Time to Equilibrate 0C Copper & 75C Water

[kendreaditya]

I have a copper cylinder that is at 0C, how much time will be needed for it to equilibrate with 10kg of water at 75C?
Copper cylinder:
Mass: 105.7 g
Specific heat: 0.385 J/gC
Initial Temp: 0C
Final Temp: 75C ish
SA: 25 cm^2

Water:
Mass: 10000g
Specific heat: 4.184 J/gC
Initial Temp: 75C
Final Temp: 75C ish

 

[Chestermiller]

Welcome to Physics Forums!

Physics Forums rules and guidelines require the person initiating a thread to summarize his thinking about the problem so far, so that other members can properly lead him to an answer. Please give us a brief summary of your current thinking.

I can help you get an answer to this problem.

Chet

 

[Chestermiller]

That surface area of 25 cm^2 is not compatible with 105.7 g of copper (density 8.96 gm/cc); the minimum surface area for a copper cylinder of that density is 28.7 cc.

 

[kendreaditya]

That surface area of 25 cm^2 is not compatible with 105.7 g of copper (density 8.96 gm/cc); the minimum surface area for a copper cylinder of that density is 28.7 cc.

That is correct, I estimated a little less because when I use it in the environment, the surface area will decrease. That 28.65 cm^2 is the actual SA.

 

[kendreaditya]

Welcome to Physics Forums!

Physics Forums rules and guidelines require the person initiating a thread to summarize his thinking about the problem so far, so that other members can properly lead him to an answer. Please give us a brief summary of your current thinking.

I can help you get an answer to this problem.

Chet

What I am thinking is to use Newtons law of cooling.

where t (time constant) is C/Ah (3.3595)
C = heat capacity (385 J/kgC)
A = heat transfer surface area (28.65 cm^2)
h = heat transfer coefficient (400 W/(m2 K) copper to water)
deltaT(t) = 0.01C
deltaT(0) = 75C

I got 385/(400 * .2865) = 3.3595 for the time constant
solving for t, I got a negative number -74 ish, which does not make sense

 

[Chestermiller]

What I am thinking is to use Newtons law of cooling.
View attachment 253139

where t (time constant) is C/Ah (3.3595)
C = heat capacity (385 J/kgC)
A = heat transfer surface area (28.65 cm^2)
h = heat transfer coefficient (400 W/(m2 K) copper to water)
deltaT(t) = 0.01C
deltaT(0) = 75C

I got 385/(400 * .2865) = 3.3595 for the time constant
solving for t, I got a negative number -74 ish, which does not make sense

The time constant should be $\frac{mC}{Ah}$, not $\frac{C}{Ah}$. Otherwise, this equation is OK. Also, the area in square meters should be 0.002865, not 0.2865. Also, where did you get that value for the heat transfer coefficient on the water side?

 

[Chestermiller]

If I do the calculation correctly for the parameters you used, I get a value of about 5 minutes for the heating time. But the heat transfer coefficient you used seems too high to me, at least for getting an upper bound to the heating time. I estimate a value of h about 10x lower than the value you used.

 

[kendreaditya]

If I do the calculation correctly for the parameters you used, I get a value of about 5 minutes for the heating time. But the heat transfer coefficient you used seems too high to me, at least for getting an upper bound to the heating time. I estimate a value of h about 10x lower than the value you used.

I got the value from https://www.engineeringtoolbox.com/overall-heat-transfer-coefficients-d_284.html

Also, I am still getting a different number now.

 

[Chestermiller]

The calculation should look like $$\frac{thA}{mC}=\ln{(75/0.01)}=8.92$$
The heat transfer coefficients in that table are typical values for forced convection. As a worst case, we should be looking at values in the limit of no convection. For radial conductive heat transfer in water to a sphere of radius R, the conductive heat transfer coefficient is $h=k_w/R$, where $k_w$ is the thermal conductivity of water. I used the radius of a copper sphere of the same mass as our cylinder, R = 1.41 cm. This give a value of about h = 40 W/m^2 C.

