Battery test in a vacuum.

[Ken S. Tucker]

The SS2S is an expert civilian organiztion sending a rocket
into space, and conduct un-classified research as they go.
Your insights/membership would be appreciated. see...
http://sugarshot.org/

Batttery's in a vacuum.....

-Craig Peterson has completed the first round of battery vacuum
testing. The batteries tested were Energizer lithium AAA cells. Craig
reports "Using the 100 and 33 ohm loads absolutely nothing moved,
voltage and current both stayed exactly the same. This was also true
for the 10 ohm load but I did notice the batteries were slightly warm
when I removed them from the chamber. Also physical size did not
change at all during this time. Now when I used the 1.2 ohm load a
couple of things happened, first off the voltage and current changed
almost constantly throughout the test (both dropping) and then as the
chamber re-pressurized both voltage and current leveled out. When I
took the batteries out of the chamber they were very hot, to the point
it had discolored the wrapper on the batteries one in particular
looked as though it had gotten very hot and the seam where the wrapper
overlaps itself had the glue exposed. I again measured the batteries
after this test and they still came out with the same dimensions. I
attached some pictures that kind of show the discoloration and I will
get the data put into a spreadsheet. The test are 10 minutes long
total from the start of the vacuum to the end when it is re-
pressurized". The next testing will be basically a repeat, except done
at ambient pressure for comparison.
Chart indicating specifics of battery electrical load test
http://sugarshot.org/downloads/batterypower.gif
Photos of batteries:
http://sugarshot.org/downloads/picture1.jpg
http://sugarshot.org/downloads/picture2.jpg
http://sugarshot.org/downloads/picture3.jpg
http://sugarshot.org/downloads/picture4.jpg
http://sugarshot.org/downloads/picture5.jpg
http://sugarshot.org/downloads/picture6.jpg
http://sugarshot.org/downloads/picture7.jpg
(photos courtesy Craig Peterson)

[John S]

Just goes to show that, in a vacuum, you can't count on convection
cooling. Hmmmmm......

Or, did you have a question you forgot to add?

[Tim Wescott]

They should also toss 100 batteries or so in there for a long time -- at
least twice as long as they plan on depending on them -- and see how they
survive.

The "dry" in "dry cell" means "relatively dry". If the water gets let
out then I wouldn't expect the battery to work correctly.

--
www.wescottdesign.com

[a7yvm1...@netzero.com]

On Nov 16, 2:04 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> The SS2S is an expert civilian organiztion sending a rocket
> into space, and conduct un-classified research as they go.

> Your insights/membership would be appreciated. see...http://sugarshot.org/

>
> Batttery's in a vacuum.....

You've created a possessive or a contraction here, not a plural.

[Nemo]

Hmm well I've done battery short circuit testing. And occasionally put
circuitry in a vacuum. Not both together. Here are some rules of thumb.

There's no table I know of that says "overrate component power ratings
by factor X in a vacuum", though no doubt people designing satellites
know. So I use a factor of 4. Remember not to put two hot things near
each other or they can't cool as easily.

Almost all batteries have hidden safety vents. These are deliberate weak
points in their casing designed to rupture in a controlled manner if
they generate gas inside. If you look at an AA cell (I have not tested
AAA) you'll see a triangular score mark at one end on some makes... but
other manufacturers hide the release point under the pip at one end.
Your battery almost certainly has a deliberate weak point, which may be
where this discolouration is occurring for your battery, and I would be
nervous about putting batteries in vacuum for this reason. The swelling
of one in your first photo indicates a high pressure built up in it. So
the idea of putting 100 in a vacuum is an excellent one. You need more
data or you could spray hot caustic soda all over the inside of your
rocket when they rupture.

I once compared Energizer and Duracell and other brands of AA alkaline.
I found they were all within 1% of the same capacity when discharged at
about 40mA - I always laugh when I see an "Energiser Bunny" advert. They
may vary in how long they last at other loads

Where batteries DO vary is the temperature they reach when short
circuited (less than one ohm) because different manufacturers use
different constructions. For AA's, some get hottest at the tip, some in
the middle of the body. There are issues here such as how to get a
reliable contact to the ends, and being careful not to make the short
too massive (like thick copper braid) or it can cool the battery. It
takes a couple of minutes for the heat to peak. Some reach 100C, others
200C... in general the manufacturers keep improving their ampere-hour
capacity, so the peak temperature of a given brand rises as the years go by.

