May someone explain Thermal diffusivity to me? I cannot understand a velocity of cm^2 / sec.
Regards
Thermal diffusivity is the ratio of thermal conductivity to heat capacity:
α=k/(Ï*Cp)
where: k - thermal conductivity, Ï - density, Cp - specific heat (heat capacity).
Of course, you can derive the units from this equation.
Basically, thermal diffusivity is a measure of how quickly heat can move through a material.
If you look at the definition, it is basically the thermal conductivity [J/(s·m·K)] divided by what one could call the volumetric heat capacity [J/(m³·K)].
So the heat transfer rate is not a velocity by itself.
You could call the thermal conductivity a kind of velocity, I suppose.
Thank both of you for reply,
For example:
We need to heat the plate up to ÎT = 40 °C (assume plate was 0 °C)
Plate dimensions 10x10x0.4 cm
Specific heat c 0.127 cal/(g * °C)
Density Ï 7.85 g/ŃĐŒ^3
Mass of plate m = 10 * 10 * 0.4 * 7.85 = 314 g
Heat Q = m * c * ÎT = 314 * 0.127 * 40 = 1595 cal
We have a heat source of q = 200 cal/sec, and we need to know time elapsed to get to the required temperature
Heat capacity C = c * m = 0.127 * 314 = 39.88 cal/°C
Rate of heating v = q / C = 200 / 39.88 = 5.02 °C/sec
Disla calculated the time required to heat the plate up to ÎT, it is below.
All things fit nicely, but how to use thermal diffusivity in calculations or what type/kind of problems to solve with it? Or are there some occasions to use it?
Thermal diffusivity is present in Fourierâs heat equation. If you are looking for examples of problems involving this property, check âFundamentals of Heat and Mass Transferâ by Incropera. Itâs one of the best books in this field.
Thank you very much for your prompt reply.
Your analysis is simplified too much.
For example, you calculate the heat flowing in, but donât take the heat flowing out into account. As soon as the plate becomes warmer than its surroundings it will start losing heat though convection and radiation.
And as soon as your heat source starts, it will start warming up the surface of the plate. How fast that heat is conducted away from the surface depends on the thermal diffusivity. When the surface warms up, the temperature difference between the heat source and the plate will decrease, which will decrease the heat flow.
Would you be so kind to me as to show the usage of the parameter in my example, please. I cannot feel cm^2 / sec. cm / sec - speed, 1 / sec - speed of rotation, but cm^2 / sec?
Youâre thinking in terms of simple equations like the accelleration of a point mass. But the reality of heat flow is more complicated.
As for the âweirdâ units, thatâs just because other units appear both in the nominator and denominator and cancel each other out.
Letâs go back to your plate. Suppose you put one edge of the (cold) plate in an infinitely large pot of boiling water. Because of the large temperature difference, a lot of heat wants to flow from the water to the metal.
If the thermal diffusivity is low, the conductivity is low compared to the heat capacity. That means that the metal in contact with the water will heat up relatively quickly, but the heat doesnât spread out fast. When that happens, the temperature difference between the water and the plate in contact with the water drops, and so does the heat flow into the plate.
If the thermal diffusivity is high, the conductivity is high compared to the heat capacity. That means that the heat flows away from the contact edge more than it heats up the material locally. That means that the plate heats up more evenly.
Throughout the plate, the temperature and heat flow are constantly changing until a steady state is reached where the plate sheds as much heat as goes into it.
The thermal diffusivity is a parameter in the partial differential equation called the heat equation, which you can find on the Wikipedia page I linked to before. Thatâs why you need an FEA program like CalculiX to find approximate solutions for boundary value problems concerning heat flow.
ÎT = 40 °C
Plate dimensions 100x100x0.4 cm
Heat Q 120 cal ?Âż? (heat source of q = 200 cal/sec)
Specific heat c 0.127 cal/(g * °C)
Density Ï 7.85 g/ŃĐŒ^3
Mass of plate m = 10 * 10 * 0.4 * 7.85 = 314 g (?Âż?Âż dimensions are 10cm or 100cm)
Heat Q = m * c * ÎT = 314 * 0.127 * 40 = 1595 cal (Check mass)
We have a heat source of q = 200 cal/sec, and we need to know time elapsed to get to the required temperature (How does that energy enter into your plate?Sides, top,âŠ)
Heat capacity C = c * m = 0.127 * 314 = 39.88 cal/°C
Time required t = q / C = 200 / 39.88 = 5.02 sec This is not the time required check units. (°C/sec)
Time required = ÎT (40 °C) / 5.02 (°C/sec) = 8 sec
Result based on purely energy balance for a uniform heating proces of the whole plate at the same time, plate isolated without heat loost and infinite conductivity (1.000.000 W/m/K). Real life requires some time for the heat to spread into the plate (finite conductivity). Thatâs why this parameter is introduced.
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Thank you. I come to a point as of while considering a material for cooler/heater to be chosen, thermal diffusivity comes handy into play.
At the beginning I had one problem in my mind then it was appeared not the one I needed. I changed the problem in my mind and missed to clean it properly after the first one. Unreliable internet connection and lunch time constraints played their role as well. What to do, it is life.
This is mostly interesting part of all scrambled things I have in my mind. If to put my example this way. All data same but the heat source is applied into the middle of the plate 10x10 cm with a section of 1 cm^2. How to obtain the rate of heat and time elapsed to get to the required temperature?
The question is not properly stated and you will realize quickly why.
When the Thermal conductivity is finite as in real life, heat needs some time to spread.
You may reach 40ÂșC at the center in one second, but the outer plate may need 3h to reach that value. It will depend on your material properties.
To do it you need to perform a Thermal transient analysis.
Thank you. But I cannot download the video right away. I will try to figure out how to deal with thermal transient analysis with Support Center. If you know there is another good book on it I will really appreciate it.
Truth, the book does not come to my expectations.