Simulation of temperature in insulated box

Question

I have a box outside that I want to insulate such that the temperature inside the box doesn't drop below 10°C given outside temperatures of (in the most extreme case - at night) -10°C. Inside the box I have a "heater" of 5 W. The box is in direct sunlight during the day (only if the sun shines, though).

I want to simulate the temperature in the box in order to help me decide on isolation material / thickness. I'm unsure which factors I need to account for. So far I have:

• heat loss due to heat transfer through box walls
• heat "gain" due to the 5 W "heater".
• heat "gain" due to the sun shining on the box at day.
• heat capacity due to the box (isolation material as well as air volume)

Are there other things to consider? Any thing which I'm modelling too simple?

Heat loss due to heat transfer through box walls:

lambda = (Q * l) / (A * deltaT)
thus =>

Q = A * lambda * deltaT * (1/l)

where
  Q = heat transfer
  lambda = thermal conductivity of material
  deltaT = temperature difference
  l = thickness of isolating material
  A = surface area of box

Heat capacity of box:

Once for the air and once for the isolation material.

E = c_v * V * T

 where
   E = "heat energy" in box
   c_v = (volume) specific heat capacity
   V = volume
   T = temperature (in K) 

Heat gain due to the sun:

The most tricky I guess. Assuming (https://en.wikipedia.org/wiki/Sunlight) I probably get about 120 W / m^2 due to sunlight and just (for simplicity) the top side of the box (A_top) I guess I still need a "efficiency" factor? This is on the outside, though, so I guess the maximal heat transfer to the box (heating) is limited by the isolating material?

Simulation:

I'd start with the box (air and isolation material) being at outside temperature. Then adding all heat gains (heater and due to sunlight during day) and (using the now higher temperature in the box and thus a temperature difference) subtract the heat lost due to heat transfer through the walls. And all that for each second.

1. E' = E_n + (P_heater + P_sun) * 1s
2. use E' to calculate T_box and thus delta_T
3. use delta_T to calculate Q
4. E_(n+1) = E' - Q

Answer 1

General

The system will require two approaches, a non-steady state and a steady state. The transient case is of no interest for the need only to solve for a minimum steady state.

The general energy balance for the steady state is

$$ \dot{q}_R + P = \dot{q}_x $$

where $\dot{q}_R$ is the solar radiation input, $P$ is the power input, and $\dot{q}_x$ is the loss from the box. All terms are in units of Watts.

Starting Assumptions

The incident solar radiation is the most difficult to express of the three terms. You are interested in finding the case for the minimum temperature in the box. By sensible logic, we can agree that the lowest temperature in the box will be when no solar radiation is incident on the box. So, we can neglect this term.

The power is given as a constant.

The heat loss can be expressed succinctly as $\dot{q}_x = U A \Delta T$ where $U$ is the overall heat transfer coefficient (W/m$^2$ $^o$C), $A$ is the box area (m$^2$), and $\Delta T = T_i - T_a$ is the temperature difference between the inside of the box and the outside air ($^o$C). The parameters $U$ and $A$ are referenced relative to the inside or the outside of the box depending on the approach taken.

Expansions and Further Assumptions

The overall heat transfer coefficient includes convention inside the box, conduction through the box, and convection outside the box. A representative general example is given at this Wikipedia link. As you might take the internal and external areas of the box to be nearly equal, and with the assumption of constant convection coefficients, the final expression becomes as below.

$$ P = \left(\frac{1}{h_i} + \frac{\Delta L_I}{k_I} + \frac{1}{h_o}\right)^{-1} A\ \left(T_i - T_o\right) $$

Solution

The value of $P$ is given. Take reasonable estimates for the internal and external convection coefficients $h$ (W/m$^2$ $^o$C), e.g. stagnate air inside and flowing air outside. Apply the value for the thermal conductivity of the insulation $k_I$ (W/m $^o$C). You now have one equation with two variables, $\Delta L_I$ the thickness of the insulation (m) and $T_i$ the temperature inside the box. Set the minimum allowable temperature and you have the minimum required thickness of the insulation.

Other Options

When the wall of the box has its own conductivity, you can to first order expand $1/U$ with an additional term $\Delta L_b/k_b$ to account for the thickness $\Delta L_b$ and conductivity $k_b$ of the box.