Modeling a wine cooler heat loss to ambient I'm trying to model the steady state heat loss to ambient, in W, for a wine cooler similar to the following:

For the modeling, I will need the following variables/constants:


*

*$T_a$ [K]: Ambient temperature (typically 300 K)

*$T_c$ [K]: Temperature inside the wine cooler enclosure (typically 280-290 K)

*$H, W, D$ [m]: height, width and depth of the cooler (facing from the front of the cooler, about the same perspective as the picture above)

*$L_w$ [m]: thickness of the thermal isolation material on the walls

*$k_w$ [W/(m K)]: thermal conductivity of the isolation material on the walls

*$Q_w$ [W]: heat conducted through the walls

*$L_g$ [m]: thickness of the cooler's front glass window

*$k_g$ [W/(m K)]: thermal conductivity of the front glass window

*$Q_g$ [W]: heat conducted through the front glass window

*$Q_c$ [W]: total heat conducted through the walls and front glass window

*$\varepsilon$: emissivity of the wine cooler walls

*$\sigma = 5.67 \times 10^{-8}$ [W/(m$^2$ K$^4$)]: Stefan-Boltzmann constant

*$Q_r$ [W]: heat loss by radiation

*$Q$ [W]: total heat loss to ambient


First define $H' = H - 2L_w$, $W' = W - 2L_w$ and $D' = D - 2L_w$, which are the dimensions of the thermal isolation walls, disregarding corners (to simplify the analysis).
Applying Fourier's law of heat conduction to the walls of the cooler, we get:
$$
Q_w = k_w \frac{2 H' D' + 2 W' D' + H' W'}{L_w} (T_a - T_c)
$$
Again, applying Fourier's law but now to the front glass window, we get
$$
Q_g = k_g \frac{H' W'}{L_g} (T_a - T_c)
$$
The total heat loss by conduction is thus
$$
Q_c = Q_w + Q_g
$$
However, since there is a front glass window on the cooler, we should also consider the effect of heat transfer by radiation. This is where it gets hard for me. This depends on the area of emission, which I understand to be the five internal walls of the cooler. Assuming this, we get:
$$
Q_r = \varepsilon \sigma (2 H' D' + 2 W' D' + H' W') (T_a^4 - T_c^4)
$$
$$
Q = Q_c + Q_r = Q_w + Q_g + Q_r
$$
The problem is that I have a wine cooler with a TEC1-12706 thermoelectric module connected to a constant 12 V voltage source. From the datasheet (linked above), assuming a $\Delta T = 10$ K (a strict lower bound to the actual $\Delta T$), the maximum heat pumping capability, reading from the graph, would be somewhat below 50 W (I estimate 48-49 W). The actual $\Delta T$ should be a little higher due to the heatsink being at a higher than ambient temperature, and hence the heat pumping capability should be smaller than the estimate above.
However, plugging in $T_a = 300.15$ K and $T_c = 290.15$ K (values that can actually be attained in practice by this cooler), and assuming $\varepsilon = 0.9$ since the inside walls are painted black, I get $Q_r = 47.1$ W. Hence all of the TEC's heat pumping capability should be used to pump the heat loss due to radiation, with nothing left to pump the heat loss due to conduction. This is clearly not the case as the wine cooler actually works at the claimed parameters, so I've got an error somewhere in my calculations. I am assuming it has to do with the fact that the front glass window is not perfectly transparent, so it should reflect some of the incoming radiation. However, I don't know how to account for this -- just multiply $Q_r$ by a constant representing the percentage of reflected radiation?
So here are my questions:


*

*Is the modeling of heat conduction correct?

*How should I fix my model of heat loss by radiation?

 A: The main problem with your approach is that you are using the wrong area for the radiation "window". The area over which the exchange of radiative energy can take place is just the window in the door - the walls "see each other" and that part of the radiation has no net effect. So you want to use just $H\cdot W$ for the area.
Secondly the window is partially reflective which again lowers the rate of heat exchange. Calculating this precisely is quite hard, but setting the emission to 0.5 seems like a reasonable way to approximate this.
A: The radiation is not the issue here, it is the outside wall temperature that does not fit. If you touch your fridge does it feel like it stands at 17 Celsius (290 K)? I would consider the outside wall at room temperature and then it is only a conduction problem through the walls. Taking my own fridge as an example for dimensions (0.86*0.46*0.46 cm, external) and taking the conductivity of styrofoam to evaluate the loss through the walls I obtain about 30W for an inner temperature of 4C and outer temperature of 22C. Then the window is a double pane which you should evaluate according to the good model found here: https://www.engineersedge.com/heat_transfer/heat_loss_doublepane_window_13864.htm. I found about 8W through the window door for a total of 38W. So I will use a 50W unit to have some extra power, you don't want to run at the unit limit all the time. I Hope this helps!