 

[kendreaditya]

The calculation should look like $$\frac{thA}{mC}=\ln{(75/0.01)}=8.92$$
The heat transfer coefficients in that table are typical values for forced convection. As a worst case, we should be looking at values in the limit of no convection. For radial conductive heat transfer in water to a sphere of radius R, the conductive heat transfer coefficient is $h=k_w/R$, where $k_w$ is the thermal conductivity of water. I used the radius of a copper sphere of the same mass as our cylinder, R = 1.41 cm. This give a value of about h = 40 W/m^2 C.

So it would take around 9 mins to equilibrate?

 

[Chestermiller]

So it would take around 9 mins to equilibrate?

With a heat transfer coefficient of 400, I get a time constant of 35.5 seconds, and an equilibration time of (35.5)(8.92)= 317 sec = 5.3 minutes. With a heat transfer coefficient of 40, I get a time constant of 355 seconds, and an equilibration time of 53 minutes.

Usually, equilibration is considered to be achieved in practice after 4 time constants (rather than 8.92). With a heat transfer coefficient of 400, that would be 2.4 minutes, and, with a heat transfer coefficient of 40, that would be 23.7 minutes.

If the cylinder is sitting upright at the insulated bottom of a tank, then the bottom of the cylinder would be insulated. To modify the previous analysis for this, the solution would involve a reflection boundary condition. The solution to this case would be the same as that for the cylinder levitating in the water, but with double the mass, double the surface area (with the end surface area subtracted out), and with the heat transfer coefficient divided by $2^{1/3}$.

By the way, even though you are only a novice at heat transfer, you've done very nicely in your analysis of this problem.

 

[kendreaditya]

With a heat transfer coefficient of 400, I get a time constant of 35.5 seconds, and an equilibration time of (35.5)(8.92)= 317 sec = 5.3 minutes. With a heat transfer coefficient of 40, I get a time constant of 355 seconds, and an equilibration time of 53 minutes.

Usually, equilibration is considered to be achieved in practice after 4 time constants (rather than 8.92). With a heat transfer coefficient of 400, that would be 2.4 minutes, and, with a heat transfer coefficient of 40, that would be 23.7 minutes.

If the cylinder is sitting upright at the insulated bottom of a tank, then the bottom of the cylinder would be insulated. To modify the previous analysis for this, the solution would involve a reflection boundary condition. The solution to this case would be the same as that for the cylinder levitating in the water, but with double the mass, double the surface area (with the end surface area subtracted out), and with the heat transfer coefficient divided by $2^{1/3}$.

By the way, even though you are only a novice at heat transfer, you've done very nicely in your analysis of this problem.

Thank You!

I am having trouble understanding the mean of "after 4 time constants", does this mean I have to have 4-time constants? or a time constant of 4?

So what would be a good way to calculate the heat transfer coefficient or calculate an upper bound, as this equilibration time ranges from 2 - 60 mins. My goal is to have it equilibrate in less than 6 mins.

So having the cylinder paced at the bottom of the container will have a large effect on the equilibration time?

 

[Chestermiller]

Thank You!

I am having trouble understanding the mean of "after 4 time constants", does this mean I have to have 4-time constants? or a time constant of 4?

Four time constants. In 4 time constants, the temperature difference decreases a factor of 0.0183, or the final temperature rises from 0 C to 73.6 C.

So what would be a good way to calculate the heat transfer coefficient or calculate an upper bound, as this equilibration time ranges from 2 - 60 mins. My goal is to have it equilibrate in less than 6 mins.

So having the cylinder paced at the bottom of the container will have a large effect on the equilibration time?

