build any vehicle with any configuration of wind-driven electric generators.
2) disengage the generators so they they rotate freely without the friction of generating current.
3) measure how much drag is reduced by disengaging the generators.
4) compare X (the reduction in drag) to Y (the amount of energy generated when the generators are engaged).
5) Is Y greater than or equal to X?

 

[Bob S]

From Bob S

This is a very interesting point. The power loss due to air drag by a vehicle is P = ½ρAC_vv³, where C_v is the coefficient of drag, and A is the frontal area of the vehicle. The object of streamlining a vehicle is to reduce C_v. If a wind turbine could be placed in a location such that the reduction in the vehicle air drag power loss was >= the output power of the turbine, then you could theoretically have turbine "efficiency" over 100%. Any comments?

The situation we're talking about involves 0 air movement. Meaning all of the energy MUST be coming from the cars kinetic energy as it moves forward. Even if drag and friction on the turbine were minimized, the turbine still could never produce 100% of the energy that it takes to spin it to generate a select amount of energy. If it somehow manages to generate more than 100% of the energy which it took to spin it, then you just made a perpetual motion machine.

A spoiler on the rear of an automobile with some wind turbines in it may be more efficient (lower overall drag coefficient) than no spoiler at all. But then, the spoiler without the turbines would be even better.

Bob S

 

[brainstorm]

A spoiler on the rear of an automobile with some wind turbines in it may be more efficient (lower overall drag coefficient) than no spoiler at all. But then, the spoiler without the turbines would be even better.

Especially if the spoiler had neon lighting and green flames painted on it. Sorry, couldn't resist:)

 

[Andrew Mason]

A windmill on a car could be useful if you had a cross wind and wanted to take advantage of it to supplement the energy delivered by the motor. Conceivably it could be designed so as to almost eliminate drag. You could turn it to forward to supplement braking and recover some of the car's kinetic energy in a battery, say. A novel twist on dynamic braking. But you cannot use the virtual wind created by the car itself, for the reasons already given.

AM

 

[ErikD]

Using a windmill to convert the car's kinetic energy when braking seems kinda not optimal. Using a flywheel (if you want to use it to accelerate again) or some kind of spring for that is much better.

 

[russ_watters]

Try this simple intuitive thought experiment:

1) build any vehicle with any configuration of wind-driven electric generators.
2) disengage the generators so they they rotate freely without the friction of generating current.
3) measure how much drag is reduced by disengaging the generators.
4) compare X (the reduction in drag) to Y (the amount of energy generated when the generators are engaged).
5) Is Y greater than or equal to X?

Let me give a real-world example that may help you out:

All cars have a radiator and grille in the front. In a lot of cars there is an air scoop to direct air to the radiator. Flow through the radiator (and then the engine compartment) is awfully draggy. Virtually any obstruction to that flow would would reduce drag. So if you placed a turbine in the airscoop, less air would get to the radiator, so the drag of the car would decrease. But a sheet of plywood across the airscoop would do a better job of blocking the airflow and would therefore decrease drag even more: and more importantly, the sheet of plywood would save more energy than the turbine could generate.

In other words, there is nothing you can do with a turbine to reduce the drag on a car that couldn't be done better another way.

 

[RonL]

Ducted fan design using tandem heat pump principle (or triple) sucks the car forward and at the same time pushes forward using a cold air discharge thrust.

One only needs to design for a air volume flow and BTU extraction equal to power needed to sustain motion. And as ErikD mentioned a flywheel has a place in the mix.

Later

Ron

 

[russ_watters]

Sooo...use a heat pump to reduce the efficiency of a ducted fan in a sort-of inverse jet engine arrangement? Why would anyone want to waste energy in such a way?

 

[OmCheeto]

And it does appear we've discussed this https://www.physicsforums.com/showthread.php?t=325282".

 

[RonL]

Om, there was that discussion but there could have been so much more

Russ, before I embarrass myself too much, let's see if what I think is right.

