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Energy efficient, waterless, Copper strip based cooling system Answered

Energy efficient ,waterless Copper stripe based cooling system:

The presented Idea is based on Heat exchange principle.As per sketch there will be a conveyor belt system and some copper stripes will be connected with this belt.Each copper strip will have a moveable hook.The conveyor belt will move with a motor.Each copper strip will be separated with each other and with conveyor belt with the help of Insulation (wooden insulation).Each copper strip will be connected with the machine for 15 to 20 seconds to absorb the heat and after that this copper strip will be removed and next copper strip will be ready to absorb the heat.The removed strip will be cooled down after some time and will be ready to couple with the machine again.In this way each copper stripe will work one by one to absorb the heat to cool the machine and each strip will get cool down due to convection .

What I thought is that if I use a continuous water jet to cool the motor then the motor will remain cool so if I use the copper stripes then the same effect will be occured.It will work in this way that temperature will be not increased of machine if I couple the device with the machine. It will be low cost,low maintenance,waterless solution to cool the machines. It will consume almost nil energy

Discussions

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Downunder35m
Downunder35m

6 weeks ago

I might have missed something, but isn't that why heat pipes were invented?
To eliminated the mechincal parts...

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vikram_gupta11
vikram_gupta11

Reply 5 weeks ago

Heatsink cannot maintain the initial temperature of machines but this design will work in this way that the initial temperature will be not increased.
That's why it is far better than Heat sink.

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Downunder35m
Downunder35m

Reply 5 weeks ago

If you say so then just build one and test it to compare reality with your theory ;)

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vikram_gupta11
vikram_gupta11

Reply 4 weeks ago

Though, i have not build it but i have done some work on it and results are as per my expectation.

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Orngrimm
Orngrimm

Reply 5 weeks ago

You, Sir, are 100% correct.

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Jack A Lopez
Jack A Lopez

6 weeks ago

You know, for these kind of situations, for which heat is being carried away by moving mass, the rate of heat flow is proportional to the rate of mass flow, and also the temperature difference between the outgoing stream and the incoming stream, like so:

(dQ/dt) = (dm/dt)*C*(Tout - Tin)

where that number C is the so called, "specific heat"

https://en.wikipedia.org/wiki/Specific_heat_capaci...

Really what I am assuming there, is that if I have a small lump of mass, dm, the amount of heat, dQ, needed to raise its temperature from Tin to Tout is:

dQ = dm*C*(Tout - Tin)

I am also assuming C is approximately constant over that temperature range, so that C(Tout) = C(Tin) = C(every T in between Tin and Tout)

Anyway, where I am going with this, is to say that with this one simple equation,

(dQ/dt) = (dm/dt)*C*(Tout - Tin)

you can broadly compare your conveyor belt of copper lumps, to a more traditional cooling system, e.g. two pipes, one that delivers cooling water, at temperature Tin, and another that takes that water away, at temperature Tout.

Your mass flow rate, (dm/dt), for a copper lump conveyor belt, is roughly, just the mass of each lump, divided by the amount of time it stays connected to the heat source plus the time it takes to connect and disconnect.

Regarding the specific heat C, of copper, or water, or whatever, this page linked below, at EngineeringToolbox, lists specific heats for common materials, (uh, probably at temperatures close to room temp, although it doesn't say)

https://www.engineeringtoolbox.com/specific-heat-c...

By the way, if all the quantities in that equation are in SI units, then that would give:

(dQ/dt) in joule/second, or watts, J/s = W

(dm/dt) in kilogram/second, kg/s

(Tout - Tin) in degrees C

C in joule/(kilogram*degrees C), J/(kg*degC)

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vikram_gupta11
vikram_gupta11

Reply 6 weeks ago

There is need of some work on it. I want to keep it simple and energy efficient.

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vikram_gupta11
vikram_gupta11

6 weeks ago

The process is to place a copper strip in proximity to heat source for 15 sec. During this time, the copper strip will absorb via convection as much heat as possible from the raised ambient temperature. The strip will then be removed and allowed to cool before being reintroduced. Identical strips will be serially introduced to maintain a constant flow of strips.