5
$\begingroup$

I'm interested in the following problem: There is a solid sphere with radius $r$ and mass $m$ at temperature $T_{s0}$. It is moving at velocity $v_s$ through a gas of temperature $T_g$. How does the temperature of the sphere, $T_s$, change with time?

A sketch:

Sketch of a flying sphere

I think I have to consider the following effects:

  • Radiation losses
  • thermal conduction at the solid/gas interface
  • Transport of hot gas away from the sphere due to $v_s$

My Approach is:

$$ T_s(t) = T_{s0} + m c \Delta E_s(t)$$

Where $c$ is the specific heat capacity and $\Delta E_s(t)$ the change of thermal energy in the sphere. For the latter I've set up an equation which incorporates the Stefan Boltzmann law.

$$ \Delta E_s(t) = - \int_0^t \sigma A (T_s(t)^4-T_g^4) dt + \Delta E_\text{cond}(t) + \Delta E_\text{conv}(t)$$

So, assuming this equation and though process is correct, my question boils down to how to take the thermal conduction (and convection?) represented by $\Delta E_\text{cond}(t)$ and \Delta E_\text{conv}(t) in the upper equation into account.

Edit: To clarify the magnitude of interest for $v_s$, $T_s$ and $T_g$: $v_s$ is about 0,5 to 5 m/s, the $T_s - T_g$ is in the range of 100 to 300 °C.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ If the speed is low enough not to heat the sphere but high enough to keep the gas in contact with the sphere at roughly a constant temperature (i.e., new gas at T$_{g}$ continuously moving past surface), then isn't the problem similar to a source in contact with an "infinite" heat sink? If T$_{s}$ is low enough, you could probably ignore radiative effects and this might become a simple conduction problem. $\endgroup$
    – honeste_vivere
    Commented Jan 5, 2015 at 21:00
  • $\begingroup$ Yes, I also think that would be the case. However, im interested in values of $T_s$ which are as high as 200°C. I'll edit this into the question. $\endgroup$
    – JHK
    Commented Jan 5, 2015 at 21:05
  • 1
    $\begingroup$ Then you might emit some infrared for a bit, but it should be easy to compare conduction to radiation rates. Radiation dominates when temperatures are high because it's $\propto$ $\Delta \ T^{4}$, but I think conduction could dominate here if the conduction coefficient is large enough. $\endgroup$
    – honeste_vivere
    Commented Jan 5, 2015 at 21:09
  • $\begingroup$ If I took your assumption, would't I have to set the temperature of the spheres border to $T_g$? Than I would have no radiation and only thermal conduction from the innards of the sphere outwards. $\endgroup$
    – JHK
    Commented Jan 8, 2015 at 6:16
  • $\begingroup$ Heat generally moves towards lower energies (kind of like water finding valleys between mountains), so if $T_{s}$ < $T_{g}$, then yes. The sphere would slowly lose heat to the gas. Whether it developed an internal $\nabla T$ depends upon the conduction coefficient for the sphere material. If the sphere can conduct heat quickly, you could approximate it as having the same T uniformly throughout. $\endgroup$
    – honeste_vivere
    Commented Jan 8, 2015 at 14:30

1 Answer 1

Reset to default
3
+50
$\begingroup$

So first off, i'm going to have to break to you that this $\Delta E_\text{cond}(t) + \Delta E_\text{conv}(t)$

is by far the most complicated bit. There is a simple equation $\frac{q}{A} = h\Delta T$ where $q$ is the heat flux (W/m^2), $A$ is the surface area of the object, $\Delta T$ is the temperature difference between the bulk of the fluid and the object and $h$ is the heat transfer coefficient. Unfortunately $h$ is quite difficult to determine as it depends on the properties of the fluid and its flow regime as well as the geometry of the problem.

In order to determine the heat transfer coefficient we define a number of dimensionless quantities such as $Pr$; the Prandtl number. This is the ratio of momentum diffusivity to thermal diffusivity for the fluid. Effectively this describes the relative size of the thermal and velocity boundary layers for the fluid. For low $Pr$ conduction through the fluid dominates (Mercury has Pr of around 0.015). In the opposite case of engine oil momentum diffusivity is much larger so convection dominates (Pr of 100 to 40000). The Prandtl number for air and most other gases is around 0.7-0.8.

Obviously the Reynolds number for the fluid, $\Re _\text{L}$ is going to be important and is defined as $\Re = \frac{v L}{\nu}$

This describes the flow regime around the object where $L$ is a characteristic length and can be taken to be the diameter of the sphere in this case. $\nu$ is the kinematic viscosity and $v$ is the velocity of the object relative to the bulk of the fluid. For the appropriate substance and temperature you can find the kinematic viscosity of your gas and then calculate the appropriate Reynolds number.

Lastly we define the Nusselt number. $Nu _\text{L} = \frac{h L}{k}$

where $L$ is again the characteristic length, $h$ is the heat transfer coefficient and $k$ is the thermal conductivity of the fluid which is evaluated at the film temperature. This is defined as the arithmetic mean of the temperature of the bulk of the fluid $T _\text{g}$ also known as $T _\infty$ and the temperature at the solid boundary.

As a small aside I will mention the Biot number. $Bi = \frac{h L}{k _\text{b}}$

where the symbols have their usual meanings except $k _\text{b}$ is the thermal conductivity of the body (your sphere in this case). If this is 0.1 or less then the conduction throughout the solid body is sufficient to assume that the body has no internal temperature gradient (which simplifies your problem).

Now on to solving the problem. We want to express the Nusselt number in terms of the Prandtl and Reynolds numbers in order to find an expression for $h$. Unfortunately this is very difficult but luckily large cohorts of experimentalists have slaved away at this problem for us.

For external flow over a sphere, T. Yuge (I don't have an explicit reference for this see textbook link at the end) found the following correlation $Nu = 2 + 0.43 Re ^\frac{1}{4} $.

This is for Pr approximately equal to 1 and Reynolds number between 1 and 10000.

(I could not find another correlation when Pr is closer to the range expected for air and most other gases. You will have to either accept this approximation or find your own information sorry.)

Decide on appropriate characteristics of your fluid and solid. Calculate the Reynolds and hence Nusselt numbers and then the heat transfer coefficient. Use this to solve for the heat flux for an appropriate temperature difference and then put this in your integral with your radiation term to solve for the total change in energy. I think that about does it. Here is a textbook that goes into a lot of detail:

http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf

I hope this is helpful.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.