A 1500 watt heat strip will radiate 1500 watts of heat, giving a 1:1 ratio, now if a heat pump with a COP of 10 will receive a 1500 watt input and return a heating value of 15,000 watts (in principle) is this basically correct ?
If that heat pump draws heat from an atmospheric temperature of 90 or 100 degrees will it's efficiency be better or is that COP based on quantity of refrigerant and cycle time and system size ?

Is the 15,000 watt equivalent heat from the COP 10 heat pump the same as a 15,000 watt heat strip of 1:1 ratio ?

If I'm correct or even close, I'll say a little more about my comment above.

Thanks
Ron

 

[russ_watters]

A 1500 watt heat strip will radiate 1500 watts of heat, giving a 1:1 ratio, now if a heat pump with a COP of 10 will receive a 1500 watt input and return a heating value of 15,000 watts (in principle) is this basically correct ?

Yes, though a typical heat pump for home heating will only give you a COP of about 4:1.

If that heat pump draws heat from an atmospheric temperature of 90 or 100 degrees will it's efficiency be better or is that COP based on quantity of refrigerant and cycle time and system size ?

The COP is based on the temperature difference between the hot and cold sides. For a residential heat pump, a 4:1 COP is based on about a 90F hot side and 40F cold side for a 50F delta-T.

Is the 15,000 watt equivalent heat from the COP 10 heat pump the same as a 15,000 watt heat strip of 1:1 ratio ?

No. The 15,000 watts from a "heat strip" (assuming you mean a resistive heater) will be vastly hotter.

I'm just guessing here, but I think you're probably operating on misconceptions about how both heat pumps and jet engines work. This sounds like an attempt to violate conservation of energy by making a heat pump provide the input heat for a turbine to generate enough power to drive the heat pump and provide excess energy.

 

[RonL]

Yes, though a typical heat pump for home heating will only give you a COP of about 4:1. The COP is based on the temperature difference between the hot and cold sides. For a residential heat pump, a 4:1 COP is based on about a 90F hot side and 40F cold side for a 50F delta-T. No. The 15,000 watts from a "heat strip" (assuming you mean a resistive heater) will be vastly hotter.

I'm just guessing here, but I think you're probably operating on misconceptions about how both heat pumps and jet engines work. This sounds like an attempt to violate conservation of energy by making a heat pump provide the input heat for a turbine to generate enough power to drive the heat pump and provide excess energy.

Not an attempt to violate conservation of energy, No design can use or store more than 100% of it's design capacity. Thermal energy can be taken in and transformed into mechanical and/or electrical or a combination of both.

We'll assume an electric car and battery storage for getting the car up to speed, now to hold that speed steady an exact amount of energy is drawn from the battery and when reducing speed some regeneration can restore a small amount of the energy used.
Now we can add different types of energy storage, such as flywheel, compressed air, wind driven generator. they all have a capacity that when full cannot be added to and they all will return less energy than what was stored.

Now to recover waste heat or bring in thermal energy to add to the needs of moving the car, it is important (at least in my mind) to consider, thermal and electrical energy can be moved in and out of system boundaries, with no consideration given to PRESSURE.

Compressing air will return high pressure gas flow capable of doing work, at the same time if the compressor is inside a refrigeration unit, the normally wasted heat will build pressure and produce work.
To make use of thermal transfer between systems, a combination of fans with related size and speeds can be designed to produce thrust and at the same time serve as heat exchange condensers or evaporators. This allows the removal of those wind brake radiators.
The fan design is a compromise of efficiency in that maximum thrust is reduced by allowing air to flow between vertical tubes spaced in such manor as to form an almost solid surface and positioned from hub to outer ring in a contour like a solid blade is shaped.
A double wall axle allowing cross flow of liquid/gas, the tubes and outer rim, comprise a complete closed loop heat exchanger and if two or three are working in a cascade fashion will result in a heat transfer from the air into the cold sinks of the A/C units or heat pumps.

This is more like a high volume low speed electric ducted fan model airplane or canister vacuum cleaner, but not so much a jet engine.

If a car is moving at 55 mph, having a slight suction in front and a slight positive thrust in the rear, the total volume of air will bring in heat far in excess of the needs of the demands. At 1 hp per ton of refrigeration, the volume of air holds more heat than a 50F Delta-T would extract. Once the car is steady the Delta would drop and my guess without lots of math, would be in the 10-20 range

If a car is at steady speed and storage systems are full, it will be impossible to extract more energy or produce any extra waste heat.