Beware battery contacts. Most are cheap and make poor contact. I've
designed robust instruments where we drop-tested them. Rockets have lots
of vibration so beware. A normal battery holder with e.g. Keystone
contacts allows the batteries to disconnect for a few milliseconds as
they bounce around. This resets processors and other circuits... you may
be able to get around this with a huge reservoir capacitor.

Suggest you wear safety goggles when you retest at ambient to avoid hot
goo etc in your eyes.

Nemo

On 16/11/11 19:04, Ken S. Tucker wrote:
> The SS2S is an expert civilian organiztion sending a rocket
> into space, and conduct un-classified research as they go.
> Your insights/membership would be appreciated. see...
> http://sugarshot.org/
>
> Batttery's in a vacuum.....
>
> -Craig Peterson has completed the first round of battery vacuum
> testing. The batteries tested were Energizer lithium AAA cells. Craig
> reports "Using the 100 and 33 ohm loads absolutely nothing moved,
> voltage and current both stayed exactly the same. This was also true
> for the 10 ohm load but I did notice the batteries were slightly warm
> when I removed them from the chamber. Also physical size did not
> change at all during this time. Now when I used the 1.2 ohm load a
> couple of things happened, first off the voltage and current changed
> almost constantly throughout the test (both dropping)

That's normal, but the battery temperature will be rising over minutes

and then as the
> chamber re-pressurized both voltage and current leveled out.

Because the battery's internal resistance is now maxed, and the charge
carriers can only move and recombine in the chemical gloop at a certain
max rate

When I
> took the batteries out of the chamber they were very hot, to the point
> it had discolored the wrapper on the batteries one in particular
> looked as though it had gotten very hot and the seam where the wrapper
> overlaps itself had the glue exposed.

This may have been the internal alkali leaking, it's a clear goo

[Ken S. Tucker]

Battery is in a vacuum ;-), with other batteries.
Ken

[Tim Williams]

"Nemo" <Ne...@nocannedmeatproducts.nosirree> wrote in message
news:M8Ywq.9452$Rz....@newsfe29.ams2...

> There's no table I know of that says "overrate component power ratings
> by factor X in a vacuum", though no doubt people designing satellites
> know. So I use a factor of 4. Remember not to put two hot things near
> each other or they can't cool as easily.

I don't know either, although I've seen ratings on individual components
before. Take this for example:
http://www.mif.pg.gda.pl/homepages/frank/sheets/131/6/6384.pdf
I have one of these laying around somewhere. Built like a tank.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

[Bitrex]

On 11/16/2011 11:23 PM, Tim Williams wrote:
> "Nemo" <Ne...@nocannedmeatproducts.nosirree> wrote in message
> news:M8Ywq.9452$Rz....@newsfe29.ams2...
>> There's no table I know of that says "overrate component power ratings
>> by factor X in a vacuum", though no doubt people designing satellites
>> know. So I use a factor of 4. Remember not to put two hot things near
>> each other or they can't cool as easily.
>
> I don't know either, although I've seen ratings on individual components
> before. Take this for example:
> http://www.mif.pg.gda.pl/homepages/frank/sheets/131/6/6384.pdf
> I have one of these laying around somewhere. Built like a tank.
>
> Tim
>

In a perfect vacuum there's only radiative cooling. If you have a
component mounted to a heatsink, say sort of like this one:

http://rocky.digikey.com/weblib/Wakefield%20Engineering/Web%20Photo/637-10ABP.jpg

and there's negligible conduction down the leads, you should be able to
approximate the power dissipation of the sink for a certain sink
temperature, assuming the heatsink is a perfect blackbody and using the
Stephan-Boltzmann law. Then you can rearrange the equation using the
various thermal resistances to find how much power could be dissipated
from the sink at the maximum die temperature.

