Can you use a fridge to cool a modern gaming computer, for 24 / 7 use?
Mini / Dorm Refrigerator = NO
Standard Size Two Door = NO
Large Side By Side = No
Commercial Reach In = Maybe, See Link, Specs in PDF
http://www.ckitchen.com/kitchen/restaurant-equipment/reach-in-refrigerators/randell/randell-2010-reach-in-rerigerator.html)
The concept of cooling a computer with a fridge has surfaced in this forum on a regular basis. Search the word “fridge” you’ll get a pretty good idea why most of the replies are negative or sarcastic, Search First, Ask Second.
Basing the answer on Btu rating and covering variation of the question.
1 Watt = 0.0009485 Btu (s)
http://www.ehow.com/how_6038931_calculate-btu-output-watts.html
Finding the BTU rating for refrigerators proved a little difficult, most folks are more interested in cubic feet, door location, ice makers, color, etc. I did find one for a Frigidaire refrigerator compressor kit, rated at 750 Btu*. What little information found for mini / dorm-fridges put the rating around 100-150 and large side by sides at 1000– 1200 Btu. Home style freezers are usually lower on the Btu rating than fridges due to the fact that freezers don’t see the action on the door a refrigerator full of beer / soda does. I’m sure if more time was spent a little more data could be found, but the 750 from a full size Frigidaire is more than adequate for this explanation.
Computer in the fridge, air cooled
Taking 350 watts out of the air to represent a loaded computer (Comes from the Guru3d site for a lightly clocked i7 920 with a strong video card, running business apps).
350 x 0.0009485 = .331975 Btu (s) x 3600 = 1195.11 Btu (h)
Subtracting the 750 from the 1195 creates (445 Btu (h)) of surplus heat, quickly turning the inside of the fridge into an easy bake oven.
Computer in the fridge, water cooled, large reservoir.
One Btu equals the energy need to raise the temperature of one pound of water one degree Fahrenheit.
The first example establishes 445 Btu’s (h) of extra heat or 130.4 watts.
130.4 x 0.0009485 = 0.1236844 Btu (s)
1 / 0.1236844 = 8.085 seconds to raise 1 pound of water 1 degree Fahrenheit.
A gallon of water weights 8.35 pounds +/-, for this example five gallons will be used for a total of 41.35 pounds of water. My own system uses about a 1.5 gallons or 5.68 liters and in general terms excessive compared to some water cooled systems.
8.085 x 41.35 = 334.31 seconds to raise 5 gallons 1 degree Fahrenheit.
Now let’s see how long it would take to go from 0c to 22c (32f to 71.6f)
71.6 – 32 = 39.6 x 334.31 = 13,238.9 seconds, (3hr 40m 38.9s)
The two examples are hypothetical, in both, based on 750 Btu being removed by the fridge, and the latter, all the heat being past to the water. Neither of which is realistic. The water cooled example biggest problem is the heat sources not cooled by the water, warming the trapped air creating the same condition in example one but at a slower rate.
Using as a water chiller, large reservoir, computer outside of fridge.
Taking information from the supplied link below.
http://www.electronics-cooling.com/2008/05/estimating-dew-point-temperature-for-water-cooling-applications/
To keep from prepping against condensation problems, from the chart, 20c ambient and 30% humidity, gives a dew point of 1.9c (36f give or take)
72f -36f = 36 degrees of difference, the 72f would be inside the fridge
Fridge at 750 Btu = 220 watts (give or take)
CPU & Video Card = 250 watts (Assumed, I’ve past this w/ a Q6700 & GTX 280)
30 watts of excess heat
30 x 0.0009485 = 0.028455 Btu (s)
1 / 0.028455 = 35.14 seconds to raise one pound of water one degree
35.14 x 41.35 (lb) = 1,453.04 seconds for five gallons of water
1,453.04 x 36 = 52,309.44 seconds or 14h 31m 49s to get the water to 72f.
This example assumes the fridge is removing 750 Btu.
In all three examples the fridge would be working continuously trying to keep up with the heat load. Chances are the thing is going to overheat causing premature failure.
Moisture would be a problem in the first two examples. The computer would be warming the inside of the fridge quickly melting any water vapor frozen to the colder parts.
Convert an AC unit, a $100 unit at 5000 Btu is quite capable of handling a gaming PC.
Try something like this.
http://www.overclock.net/t/1264939/final-oc-results-window-air-con-chill-box-build
Pulling an older 1 ton unit apart in my garage to try this.
Power Supply wattage rating to Btu (h), in reference post #3, sizing.
500 = 1710
750 = 2560
1000 = 3410
1250 = 4270
Notes
Added side by side information.
*, The 750 Btu used in the example is a little low, found a general answer of 800 Btu average for standard two door fridges.
Since a lot of refrigerators’ hot side is not actively cooled, Removed, the three dorm size fridges I’ve owned were passive on the hot and cold side. Looking at schematics for newer larger units the statement was not true. Side Note, would you use one 120mm fan to cool your computer?