If I had all the resources needed, I would do it like this and the flywheels can be designed for 5 other options besides kinetic storage. Pulling heat from the air is not a violation, It's just more complicated than the combustion of a charge of fuel.

Ron

 

[russ_watters]

To make use of thermal transfer between systems, a combination of fans with related size and speeds can be designed to produce thrust and at the same time serve as heat exchange condensers or evaporators. This allows the removal of those wind brake radiators.

No, a fan is not an evaporator or condenser, the heat exchangers are evaporators and/or condensers. You're removing one wind-brake radiator and adding two in its place!

All of the rest of the post is incoherent. I'm still not seeing the point.

 

[OmCheeto]

Om, there was that discussion but there could have been so much more

There was https://www.physicsforums.com/showthread.php?t=180442". That thread convinced me that improving the aerodynamics of a vehicle would yield more energy savings vs. any type of fan/turbine addition.

And there would appear to be more of these threads:

Aug14-07, 05:35 PM
Search the forum on this. There was a lengthly thread about it.

They should put an index at the back of the book, as I cannot find the thread ASN was talking about.

Vehicles
...Cars
...Adding wind turbines(see also: crackpottery, but tolerated due to educational value)

 

[RonL]

No, a fan is not an evaporator or condenser, the heat exchangers are evaporators and/or condensers. You're removing one wind-brake radiator and adding two in its place!

All of the rest of the post is incoherent. I'm still not seeing the point.

Sorry the fan design did not register in your mind, as it is critical to the operation.

I'm sure you have seen a six bladed paint booth exhaust fan, now imagine a tube ring around the six tips, each blade is hollow inside, a circular hole at the root then changing to a flatter rectangular shape where it contacts the outer ring. liquid refrigerant flows outward through three and becomes a gas in the outer tube ring and gas flow returns through the other three, back to the double core hub axle.

If this is spinning at some rate, it will flow the same amount of air as it's original design called for, but if refrigerant flows through the insides of the blades and tube, it will transfer heat out or draw heat in, based on the temperature of the liquid or gas. This in no way retards or changes the performance of the designed air flow, in fact the outer ring will allow for much higher RPM and prevent any distortion of the blade tips. that means much higher thrust can be produced.

Now what I described in that last post is an invention of my mind and you won't find it on any manufacturer's parts list (at least I don't think so) One patentable idea ruined for profit.

The typical A/C system has the generic 1/2 hp motor drawing air through a mass of tubing, using a typical three blade flat, solid sheet paddle, these blades are twist formed to a shape that by design will move a calculated volume of air at a certain speed.

Instead of a stationary mass of tubing flowing refrigerant and having air forced through, I have elected to have sections of tubing standing in-line (or slightly curved) along the hub and as they extend outward to the outer ring, they might be flattened a little (for taking up the space) and formed into a fan like shape.
Due to speed of rotation they will perform as well as a solid blade, but because of the slight space between each tube maximum thrust is compromised in exchange for heat exchange. For a visual image it would look much like a stator fan in the torque converter section of an automatic transmission.

This should have huge value in home or commercial design of A/C systems. (IMHO)

Seals for high speed rotation and high pressure gas are what I think will be hardest to design, but this is not a worry so much as I know this is already a well tested area in design engineering.

If this description is not understood by anyone, then I'll know that I do live in my own little world.

Back to the original topic, no positive energy from air movement due to motion of a car can be obtained, unless there is a thermal reduction of that air.
Cold air requires less energy to move than hot air, does anyone disagree ?
Heat increase in any closed system will cause an increase in pressure ?
Liquid and gas volume will define maximum energy movements within a closed system ?
Heat can be moved in and out of a closed system ?
Electricity can be moved in and out of a closed system ?
Mechanical energy can be moved in and out of a closed system with magnetic couplings?
Kinetic energy can be stored in a flywheel, which can reside inside or outside a closed system?
There are a number of things that can still be brought in, but this list of statements represent a group of actions that can interact in one harmonious motion with one basic thermal conclusion.

