Compressed air for rapidly cooling motors?

[Jon Stratis]

Originally Posted by BJC

First off, Aren Hill deserves all credit for this idea.

Basically, the mounting holes for a CIM type motor are the only access points to the inside of the motor. For reasons described earlier cooling is much more effective when air is forced directly past the armature. When you put 2 and 2 together you end up with hollow fasteners. I'll leave the rest as an exercise to the reader.

Cheers, Bryan

If you mean replacing the bolts holding the motor together with different ones... I would think this is a violation of R30:

The integral mechanical and electrical system of any motor must not be modified.

Those bolts are pretty integral to the mechanical system of the robot... without them the whole thing falls apart and stops working!

[Aren_Hill]

Originally Posted by Jon Stratis

If you mean replacing the bolts holding the motor together with different ones... I would think this is a violation of R30:



Those bolts are pretty integral to the mechanical system of the robot... without them the whole thing falls apart and stops working!

Nope, a CIM motor has two openings in it that are entirely legal to blow air into without modifications.

-Aren

[Richard Wallace]

Originally Posted by Aren_Hill

Nope, a CIM motor has two openings in it that are entirely legal to blow air into without modifications.

Very cool, Aren! 91979A605 or something similar?

[Aren_Hill]

Originally Posted by Richard Wallace

Very cool, Aren! 91979A605 or something similar?

Yup, you can even get 10-32 to 1/4" tube push to connect fittings to make it easier to hook up air.

Feel free to come by 1296 at champs, we'll also be running it.

-Aren

[Thad House]

Originally Posted by Aren_Hill

Yup, you can even get 10-32 to 1/4" tube push to connect fittings to make it easier to hook up air.

Feel free to come by 1296 at champs, we'll also be running it.

-Aren

So is the air actually flowing during a match or only being hooked up afterwards between matches?

[Aren_Hill]

Originally Posted by Thad House

So is the air actually flowing during a match or only being hooked up afterwards between matches?

Between.

Nothing hooked up to the robots pneumatics system.
It's mainly about air flow, and the robot system isn't the greatest for that anyway.

The testing I did on some of this helped lead to motors.VEX.com
Mainly the "peak power duration test"

-Aren

[runneals]

Originally Posted by ctt956

DO NOT use a canned "air" duster to cool it, especially the liquid. Spraying highly flammable liquid that gives off highly flammable fumes onto a burning hot motor near electrical stuff will not end well! Not to mention that the fumes are also highly toxic. Regular compressed air, as in what you get out of a compressor that's used for riveting/nailing/etc, will cool the motors. If the pressure is high enough, it will be cold enough to cause frostbite. Even if it's not cold, passing the cool air over the CIMs will cool them down. It's like a fan blowing on a heat sink.

We had our pit partners ask for an ice pack during the finals to which we gave them our ice pack from our first aid kit I guess that works lol

[Ian Curtis]

Originally Posted by Al Skierkiewicz

Yes teams also spray the main breaker, but in this case there simply is not enough thermal mass inside the main breaker for this method to be effective.

Al,

I thought these breakers were thermal (too much heat and a bi-metallic strip trips). What does it mean that there isn't enough thermal mass? Just that the electric heat builds up too quickly for the temperature at the start of the match to matter much?

That doesn't really jive with my (anecdotal) experiences. The only time I've ever tripped a main breaker is after consecutive matches where I assume heat would be building up in the breaker. Is it that that the heat building up in every motor makes everything less efficient resulting in more current draw and more heating? Or something else?

[Al Skierkiewicz]

Ian, There isn't enough metal to hold the cold long enough to be effective. It likely will reach ambient before being placed on the field. The bimetal is the silver part in the picture. It is roughly .020 to .040 thick. The terminals connect to the left and right side of the bimetal. The terminals are connected to the #6 on both sides which will transfer temperatures and draw the parts close to ambient. The handle parts are removed in this drawing. Even spraying directly into the handle opening is unlikely to take this part down below ambient all things considered. The base is Bakelite.

