Help with Newton's Law of Cooling

In summary: Hello!I am trying to figure out how long an object takes to cool and I am just banging my head on the desk.Info:Metal object is placed into an oven set to 302F (150C).The object is 3' long and 2.75" diameter, and is solid metal (steel).The object is "baked" in the oven for 30 minutes at 302F and then the oven is turned off, door closed.Oven is of industrial type, and is approx. 3ft by 3ft big (oven cavity).Once the oven is turned off, how long would it take the object to cool to ambient (72F) temperature?
  • #1
Mark Nez
5
1
Hi - I have been trying to figure out how long an object takes to cool and I am just banging my head on the desk. Info:

Metal object is placed into an oven set to 302F (150C).
The object is 3' long and 2.75" diameter, and is solid metal (steel).
The object is "baked" in the oven for 30 minutes at 302F and then the oven is turned off, door closed.
Oven is of industrial type, and is approx. 3ft by 3ft big (oven cavity).

Once the oven is turned off, how long would it take the object to cool to ambient (72F) temperature?

If anyone could help I wold be overwhelmed with happiness :)

Also - educated estimates are fine!

Thanks!
Mark
 
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  • #2
Mark Nez said:
Hi - I have been trying to figure out how long an object takes to cool and I am just banging my head on the desk. Info:

Metal object is placed into an oven set to 302F (150C).
The object is 3' long and 2.75" diameter, and is solid metal (steel).
The object is "baked" in the oven for 30 minutes at 302F and then the oven is turned off, door closed.
Oven is of industrial type, and is approx. 3ft by 3ft big (oven cavity).

Once the oven is turned off, how long would it take the object to cool to ambient (72F) temperature?

If anyone could help I wold be overwhelmed with happiness :)

Also - educated estimates are fine!

Thanks!
Mark
Welcome to the PF. :smile:

How well insulated is the oven? That will make a big difference.

Is there a window so you can see the object? If so, do you have an IR thermometer that you could use to take some data? That may be the quickest way to figure this out.

Why are you leaving the oven door closed if you want the object to cool?
 
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  • #3
Thank you Sir! Great bike BTW!

The oven will be insulated very well, no window, and will have a thermometer in there eventually...I am in a meeting and was just trying to get a "ballpark" on timing for a prototype assembly we are working on.
 
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  • #4
Mark Nez said:
The oven will be insulated very well, no window, and will have a thermometer in there eventually.
So the intention it to maximize how long it takes to cool off? If I had that object in one of my ovens, it would probably take 12-24 hours to cool most of the way back to ambient. Can you say what the application is, and why you want to maximize the cooling time?
 
  • #5
Mark Nez said:
Once the oven is turned off, how long would it take the object to cool to ambient (72F) temperature?

The oven and its contents will cool, but how fast depends on how well heat energy is transferred from the oven to the surroundings. Insulation inhibits the transfer of heat energy by conduction, and there are formulas that can be used to calculate the magnitude of the effect. They depend on things like the R-value of the insulation and the surface areas of the oven walls. But heat energy is also transferred via convection and radiation. Again, there are formulas that can be used to calculate these effects. But all you'll get from that is an estimate. You really need to measure the time if you want to get an accurate value.
 
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  • #6
Mister T said:
The oven and its contents will cool, but how fast depends on how well heat energy is transferred from the oven to the surroundings. Insulation inhibits the transfer of heat energy by conduction, and there are formulas that can be used to calculate the magnitude of the effect. They depend on things like the R-value of the insulation and the surface areas of the oven walls. But heat energy is also transferred via convection and radiation. Again, there are formulas that can be used to calculate these effects. But all you'll get from that is an estimate. You really need to measure the time if you want to get an accurate value.
Just a rough estimate would work. I am going to have test run and proper calculations done by the developing engineers, again, this is just a prototype design meeting and trying to set estimates on timing so we can get some ballpark figures for the prototype 1 build. Thank you for the info and help!
 
  • #7
berkeman said:
So the intention it to maximize how long it takes to cool off? If I had that object in one of my ovens, it would probably take 12-24 hours to cool most of the way back to ambient. Can you say what the application is, and why you want to maximize the cooling time?
not really maximize, just get an idea of what we are looking at in terms of time for scheduling of personnel. I was in the ballpark of 6-8 hours based on past tools but wasn't sure if there was a calculator that I could "fudge" to get estimates. In this rare instance (at this point of time) just an educated guess is all I was hoping for. But, sounds like there are too many variables and I understand the hesitation of giving an answer based on the lack of hard numbers needed to calculate. Thank you for your fast replies!

I am actually thinking we are going to get cooling ovens for manufacturing at this point...

And no sorry, I can't discuss any of the project details...just a direction drilling tool for one of the top O&G companies :)
 
  • #8
Is the object taken out of the oven after the heating, or is it left in the oven?
 
  • #9
Chestermiller said:
Is the object taken out of the oven after the heating, or is it left in the oven?
Left in
 
  • #10
I suggest you get an oven thermometer and first measure temperature vs time as the oven cools without the object present. This will give you the Newton's cooling coefficient exclusively from the oven to the surrounding room. We can then set up a model for the case with the object present within the oven.
 
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FAQ: Help with Newton's Law of Cooling

What is Newton's Law of Cooling?

Newton's Law of Cooling is a mathematical equation that describes the rate at which an object cools down in a given environment. It states that the rate of change of the temperature of an object is directly proportional to the difference between the object's temperature and the temperature of its surroundings.

How is Newton's Law of Cooling used in real life?

Newton's Law of Cooling is used in various fields such as meteorology, engineering, and biology. It is used to predict the cooling rate of hot objects in different environments, to design efficient cooling systems, and to study the metabolism of living organisms.

Can you give an example of Newton's Law of Cooling in action?

One example of Newton's Law of Cooling is when a hot cup of coffee is left on a table. The coffee will cool down faster in a colder room compared to a warmer room because the temperature difference between the coffee and the surroundings is greater in the colder room.

How is the rate of cooling affected by different factors?

The rate of cooling is affected by factors such as the temperature difference between the object and its surroundings, the surface area of the object, and the type of material the object is made of. Objects with a larger surface area or made of materials with higher thermal conductivity will cool down faster.

Can Newton's Law of Cooling be applied to objects that are heating up?

Yes, Newton's Law of Cooling can also be applied to objects that are heating up. In this case, the rate of change of temperature is directly proportional to the difference between the object's temperature and the temperature of its surroundings, but the temperature of the object is increasing instead of decreasing.

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