Yes it's a LOAD to comprehend! and I know my writing skills suck
But for now it's my best effort.

Ron

PS If this is considered crackpot thinking then I definitely am in the wrong place, but please state what action violates anything and what, why ?

 

[RonL]

There was https://www.physicsforums.com/showthread.php?t=180442". That thread convinced me that improving the aerodynamics of a vehicle would yield more energy savings vs. any type of fan/turbine addition.

And there would appear to be more of these threads:



They should put an index at the back of the book, as I cannot find the thread ASN was talking about.

Vehicles
...Cars
...Adding wind turbines(see also: crackpottery, but tolerated due to educational value)

Hi Om,
I went back and looked at that thread that you under lined "much more", unless I missed a post, It's interesting to me that Ranger Mike was the first to mention the thought of HEAT, a few post later the thread was locked.
Being able to recover waste heat from a hot engine but not from a mass of warm air, seems a little hypocritcal to me.

Guess I was out working and missed that thread, thanks for linking to it.

A little of what I have been trying to explain, is what I offered to try and pass on to Mike for use on his race car, figured he might be able to calculate things much better and run numbers that are far above my ability.

Thanks
Ron

 

[OmCheeto]

Hi Om,
I went back and looked at that thread that you under lined "much more", unless I missed a post, It's interesting to me that Ranger Mike was the first to mention the thought of HEAT, a few post later the thread was locked.
Being able to recover waste heat from a hot engine but not from a mass of warm air, seems a little hypocritcal to me.

Guess I was out working and missed that thread, thanks for linking to it.

A little of what I have been trying to explain, is what I offered to try and pass on to Mike for use on his race car, figured he might be able to calculate things much better and run numbers that are far above my ability.

Thanks
Ron

I'm hesitant to paraphrase what I think you are describing, but I'll do it anyways.

Are you saying that cooling the air in front of the vehicle via some sort of heat exchanger with the back of the vehicle is going to improve the efficiency of the vehicle?

In a closed system, a thermal gradient can be a source of energy, but in this open system?

If this were true, then you could propel a vehicle in this way. I seriously doubt you could get anything resembling a car to move in this manner without a [STRIKE]gigawatts[/STRIKE] terawatts worth of heat exchangers.

This idea isn't something you've transferred from hot air balloon technology is it?

 

[RonL]

I'm hesitant to paraphrase what I think you are describing, but I'll do it anyways.

Are you saying that cooling the air in front of the vehicle via some sort of heat exchanger with the back of the vehicle is going to improve the efficiency of the vehicle?

In a closed system, a thermal gradient can be a source of energy, but in this open system?

If this were true, then you could propel a vehicle in this way. I seriously doubt you could get anything resembling a car to move in this manner without a [STRIKE]gigawatts[/STRIKE] terawatts worth of heat exchangers.

This idea isn't something you've transferred from hot air balloon technology is it?

Om,
Nearly all my thoughts have evolved from the 60s' dream of a submersible, provoked by J. Cousteau, designed to be free of any surface support in any ocean or body of water that might need some kind of attention. A design that draws it's energy from the water. A design that involves no combustion of any kind that would have a polluting discharge. When R. Ballard showed us Hot Vents at close to 800F degrees, I really got a renewed fever as I had just started study about A/C and heat pumps (recharge points in any ocean who knows where).
Years of working with mechanical things has given a gift (or curse ??) of mixing and matching anything and not be concerned at the start about it being practicable or will it work in a positive way, the answer to failing conditions almost always opens a door to other lines of thought. The thermal exchanging fan is actually a spin-off idea that started as a propeller design for the submersible, A sealed system that moved refrigerant through cross flow shaft and tube, into the hub area then circulating through copper tubing curved to form a shape like a typical propeller. Only problem might be ice buildup, but under what conditions ??

I did just figure a way to phrase a question that will determine the validity of my thinking.

Without any detail of mechanics involved, if we consider a heat pump with a COP of 4:1 and the wattage input is 1,000 required by the motor and if the heat transferred is (equal ??) to 4,000, then if the coils move that heat into a second unit's cold liquid section will the pressure gain in the second unit be based on 1,000 or the 4,000 figure ?