[Nuttyman54]

Originally Posted by Al Skierkiewicz

Ian, There isn't enough metal to hold the cold long enough to be effective. It likely will reach ambient before being placed on the field. The bimetal is the silver part in the picture. It is roughly .020 to .040 thick. The terminals connect to the left and right side of the bimetal. The terminals are connected to the #6 on both sides which will transfer temperatures and draw the parts close to ambient. The handle parts are removed in this drawing. Even spraying directly into the handle opening is unlikely to take this part down below ambient all things considered. The base is Bakelite.

Al - my understanding was that the practice of cooling of breakers WAS to get them back to ambient, as opposed to potentially starting the match at higher than ambient temperatures and thus biasing the breaker to trip earlier than may be warranted.

[Road Rash]

Originally Posted by runneals

We had our pit partners ask for an ice pack during the finals to which we gave them our ice pack from our first aid kit I guess that works lol

As someone that focuses on safety, I don't recommend using an ice pack from a first aid kit to cool a motor on a robot unless you have plenty of ice packs in your kit to spare. An actual person may need that ice pack due to an injury. Just a friendly observation...

[Al Skierkiewicz]

Evan, there is enough heat sink in the wiring and mounting that I would bet the breaker is back to ambient in a minute or two.

[Mark Sheridan]

I have not tested the main breaker yet but I did get an opportunity to test the 30 amp breaker when we were debugging our current control software. We found it was difficult to get 30 amp breakers back down to an ambient temperature in a minute or two. We did get the 30 amp breakers to still open below 30 amps. Bare in mind we got them really hot (painful to touch) and the adjacent wires were warm to touch but it prompted us to cool these if we have back to back matches. However we only bother to do this under those circumstances and only particular circuits.

Not sure about the main breaker because we are careful with managing our current draw but in rapid practice matches we have gotten our battery lead wires quite warm. I hypothesize that this circumstance would limit the cooling of the breaker or at the very least have a temperature elevated than ambient. Maybe we will try to cook the practice robot this summer to see what happens.


Also to correct some misconceptions about using can air. You can buy non-flammable varieties and they come with lovely little set of warnings explaining the operating temperature. Those are the ones I buy, you can get them on mcmaster even. You even buy varieties designed for rapid cooling so you don't even have to turn the can upside down. The one I have has an operating temperature up to 450 degrees F. Paper burns at 480 F so if your robot is over 450F in any area, you probably should be using a fire extinguisher instead.

For the propellant HFC-134a, its non-flammable.
https://www.chemours.com/Refrigerant...c134a_push.pdf
It should be noted its non-flammable in air in temperatures up to 212F in normal atmosphere conditions. It can be ignited in the presence of oxygen or chlorine.

[Ryan_Todd]

Originally Posted by Ian Curtis

I thought these breakers were thermal (too much heat and a bi-metallic strip trips). What does it mean that there isn't enough thermal mass? Just that the electric heat builds up too quickly for the temperature at the start of the match to matter much?

One way to find out:
MATH!
(Hold on to your hats, people; this is gonna be a bumpy ride.)


The internal temperature required to trip these breakers is quite high relative to their size: a bit over 300 degrees Fahrenheit (149 C, or 422 K), to be precise (citation). At an ambient temperature of 77 F (25 C, or 298 K), they are still guaranteed to permit at least 120 A of continuous current for an indefinite period of time without tripping; that means that the steady-state internal temperature will never exceed ~270 F under those circumstances.

Given this, we can find that for a 120 A load, the approximate steady-state temperature rise between the ambient air and the breaker snap element is...

delta T = (270 F - 77 F)
delta T = 193 degrees Fahrenheit (or 89.4 C)

From thermodynamics, we also know that the rate of heat dissipation through a solid object is directly proportional to the temperature difference between two points (in this case, one point is inside the breaker switch and the other is in the air directly next to the surface of the breaker). From this, we can find a good approximation for the rate that the breaker in question can dissipate heat energy into the surrounding air.


* For the sake of my own sanity, I’ll use SI units for the rest of the math: degrees Kelvin for temperature, Joules for energy, et cetera. *


Looking around, I can't seem to find a stat for the main breaker's electrical resistance. In the absence of a known value, I'll just roll with 0.05 ohms as a fairly reasonable guess. This gives us a thermal conductance of...

