How to Calculate Heat Transfer to Coolant in a Hybrid Lighting System?

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Discussion Overview

The discussion revolves around calculating the heat transfer to coolant in a hybrid water/forced air heat transfer system for a prototype lighting solution. Participants are exploring the calculations necessary to determine how much of the heat generated by the LEDs is transferred to the coolant (water) as it passes through the system's radiators.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that with 12% efficiency, 88% of the 750 watts input becomes heat, suggesting that all of this heat would be transferred to the coolant.
  • Another participant clarifies that they are specifically interested in how much heat is transferred to the coolant versus how much is lost to the surrounding area, emphasizing the evaluation of the cooling system's effectiveness.
  • A different participant suggests that to calculate heat transfer, one needs to know the heat capacity of the fluid, flow rate, and inlet/outlet temperatures, noting that phase changes could complicate the calculations.
  • Another contribution mentions that if the LEDs are mounted on a ceramic substrate bonded to a water-cooled copper heat sink, most of the heat would be transferred, but if they are screw-in types, practically none would be transferred.
  • One participant proposes calculating the maximum allowed thermal resistance from the LED to the coolant based on the energy to be removed and the desired LED temperature.
  • Another participant suggests assuming the entire power input goes to the cooling system as a safety margin against temperature rises.
  • One participant raises a question about the confirmed efficiency of the LEDs, suggesting that modern LEDs might achieve better efficiency.

Areas of Agreement / Disagreement

Participants express differing views on the efficiency of heat transfer to the coolant and the factors affecting it. There is no consensus on the exact calculations or methods to be used, and multiple competing views remain regarding the effectiveness of the cooling system.

Contextual Notes

Participants mention various factors that could influence heat transfer calculations, including the design of the LED mounting, the characteristics of the coolant, and the potential for phase changes in the fluid. These aspects introduce complexity and uncertainty into the calculations.

craserv
My colleague has been developing a hybrid water/forced air heat transfer system for a prototype lighting solution we are developing.

I am struggling to get my head around the calculations required to figure out exactly how many watts of energy (heat in this case) are being transferred to the coolant (water) as it passes through it's radiators.

We are driving the at 750 watts, the fixture's LEDs are ~12% efficient--- so how would we calculate exactly how much of the remaining 660 watts radiant heat is being transferred to the coolant via our cooling system?

Thanks for in advance, any help would be greatly appreciated.
 
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Here's an article that talks about lighting and heat.

So based on it your 12% efficiency means 88% becomes heat so that would 88% of 750 watts becomes heat so I would suspect all of it would be transferred.
 
jedishrfu said:
Here's an article that talks about lighting and heat.

So based on it your 12% efficiency means 88% becomes heat so that would 88% of 750 watts becomes heat so I would suspect all of it would be transferred.
Where is the article?
 
Jedishrfu, Thank you very much for the reply- while you would be correct that all of the 660 watts of energy converted to heat would eventually be transferred out of the system- I am looking to calculate specifically how much is transferred into the coolant medium itself versus how much is transferred to the surrounding area. This is an evaluation of the effectiveness of the cooling system, not the efficiency of the LEDs.

Thanks very much,

-A
 
craserv said:
Jedishrfu, Thank you very much for the reply- while you would be correct that all of the 660 watts of energy converted to heat would eventually be transferred out of the system- I am looking to calculate specifically how much is transferred into the coolant medium itself versus how much is transferred to the surrounding area. This is an evaluation of the effectiveness of the cooling system, not the efficiency of the LEDs.

Thanks very much,

-A
You need to know the heat capacity of the fluid, flow rate, inlet and outlet temperatures. This allows you to calculate how much heat is transferred to the fluid itself, the rest goes elsewhere. It gets more complicated if you have phase change (boiling) of the fluid.
 
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Thanks so much for the reply Vitro. The liquid used for cooling is water origining from a ground source with ~300 ppm TDS.
 
craserv said:
My colleague has been developing a hybrid water/forced air heat transfer system for a prototype lighting solution we are developing.

I am struggling to get my head around the calculations required to figure out exactly how many watts of energy (heat in this case) are being transferred to the coolant (water) as it passes through it's radiators.

We are driving the at 750 watts, the fixture's LEDs are ~12% efficient--- so how would we calculate exactly how much of the remaining 660 watts radiant heat is being transferred to the coolant via our cooling system?

Thanks for in advance, any help would be greatly appreciated.
I gather that this is still in the design phase and you are trying to size the cooling system. If the LEDs are surface mounted on a ceramic substrate that is bonded to a water-cooled Copper heat sink, the answer is 'most of it' (assuming adequate water flow). If the LED configuration is screw-in as the household consumer ones are, then the answer is 'practically none.'

The general approach is to define the amount of energy to be removed (660 Watts), the maximum desired LED temperature, and the temperature of what ever you are dumping the heat to. From that you calculate the temperature difference you have available to transfer the heat. Divide the temperature difference by your 660 Watts to yield the maximum allowed Thermal Resistance (Degrees C per Watt, °C/W) from LED to Coolant. (Another way to show that is as Watts/°C, the reciprocal of what I calculated here. Use which ever you are comfortable with.)

With a fluid coolant, water in this case, you need the incoming temperature and the heat capacity of the fluid, (generally given as Watts per °C per unit weight.) With the heat capacity of the fluid, the fluid flow rate, and the thermal energy entering the fluid, you calculate the fluid temperature rise. The temperature rise plus the temperature of incoming water gives the exit temperature. For initial calculations, you can use the average of the incoming and exit temperatures.

Since you are using forced air to a radiator to water, you have the extra stage with Air as the initial working fluid and Water as the second working fluid.

That is the easy part. The Engineering part is creating an LED mounting method with a low enough thermal resistance to keep them cool.

Well, that is my two cents worth. Others here are much better at walking you thru the fine details and finding the material constants you will need.

Hope it helps, and good luck.

p.s. I'm curious. Those are some awfully bright LEDs, what are they being used for?
 
IMHO, I'd assume the entire power input was going to the cooling system, and keep that 12% as part of your safety margin against ground-water temperature rises, filter blinding etc etc...

Um, have you had a look at fluid-cooled CPU & PC systems ? IIRC, they can be similar power levels, and have some scary solutions.
Including total immersion...
 
To give and idea of how much heat you are talking about, that 750W is 2558BTU, a small window are conditioner running at 50%.
 
  • #10
Have you got definite confirmation about that 12 % efficiency ? You should be able to do better than that with modern LED lighting .

You certainly could with other forms of electric lighting . You could probably do better with a paraffin lamp for that matter .
 
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