Everything I think I understand is that the 4,000 number is now heat energy available to drive the second unit in some form of conversion.

The numbers don't have to be exact just simply put is this example close to correct ?

Thanks for your help

Ron

PS In answer to your first question, The warm air is drawn in by the spinning fan, heat is taken and adds to the driving system which is composed of several things, as air is moved from intake to cold discharge it might be pulled through even more fans until it's final discharge. The total heat that can be removed will equal the total work required to sustain motion.

 

[OmCheeto]

Cold air requires less energy to move than hot air, does anyone disagree ?

Since http://en.wikipedia.org/wiki/Density_of_air" , it seems to me it would require more energy to move.

ρ=P/(RT): if T goes down, rho goes up

F = ½ρC_dAv²

When rho goes up, so does drag force.

Hmmm... Perhaps we should put a spray mister on the end of a long tube in front of the car, bombard it with microwaves, turning the spray into steam, which when it cools, will produce a vacuum in front of the car, and atmospheric pressure at the rear of the car will provide the propulsive force.

Oh! And put a wind driven fan in there somewhere to power the whole thing. Otherwise I'll be off topic.

 

[stevenb]

There is one practical application—a turbine mounted on a stationary vehicle in the presence of a significant breeze. You could, for instance, stick one on the roof of a parked RV to power your TV or computer. In that case, there would be no parasitic losses to the vehicle.

There is also the case of going down a steep hill. This does not violate the conservation of energy.

 

[RonL]

Since http://en.wikipedia.org/wiki/Density_of_air" , it seems to me it would require more energy to move.

ρ=P/(RT): if T goes down, rho goes up


When rho goes up, so does drag force.

Hmmm... Perhaps we should put a spray mister on the end of a long tube in front of the car, bombard it with microwaves, turning the spray into steam, which when it cools, will produce a vacuum in front of the car, and atmospheric pressure at the rear of the car will provide the propulsive force.

Oh! And put a wind driven fan in there somewhere to power the whole thing. Otherwise I'll be off topic.

First let me make a most important point for any young mind reading this thread. "GET THE EDUCATION" you might have the greatest idea the world has ever seen, but if you can't articulate what's in your mind to other people, it likely will never see the light of day...Again "GET THE EDUCATION".

Om,
I think you are right, but let me see if my "mirror image, reverse logic" can make sense and you still be right.

I found a prop calculator and filled in all the numbers and at the bottom the standard air density was set at 29.92, lowering the number produced less thrust and then raising the number increased thrust, I think we will both agree that a thrust will always have an equal input requirement of power from some source.

But looking at power to compress air, each scf (standard cubic foot)compressed to 200 psi in a single stage compressor will require .250 hp, a two stage will require .230 hp, and a three stage will require .212 hp. At each stage heat has been removed and as a result horse power needs have gone down.

What I have described is a way to remove heat in a volume of air as it is continually reducing in volume in a duct that reduces in size and at the rear the same mass of air is thrusted rearward at a colder temperature than it entered. "if there had been no temperature reduction, more power would have been required to move the mass and a higher temperature discharge would have resulted".

The heat is used to produce pressure in the gases that help drive the primary "power group" that turns the fans and puts work into the motion of the car.

Did the fan design that provides thrust and at the same time removes heat (or if needed adds heat) go unnoticed ? (Russ)? anyone?

Maybe my mind has been inside that little submarine tooo long!

Ron

 

[russ_watters]

Sorry the fan design did not register in your mind, as it is critical to the operation.

A shrouded fan which you pump refrigerant through in order to allow the fan to double as an evaporator. Got it. I highly doubt that the idea would be workable as a commercial product (reliably sealing the tubes in a rotating hub would be near impossible), but even if it could be made to work, I can't see it being efficient at either heat transfer or moving air. For one thing, the vast majority of HVAC evaporator fans aren't axial, but centrifugal. Anyway...

Back to the original topic, no positive energy from air movement due to motion of a car can be obtained, unless there is a thermal reduction of that air.
Cold air requires less energy to move than hot air, does anyone disagree ?

