How Do You Calculate Cooling Time Without the Convection Coefficient?

[JSBeckton]

A 20-20-5 cm slab of copper at a uniform temp of 200c suddenly has its surface temp lowered to 35c. Using the concepts of thermal resistance and capacitance and the lumped-capacity analysis, find the time at which the center temp becomes 90c.
density =8900 kg/m^3
specific heat = .38 kJ/kg-k
k=370 w/m-k

My question is how I begin to solve this w/o knowing h? All of the formulas in this section have a h term that I am not given. Is there a way to find h from this info or is there another formula tha I am not seeing?

Thanks in advance

 

[OlderDan]

A 20-20-5 cm slab of copper at a uniform temp of 200c suddenly has its surface temp lowered to 35c. Using the concepts of thermal resistance and capacitance and the lumped-capacity analysis, find the time at which the center temp becomes 90c.
density =8900 kg/m^3
specific heat = .38 kJ/kg-k
k=370 w/m-k

My question is how I begin to solve this w/o knowing h? All of the formulas in this section have a h term that I am not given. Is there a way to find h from this info or is there another formula tha I am not seeing?

Thanks in advance

I can't speak for everyone of course, but in a physics forum you probably need to be a bit more specific about the parameters for engineering topics.

h = height above surface of the earth? Planck's constant?

If h has something to do with heat transfer from one material to another, it is not relevant to this problem.

What is k? In one place it looks like k is a Kelvin temperature and in another perhaps a thermal conductivity

k=370 w/m-k

Does this mean k^2 = 370 w/m?

Are these units correct. I would gues there would be an area involved in heat flow. Is the m an m^2 perhaps?

 

[JSBeckton]

Sorry, I have never seen h used as anything other than a heat transfer coeficient in a heat transfer book. I should be more specific.

the k=370W/m-k is the thermal conductivity

I do not understand how to use latex in this forum or I would.

 

[Cyrus]

Is this from the book or did the teacher write it?

I'm not seeing how to find the h value yet. T_infinity is not given either, hmmm.


*In the future, you should post engineering questions in the engineering section, and not the advanced physics section.

 

[OlderDan]

Sorry, I have never seen h used as anything other than a heat transfer coeficient in a heat transfer book. I should be more specific.

the k=370W/m-k is the thermal conductivity

I do not understand how to use latex in this forum or I would.

I interpret heat transfer (h) to be related to heat flowing from one body to another. If that is the correct interpretation, I do not think it is needed in this problem. The problem says that the surface temperature of the plate is suddenly changed. While this may be unrealistic, I take it to mean what it says. That makes this a heat conduction problem within a single material. I don't know what additional assumptions, if any, you can make. If the temperature is everywhere 200c except at the surface, there would initially be a uniform temperature within the copper except for an infinite temperature gradient at the surface. That would imply infinite heat conduction. Very quickly there would be some temperature gradient established between the hot core and the cool surface that would diminish the core temperatrure as heat flowed to the surface. Eventually the core temperature would reach the final 90c.

I see now why the conductivity units are what you stated. There is an area involved, but there is also a length between the points of temperature difference.

I have an intuitive idea of how the process should go, but I do not know the theoretical foundation you have to work with. Since the temperature of the plate is initally uniform, the sudden drop in surface temperature should result in a huge rate of heat flow from leyers near the surface. The deeper layers will not respond initially because there is zero temperature gradient in the interior. As the heat flows out of the layers near the surface their temperature drops and the gradient increases at deeper levels. My guess is the temperature profile within the material would take on a shape somewhat like an ellipse, with a gradual temperature variation near the center and a steeper variation near the surface.

I can't do any more with this right now. I think with some effort I might be able to come up with a DE for the temperature change with distance profile for this situation, but I would probably be reinventing the wheel. I would think yoiur text has already worked it out for similar situations.

 

[Cyrus]

I'm sorry to say your interperation of the problem is wrong OlderDan.

Lumped Capacitance means the Biot number is <<0.1.

 

[OlderDan]

I'm sorry to say your interperation of the problem is wrong OlderDan.

Lumped Capacitance means the Biot number is <<0.1.

I have no idea what the Biot number is. I have no problem with being wrong about the interpretation, but I'd like to know what is wrong with it, or what assumptions the engineers make when dealing with such problems.

From a quick google it still appears that h is related to a transfer of heat from one object to another. If the problem is one of bringing the plate in contact with a heat sink at 35c, then that is what it should say. That is not the same thing as suddenly changing the surface temperature to 35c, though as I said earlier, a sudden change in surface temperature is unrealistic.

 

[Cyrus]

The lumped capacitance is a simplification that that you can make when Bi<<0.1. In other words, the thermal resistance of the material is extremely small in comparison to the thermal resistance of the convection (h). Therefore, the body acts as an isothermal body at every instant in time.

If the Biot number is NOT satisfied, then you have to solve for the erf function of the differential equation using the non-dimensional charts.

 

[JSBeckton]

I'm sorry to say your interperation of the problem is wrong OlderDan.

Lumped Capacitance means the Biot number is <<0.1.

My book states that Biot number must be <<1, it says to use 0.1 as the cutoff.