C = (power dissipated) / (temperature rise)
C = (I^2 * R) / (dT)
C = ((120 A)^2 * (.05 ohms)) / (89.4 C)
C = 8 Watts per degree Celsius

...Which means that each additional degree of difference between internal and ambient temperature will increase the rate of heat energy transmission by 8 Watts.


Next, we can find a full equation for temperature (big T) as a function of time (little t), given the following constants: current (I), resistance (R), thermal conductance (big C), and initial temperature (Ti). We will also need the main breaker's specific heat capacity (little c), which will be a constant representing how much energy (in Joules) is required to raise the breaker's temperature by 1 degree Celsius. Here's how we work that out:

Temperature = (specific heat capacity) * (total heat energy)
T = c * E

Now to find the change in temperature over time, we take the derivative:

(rate of change in temperature) = c * (rate of change in heat energy)
dT/dt = c * dE/dt

Expand:

dT/dt = c * ((power usage) – (C * (T - Tambient))
dT/dt = c * (I^2 R – CT + CTambient)
dT/dt = c * (I^2 R – CT + CTambient)

Integrating this over the time range (0-t) lets us find the final temperature with respect to time:

T(t)= c*∫(0-->t)〖(I^2 R–CT+CTambient)dt〗

Well, this is fun. Looks like we've got ourselves a full-blown differential equation, which means that our dependent variable can't be extracted completely from the independent variable. To avoid the tedium of university-level DiffEq, then, we'll use a computer to help us from here on out.

T(t)=(I^2 R)/C+k1 e^(-Cct)+Tambient

Here, k1 is a constant of integration; that means that we need to solve for that value before we can proceed any further. If we set t = 0, that should help:

T(0) = (I^2 R)/C+k1 e^(-Cc(0))+Tambient
Ti = (I^2 R)/C+k1 e^0+Tambient
Ti = (I^2 R)/C+k1 (1)+Tambient
Ti = (I^2 R)/C+k1+Tambient

Solving for k1, we get:
k1 = Ti-(I^2 R)/C-Tambient

That’s pretty ugly, but we’ll plug it back into the big equation anyways:

T(t)=(I^2 R)/C+(Ti-(I^2 R)/C-Tambient ) e^(-Cct)+Tambient
T(t)=(I^2 (.05))/8+(Ti-(I^2 (.05))/8-Tambient ) e^(-8ct)+Tambient
T(t)=I^2/160+(Ti-I^2/160-Tambient ) e^(-8ct)+Tambient

Next step: solve for the specific heat constant (c)!
Consulting the main breaker's data sheet again, we find that 5 seconds is the time it takes for the breaker to trip if you start from room temperature (defined as 25 C) and run about 3.66 times the rated current. Our battery can't actually deliver this kind of current, but that doesn't matter just yet. These numbers still give us a useful data point (Ti = Tambient= 298 K, I = 439 A, t = 5 s, Tfinal = 422 K, ), which is precisely the sort of information we need to find (c)! Feeding this into our equation, we get:

T(t)=I^2/160+(Ti-I^2/160-T_ambient ) e^(-8ct)+Tambient
422=〖439〗^2/160+(298-〖439〗^2/160-298) e^(-8c5)+298
c = 1/40 (-3 log(3)-log(19)-log(337)+2 log(439))
c = 0.002716 degC / Joule

Plugging that result back in, we get our final equation for the breaker's internal temperature with respect to time, given the starting conditions (Ti), (Tambient), and (I):

T(t)=I^2/160+(Ti-I^2/160-Tambient ) e^(-8(0.002716)t)+Tambient
T(t)=I^2/160+(Ti-I^2/160-Tambient ) e^(-0.02173t)+Tambient

Now we consider the case of a 6-CIM drivetrain, running on 40 A snap-action breakers. If you manage to stall out all 6 motors, you'll approach the battery's maximum discharge current of 270 A (citation) for a few brief, glorious moments before something gives out. Let's leave aside the guaranteed RoboRIO brownout condition after only a couple seconds of this (citation), and also ignore the fact that the battery can only handle that kind of current draw for up to 5 seconds before it starts to sustain permanent damage. We'll even neglect the fact that (depending on manufacturing tolerances) the snap-action breakers might start to cut you off as soon as 4 seconds into the stall (citation); all this, so that we can focus on the main breaker by itself.
Subjected to 270 amps' worth of abuse, the main breaker is rated to hold out for anywhere from 6-25 seconds (there’s your manufacturing tolerance again) if you start from room temperature. For a quick sanity check, let’s throw this data point into our new equation and make sure it jives:

