### Author Topic: Aluminum Block and/or chassis as heatsink?  (Read 1611 times)

0 Members and 1 Guest are viewing this topic.

• Supreme Robot
• Posts: 611
##### Aluminum Block and/or chassis as heatsink?
« on: September 30, 2013, 12:34:46 PM »
Hi all,
Quick question for those of you who know more about thermal modeling than me. I'm building a very simple device for some research. I'm using a solid state relay with an internal resistance of ~ 8000 Ohms. It's switching 240VAC therefore will be dissipating at max 7.5W of power in terms of heat. I was debating just buying a \$5.00 10 W heatsink from digikey then I remembered I have a large 3/8 inch plate of aluminum. I was thinking that I could simply use a block of aluminum as my heatsink and attach it to the chassis of my device for more heat dissipation, but I have no idea how to determine how large my block of aluminum would have to be to make the device safe.

I've looked at a few websites and I can calculate the temp of the aluminum block vs. time, but I don't know how to factor in the transfer of heat to the air. I've seen very varied results for the heat transfer coefficient of aluminum that either make my device work just fine, or make it way too hot. (I've seen anywhere from 6 W/m^2*K to 35 W/m^2*K) (And, in different units 205-250 W/m*K)

Sure I can just plug these numbers into solidworks and find out that with ~10 W/m^2*K, the temp of a 5x6x2 aluminum block will raise about 80K with 8W of power.
Hm.... those numbers look suspiciously easy...

Surface area of the block: 104 cm^2
.001 W/cm^2*K
80K
8W

(.001W/cm^2*K * 104cm^2)/8W = .013 T^-1 ~~ 77K.

Ok, so that's how you do it, but what's the correct number for the heat transfer coefficient for aluminum? If it's 30 W/m^2*K then it raises only 26K, but if it's 10 like in the example above, it's way too hot. And what if I add a fan blowing across the large face of the heatsink what difference does that make?

#### waltr

• Supreme Robot
• Posts: 1,944
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #1 on: September 30, 2013, 01:36:40 PM »
Yep, I've run across this also. Best way is to actually measure the temperature rise with a known amount of heat then calculate the transfer.

The transfer values will differ pending several factors including (but not limited to):
The surface finish, smoothness and color.
The actual alloy used
Is the surface horizontal or vertical?

Forcing air across the heat sink increase the heat transfer. The greatest increase is with the smallest amount of air and the increase diminishes as the amount of air increases.

• Supreme Robot
• Posts: 611
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #2 on: September 30, 2013, 02:13:46 PM »
Thanks waltr, I'll cut out a piece of the aluminum and run some quick tests while monitoring the temp.

#### jwatte

• Supreme Robot
• Posts: 1,345
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #3 on: October 01, 2013, 10:28:57 AM »
Even more important is airflow. Just a little bit of air moving across it will make a huge difference.
If you have a mill nearby, slotting the block to make fins may also significantly increase the surface area, and thus make for better heat transfer to air (but less thickness to transfer heat out of the original contact.)

By the way: What is the 8 kOhm solid state relay value? Are you saying it introduces 8 kilo-ohms of resistance to the 240V signal? That's... really weird, and I don't understand what that would be useful for. Could you elaborate?
In general, heat generated equals resistance times current squared, so how much current are you controlling?

• Supreme Robot
• Posts: 611
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #4 on: October 01, 2013, 03:44:30 PM »
By the way: What is the 8 kOhm solid state relay value? Are you saying it introduces 8 kilo-ohms of resistance to the 240V signal? That's... really weird, and I don't understand what that would be useful for. Could you elaborate?
In general, heat generated equals resistance times current squared, so how much current are you controlling?

Yeah I was thinking a bit about that more the other day and it doesn't seem right. I'm using this solid state relay.
http://www.coleparmer.com/Product/Solid_state_relay_40_amps_with_3_to_32_VDC_control_input/EW-94469-40

If I switch it on using 12VDC, the resistance between the two other contacts (the load) is about 8 kohms. I am driving a furnace that's made up of some resistance wire. The resistance is on the order of 10s of ohms.

Just for the record, this relay was recommended to me for this application by the people at cole palmer.  (And setups like this have been used many times in the past, in fact I replaced a system with pretty much the exact same components to build this one (mine can handle more current.))

So if I connect a furnace with resistance on the order of 10s of ohms in series with this relay, most of the heat will be dissipated within the relay itself, with almost none going to my furnace.

The load itself is huge, since the furnace is tiny the current running through it has to be relatively high (up to a point) to get the temp up to where I need it.

I haven't had time to think about it this week, too much to do, but I'll be working on it more next week.

(I'm assuming/hoping as the relay heats up the resistance drops or it's supposed to be wired up in parallel... but I'm not quite sure how that'd work.)

#### Roman505

• Jr. Member
• Posts: 44
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #5 on: October 01, 2013, 04:29:54 PM »
Are you able to discover how much current, exactly, or the exact resistance of the furnace load?

Where did you get 8 kOhm internal resistance for the SSR please? I saw 1.5kOhm on the spec sheet but may have missed something.

Looking at the Derating Curves, I would choose a proper heatsink as proposed on that sheet, not hope to manage it with an aluminium case or plate. Doing that (a proper heatsink) makes the rest of this rather moot.

• Supreme Robot
• Posts: 611
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #6 on: October 01, 2013, 05:28:45 PM »
Are you able to discover how much current, exactly, or the exact resistance of the furnace load?

Where did you get 8 kOhm internal resistance for the SSR please? I saw 1.5kOhm on the spec sheet but may have missed something.

Looking at the Derating Curves, I would choose a proper heatsink as proposed on that sheet, not hope to manage it with an aluminium case or plate. Doing that (a proper heatsink) makes the rest of this rather moot.

The resistance of one of the furnaces is 10 ohms. (As of right now, I may have to use different nichrome wire for more resistance.) The 8kOhm resistance I got from activating the relay with a 12VDC signal and measuring the resistance across the other two terminals with a multimeter...

Yes, I could buy the \$30-40 heatsink (on the website I bought it from), or I could go to digikey and find one that'll work for much cheaper. I expect it not to be possible to use a block of aluminum with this relay, I was asking the question in general during the original post.
« Last Edit: October 01, 2013, 05:36:06 PM by corrado33 »

#### waltr

• Supreme Robot
• Posts: 1,944
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #7 on: October 01, 2013, 06:39:46 PM »
That was a DC resistance with no load.
Try measuring the Voltage drop with a know load an AC power.

http://en.wikipedia.org/wiki/Solid-state_relay

I see in the specs an input impedance of 1.5k Ohm. and a Voltage drop at rated load of 1.5V.
http://www.coleparmer.com/Virtual-Catalog/US/2136

Using Ohms Law that means the resistance is 1.5V/40Amp = 37.5mOhm.
This makes sense.

Have you tried the SSR yet?

• Supreme Robot
• Posts: 611
##### Re: Aluminum Block and/or chassis as heatsink?
« Reply #8 on: October 01, 2013, 07:39:49 PM »
That was a DC resistance with no load.
Try measuring the Voltage drop with a know load an AC power.

http://en.wikipedia.org/wiki/Solid-state_relay

I see in the specs an input impedance of 1.5k Ohm. and a Voltage drop at rated load of 1.5V.
http://www.coleparmer.com/Virtual-Catalog/US/2136

Using Ohms Law that means the resistance is 1.5V/40Amp = 37.5mOhm.
This makes sense.

Have you tried the SSR yet?

I figured as much for the resistance after I saw the data sheet. I still haven't had time to play with it, very busy week.