6
$\begingroup$

Let water (liquid) be in a chamber (with no heat transfer to the walls). The gas is the residual space above the liquid, meaning that the contact surface is the surface of the water.

Given the gas and the liquid are at a certain pressure, but the temperature of the gas and the liquid differ. How can I find the heat transfer rate for this problem?

$\endgroup$
  • 4
    $\begingroup$ In general, there is no easy answer to this, because the total heat transfer will be a complicated function of different effects like conduction, convection, radiation, evaporation etc.. Can you restrict the problem to a much more specialized case? $\endgroup$ – CuriousOne Sep 5 '14 at 8:17
1
$\begingroup$

Radiation:

The power of energy emitted by a body through radiation is given by Stefan–Boltzmann law:

$$P = eA\sigma T^4$$

where $e$ is the emissivity of the body, $A$ is the surface area of the body and $\sigma$ is the Stefan-Boltzmann's constant and $T$ is the temperature of the body.

Conduction:

For a conductor, the rate at which heat is transferred is calculated using the following formula:

$$\frac{dQ}{dt} = kA\frac{dT}{dx}=kA\frac{\Delta T}{\Delta x}$$

where $k$ is the thermal conductivity of the conductor, $A$ is the surface area through which heat is being transferred, $\Delta T$ is the temperature difference between the ends of the conductor and $\Delta x$ is the length of the conductor.

Convection:

The gas surrounding the liquid is not in equilibrium until it reaches a certain distance away from the surface of the liquid. It is quite difficult to quantitatively work with heat transfer through convection. The best approach to this problem is by treating the convective heat transfer to be a conductive heat transfer (or solve for conductive heat transfer and then adjust parameters to account for convection). The exchange of energy between the gas and the liquid can hence, be modeled as:

enter image description here

The region colored in blue is the liquid and the thick black line indicates the surface of the liquid. The region colored in yellow represents the gas which is at equilibrium. The orange region represents the gas which is not in equilibrium. In this region, we assume that the gas behaves as a thick conductor. The width of this region is called effective wall thickness and is denoted by $d$. The thin black line represents a line separating the orange region, where active conduction is taking place, and the yellow region where the gas has already reached the equilibrium temperature.

We can make adjustments to the heat transfer equation for conduction to account for convection. In cases where conduction is not negligible, we add a correction factor to the thickness of the conductor to account for the heat transferred through convection. In cases where conduction is negligible and heat transfer overwhelmingly happens through convection, we assume that there exists a hypothetical conductor separating the two surfaces; in our case, that is the orange region of air.

The value of $\Delta T$ is taken as the equilibrium temperature of the gas (yellow region) and temperature of the liquid (blue region). The value of $\Delta x$ is taken as the effective wall thickness $d$. The value of $A$ is taken as the total surface area of the liquid that is in contact with the air. The value of $k$ is taken as the thermal conductivity of air.

The value of effective wall thickness depends on several factors such as wind speed (turning on an air blower helps disperse the hot air near the surface of the liquid and hence, improves the conduction by reducing the effective wall thickness). This value is in most cases derived empirically or calculated through experiments.

Comments on the original question:

The above explanation dealt with a general case. In your question, you have the gas enclosed in a container. As the hot air rises, the temperature at the top of the container will be the highest. You can take nearly the entire height of the gas as the effective wall thickness.

References:

Conduction and convection mechanism for the human body

$\endgroup$
0
$\begingroup$

You could either use:

$ dS = \frac{dQ}{T} $

with $ T dS = dQ$

First Law: $T dS = dE - dW $

so $ dE = T dS + dW $

alternatively, you could use

$ Q = mc \Delta \theta $

Where $ \Delta \theta $ is the temperature difference between the gas and the liquid.

To use this in regard to both systems, the formula gets a bit more complex.

Firstly you have to compute the final temperature of the lower temperature substance using, lets say the gas is:

$ Q = M_{g}c_{g}(T_{f} - T_{i}) = M_{l}c_{l}(T_{i} - T{f}) $

so $ T_{f} = \frac{M_{l}c_{l}T_{i} + M_{g}c_{g}T_{g}}{M_{g}c_{g} + M_{l}c_{l}} $

I.e the joint heat capacity of the liquid and gas would be:

$ c_{lg} = \frac{c_{g} + c_{l}}{c_{l}c_{g}} $

Then you can again use $ Q = mc \Delta \theta $

and iterate for an arbitrary addition of systems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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