Added PS Info
Edited by Aleslammer - 8/28/12 at 9:27am
Mini / Dorm Refrigerator = NO
Standard Size Two Door = NO
Large Side By Side = No
Commercial Reach In = Maybe, See Link, Specs in PDF
http://www.ckitchen.com/kitchen/restaurant-equipment/reach-in-refrigerators/randell/randell-2010-reach-in-rerigerator.html)
The concept of cooling a computer with a fridge has surfaced in this forum on a regular basis. Search the word “fridge” you’ll get a pretty good idea why most of the replies are negative or sarcastic, Search First, Ask Second.
Basing the answer on Btu rating and covering variation of the question.
1 Watt = 0.0009485 Btu (s)
http://www.ehow.com/how_6038931_calculate-btu-output-watts.html
Finding the BTU rating for refrigerators proved a little difficult, most folks are more interested in cubic feet, door location, ice makers, color, etc. I did find one for a Frigidaire refrigerator compressor kit, rated at 750 Btu*. What little information found for mini / dorm-fridges put the rating around 100-150 and large side by sides at 1000– 1200 Btu. Home style freezers are usually lower on the Btu rating than fridges due to the fact that freezers don’t see the action on the door a refrigerator full of beer / soda does. I’m sure if more time was spent a little more data could be found, but the 750 from a full size Frigidaire is more than adequate for this explanation.
Computer in the fridge, air cooled
Taking 350 watts out of the air to represent a loaded computer (Comes from the Guru3d site for a lightly clocked i7 920 with a strong video card, running business apps).
350 x 0.0009485 = .331975 Btu (s) x 3600 = 1195.11 Btu (h)
Subtracting the 750 from the 1195 creates (445 Btu (h)) of surplus heat, quickly turning the inside of the fridge into an easy bake oven.
Computer in the fridge, water cooled, large reservoir.
One Btu equals the energy need to raise the temperature of one pound of water one degree Fahrenheit.
The first example establishes 445 Btu’s (h) of extra heat or 130.4 watts.
130.4 x 0.0009485 = 0.1236844 Btu (s)
1 / 0.1236844 = 8.085 seconds to raise 1 pound of water 1 degree Fahrenheit.
A gallon of water weights 8.35 pounds +/-, for this example five gallons will be used for a total of 41.35 pounds of water. My own system uses about a 1.5 gallons or 5.68 liters and in general terms excessive compared to some water cooled systems.
8.085 x 41.35 = 334.31 seconds to raise 5 gallons 1 degree Fahrenheit.
Now let’s see how long it would take to go from 0c to 22c (32f to 71.6f)
71.6 – 32 = 39.6 x 334.31 = 13,238.9 seconds, (3hr 40m 38.9s)
The two examples are hypothetical, in both, based on 750 Btu being removed by the fridge, and the latter, all the heat being past to the water. Neither of which is realistic. The water cooled example biggest problem is the heat sources not cooled by the water, warming the trapped air creating the same condition in example one but at a slower rate.
Using as a water chiller, large reservoir, computer outside of fridge.
Taking information from the supplied link below.
http://www.electronics-cooling.com/2008/05/estimating-dew-point-temperature-for-water-cooling-applications/
To keep from prepping against condensation problems, from the chart, 20c ambient and 30% humidity, gives a dew point of 1.9c (36f give or take)
72f -36f = 36 degrees of difference, the 72f would be inside the fridge
Fridge at 750 Btu = 220 watts (give or take)
CPU & Video Card = 250 watts (Assumed, I’ve past this w/ a Q6700 & GTX 280)
30 watts of excess heat
30 x 0.0009485 = 0.028455 Btu (s)
1 / 0.028455 = 35.14 seconds to raise one pound of water one degree
35.14 x 41.35 (lb) = 1,453.04 seconds for five gallons of water
1,453.04 x 36 = 52,309.44 seconds or 14h 31m 49s to get the water to 72f.
This example assumes the fridge is removing 750 Btu.
In all three examples the fridge would be working continuously trying to keep up with the heat load. Chances are the thing is going to overheat causing premature failure.
Moisture would be a problem in the first two examples. The computer would be warming the inside of the fridge quickly melting any water vapor frozen to the colder parts.
Convert an AC unit, a $100 unit at 5000 Btu is quite capable of handling a gaming PC.
Try something like this.
http://www.overclock.net/t/1264939/final-oc-results-window-air-con-chill-box-build
Pulling an older 1 ton unit apart in my garage to try this.
Power Supply wattage rating to Btu (h), in reference post #3, sizing.
500 = 1710
750 = 2560
1000 = 3410
1250 = 4270
Notes
Added side by side information.
*, The 750 Btu used in the example is a little low, found a general answer of 800 Btu average for standard two door fridges.
Since a lot of refrigerators’ hot side is not actively cooled, Removed, the three dorm size fridges I’ve owned were passive on the hot and cold side. Looking at schematics for newer larger units the statement was not true. Side Note, would you use one 120mm fan to cool your computer?
Added PS Info
Edited by Aleslammer - 8/28/12 at 9:27am
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I can honestly tell you the answer is yes.
(fremantle DOCKERS!)