Take the reciprocal of the derivative of the Stephan-Boltzmann equation
multiplied by the total surface area of the sink and turn it into a
thermal resistance as a function of power: dT/dP =
1/[area*alpha*4*(P/2*area)^3/4]. That should give you an approximation
of the thermal resistance from the sink to ambient. The add in the
other thermal resistances to the equation, like junction to case, etc.
Set the equation equal to the max die temperature and solve for P and
that should give an approximation to the maximum allowable power
dissipation. Of course this doesn't account for second-order effects,
such as if the component is in a tiny compartment and the radiated
energy is then radiated back at the component until the whole system is
in some kind of thermodynamic equilibrium.

It doesn't seem like some heatsink manufacturers, such as the
aforementioned Wakefield, give figures for the surface area of their
sinks, but it should be easy enough to estimate from the mechanical
drawings in the datasheets.

[Bitrex]

On 11/16/2011 11:23 PM, Tim Williams wrote:

> "Nemo" <Ne...@nocannedmeatproducts.nosirree> wrote in message
> news:M8Ywq.9452$Rz....@newsfe29.ams2...
>> There's no table I know of that says "overrate component power ratings
>> by factor X in a vacuum", though no doubt people designing satellites
>> know. So I use a factor of 4. Remember not to put two hot things near
>> each other or they can't cool as easily.
>
> I don't know either, although I've seen ratings on individual components
> before. Take this for example:
> http://www.mif.pg.gda.pl/homepages/frank/sheets/131/6/6384.pdf
> I have one of these laying around somewhere. Built like a tank.
>
> Tim
>

In a perfect vacuum there's only radiative cooling. If you have a
component mounted to a heatsink, say sort of like this one:

http://rocky.digikey.com/weblib/Wakefield%20Engineering/Web%20Photo/637-10ABP.jpg

and there's negligible conduction down the leads, you should be able to
approximate the power dissipation of the sink for a certain sink
temperature, assuming the heatsink is a perfect blackbody and using the
Stephan-Boltzmann law. Then you can rearrange the equation using the
various thermal resistances to find how much power could be dissipated
from the sink at the maximum die temperature.

Take the reciprocal of the derivative of the Stephan-Boltzmann equation
multiplied by the total surface area of the sink and turn it into a
thermal resistance as a function of power: dT/dP =

1/[area*alpha*4*(P/2*area)3/4]. That should give you an approximation

of the thermal resistance from the sink to ambient. The add in the
other thermal resistances to the equation, like junction to case, etc.
Set the equation equal to the max die temperature and solve for P and
that should give an approximation to the maximum allowable power
dissipation. Of course this doesn't account for second-order effects,
such as if the component is in a tiny compartment and the radiated
energy is then radiated back at the component until the whole system is
in some kind of thermodynamic equilibrium.

Even so, given that alpha is 5.67*10^-8, it looks like you're going to
need very large sinks to dissipate even modest amounts of power if you
want to keep a die under its max temperature in a total vacuum.

[Bitrex]

Just to clarify you'd set the equation up like this:
1/[area*alpha*4*(P/2*area)^3/4]*P + sum(thermal_resistances)*P =
die_temperature. Solve using your numerical method of choice.

[George Herold]

On Nov 17, 2:44 am, Bitrex <bit...@de.lete.earthlink.net> wrote:
> On 11/16/2011 11:23 PM, Tim Williams wrote:
>

> > "Nemo" <N...@nocannedmeatproducts.nosirree> wrote in message

> >news:M8Ywq.9452$Rz....@newsfe29.ams2...
> >> There's no table I know of that says "overrate component power ratings
> >> by factor X in a vacuum", though no doubt people designing satellites
> >> know. So I use a factor of 4. Remember not to put two hot things near
> >> each other or they can't cool as easily.
>
> > I don't know either, although I've seen ratings on individual components
> > before.  Take this for example:
> >http://www.mif.pg.gda.pl/homepages/frank/sheets/131/6/6384.pdf
> > I have one of these laying around somewhere.  Built like a tank.
>
> > Tim
>
> In a perfect vacuum there's only radiative cooling. If you have a
> component mounted to a heatsink, say sort of like this one:
>

> http://rocky.digikey.com/weblib/Wakefield%20Engineering/Web%20Photo/6...

