An Idea for a load tester

[pythagoras]

How do you people think a load tester based on those small halogen bulbs would work?

I was thinging about a block made out of a real bad conductor, maybe some wood or plastic. Imbedded in the top would be a parabolic reflector, into which you could wire a number of these bulbs at different wattages.

Over the reflector would be a thin copper plate, black on the inside and lapped on top. Mounting holes would be easy to place in the top of the block.

The load could be varied by simply switching the required number of bulbs on or off.

Thoughts?

Regards

John.

[wdrzal]

That will work.how about a dimmer switch(reostat) to adjust the load.

walt

[LostInSpace278]

I think it would easier just to change the bulb. This way you would not need a DMM. A rheostat would work as well, if someone was really particular about the load.

[berkut]

Imo - a resistor is a better idea

Or a heatplug off a diesel engine

[bxa121]

cant you guys use a tec???

[pythagoras]

The one problem I see with a resistor is that it needs to be placed in a conductive box, and the heat travels in all directions, not just towards the evaporator.

Regards

John.

[craig588]

I think using a TEC is fine, that is what I have always done, it puts out the TECs wattage, plus the heat it is removing from the cold side.

[Redwolf]

Like berkut said... resistors. Personaly I like cartriage heaters. Basicly the same thing.

[blinky]

that would work but i think TECs are more simple to set up

[GuGaCoSa]

Where can I buy like 50 or 100w resistor to make a load tester?Is there special resistor?Its originally made for what funcion?
Thanks

[total_assault]

just use a tec and be done with it, they are like $5 on ebay pick one that represents the wattage output of your cpu when overclocked...

[Malachia]

guys a "paint" for a resistor heatload creator???

[pythagoras]

I want a variable load, the range must be from 50 watts to 250watts, and when I want a 50 watt load I want the power consumption of the load tester to be 50watts.

Lets say I use a resistor in a cuboid of copper, no a cube to make the calcs easier. If the resistor is 200watts, the cube has 6 faces therefore the heat on anyone face is going to be 200/6=33 1/3 watts. Or to load the evap connected to one face with 200 watts I would need a 1.2kw resistor. Maybe I'm missing something here.

I'm not sure on how to calculate the heatoutput of the hotside of a tec, given its nominal rating also the ambient temp on the coldside would alter the hotsides heat output.

I would like an accurate heat output, i.e. one that I can be confident is within 1 or 2 %.

More input please :D

Regards

John.

[blinky]

cant u buy one of those cpu load testers, and alter the watts accurately like they use to review heatsinks on tech sites?

[pythagoras]

Have you a link to what the review sites use, or a review that uses one of those cpu load testers?

Regards

John.

[wdrzal]

you are right and wrong with your cube theory ,if a resistor was in the direct center all 6 faces would heat equally. if a cold evap was applied to one surface, the other 5 surfaces would change ,but not equally. remember the rate at which heat flows depends on temp. differance & thermal resistance. the transfer would be quicker between the resistor and the cold face.the 4 faces that intersect the face with the different load will vary in temp across their faces.colder closer to the intersection,warmer farther away. only the oppisite face will have a uniform temp(this is not exactly true) but lets say that. the calculations to solve this simple thermodynamic problem could take days that where thermal moldaling software comes in handy.

as to the tec your are right you will have to add the load that the cold side absorbs to the wattage,I don;t remember what that formula is.

walt

[pythagoras]

I was just about to correct my assumption, though not on your observations, which are correct. The load would actually spread to a geater extent over just 4 faces not 6 :( .

But, your correct what seems simple, i.e. a resistor in a block is actually quite complex, when it comes to calculating the load on the evap face. Hence my idea of the halogen bulbs, who's heat can be focused on just an area equivalent to a cpu area.

I'm sure my idea has some flaws, that I'm blinded too though, any negative points to the idea?

Regards

John.

[wdrzal]

cube or reflector,insulate where you do not want the heat to go,that minamizes gains or losses. walt

[EMC2]

Use a cartridge heater, do it for setting up thermal testing all the time, can be accurately controlled. Regarding the issue of heat loss on the other faces... insulated all faces except the test side. If you want really high accuracy, get it calibrated, a lab with a calorimeter is the key.

[saratoga]

I was just about to correct my assumption, though not on your observations, which are correct. The load would actually spread to a geater extent over just 4 faces not 6 :( .

But, your correct what seems simple, i.e. a resistor in a block is actually quite complex, when it comes to calculating the load on the evap face. Hence my idea of the halogen bulbs, who's heat can be focused on just an area equivalent to a cpu area.


Regards

John.

Insulate the sides of the cube and the problem becomes trivial.

I'm sure my idea has some flaws, that I'm blinded too though, any negative points to the idea?


Very compicated compared to resistive heating.
Possibility of damaging bulbs due to excessive ambient temperature.
Nonconstant thermal load (bulb resistance is not a constant, will drift as the system runs).
Much more expensive.

