1
$\begingroup$

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.

The box in question

$\endgroup$
  • $\begingroup$ Material and thickness of the box? r value? $\endgroup$ – Solar Mike Aug 10 '17 at 16:38
  • 1
    $\begingroup$ What is sensible heat? What happens to the insensible, secret heat? :) $\endgroup$ – Gürkan Çetin Aug 10 '17 at 19:54
  • $\begingroup$ @GürkanÇetin latent heat due to moisture loading $\endgroup$ – Solar Mike Aug 10 '17 at 21:23
  • $\begingroup$ @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. $\endgroup$ – Charlotte Abbott Aug 11 '17 at 12:47
  • $\begingroup$ If solving this problem accurately were easy, a lot of Finite Model experts would be out of a job. $\endgroup$ – Carl Witthoft Aug 11 '17 at 13:12
1
$\begingroup$

Take a system picture as below.

picture of hen house

  • $\dot{V}$ - 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.

| improve this answer | |
$\endgroup$

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.