In my judgment, the heat transfer coefficient of 400 is excessive. Based on using 4 time constants (final temperature 73.6 C) and a worst case heat transfer coefficient 40, we calculated a heating time of 24 minutes. Realistically, the heating time is probably going to be about half of this, or about 12 minutes. But, to check out 6 minutes, the calculation has to be done more accurately, by solving the detailed partial differential equations for heat transfer. I would look at the case of an infinitely long cylinder of the same diameter, so that the problem could be reduced to 2D. So, you would be solving the transient heat conduction equation in cylindrical coordinates for the temperature of the cylinder and the temperature profile in the water as a function of radial position and time. The problem is that, using the worst case scenarios we have looked at so far would not be adequate for determining whether the actual heating time would be less than 6 min.

 

[Chestermiller]

You do realize that, if it is a very long thin cylinder or a very short flat cylinder (pancake/disk), the heating will be much faster than if it is close to minimum surface area, right?

 

[kendreaditya]

You do realize that, if it is a very long thin cylinder or a very short flat cylinder (pancake/disk), the heating will be much faster than if it is close to minimum surface area, right?

Yes, I understand that because increasing the SA is inversely proportional to the equilibration time.

This being the work case scenario, and having a heating time of 12 min is not terrible, but it may be an area of concern.

Background/purpose: The reason I need to know this because I am trying to calibrate a thermistor with another probe. My only source of heat is water, which is not insulated, thus it is always trying to equilibrate with the ambient temp. This decrease is temp at 75C is limiting the accuracy of my thermistor when I take the resistance readings of the thermistor, the temp displayed on the calibration probe is different because of the difference in the thermal mass. To resolve I insert both probes into a copper block so the temp is more stable, in a sense acting as a heat reservoir. Basically, I have to do this process mentioned above 3 times in less than 20 mins. The ideal calibration temperatures would be 0C, 37.5C, 75C. So which order in placing the copper block do you think would take the least amount of time?

My thought was the 75C, 37.5C, 0C. But originally I would have the heat the copper block from 22C ish (room temp) to 75C. Would this be faster than doing 37.5C first since its closer to the room temp, and then doing 75C, and 0C?

Let me know if you have ideas in trying to make a more stable temperature environment for calibration.

 

[Chestermiller]

Yes, I understand that because increasing the SA is inversely proportional to the equilibration time.

This being the work case scenario, and having a heating time of 12 min is not terrible, but it may be an area of concern.

Background/purpose: The reason I need to know this because I am trying to calibrate a thermistor with another probe. My only source of heat is water, which is not insulated, thus it is always trying to equilibrate with the ambient temp. This decrease is temp at 75C is limiting the accuracy of my thermistor when I take the resistance readings of the thermistor, the temp displayed on the calibration probe is different because of the difference in the thermal mass. To resolve I insert both probes into a copper block so the temp is more stable, in a sense acting as a heat reservoir. Basically, I have to do this process mentioned above 3 times in less than 20 mins. The ideal calibration temperatures would be 0C, 37.5C, 75C. So which order in placing the copper block do you think would take the least amount of time?

My thought was the 75C, 37.5C, 0C. But originally I would have the heat the copper block from 22C ish (room temp) to 75C. Would this be faster than doing 37.5C first since its closer to the room temp, and then doing 75C, and 0C?

Let me know if you have ideas in trying to make a more stable temperature environment for calibration.

Sorry. I really don't follow what you are trying to do.

Regarding the insulated bottom boundary condition, this can be handled by doubling the mass of the cylinder (keeping the aspect ratio the same: diameter equal to length), and taking the cylinder to be in an infinite ocean of water (no insulating boundary). So, in this case, the equivalent cylinder would be 3.1 cm in length and diameter, and have a mass of 211.4 grams.

Based on all this, the time constant with an insulated boundary on the bottom would increase by a factor of $2^{2/3}=1.58$. That would result in an upper bound heating time (with 4 time constants) of 37.4 minutes. In my judgment, realistically, the required heating time would be about 30 minutes. Sorry that this is not the 6 minutes you were hoping for.