How To Ice Up Your Motor? (Cooling Ideas)

[TylerDurden]

Basically in the heat of the day in the summer a fan blowing hot outside air into the motor doesn't do any good... you couldn't increase the airflow any more and expect to achieve anything.

Oh please.

If it's 30C ambient and 60C inside your motor, you can be sure more air is a good thing.

As you point out, air conducts heat more slowly than water, so you want to move it 4x faster to keep the greatest differential between the air and the motor for the greatest absorbtion.

Refrigeration systems (condenser/evaporator) are quite lossy. You throw a lot of energy away as heat in the condenser.

I wouldn't recommend using your valuable batteries to make more heat, when the simplest, most effective way to cool your motor is a fan.

 

[fechter]

Check this out:
http://en.wikipedia.org/wiki/Specific_heat_capacity

I think most of the formulas you need are there.

You can take a tire pump compressor and run it through a small restriction to model what happens. The hose gets hot, and the air coming out of the restriction is cooler than ambient. It's just not a lot of air.

 

[safe]

Refrigeration systems (condenser/evaporator) are quite lossy. You throw a lot of energy away as heat in the condenser.

That's because they are a loop. You first compress the refrigerant, then cool it, then expand it again and allow the cooled expanded air to absorb heat which creates the refrigeration process.

This is a short circuit of that...

The compression phase creates heat which is dissipated in the radiator, then it's directly dumped into the motor where it expands and comes into contact with the motors insides. The motors insides effectively become the "condensor" part of the system. Since all the cooling effect goes right into the motor it transfers heat at whatever rate the compression phase determines. The MORE boost, then the MORE cooling, but more motor strain. Less boost means less cooling and less drain. So you could have an ADJUSTABLE system rather than something that is just a simple fan.

In a system like this you don't need high flow rates... you're focus is on COOLING and not air flow...

 

[TylerDurden]

This is a short circuit of that...

Not really.

It's just an open system where you recover your refrigerant from the atmosphere.

You're still losing lots of energy in the compression phase.

 

[TylerDurden]

In a system like this you don't need high flow rates... you're focus is on COOLING and not air flow...

You're focusing on compression and not cooling.

The temperature differential of your released air compared to your motor temperature will still require massive amouts of airflow to achieve any significant reduction in motor temps.

Plus, you can't sustain high pressure without high volume. So, you're back to a pressure-vessel of something like 3000psi to provide 30 min of moderate airflow... CO2 would be better... dry-ice even better than that.

 

[safe]

Formal Mathematical Proof

At some point idle opinions have to make way for the formal proof. Time to find out what the numbers really say. At this point I have not done the "full proof" so you will see the discovery as it happens... 8)

First I need to nail down EXACTLY what we are dealing with. If you are dealing with air you have to deal with all kinds of temperature and pressure issues throughout the system. The only logical way to go is to start with a baseline which is the actual number of molecules you are dealing with. With gases that standard is the mole (mol) which is 6.022×10^23 molecules of a gas.

But we don't want to deal with one full "mol", but some size that makes sense to us. The easiest size for us to handle is the Liter since that's about the size of the motor.

Some things that I looked up about air:

The mean molar mass of air is 28.97 g/mol.

The density of air at sea level is about 1.2 g/L.

So if we divide 1.2 g/L by 28.97 g/mol we get:

(1.2 g/L) / (28.97 g/mol) = 0.041 mol/L (at sea level)

I'll stop this posting here to give a single issue to respond to. If you think that my calculation is wrong be sure to explain where it went wrong.

 

[safe]

PV = nRT

P = the gas absolute pressure, in Pa
n = number of moles, in mol
V / n = the gas molar volume, in m³/mol
T = the gas absolute temperature, in K
R = the universal gas law constant of 8.3145 m³Â·Pa/(mol·K)

So let's get our 0.041 mol/L (at sea level) into this equation so that we can begin to use this equation.

n = number of moles, in mol = 0.041 mol

V / n = the gas molar volume, in m³/mol... this one requires some work:

If we have one Liter for 0.041 mol (L/mol) and we need to convert that to m³/mol then since "1 liter = 0.001 cubic meter" the value for V will be 0.001.

So:

n = 0.041 mol

V = 0.001 m³

So far so good?

 

[TylerDurden]

*yawn*

Sure.

 

[safe]

Does it Balance?

P = standard atmosphere = 101325 Pa = 14.67 PSI

V = 0.001 m³.

n = 0.041 mol

R = 8.3145 m³Â·Pa/(mol·K)

T = 297.23 K = 75.34 F

So:

101325 * 0.001 = 0.041 * 8.3145 * 297.23

101.325 = 101.324

Close enough! :wink:

 

[safe]

Raise the Pressure?

