Calculate Cooling Time: Aluminum Block in Steel Crate at 90°C

In summary: After determining the maximum possible time, I would then need to account for the fact that the object will slowly warm up during the cooling process. This would require estimating the average temperature over the course of the cooling process. After estimating the average temperature, I would then use that information to determine the amount of time that it would take for the average temperature to reach the desired temperature. In summary, an aluminum pipe with an ID of 0.3m and an OD of 1m would require about 60 minutes to cool down to 30 degrees Celsius.
  • #1
Shady99
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0
How to calculate cooling time for a aluminum block in a steel crate placed in a room...both at 90°c ...how much time aluminum block will tak take to cool down to 25°c?
 
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  • #2
What is the exact geometry of the block and the crate, the thickness of the crate wall, the heat transfer coefficient between the crate and the outside air, the thermal conductivity of the crate wall, and the geometric arrangement of the block and crate relative to one another?
 
  • #3
...and the ambient temperature of the room?
 
  • #4
Chestermiller said:
What is the exact geometry of the block and the crate, the thickness of the crate wall, the heat transfer coefficient between the crate and the outside air, the thermal conductivity of the crate wall, and the geometric arrangement of the block and crate relative to one another?
Aluminum block is of brick shape with dimensions 10x 10 x 10 mm.

The steel crate wall is 5 mm.

The heat transfer coefficient is 25 w/m2 k

The thermal conductivity of crate wall is 40 w/m k

The block and crate are arranged in such a way that there is no gap between both . The crate covers the block from all side except from the top face which is exposed to environment/room

The ambient temperature is 25 deg c
 
  • #5
russ_watters said:
...and the ambient temperature of the room?
The ambient temperature is 25 deg c
 
  • #6
About 20 minutes if you sit the box on a larger steel block in the open air .

About 40 minutes if you sit the box on a brick and put it in a cupboard .

About 60 seconds if you drop the box in water .
 
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  • #7
Nidum said:
About 20 minutes if you sit the box on a larger steel block in the open air .

About 40 minutes if you sit the box on a brick and put it in a cupboard .

About 60 seconds if you drop the box in water .
Can you please share the formulas u used sequence wise
 
  • #8
From the data you provided, it looks like the dominant resistance to heat transfer is on the outside of the crate, assuming that the heat transfer resistance between the crate and the block is negligible. If the crate are assumed to be levitated in the air (and neglecting the internal heat transfer resistance of the block and crate), the equation for heat transfer would be:
$$(M_BC_B+M_CC_C)\frac{dT}{dt}=-hA(T-T_a)$$where the M's are the masses of the block and crate, the C's are the heat capacities of of the block and crate, h is the heat transfer coefficient, A is the total outside surface area of the block and crate, and ##T_a## is the ambient temperature
 
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  • #9
Chestermiller said:
From the data you provided, it looks like the dominant resistance to heat transfer is on the outside of the crate, assuming that the heat transfer resistance between the crate and the block is negligible. If the crate are assumed to be levitated in the air (and neglecting the internal heat transfer resistance of the block and crate), the equation for heat transfer would be:
$$(M_BC_B+M_CC_C)\frac{dT}{dt}=-hA(T-T_a)$$where the M's are the masses of the block and crate, the C's are the heat capacities of of the block and crate, h is the heat transfer coefficient, A is the total outside surface area of the block and crate, and ##T_a## is the ambient temperature
was viewing it on cellphone so it wasnt showing the post properly.Now its clear on the mac
 
  • #10
Shady99 said:
was viewing it on cellphone so it wasnt showing the post properly.Now its clear on the mac
$$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{hAt}{(M_BC_B+M_CC_C)}$$
 
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  • #11
hello again chester...Sir this discussion we had about the heat transfer we considered the our objects as lumped system...what if the mass is quite large and we can't use lumped approach as there are temperature difference over along the length itself. For example i have an aluminum pipe with ID=0.3m diameter and OD=1 m diameter and length=6m...how would i find out time to cool it down to 30 c...from my understanding now due to large size of object,there will be conduction inside the pipe and then convection from the surface of the pipe exposed to environment...how can i find cooling time for this pipe...initial temperature that the pipe was heated to is 90 c and then it was left to cool down to a final temperature that is 30 c and ambient is 25 c...please help
 
  • #12
You are correct in what you are saying. To take the conduction within the solid into consideration, you need to learn about transient solutions to the heat conduction equation. This is not the proper format for a complete tutorial on transient heat conduction within solids. But, if I were solving the specific problem you have posed, as an experienced heat transfer guy, this is how I would proceed:

I would focus on trying to determine the maximum possible time that the cooling would require, by bounding the answer. (I assume that you want to find out the maximum possible time you would have to wait for the object to cool.) The approach I would use would deliver not only the maximum, but also a close approximation to the actual time. I would start out by neglecting the heat transfer at the inner diameter (because of the small amount of surface there) and at the ends of the cylinder (because of the relatively large aspect ratio). So the heat transfer would then be purely radial. I would also mathematically fill in the hollow part with metal, so that I now have a solid cylinder rather than an annulus. So now I have a problem of transient radial heat conduction in an (essentially) infinite cylinder.