Tfinal = 300 F = 422 K
Ti = Tambient = 77 F = 298 K
I = 270 A
422=〖270〗^2/160+(298-〖270〗^2/160-298) e^(-0.02173t)+298
0.7278=e^(-0.02173t)
t = 14.6 seconds before the breaker trips

Cha-CHING! This number is squarely in the middle of the range where we expected to find it.


With that success in hand, we can find out exactly how long it takes for the inside of the breaker to cool down after a hard match. Just plug in the following values:

Ti = 270 F = 405 K
Tambient = 77 F = 298 K
I = 0 A (robot is off)

…And since we know that temperature changes are asymptotic (the final value is never quite equal to ambient; it just gets really, really close), we’ll ask for a Tfinal of 80 F = 300 K. Plug that all in, and solve for t:

300=0^2/160+(405-0^2/160-298) e^(-0.02173t)+298
2=107e^(-0.02173t)
t=183 seconds

All it takes is 2.5 minutes after a hard match, and the innermost parts of the main breaker are already back down to room temperature.


Now let’s investigate what happens if you chill the main breaker down to, say, -60 degrees Fahrenheit (source) by spraying it with a can of gas duster propellant. Will it still be cold by the time the match starts?

Ti = -60 F = 222 K
Tambient = 77 F = 298 K
I = 0.8 A (idle current)
Tfinal = 74 F = 296 K
296=〖0.8〗^2/160+(222-〖0.8〗^2/160-298) e^(-0.02173t)+298
-2=-76e^(-0.02173t)
t=167 seconds

Once again, it only takes a couple of minutes for the innermost parts of the main breaker to get back up to room temperature.

Okay, so we know that the extra chill doesn’t last all that long, but I’m still not quite convinced that there isn’t any benefit to be had. 167 seconds is still longer than a match, after all! What happens if we chill the main breaker at the last second before loading the bot onto the field, then start the match by immediately ramming an opponent and stall out our 6-CIM drivetrain? That’s the real situation where we might want the extra chill, after all. Exactly how much longer can we keep drawing the max current before the main breaker trips?
Part 1: load the bot onto the field

Ti = -60 F = 222 K
Tambient = 77 F = 298 K
t = 20 seconds
I = 0.0 A (robot is off)
Tfinal =0^2/160+(222-0^2/160-298) e^(-0.02173(20))+298
Tfinal =248.8 K=-11.8 F

Part 2: wait for match to start

Ti = 248.8 K
Tambient = 77 F = 298 K
t = 40 seconds
I = 0.8 A (idle current)
Tfinal =〖0.8〗^2/160+(248.8-〖0.8〗^2/160-298) e^(-0.02173(40))+298
Tfinal =277.4 K=39.7 F

Part 3: stall the drivetrain

Ti = 277.4 K
Tambient = 77 F = 298 K
Tfinal = 300 F = 422 K
I = 270 A (max possible current from battery)
422 =〖270〗^2/160+(277.4-〖270〗^2/160-298) e^(-0.02173t)+298
t=16.7 seconds

Our previous max stall time was 14.6 seconds, so the most benefit that we can possibly expect to get from chilling the main breaker is 2.1 extra seconds of stalling out the drivetrain once the match starts.

So now it seems like this could make the difference between dying in auto and surviving until teleop...but wait! Remember that the RoboRIO would brown out long before that happens, and that you'd probably also start tripping the snap-action breakers before you get to that point as well? That means that it's practically impossible to trip the main breaker before teleop, regardless of whether or not we chill it first...

So, what do you think?
Is it really worth risking the possibility of damage to your robot and/or yourself, or is this just a solution in search of a problem?

[techhelpbb]

I really thought this topic was going to about pneumatic vortex coolers.