Yes. The energy required to move air - by mass - is independent of temperature. It has to be. The kinetic energy equation is just E=.5MV^2.

Heat increase in any closed system will cause an increase in pressure ?

Temperature increase, yes.

Liquid and gas volume will define maximum energy movements within a closed system ?

That's incoherent. Word salad. Many of those words don't belong in the same sentence with each other. Volume of a closed system has nothing to do with energy movement.

Heat can be moved in and out of a closed system ?
Electricity can be moved in and out of a closed system ?
Mechanical energy can be moved in and out of a closed system with magnetic couplings?

Yes.

Kinetic energy can be stored in a flywheel, which can reside inside or outside a closed system?

No.

Without any detail of mechanics involved, if we consider a heat pump with a COP of 4:1 and the wattage input is 1,000 required by the motor and if the heat transferred is (equal ??) to 4,000, then if the coils move that heat into a second unit's cold liquid section will the pressure gain in the second unit be based on 1,000 or the 4,000 figure ?

What pressure gain? What is a "second unit"? You haven't described the thermodynamic process you are trying to drive! That's the problem that I asked about in my previous post! That said, what it looks like to me is that you think you can use a heat pump to drive power production, in a way that will produce more energy than is consumed. That would be an obvious violation of conservation of energy.

What I need is a concise, coherent description of the process, in single-sentence steps. For a gas turbine, for example, the process looks like this:

1. Incoming air is compressed (and it heats up).
2. Heat is added via a burner.
3. Energy is extracted via a turbine (and the exhaust cools).
4. Exhaust/heat rejection.

It should be possible to describe your process in that short/simple of a format. Please try! Let me help get you started by describing what I see so far:
1. Incoming air is pulled-in by a fan.
2. Incoming air is heated by a heat pump (it may be in the same device, but it is two separate thermodynamic processes).
3. Compress the air (it heats up more).
4. Cool the air using a the other side of the heat pump in #2 (pressure drops).
5. Extract electrical energy via a turbine.
6. Low temperature heat rejection.

Does this come anywhere close to what you are proposing?

 

[RonL]

A shrouded fan which you pump refrigerant through in order to allow the fan to double as an evaporator. Got it. I highly doubt that the idea would be workable as a commercial product (reliably sealing the tubes in a rotating hub would be near impossible), but even if it could be made to work, I can't see it being efficient at either heat transfer or moving air. For one thing, the vast majority of HVAC evaporator fans aren't axial, but centrifugal. Anyway... Yes. The energy required to move air - by mass - is independent of temperature. It has to be. The kinetic energy equation is just E=.5MV^2.
Temperature increase, yes. That's incoherent. Word salad. Many of those words don't belong in the same sentence with each other. Volume of a closed system has nothing to do with energy movement. Yes.
No. What pressure gain? What is a "second unit"? You haven't described the thermodynamic process you are trying to drive! That's the problem that I asked about in my previous post! That said, what it looks like to me is that you think you can use a heat pump to drive power production, in a way that will produce more energy than is consumed. That would be an obvious violation of conservation of energy.

What I need is a concise, coherent description of the process, in single-sentence steps. For a gas turbine, for example, the process looks like this:

1. Incoming air is compressed (and it heats up).
2. Heat is added via a burner.
3. Energy is extracted via a turbine (and the exhaust cools).
4. Exhaust/heat rejection.

It should be possible to describe your process in that short/simple of a format. Please try! Let me help get you started by describing what I see so far:
1. Incoming air is pulled-in by a fan.
2. Incoming air is heated by a heat pump (it may be in the same device, but it is two separate thermodynamic processes).
3. Compress the air (it heats up more).
4. Cool the air using a the other side of the heat pump in #2 (pressure drops).
5. Extract electrical energy via a turbine.
6. Low temperature heat rejection.

Does this come anywhere close to what you are proposing?

Russ,
Thanks for being patient with me and thanks for the suggestion about how to keep my thoughts in a short, simple, step by step method.

I think you have misinterpreted a few things that I have tried to imply, but that's OK for now, I'll try to address each and every one as this progresses.