This question is straight out of the book, and its one of the first for this section so i would not expect it to be too involved.

Woud it be wrong to assume T infinity is 35?

 

[Cyrus]

Woud it be wrong to assume T infinity is 35?

Yes, T infinity is not 35. The surface temperature T_s, would be 35.

 

[JSBeckton]

Any ideas where to start? Is there a way that I can find h with the given info?

 

[OlderDan]

Any ideas where to start? Is there a way that I can find h with the given info?

Have you tried following the suggestion from cyrusabdollahi

*In the future, you should post engineering questions in the engineering section, and not the advanced physics section.

As for

The lumped capacitance is a simplification that that you can make when Bi<<0.1. In other words, the thermal resistance of the material is extremely small in comparison to the thermal resistance of the convection (h). Therefore, the body acts as an isothermal body at every instant in time.

If the Biot number is NOT satisfied, then you have to solve for the erf function of the differential equation using the non-dimensional charts.

To my mind, the notion of an isothermal assumption is inconsistent with the problem. After all you are trying to find the time for a temperature change in the center of the plate when the surface is at another temperature. The temperature is neither constant nor uniform. I am not familiar with the other approach he speaks about, but maybe you are.

Interstingly to me, the assumption of a uniform temperature in the metal is what I was trying to make in an earlier problem we talked about for a metal cylinder exposed to air. You decided that was probably not valid. Did you ever resolve that problem?

 

[Cyrus]

To my mind, the notion of an isothermal assumption is inconsistent with the problem.

No, this is exactly what the problem is about. Bi<<0.1

The temperature is neither constant nor uniform.

No, the temperature is not constant, but it is uniform.

 

[Astronuc]

JS, are there no examples on lumped parameter model/analysis.

OlderDan is correct, heat transfer factor is not needed. One would just assume the heat transfer coefficient is equal to the thermal gradient at the surface, which comes from the T_s and the temperature inside.

The key is

Using the concepts of thermal resistance and capacitance and the lumped-capacity analysis

which gives a crude estimate.

 

[Cyrus]

OlderDan is correct, heat transfer factor is not needed. One would just assume the heat transfer coefficient is equal to the thermal gradient at the surface, which comes from the Ts and the temperature inside

I'm afraid this too, is wrong. That is not what the problem is asking. There is no such thing as a 'heat transfer factor' either The problem is finding the heat transfer coefficient, h, due to convection. $$q''_{conv} = h(T_0 - T_\infty)$$.

The Biot number is defined as: $$Bi = \frac{hL_c}{k}$$

I appreciate you guys trying to help, but in this case you are just making more hurt than help.

I also think this author did a piss poor job writing the problem.

the heat transfer coefficent, h, is VITAL to solving this problem, and the body is ISOTHERMAL when using the lumped capciatance method.

 

[OlderDan]

I'm afraid this too, is wrong. That is not what the problem is asking. There is no such thing as a 'heat transfer factor' either The problem is finding the heat transfer coefficient, h, due to convection. $$q''_{conv} = h(T_0 - T_\infty)$$.

The Biot number is defined as: $$Bi = \frac{hL_c}{k}$$

I appreciate you guys trying to help, but in this case you are just making more hurt than help.

I also think this author did a piss poor job writing the problem.

the heat transfer coefficent, h, is VITAL to solving this problem, and the body is ISOTHERMAL when using the lumped capciatance method.

I will happily defer to you on this. You are obviously far more in tune with the terminlogy than I am.

Just for my own understanding, what led me in the wrong direction was the statement about the surface temperature suddenly changing to 35c. If I understand what you are saying, that is impossible. I believe you are saying that the internal conductivity is so high compared to the rate at which heat leaves the surface that the copper essentially has a uniform time-dependent temperature throughout that is gradually falling toward 90c and that the rate of temperature change is dependent on the temperature difference between the copper and the environment. Is that about right?

 

[Cyrus]

Just for my own understanding, what led me in the wrong direction was the statement about the surface temperature suddenly changing to 35c.

And rightly so, as it makes little to no sense to me either...

I believe you are saying that the internal conductivity is so high compared to the rate at which heat leaves the surface that the copper essentially has a uniform time-dependent temperature throughout that is gradually falling toward 90c and that the rate of temperature change is dependent on the temperature difference between the copper and the environment. Is that about right?

Yes,

 

[JSBeckton]

Cyrus, I asked the TA about this and this is what he had to say

You do not need h. You only need to evaluate hA/(rho.Vc)= 1/((1/hA)rho.Vc)

let g = rho.Vc and
1/hA = Rth (your total resistance) then your

hA/(rho.Vc)t = t/(Rth.g)

Does he mean to solve the first equation for h and then use that value to solve for time?

Thanks for your help by the way. Also last semester I posted a thermo question in the engineering forum and you said to post in the advanced physics forum, it heat transfer considered engineering?

 

[Cyrus]

So how does he expect you to calculate the thermal resistance without knowing the the h value or the R_thrm not being given? You need to know 1 of the 2.

 

[JSBeckton]

He said that Rth=1/2 (deltaX/kA)

I used that the solve for h and got a huge number, 44326.2

I used that value for h and of course got a very small time~.087 seconds

Do you see what he means? Am I doing this right?