Okay so now we have a model of our one Liter of air. We are now going to compress that air by adding 10 PSI of pressure to it.

Using one of those conversion tools:

14.67 PSI + 10 PSI = 24.67 PSI = 170093 Pa

PV = nRT

T = PV/nR

T = (170093 * 0.001) / (0.041 * 8.3145) = 498.96 K = 438.46 F

Which is totally wrong... there's no way that 10 PSI of added pressure is going to raise the air temperature to over 400 degrees F!!!!

So air obviously doesn't behave anything like an "ideal gas"... it looks like the simple PV = nRT equation won't help us very much. :? (or I've already done something very wrong)

It's supposed to look about like this... so we would expect the temperature to be about 185 degrees F based on this chart.

 

[ngocthach1130]

i don't think increasing the boost into the motor will be the same as using a turbo. in ICE the air is added so more fuel can be dump in. with electric motor i think there might be a limit on how fast it can dissipate heat.

On another note sureelectrics on ebay have chinese made peltier in pack of 10 and for 12v at 29.99+shipping. I guess all that's left to add is some heat sink and that should help. Although they are pretty high wattage. I think they varied from 136W to 46.5W.

 

[safe]

Yeah, I've already passed over the electrical approach as being too inefficient. The peltier cooler isn't a good way to go.

Back to the math...

This website agrees with my numbers when I plug them in (so it wasn't a calculation error) so from the standpoint of an "ideal gas" the temperature does indeed rise very fast with boost pressure.

Maybe the air is really at 400 degrees, but it doesn't transfer the heat instantly?


http://www.ausetute.com.au/idealgas.html

 

[safe]

Monatomic vs Diatomic

I think I remember this from a class long ago...

There's a difference between the simple Monatomic gasses and the more complex gasses like air that behave like a Diatomic gas:

http://en.wikipedia.org/wiki/Heat_capacity_ratio

The terrestrial air is primarily made up of diatomic gasses (~78% nitrogen (N2) and ~21% oxygen (O2)) and, at standard conditions it can be considered to be an ideal gas. A diatomic molecule has five degrees of freedom (three translational and two rotational degrees of freedom, the vibrational degree of freedom is not involved except at high temperatures). This results in a value of

As temperature increases, higher energy rotational and vibrational states become accessible to molecular gases, thus increasing the number of degrees of freedom and lowering γ. For a real gas, CP and CV usually increase with increasing temperature and γ decreases. Some correlations exist to provide values of γ as a function of the temperature.

 

[safe]

Here's a thought...

What if you simply charged the tubes of your frame up with 2,000 PSI of regular air in advance? It will initially heat your frame a little, but it would cool down before your rode. You then begin to release that air into the motor to cool it (since expansion means cooling) and by the time the batteries are done (about a half hour to hour) the air runs out. You control the rate of release and cooling to your expected ride time.

Net loss to motor performance - Zero.

Net gain in cooling - Potentially significant.

Hardware required - Some sort of nozzle to release the air into the motor and a high PSI air charger.

There are people that try to power their entire bikes on air alone, why not just use a little to supply the cooling?

8)


I like this one...

 

[Deepkimchi]

Jeeesh - Safe

Weld some cooling fins on the casing, maybe even extend the casing and mount fan blades on the motor shaft and ride on a cool day...and leave it alone

 

[safe]

Weld some cooling fins on the casing, maybe even extend the casing and mount fan blades on the motor shaft and ride on a cool day...and leave it alone

No you don't understand... I'm planning to take a 350 watt motor and make it perform like a big motor. My goal is to prove the potential of lightweight machines being able to outperform expectations.

If I'm going to produce 3-4 times the rated power I need to do stuff like this. The idea of using the frame as a air bottle seems very good. I might as well, if the frame is there to be used, why not?

Besides... it's snowing outside... I might as well think up new ideas...

 

[vanilla ice]

Do it man. I want to see this thing in action.

Have you ever watched semi truck racing? They spray a water mist at the brake rotors. Makes big clouds of steam at the end of a strait. Scaled way down, a small water mist focused at the outside of the motor housing could work. With plastic ducting to direct it and a temp probe closed loop system. Just be sure to keep the mist away from your rear tire...

 

[safe]

I've got welding bottles and that means I have compressed air already. It wouldn't take much to connect the oxygen bottle to the frame and simply charge it up with air. (as high as 2000 psi) Assuming the frame main tube doesn't leak (which would be a real problem to fix) it could take very little effort to fill. The only issue is then just a valve... which like the welding regulators could show the total pressure as well as the active pressure. (the flow rate) The easy thing to do would be to just use a oxygen welding regulator. I looked on Ebay and saw some that looked like about $20.

