Basic Problem with Heat Transfer

In summary, the conversation discussed the loss of heat from a glass aquarium tank maintained at a temperature above ambient. It was determined that when a glass pane is placed between the water and surrounding air, heat transfer is due to conduction, while without the pane, heat transfer is due to radiation and convection. The conversation also considered a calculation for heat loss in a specific scenario and discussed the impact of thickness and material of the glass pane on heat loss. It was concluded that glass is not a good insulator for heat loss.
  • #1
TheH
12
0
I am trying to understand how heat is lost from a glass aquarium tank maintained at some temperature above ambient.

If we imagine a direct interface between the body of water and the surrounding air, I think heat transfer would be entirely due to radiation and convection.

When a glass pane is placed in-between the two, heat transfer is due (entirely?) to conduction.

What I am having trouble understanding, is how the heat lost by conduction through the glass can be so significant but as soon as that glass is removed the heat loss by radiation and convection is so trivial. It seems that the presence of the glass pane encourages heat-loss?

Here is what I am considering:

A body of water with surface area 1m x 0.25m , or 0.25 m^2 at temperature 29C.
The surrounding environment temperature of 25C.
A glass pane of thickness 0.01 m.

Heat loss by conduction:
Watts = (k * A * (Th - Tl)) / d , where k for glass is 1.1 Watts/m*K

= (1.1 * 0.25 * (29 - 25)) / 0.01
= 110 Watts of heat flow.

When the pane is removed:

Heat loss by radiation:
Watts = (em * sbc * A * (Th^4 - Tl^4)), where em for water is 0.95 and sbc = 5.6703x10^(-8) W/m^2*K^4

= 0.95 * 0.000000056703 * 0.25 * ( (29^4) - (25^4) )
= 0.0043 Watts

Heat loss by convection:
Watts = (h * A * (Th - Tl)), where h for water is taken to be 20 W/m^2*K

= 100 * 0.25 * (29 - 25)
= 20 Watts

Here it appears that when a glass pane is present 110 watts of heat are lost, but when it is removed only 20.0043 watts are lost.

Can anyone explain this to me?

Thank you for your time.
 
Science news on Phys.org
  • #2
It may not be clear from the above, but I am considering heat loss from each side of a rectangular aquarium turn-by-turn. In the example I gave, I am only considering heat loss from one side.

Also, is this perhaps the wrong sub-forum for this kind of question?
 
  • #3
In your calculation, you assumed that the temperature difference in case 1 is directly 29-25 = 4 degrees. However, temperature varies from the water surface, whose Th=29C, to the outside air, whose Tl=25C. So the temperature difference between the 2 faces of the pane is not necessarily 4 degrees, but smaller. Denote T the temperature of the face which faces the air, we have:
P = (k_glass / d) * A * (Th-T)
P = h_air * A * (T - Tl)
We have (k_glass / d) = 110; h_air = 20. Thus: T = 28.4C, P = 16.5W!
Now you may rethink about your conclusion :smile:
 
  • #4
hikaru1221, thank you for your reply.

I now see that the temperature will not be a full four degree difference as you pointed out, but I am not quite clear how you settled on T = 28.4C ? Will this value be a function of the thickness and material of the pane and the overall difference in temperature?

Also, in the convection equation:
P = h_air * A * (T - Tl)
Shouldn't this remain as:
P = h_air * A * (Th - Tl), unless we are convecting heat off of the panel (rather than directly from the surface of the water)?

And just one last question, I noticed that you used h_air for the convection equation whereas I used h_water - was I mistaken in doing so?

Thank you for your time.
 
  • #5
TheH said:
hikaru1221, thank you for your reply.

I now see that the temperature will not be a full four degree difference as you pointed out, but I am not quite clear how you settled on T = 28.4C ? Will this value be a function of the thickness and material of the pane and the overall difference in temperature?
Yes. Plug in the values of those two and you will get T=28.4C.

Also, in the convection equation:
P = h_air * A * (T - Tl)
Shouldn't this remain as:
P = h_air * A * (Th - Tl), unless we are convecting heat off of the panel (rather than directly from the surface of the water)?
I was considering the 1st case, where the pane is present. Convection doesn't happen inside the pane. Heat flows from the water through the pane in conduction, and then through the air in convection.