From transient solutions to the heat conduction equation in a cylinder, I would know that the heat transfer coefficient on the solid side of the interface (based on the temperature difference between the interface and the bulk mean temperature) starts out very high at short times, and then decreases monotonically to a lower limiting asymptotic value at very long times. This asymptotic value is given by:$$h=8\frac{k_m}{D}$$where ##k_m## is the thermal conductivity of the solid and D is the cylinder diameter. So the heat flux at the cylinder boundary is given by:
$$q=8\frac{k_m}{D}(T_i-\bar{T})$$where ##\bar{T}## is the average temperature of the cylinder at time t.

Beyond this, you just solve the problem as you did for the lumped model, of course based on the overall heat transfer coefficient from the combination of conduction and external convection.

You can see from all this that being able to solve a problem like this requires considerable experience with solution to the transient heat conduction equation, including (a) recognizing the kinds of approximations that you can make and (b) familiarity with the asymptotic limiting behavior of the solutions (for different kinds of boundary conditions). If you wanted to be even more accurate, you could use graphs in McAdams (Heat Transfer) showing the dimensionless temperature change as a function of the dimensionless time (Fourier Number) and the dimensionless external heat transfer coefficient (Biot number), or solve the transient heat conduction equation yourself. But, for what you are asking, I think that the bounding approach would be adequate.
 
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  • #13
Chestermiller said:
You are correct in what you are saying. To take the conduction within the solid into consideration, you need to learn about transient solutions to the heat conduction equation. This is not the proper format for a complete tutorial on transient heat conduction within solids. But, if I were solving the specific problem you have posed, as an experienced heat transfer guy, this is how I would proceed:

I would focus on trying to determine the maximum possible time that the cooling would require, by bounding the answer. (I assume that you want to find out the maximum possible time you would have to wait for the object to cool.) The approach I would use would deliver not only the maximum, but also a close approximation to the actual time. I would start out by neglecting the heat transfer at the inner diameter (because of the small amount of surface there) and at the ends of the cylinder (because of the relatively large aspect ratio). So the heat transfer would then be purely radial. I would also mathematically fill in the hollow part with metal, so that I now have a solid cylinder rather than an annulus. So now I have a problem of transient radial heat conduction in an (essentially) infinite cylinder.

From transient solutions to the heat conduction equation in a cylinder, I would know that the heat transfer coefficient on the solid side of the interface (based on the temperature difference between the interface and the bulk mean temperature) starts out very high at short times, and then decreases monotonically to a lower limiting asymptotic value at very long times. This asymptotic value is given by:$$h=8\frac{k_m}{D}$$where ##k_m## is the thermal conductivity of the solid and D is the cylinder diameter. So the heat flux at the cylinder boundary is given by:
$$q=8\frac{k_m}{D}(T_i-\bar{T})$$where ##\bar{T}## is the average temperature of the cylinder at time t.

Beyond this, you just solve the problem as you did for the lumped model, of course based on the overall heat transfer coefficient from the combination of conduction and external convection.

You can see from all this that being able to solve a problem like this requires considerable experience with solution to the transient heat conduction equation, including (a) recognizing the kinds of approximations that you can make and (b) familiarity with the asymptotic limiting behavior of the solutions (for different kinds of boundary conditions). If you wanted to be even more accurate, you could use graphs in McAdams (Heat Transfer) showing the dimensionless temperature change as a function of the dimensionless time (Fourier Number) and the dimensionless external heat transfer coefficient (Biot number), or solve the transient heat conduction equation yourself. But, for what you are asking, I think that the bounding approach would be adequate.
thakyou so much chester. after finding the conductive heat flux...how can i combine the two (bounded + lumped approah)for finding time
 
  • #14
Shady99 said:
thakyou so much chester. after finding the conductive heat flux...how can i combine the two (bounded + lumped approah)for finding time
The equation is now: $$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{UAt}{(MC)}$$where $$\frac{1}{U}=\frac{1}{h_c}+\frac{D}{8k_m}$$where T is the average temperature of the solid and ##h_c## is the convective heat transfer coefficient.
 
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  • #15
Thank you chester
 
  • #16
Chestermiller said:
The equation is now: $$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{UAt}{(MC)}$$where $$\frac{1}{U}=\frac{1}{h_c}+\frac{D}{8k_m}$$where T is the average temperature of the solid and ##h_c## is the convective heat transfer coefficient.
 

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  • #17
Chestermiller said:
The equation is now: $$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{UAt}{(MC)}$$where $$\frac{1}{U}=\frac{1}{h_c}+\frac{D}{8k_m}$$where T is the average temperature of the solid and ##h_c## is the convective heat transfer coefficient.
Please see the hollow cylinder.The red colored sides are insulated and the only exposed surface is the hollow pipe surfaces
 
  • #18
hello chester. I have posted a picture of the cylinder i was discussing with you. Please see the cylinder is insulated from the sides with red color (the bottom side isn't visible in the pic but that is also insulated similar to top side) and the yellow color surfaces in the hollow pipe area is exposed to room conditions. The length of the steel cylinder is 6 m, OD is 1m,ID is 0.3 m,To is 333k ,Ta is 298k, T is 299k , specific heat is 480 j/kg k, Thermal conductivity is 0.4 w/m k...the insulation is of NBR with specific heat is 1430 j/kg k, Thermal conductivity is 0.4 w/m k...can you help me find the cooling time...I m getting too many days which seems wrong
 
  • #19
correction: thermal conductivity of steel cylinder 51.9 w/m k
 
  • #20
Shady99 said:
correction: thermal conductivity of steel cylinder 51.9 w/m k
OK. What do you calculate for the overall heat transfer coefficient U?
 