Compressed air is the perfect coolant.

They had a story about the Belagio on one of the science shows and apparently the compressed water they use to make the fountains create so much cold temperature that they were initially having troubles with the valves FREEZING and getting stuck.

So the process of releasing compressed air is a great way to cool off the motor... 8)

This thing only costs $12.

 

[TylerDurden]

The idea of using the frame as a air bottle seems very good. I might as well, if the frame is there to be used, why not?

Boom.

 

[vanilla ice]

I knew a guy who charged up the tubular bumper on his off road truck to like 200psi as a compressor tank, but 2000??! And people worry about lithium explosions...

 

[OneEye]

Raise the Pressure?

Okay so now we have a model of our one Liter of air. We are now going to compress that air by adding 10 PSI of pressure to it.

Using one of those conversion tools:

14.67 PSI + 10 PSI = 24.67 PSI = 170093 Pa

PV = nRT

T = PV/nR

T = (170093 * 0.001) / (0.041 * 8.3145) = 498.96 K = 438.46 F

Which is totally wrong... there's no way that 10 PSI of added pressure is going to raise the air temperature to over 400 degrees F!!!!

I haven't looked at your calculations too much, but it appears what you have calculated is the temperature you would have to heat the same volume of air to exert 10psi relative (24.67 psi absolute) if confined to the same .001m^3 volume. If you are compressing the air, there is a big volume change along with a much smaller temperature change. There are curves for the type of compression you are trying to do. I think the term is "adiabatic", but it's been a more than a decade since my only thermodynamics course.

 

[OneEye]

More info from wikipedia

P^(gamma-1) * T ^(-gamma) is constant in an adiabatic process.

"Gamma" for dry air is about 1.4 (@ 20 degrees C)

P1=14.67
P2=24.67

T1=20 degrees C = 293.15

therefore T2 = 67 degrees C

or about 150 degrees F

 

[Deepkimchi]

If you are going to that trouble then use liquid nitrogen - controlled time release :wink:

 

[safe]

It's starting to come back to me a bit... it's been 25 years or so since I studied this stuff in school. I've thrown up a quickie chart of uncertain value, but it's kind of getting me there.

You have two things to deal with...

There's the actual temperature drop that you get from expansion which you can calculate and in the end have a "watt hour cooling effect" value.

Then there's just the natural cooling of natural hot outside (not air conditioned air) across even hotter electrical coils.

So before all this is done I'll need to integrate all that into a mathematical model (using a spreadsheet of course ) that matches my motor. In other words... it will take a little more study.

This chart deals with the TEMPERATURE DROP cooling energy and its expressed in watt hours. This is the "refrigeration effect" above and beyond natural cooling.

As you can see you have to pump more and more volume at the lower pressures in order to get any cooling. So my basic premise that lower volume and higher pressure still seems viable. (so far)

I just grabbed some numbers for volume and pressure to see how they look... there's no formal pattern involved, but the relationship of the data points are correct. (it's a three dimensional data space it seems)

 

[safe]

Ice, Ice, Baby?

What about using a one liter water bottle filled with frozen water and contained in a vessel that has air passing over it and then into the motor?

How many watt hours of refrigeration would we get?

How does this compare to mechanical cooling?

It's turns out that if you look up how many moles of water are in a liter of water than you get 55.55 mol/L. The cooling that you get (through warming the ice) operates at a rate of 6.02 kJ/mol.

So we do the very simple math operation:

55.55 mol * 6.02 kJ/mol = 334.411 kJ = 92.89 Wh

93 Watt Hour

That pretty much dwarfs the refrigeration effect that mechanical compression can achieve (unless you are willing to expend large amounts of energy to compress at high levels) since the typical watt hour of refrigeration for the mechanical tends to be around 15 Wh or less.

So the simple answer might be to use a 1 PSI fan that runs it's air first past a frozen water bottle (thus becoming cooled) THEN passing that cool air into the motor. This is in effect still doing "external" refrigeration and also allows for whatever flow rate you can manage. A liter of water weighs about 2.2 lbs. (1000 grams)

And when the water melts (on a hot day) you have some water to drink.

Just to get a sense of perspective on this...

My little motor is supposed to create (@ 30 amps) about 181 Wh of total heat at full throttle. So if you apply 93 Wh of cooling to that (on top of the airflow which also helps) you would drain your batteries before the ice melts and your motor would stay cool... assuming you ride right away. The quicker the ice melts the shorter your cooling will work. (so taking a break under that shady tree means that you lose the effect)

Aaaaah... that summer heat....