And just one last question, I noticed that you used h_air for the convection equation whereas I used h_water - was I mistaken in doing so?

Thank you for your time.

First, have a look at the last 2 lines :smile:
http://en.wikipedia.org/wiki/Heat_transfer_coefficient#Overall_heat_transfer_coefficient

Again, I was considering the 1st case (sorry for not specifying this). The temperature at water surface is 29C, which is my assumption. In your solution, in case 1, you only consider the heat loss due to the pane only, which means you ignored the convection in the water (or assumed that the temperature was uniform in the water) and this is my reason for the assumption. So convection we should consider here is due to the air.

Actually convection in the water does happen. You may recalculate heat loss rate by including it besides conduction in the pane and convection in the air, just as I did.
 
  • #6
I see, so the heat transferred by conduction through the pane must be equal to the heat transferred by convection from the pane into the air? I did not realize you were stating a system of equations, I thought they were two independent equations.

I see that 0.1m of glass doesn't offer the water much insulation from heat loss - 16.5Watts versus 20Watts. Glass really is not much of an insulator.

Thank you so much for your help here hikaru1221. Do you mind if I ask you another question? Does my estimation for heat loss by radiation look reasonable? Just 0.0043 watts?
 
  • #7
TheH said:
I see that 0.1m of glass doesn't offer the water much insulation from heat loss - 16.5Watts versus 20Watts. Glass really is not much of an insulator.

It's 0.01m, or 1mm, too thin to insulate anything. If you consider another more practical case, a glass pot of thickness 5.5mm containing boiling water (at 100C)when the surrounding is at 20C, you will see that T = 60C! They don't produce glass pot for no reason :smile:

Do you mind if I ask you another question? Does my estimation for heat loss by radiation look reasonable? Just 0.0043 watts?

Well I was kind of surprised when seeing that em of water = 0.95. Is water that close to black body? I did a little search and found that em of water at 38C = 0.67, so you should check your source.
Anyway your calculation is fine. Even if em=1, it doesn't matter much.
 
Last edited:
  • #9
TheH said:
hikaru, thanks again.

I can't remember where I got 0.95 from but most the stuff I was seeing on the web was in that range. For example on this page they say 0.99: http://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html

Wow, I again did a search then found, just like you said in most sites, it's around 0.95 at 300K or 27C. I didn't expect the emissivity coefficient of water to vary that much from 0.95 at 27C to 0.67 at 38C. I had better checked my source :biggrin:
Anyway even if emissivity coefficient = 1, the heat loss rate due to radiation is still infinitesimal, right?

From this page: http://physics.info/convection/ he specifies a formula on the bottom for estimating the convection heat-transfer coefficient of air as a function of air-speed. He doesn't cite any source, does it look reasonable to you?

I don't know much about this topic, so I cannot tell if it's right or not. The formula is empirically derived, and as we can see, h only depends on v. I think it's much more complicated, depending on the conditions we conduct the experiment. For example, in the extreme case where the density of the air is infinitesimal (i.e. ~ vacuum), convection hardly happens. However it does agree with Wikipedia about the range h should be under normal conditions, i.e. from around 10 to 100.

I guess you want to do some exact calculations for some practical purpose. However this is the best I can do - to give qualitative opinions :smile:

EDIT:
I'm sorry, I made a big mistake. Your calculation for heat loss rate due to radiation is wrong. You must use the absolute temperature instead. The result is about 6W, not so small at all!
 
Last edited:
  • #10
This is quite embarrassing, I can see my mistake now.

Actually the first time I used the equation, I thought it was (Th - Tl)^4 and so I thought to myself celsius/kelvin wouldn't matter. I forgot to re-consider that when I corrected it to (Th^4 - Tl^4). Thanks for the correction.
 
  • #11
So since the heat loss by radiation is non-trivial, should we revise the system of equations to:

P = (k_glass / d) * A * (Th-T)
P = ( h_air * A * (T - Tl) ) + (em * sbc * A * (T^4 - Tl^4))

and use this to solve for P?
 