>
> and there's negligible conduction down the leads, you should be able to
> approximate the power dissipation of the sink for a certain sink
> temperature, assuming the heatsink is a perfect blackbody and using the
> Stephan-Boltzmann law.  Then you can rearrange the equation using the
> various thermal resistances to find how much power could be dissipated
> from the sink at the maximum die temperature.
>
> Take the reciprocal of the derivative of the Stephan-Boltzmann equation
> multiplied by the total surface area of the sink and turn it into a
> thermal resistance as a function of power: dT/dP =
> 1/[area*alpha*4*(P/2*area)^3/4]. That should give you an approximation
> of the thermal resistance from the sink to ambient.  The add in the
> other thermal resistances to the equation, like junction to case, etc.
> Set the equation equal to the max die temperature and solve for P and
> that should give an approximation to the maximum allowable power
> dissipation.  Of course this doesn't account for second-order effects,
> such as if the component is in a tiny compartment and the radiated
> energy is then radiated back at the component until the whole system is
> in some kind of thermodynamic equilibrium.
>
> It doesn't seem like some heatsink manufacturers, such as the
> aforementioned Wakefield, give figures for the surface area of their
> sinks, but it should be easy enough to estimate from the mechanical
> drawings in the datasheets.

Oh, he could paint his batteries black. That would help radiate
more. The silver wrapper around the batteries won't help.

It'd be fun to compare blackened to unblackened.

George H.

[Bitrex]

A reflective silver coating is about the worst choice possible!

I remember a sci-fi novel where one of the alien species involved had
modified their bodies to live off-planet, directly in the vacuum of
space. They were called "Silver Ghosts" because they packed themselves
into silver spheres to conserve heat energy; the silver surface
reflected most incident heat energy but it also acted as a near perfect
insulator.

[Uwe Hercksen]

Bitrex schrieb:

>
> In a perfect vacuum there's only radiative cooling.

Hello,

no, there is also some conductive cooling too, the battery must be
supported by something solid and this will conduct some heat to the
walls of the vacuum chamber.

Bye

[patricia herold]

> insulator.- Hide quoted text -
>
> - Show quoted text -

Yeah, super insulation is just many many layers of aluminum coated
mylar.

Oh here's an easy way to guestimate the heat loss from a perfect
'black' radiator.

It goes as temperature (T) to the fourth. T**4, and 300K to 0K is
close to 0.5 kW/m**2
Which is easy to remember since it's 1/2 the solar radiation. (1kW/
m**2)

George H.

[Tim Williams]

"patricia herold" <pmdh...@gmail.com> wrote in message
news:5b626cc0-b9f1-4188...@y12g2000vba.googlegroups.com...

> Oh here's an easy way to guestimate the heat loss from a perfect
> 'black' radiator.
>
> It goes as temperature (T) to the fourth. T**4, and 300K to 0K is
> close to 0.5 kW/m**2
> Which is easy to remember since it's 1/2 the solar radiation. (1kW/
> m**2)

Note:

Loss, yes.

But not gain.

Radiation is an equilibrium thing, so when you're doing heat flow, you
need the difference, which is naturally the difference of quartics --
t_hot^4 - t_cold^4, where the temperatures are absolute. For sufficiently
large differences (like 1000K vs room temp), it doesn't matter a lick, but
for small differences it's critical.

[Ken S. Tucker]

On Nov 18, 7:58 pm, "Tim Williams" <tmoran...@charter.net> wrote:
> "patricia herold" <pmdher...@gmail.com> wrote in message

>
> news:5b626cc0-b9f1-4188...@y12g2000vba.googlegroups.com...
>
> > Oh here's an easy way to guestimate the heat loss from a perfect
> > 'black' radiator.
>
> > It goes as temperature (T) to the fourth. T**4, and 300K to 0K is
> > close to 0.5 kW/m**2
> > Which is easy to remember since it's 1/2 the solar radiation. (1kW/
> > m**2)
>
> Note:
>
> Loss, yes.
>
> But not gain.
>
> Radiation is an equilibrium thing, so when you're doing heat flow, you
> need the difference, which is naturally the difference of quartics --
> t_hot^4 - t_cold^4, where the temperatures are absolute. For sufficiently
> large differences (like 1000K vs room temp), it doesn't matter a lick, but
> for small differences it's critical.
> Tim

That could be an issue, especially in a small nosecone that has
been heated by aerodynamic friction.
That radiation factor depends on the interior infared radiation flux.
Ken