# How do I calculate the temperature change inside a box with a heat source? What if I ventilate the box?

I have a box (as shown below). I know the temperature outside of the box (for example you could use 293K) and I would assume the temperature inside the box is initially the same as outside of the box.

I have 6000 chickens in the box. They produce a total of 15.8 W of heat with 8 W of that being sensible heat.

If the box is sealed and there is no ventilation how would I calculate the change in temperature over time? Would it be a simple case of $Q=mcΔT$?

Obviously in the real world I would not leave 6000 chickens in a box to overheat. How could I calculate the temperature change over time if I ventilated the box at a rate of 4m3/s?

Edit: Material would be PVC coated material (like on the side of lorries/trucks). I have an found an R value of 0.16 m²K/W for this type of material and it's thickness would be around 0.75mm. • Material and thickness of the box? r value? Aug 10 '17 at 16:38
• What is sensible heat? What happens to the insensible, secret heat? :) Aug 10 '17 at 19:54
• @GürkanÇetin latent heat due to moisture loading Aug 10 '17 at 21:23
• @Solar Mike: Material would be PVC coated material (like on the side of lorries/trucks). I have an found an R value of 0.16 m²K/W for this type of material and it's thickness would be around 0.75mm. Aug 11 '17 at 12:47
• If solving this problem accurately were easy, a lot of Finite Model experts would be out of a job. Aug 11 '17 at 13:12

Take a system picture as below. • $\dot{V}$ - air flow (m$^3$/s)
• $\rho$ - air density (kg/m$^3$)
• $\tilde{C}_p$ - air specific heat (J/kg $^o$C)
• $h_a$ - air convection coefficient (W/m$^2$ $^o$C)
• $T_a$ - air temperature (K)
• $A$ - area of container (m$^2$)
• $w$ - wall thickness (m)
• $k$ - wall thermal conductivity (W/m K)
• $\hat{\dot{q}}_I$ - internal heat generation (W/m$^3$)
• $V$ - volume of container (m$^3$)
• $T_f$ - unknown final equilibrium temperature

Assume that heat generation is uniform throughout the container, the container is (mostly) air, and the air is well mixed. The energy balance equation gives

$$\hat{\dot{q}}_I V = \left(\dot{V}\rho\tilde{C}_p + (A/R_T)\right)\left(T_f - T_a\right)$$

where $R_T$ is a thermal resistance defined as

$$R_T = \left(\frac{w}{k} + \frac{1}{h_a}\right)$$

Cast this into the following form

$$\left(\frac{T_f}{T_a} - 1 \right) = \left[\alpha\ \hat{\dot{q}}_I\right] \left[1 + \alpha \left(\frac{A}{V}\right)\left(\frac{1}{R_T}\right) \right]^{-1}$$

with

$$\alpha = \frac{\tau}{\rho \tilde{C}_p T_a}$$ $$\tau = \frac{V}{\dot{V}}$$

Consider the left side as a relative difference in the temperature inside the container versus the air (this can be expressed as a percentage when multiplied by 100). Consider $\tau$ as a turnover time (how often does one entire volume of air in the container get exchanged).

Now take two extremes.

# Case A

In the limit that you do not ventilate, $\tau \rightarrow \infty$. For large $\alpha$, the expression approximates as

$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\frac{A}{V}\right]^{-1}\left[R_T\right] \left[\hat{\dot{q}}_I\right]$$

To keep the inside of the container closest to the external air temperature, have the greatest wall area to container volume and have a low thermal resistance. As an extreme analogy, using a thin walled metal sphere that houses all the chickens would be the best option.

# Case B

In the limit that you ventilate well, $\tau \rightarrow 0$. For small $\alpha$, the expression approximates as

$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\tau\right]\left[\rho\tilde{C}_p T_a\right]^{-1} \left[\hat{\dot{q}}_I\right]$$

To keep the inside of the container closest the external air temperature, have a high air flow (low $\tau$) and replace the air with a denser gas that has a higher heat capacity. So again by extreme analogy, do not ventilate the container with helium (for other reasons as well I can imagine :-)).

# Summary

• Use a container with the greatest area to volume possible
• Use a container that has low thermal resistance (alternatively blow the air around the outside of the container as well to increase $h_a$)
• Ventilate well

Interestingly, Case B suggests that, all else being equal, cooling will be more efficient on cooler days than on hot days.