  • #12
Actually there is problem with the heat loss rate by radiation. The exact equation is:

P = P(29) - Ps = em(29)*sbc*A*Th^4 - Ps(25/29)

Ps(25/29) is the power of radiation that is emitted by the surrounding at 25C and absorbed by the water at 29C. However I don't think we can find the exact formula for Ps(25/29) for the following reason. The power of the radiation coming in the area A, emitted from the surrounding may be written as:

P' = e(surrounding) * sbc * A * Tl^4

However, since water doesn't absorb all the radiations, Ps(25/29) < P', or Ps(25/29)=kP', where k<1. We may luckily find e(surrounding) on some website, but not k as k depends on the material (water), the wavelength of the radiation (we don't know at which range of wavelength the air radiates), the dimensions of the tank and water's temperature. It's too specific and thus, most likely unavailable on the internet.

At first, I thought of some way to go around. If the water is at 25C, then it's in thermal equilibrium with the surrounding, which means:

the radiation power it emits P(25) = the one it absorbs from the surrounding = em(25)*sbc*A*Tl^4.

P(25) = the power the water absorbs when it's at 25C. It should be different from the power it absorbs when it's at 29C, though the same surrounding's temperature. However as emissivity = 0.95, very close to 1, we may expect that it changes slightly from 25C to 29C, i.e. em(25)=em(29). The same idea arose in my mind. We may expect that the ability to absorb radiation of water (or its absorptivity) changes slightly from 25C to 29C. Thus: Ps(25/29) = P(25), and that leads to your equation.

I'm not really sure about my reasoning. I'll think of some other way or search on the internet; hopefully I can find something. Any other opinion would be appreciated.
 
  • #13
hikaru, I don't completely understand the issue.

The equation I am using for heat-loss by an object via radiation to it's surroundings is:

q = ε σ (Th^4 - Tc^4) A

from: http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

They do not indicate on that page anything about wavelengths? Is this a simplified form the of the equation, some sort of approximation?
 
  • #14
I'm not really sure about the issue. Both my thought and the website lead to the same equation, i.e. your equation. However I still think it's just an approximation. Anyway, since I'm not quite sure and the website seems to be reliable, you may apply the equation :smile:
 
  • #15
Alright then :)
 

1. What is heat transfer and why is it important?

Heat transfer is the movement of thermal energy from one object to another. It is important because it plays a crucial role in many natural and industrial processes, such as maintaining the Earth's climate and allowing for the operation of heating and cooling systems.

2. What are the three modes of heat transfer?

The three modes of heat transfer are conduction, convection, and radiation. Conduction is the transfer of heat through a solid material, convection is the transfer of heat through a fluid, and radiation is the transfer of heat through electromagnetic waves.

3. What is the difference between thermal conductivity and thermal resistance?

Thermal conductivity is a measure of a material's ability to conduct heat, while thermal resistance is a measure of a material's resistance to heat flow. In other words, thermal conductivity is the inverse of thermal resistance.

4. How does heat transfer affect energy efficiency?

Heat transfer can impact energy efficiency in buildings and other systems by causing unwanted heat loss or gain. By understanding the principles of heat transfer, engineers and designers can create more efficient structures and products that minimize heat transfer and reduce energy consumption.

5. How do you calculate heat transfer?

The calculation of heat transfer depends on the specific situation and the mode of heat transfer involved. For example, the heat transfer between two objects in contact can be calculated using the equation Q = kA(T2-T1)/d, where Q is the heat transferred, k is the thermal conductivity, A is the contact area, T2 and T1 are the temperatures of the two objects, and d is the distance between them. Other equations exist for calculating heat transfer through different modes, such as convection and radiation.

Similar threads

  • Thermodynamics
Replies
2
Views
997
Replies
2
Views
1K
  • Thermodynamics
Replies
5
Views
945
  • Thermodynamics
Replies
2
Views
623
Replies
4
Views
1K
  • Materials and Chemical Engineering
Replies
7
Views
1K
Replies
10
Views
2K
Replies
31
Views
4K
Replies
16
Views
12K
  • Thermodynamics
Replies
20
Views
9K
Back
Top