Wow! the flywheel is the last thing I would have expected you to disagree with, if you can give the reason I would be thankful.

My next few days are going to be pretty full, but I will carry a note pad and try to jot down little bits at a time and maybe by Thursday of the coming week I can make a decent post.

I do hope you understand that even if I have not worded things well, I have always maintained staying within the first two laws.

Guess I better not try to say anything else, except THANKS.

Ron

 

[russ_watters]

Wow! the flywheel is the last thing I would have expected you to disagree with, if you can give the reason I would be thankful.

Actually, I'm not sure how that got in there - it must have been an editing mistake. A flywheel is a flywheel - it is an energy storage device and it has no relevance whatsoever to the discussion.

 

[RonL]

Actually, I'm not sure how that got in there - it must have been an editing mistake. A flywheel is a flywheel - it is an energy storage device and it has no relevance whatsoever to the discussion.

I might be wrong to think it's needed, but in my mind I see a need to use a storage of some kind to carry a cycle above and below it's steady state flow. Hope that's not word salad.

 

[OmCheeto]

Om,
I think you are right, but let me see if my "mirror image, reverse logic" can make sense and you still be right.

I found a prop calculator and filled in all the numbers and at the bottom the standard air density was set at 29.92, lowering the number produced less thrust and then raising the number increased thrust,

Yes.

I think we will both agree that a thrust will always have an equal input requirement of power from some source.

Power = Thrust * Velocity (according to wiki)

But looking at power to compress air, each scf (standard cubic foot)compressed to 200 psi in a single stage compressor will require .250 hp, a two stage will require .230 hp, and a three stage will require .212 hp. At each stage heat has been removed and as a result horse power needs have gone down.

This is where I start having a problem.
As soon as you start incorporating heat transfer, you are going to have to deal with the second law of thermodynamics, and the dreaded http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html#c1".

What I have described is a way to remove heat in a volume of air as it is continually reducing in volume in a duct that reduces in size and at the rear the same mass of air is thrusted rearward at a colder temperature than it entered. "if there had been no temperature reduction, more power would have been required to move the mass and a higher temperature discharge would have resulted".

The heat is used to produce pressure in the gases that help drive the primary "power group" that turns the fans and puts work into the motion of the car.

Without ever inserting numbers into your theory, you will never know how inefficient it would really be.

And without having a clue as to what your contraption looks like, aka a schematic, we can't plug the numbers in for you.

 

[kmarinas86]

It seems like a wind turbine would be used to generate electricity.

So why not just have the wind turbine where it belongs and just plug in the vehicle to the grid to recharge?

Far better improvements can be made by making cars as aerodynamic as airplanes.

Adding windmills to cars to generate energy is a stupid idea.

 

[RonL]

It seems like a wind turbine would be used to generate electricity.

So why not just have the wind turbine where it belongs and just plug in the vehicle to the grid to recharge?

Far better improvements can be made by making cars as aerodynamic as airplanes.

Adding windmills to cars to generate energy is a stupid idea.

Everything you say is "precisely true". The question is, what if anything, can move it beyond the stupid point??

I say yes.

There is only time for this short answer.

Ron

 

[RonL]

Yes.

Power = Thrust * Velocity (according to wiki)

This is where I start having a problem.
As soon as you start incorporating heat transfer, you are going to have to deal with the second law of thermodynamics, and the dreaded http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html#c1".


Without ever inserting numbers into your theory, you will never know how inefficient it would really be.

And without having a clue as to what your contraption looks like, aka a schematic, we can't plug the numbers in for you.

Hi Om,
Don't have time right now, but will get back later this evening.

I really have never fully understood the Carnot cycle, but isn't it describing a single machine or process cycle ?

I'm working on Russ's suggestion, so I'll throw your posts in as well.

Thanks

Ron

 

[russ_watters]

The Carnot cycle is simply the ideal thermodynamic cycle, containing the four basic processes (there are always variants of them...) in any thermodynamic cycle. Learning it is a critical part of the beginning of learning thermodynamic analysis. If you tell us what, specifically, you don't understand about it, maybe we can help...