BTW: The pros use nicrome heater catridges because the thermal load is independent of the die temperature. they're also cheap.

[hatemi]

Just check bowmans load tester at xtremerecources if you want an adjustable and more than acurate enough tester.

[wdrzal]

what about using the evaporator itself as the heat source. Its conductive right. Any body ever see the trick to thaw frozen water pipes? You just clamp on a dc welder,the resistance heats up the copper.you could pass the current from one side of base plate to the oppisite. oh well, their goes another one of my get rich patents :rolleyes: LOL. Walt

[wdrzal]

seems like a good idea,there would be no losses to consider. since energy is mutually convertable all you would have to do is measure current used.

[alexio]

What about a laser on a black surface?, that would generate some serious heat very concentrated.

It's just a thought :)

[wdrzal]

What about a laser on a black surface?, that would generate some serious heat very concentrated.

It's just a thought :)


What size is you laser? I may like to borrow it.mine are all milliwatt. walt

[gkiing]

Just check bowmans load tester at xtremerecources if you want an adjustable and more than acurate enough tester.

yes.. that was a work of art. He had machined copper plates to hold the load resistors.

You can use hot water as a load test to calculate the heat the evap can remove (in watts, eventually); time how long it takes to cool say 40C to 30C in an insulated calorimeter and do the physics.

[saratoga]

seems like a good idea,there would be no losses to consider. since energy is mutually convertable all you would have to do is measure current used.

Running electric current through your suction line would be a bad idea since its grounded via your compressor. It might not do any damage, but its not particularly safe. Its also difficult to precisely regulate load, and suffers from even worse temperature dependence then other methods. Also, secondary losses are more or less the same as with a heater catridge.

[wdrzal]

Running electric current through your suction line would be a bad idea since its grounded via your compressor. It might not do any damage, but its not particularly safe. Its also difficult to precisely regulate load, and suffers from even worse temperature dependence then other methods. Also, secondary losses are more or less the same as with a heater catridge.

current would not pass down suction line,just across evap.Could you explain what what you mean by primary and secondary losses?

[wdrzal]

what about this,2 carbon blocks clamp on pipe to solder with, would directly heat evap.adjusting current would regulate load. you could make your own.

We can call it :idea: "Direct Die Heating"

[saratoga]

current would not pass down suction line,just across evap.

Whats your suction line made out of? Metal?

Could you explain what what you mean by primary and secondary losses?

Secondary loss is any heat that doesn't go into the evap.

[pythagoras]

Wont current follow the path of least resistance, hence across the base of the evap? However, I wouldn't like to pass a high voltage across the evap with the system working when there are simpler methods.

How about just two bulbs a 50 watt and a 200watt that has a dimmer connected, that to me sounds easier and cheaper than a machined block with insulation.

Regards

John.

[pythagoras]

Opps major problem :rolleyes: My idea will rain, literally on the bulb and conections, back to the drawing board.

What did people go back to before the drawing board was invented :D

Regards

John.

[wdrzal]

current would take path of least resistance, here are the problems vs benifits I see. If you wanted too apply preciese heat load to systems with different evaporatos this would work well,It would take out of play the effiicancy of the evaporator,since the heat was not being conducted into baseplate.if you were testing to find the best evap. this idea would not work. If you were testing the rest of system it would be ok. I would like to hear more design ideas, I did like bowmans design. walt

[wdrzal]

lets see what we can come up with for pythagoras, I think he wants preciese heat load,inexpensive,can be fabricated out of easily obtained materials.Oh,and its completed before his vacation is over. ;)

[STEvil]

Current does not take the path of least resistance.

It takes the paths of least resistance.



This is why you dont want to have your feet significantly different distances away from a lightning strike within about 20 feet.

EDIT

peltiers seem the best way to go to me.. if your PSU has an adjustable voltage supply you can attach it at 5v, 7v, 12v, and adjust each rail for between 4v to 15v possibly then measure current draw at that voltage. If you have something to measure current draw from the wall at you can measure before/after and it will be within about 5 watts or so..

Gonna be testing my setup with a 326w peltier (@ 15v).. thats ~ 115w @ 5v, 155w @ 7v, 260w @ 12v... although i'll need something to measure amperage draw from the wall with to be more accurate than 325/15*xV

[hatemi]

I copy pasted my own post in another thread.

The basic principle behind all of this is ohms law. I wont get in detail on this matter since It's been awhile since I last studied it so I'll let someone else to explain it.

But you dont even need to understand the concept to use it. and This simple calculator programme helps you to determine the needed resitance at given voltage to get certain heatload: http://www.arcol.co.uk/pages/resour...calculator.html

And heres how I did mine. I use 230v so I dont need to carry around a power supply. I used 3 100ohm resistors connected in series to get 300ohm total resistance(measured 294ohm with my DMM)so teh heat imput is very close to 180w. You can use any pretty much any voltage you have and calculate what resitance you need for certain heatload. And you could even use those light dimmer thingys to lower the voltage and make the load variable.

And heres some pictures of it.