Can 100 liters of water keep a computer cool?

In summary: I'm not sure that is the whole question. It seems to me that they are interested in knowing if the reservoir can be ignored or not.In summary, the problem involves a water-cooled computer with a 100 liter copper reservoir and no radiators. The goal is to determine how long the temperature of the components can be kept below 70 degrees Celsius if the initial water and air temperature is 20 degrees Celsius. Using the equation (area x heat transmission coefficient x (water temperature - air temperature)), it is determined that the components can be kept below 70 degrees Celsius until the room temperature reaches 41 degrees Celsius. However, this calculation does not take into account the heat dissipated by the copper reservoir. To account for this
  • #1
PraAnan
77
1

Homework Statement


[/B]
If you had a water-cooled computer that puts out 500W of heat into the loop with a large copper 100 liter reservoir but no radiators, how long can you keep the temperature of the components below 70 degrees Celsius if the initial water and air temperature is 20 degrees Celsius?

Homework Equations



area x heat transmission coefficient x (water temperature - air temperature)

Cube length for 100 liters of water = 0.465m
Heat transmission coefficient for water-copper-air = 13.1W/m^2K
Room temperature = 20C & maximum water temperature allowed = 70C

The Attempt at a Solution



(0.465^2x6) x 13.1 x (70-20) = 849.8W
(0.465^2x6) x 13.1 x (70-41) = 493WSo 100 liters of water as well as the copper dissipating the heat should keep the water cooled pc components less than or equal to 70C until the room temperature reaches 41C.Is this ok?
 
Physics news on Phys.org
  • #2
Why not just simplify the problem to the time it takes to heat 100 l of water from 20 to 70 deg C with 500 W of power?

Am I missing something?
 
  • #3
That equation would be:
((volume x heat capacity x (final temperature - starting temperature) ) /3600) / 500
((100000x 4.185 x (70 - 20) ) /3600) / 500 = 11.625 hours.

I did look at this equation but it doesn't take into account the heat that would be dissipated by the copper reservoir.
 
  • #4
Is the mass of copper or anything about how it would dissipate heat into the air provided?
 
  • #5
PraAnan said:
I did look at this equation but it doesn't take into account the heat that would be dissipated by the copper reservoir

The problem statement says..

"large copper 100 liter reservoir but no radiators"

Unless there is more info in the problem statement I would take "no radiators" to mean the 100 litre reservoir is effectively insulated. That's not realistic but the problem statement as written provides no data with which to calculate the losses from the reservoir.
 
  • #6
russ_watters said:
Is the mass of copper or anything about how it would dissipate heat into the air provided?

The mass of the copper tank is not provided and I was thinking that the walls of the tank itself will dissipate the heat into the surrounding air, was this thinking wrong? Would the numbers change a lot depending on the thickness of the copper?

CWatters said:
The problem statement says..

"large copper 100 liter reservoir but no radiators"

Unless there is more info in the problem statement I would take "no radiators" to mean the 100 litre reservoir is effectively insulated. That's not realistic but the problem statement as written provides no data with which to calculate the losses from the reservoir.

The original problem wasn't written in English so I had to translate. The reservoir is not insulated, it's basically a copper tank that holds the water and the outside is exposed to the air.
 
  • #7
Is the shape provided? It doesn't sound to me like there is enough information to include the reservoir or air in the calculation.
 
  • #8
russ_watters said:
Is the shape provided? It doesn't sound to me like there is enough information to include the reservoir or air in the calculation.

The shape is a cube that has a length of 46.5cm, what other information is needed?
 
  • #9
I see that you calculated it as if it were a cube, but the problem statement doesn't say that. It also doesn't say whether it is covered, insulated, how thick it is or what it is sitting on, all of which will impact heat capacity and dissipation.
 
  • #10
I'll find out and get back to you tomorrow.
Apart from shape, insulation, wall thickness and resting position is there anything else that will be needed?
 
  • #11
I asked the teacher and he said that the tank can be any shape, there is no insulation (both the air and water have direct contact with copper), the thickness is 5mm and the resting position needs to be realistic (so a part of the area will need to be deducted).

Is the heat transmission equation that I used in my first post completely wrong?
 
  • #12
Bumping this for some advice.
 
  • #13
Ok so it looks like they want you to assume there are losses from the copper tank. They say that "both the air and water have direct contact with copper" so I would probably assume they are only interested in conducted losses not radiated or convection losses. Perhaps just assume you have a water-copper-air sandwich where the temperature of the water and air are the same as the surface of the copper.
 