  • #21
Chestermiller said:
OK. What do you calculate for the overall heat transfer coefficient U?
 

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  • #22
I thought you said that the outside heat transfer coefficient was 25.
 
  • #23
Chestermiller said:
I thought you said that the outside heat transfer coefficient was 25.
that was a random value i picked . However while exploring this issue over the internet. i came to this relationship for heat transfer co efficient h=k/l where k is the thermal conductivity and l is the thickness/length of the object
 
  • #24
Shady99 said:
that was a random value i picked . However while exploring this issue over the internet. i came to this relationship for heat transfer co efficient h=k/l where k is the thermal conductivity and l is the thickness/length of the object
This is done incorrectly. First of all, you should be using the thermal conductivity of air, not the thermal conductivity of the metal. Secondly, you should be using the thickness of the natural convective thermal boundary layer as the length l, not the length or diameter of the cylinder. Typical values of the heat transfer coefficient for natural convection heat transfer from a cylinder are on the order of 3 - 20 W/m K. So, from this range, what value would you be comfortable using?
 
  • #25
10 w/m k would be fine
Chestermiller said:
OK. What do you calculate for the overall heat transfer coefficient U?

Chestermiller said:
OK. What do you calculate for the overall heat transfer coefficient U?

Chestermiller said:
This is done incorrectly. First of all, you should be using the thermal conductivity of air, not the thermal conductivity of the metal. Secondly, you should be using the thickness of the natural convective thermal boundary layer as the length l, not the length or diameter of the cylinder. Typical values of the heat transfer coefficient for natural convection heat transfer from a cylinder are on the order of 3 - 20 W/m K. So, from this range, what value would you be comfortable using?
Sent from my Nexus 5 using Physics Forums mobile app
 
  • #26
OK. Redo the calculation with that value please.
 
  • #27
Chestermiller said:
OK. Redo the calculation with that value please.
Chestermiller said:
OK. Redo the calculation with that value please.
still seems incorrect...specific heat i have used of my specific object...rest all values as u suggested
 

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  • #28
$$\frac{1}{U}=\frac{1}{10}+\frac{1}{(8)(51.9)}=\frac{1}{10}+\frac{1}{415.2}=9.76\ \frac{W}{m^2\ K}$$
Note that, even though the cylinder diameter is very large, the heat transfer is still dominated by the outside convective heat transfer coefficient.
$$V=\frac{\pi (1)^2(6)}{4}\ m^3=4.71\ m^3$$
$$M=(7870)(4.71)=37100\ kg$$
$$MC=(37100)(480)=17800000\ J/K$$
$$A=\pi (1)(6)\ m^2=18.84\ m^2$$
$$\frac{UA}{MC}=\frac{(9.76)(18.84)}{17800000}=\frac{1}{96847}\ sec^{-1}$$
$$\ln{\frac{(30-25)}{(90-25)}}=-2.56$$
So, $$t=(96847)(2.56)=248000\ sec=2.9\ days$$
 
  • #29
thankyou chester...really appreciate your help
 

What is the purpose of calculating cooling time for an aluminum block in a steel crate at 90°C?

The purpose of calculating cooling time is to determine how long it will take for the aluminum block to reach a desired temperature within the steel crate. This information can be useful for various industrial and scientific processes that require precise temperature control.

What factors affect the cooling time of an aluminum block in a steel crate?

The cooling time of an aluminum block in a steel crate can be affected by several factors such as the initial temperature of the block, the size and shape of the block, the material and thickness of the crate, the surrounding temperature, and any insulation or heat transfer mechanisms present.

How is the cooling time of an aluminum block in a steel crate calculated?

The cooling time can be calculated using the Newton's Law of Cooling formula: t = (m x c x ΔT) / h, where t is the cooling time in seconds, m is the mass of the aluminum block, c is the specific heat capacity of aluminum, ΔT is the temperature difference between the block and the surrounding environment, and h is the heat transfer coefficient.

Can the cooling time of an aluminum block in a steel crate be shortened?

Yes, the cooling time can be shortened by increasing the heat transfer coefficient through methods such as using a more efficient insulation or increasing air circulation around the crate. The initial temperature of the block can also be lowered to decrease the temperature difference and therefore reduce the cooling time.

How accurate are the calculations for cooling time?

The accuracy of the cooling time calculations depends on the accuracy of the input parameters and the assumptions made in the calculation. It is important to have precise measurements and to consider all relevant factors in order to obtain an accurate estimate of the cooling time.

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