  • #14
PraAnan said:
Is the heat transmission equation that I used in my first post completely wrong?

No that looks ok.

One potential complication is that the temperature of the water isn't constant so the heat loss (power loss) isn't constant.
 
  • #15
Thanks for your reply.

CWatters said:
Ok so it looks like they want you to assume there are losses from the copper tank. They say that "both the air and water have direct contact with copper" so I would probably assume they are only interested in conducted losses not radiated or convection losses. Perhaps just assume you have a water-copper-air sandwich where the temperature of the water and air are the same as the surface of the copper.

Isn't that what my initial equation is about?
CWatters said:
No that looks ok.

One potential complication is that the temperature of the water isn't constant so the heat loss (power loss) isn't constant.

Does this need to be worked out to answer the question? I just need to find how much heat the copper will be dissipating into the air when the water is at 70C and if that is higher than 500W then the answer would be 'indefinitely'. I could also then work out the equilibrium temperature where the copper is dissipating 500W.

People have said that I need to give more information which I did but I'm still not sure about the equation that needs to be used.
 
  • #16
PraAnan said:
Does this need to be worked out to answer the question? I just need to find how much heat the copper will be dissipating into the air when the water is at 70C and if that is higher than 500W then the answer would be 'indefinitely'.

Yes that would be a reasonable approach to take. Your equation is the correct one to use. Give it a go.
 
  • #17
To find heat dissipation at a water temperature of 70C with one side of the cube covered.
(0.465^2 x5) x 13.1 x (70-20) = 708W

When rearranged to find the equilibrium you get a water temperature of 55.3C.

I'm curious about (not part of my schoolwork) is how the air in a room will react to this, because of air convection 'new' air will constantly be coming in contact with the copper right? I was also wondering about how the temperature of the room would change.

dry air specific heat = 1.00J/g K
dry air density = 1275g/m3
room area = 30m2
room height = 3m
temperature change = 5C

Energy = density x volume x specific heat x temperature difference = 574kJ
500Wh is 1800kJ

So does this mean that in about 20 minutes a room with a volume of 90m^3 will go up in temperature by 5C if this computer is putting out 500W?
 
  • #18
Yes but only if there are no losses through the walls of the room or to ventilation.

A common problem is working out how big (in Watts) does the boiler/furnace of a house need to be to maintain say 21C indoors when it's -10 outside. The same basic equation you used to calculate the loss through the copper also works for combinations of brick, insulation and glass etc.

Ventilation has the effect of providing new cold air that has to be heated. So if you know the rate of ventilation (usually specified in air changes per hour) you can also account for the power needed for that.
 
  • #19
CWatters said:
Yes but only if there are no losses through the walls of the room or to ventilation.

A common problem is working out how big (in Watts) does the boiler/furnace of a house need to be to maintain say 21C indoors when it's -10 outside. The same basic equation you used to calculate the loss through the copper also works for combinations of brick, insulation and glass etc.

Ventilation has the effect of providing new cold air that has to be heated. So if you know the rate of ventilation (usually specified in air changes per hour) you can also account for the power needed for that.

Thank you for all your help CWatters, it's very much appreciated.
 

1. How does water keep a computer cool?

Water has a high specific heat capacity, meaning it can absorb and release large amounts of heat without a significant change in temperature. When water is circulated through a computer's cooling system, it absorbs the heat generated by the components and carries it away, keeping the computer cool.

2. What is the recommended amount of water to cool a computer?

The recommended amount of water to cool a computer varies depending on the computer's specifications and the type of cooling system being used. In general, 100 liters of water should be enough to effectively cool most desktop computers.

3. Is it safe to use water to cool a computer?

Yes, it is safe to use water to cool a computer as long as it is done properly. The water should be distilled or deionized to prevent corrosion and mineral buildup, and the cooling system should be properly sealed and maintained to prevent leaks. It is also important to regularly monitor the water levels and temperature to ensure the system is functioning properly.

4. Can using water to cool a computer damage the components?

No, as long as the water is properly distilled or deionized and the cooling system is well-maintained, it should not damage the components. In fact, using water to cool a computer can be more efficient and effective than traditional air cooling methods.

5. Are there any alternatives to using water to cool a computer?

Yes, other alternatives to using water for computer cooling include using liquid coolants specifically designed for computer cooling, as well as other methods such as air cooling or thermoelectric cooling. However, water is still considered one of the most efficient and cost-effective options for cooling a computer.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
6
Views
7K
  • Mechanical Engineering
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
4K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Engineering and Comp Sci Homework Help
2
Replies
44
Views
6K
Back
Top