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Suppose I have a room, 10x10x10m with insulated walls (R-value: 4) and a single double-pane window 1x1m (R-value: 0.35).

Now the total thermal power required to keep the room at a desired temperature is given by:

$$ P = \left( \frac{6\cdot 10\cdot 10 -1\cdot 1 }{4} + \frac{1\cdot 1}{0.35} \right) \Delta T $$

I am wondering what would happen if the window is left open? In that case, wouldn't the R value be zero and power go to infinity? Clearly this can't be right... I should be able to assume a finite R value for this case. But which and why?

EDIT: I am somehow suspicious in this case the hole is completely ignored in this formula but taken into account as convection term?

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  • $\begingroup$ give the hole a negative R value ... maybe a -10 $\endgroup$
    – jsotola
    Commented Jan 27 at 21:05
  • $\begingroup$ Negative? That wouldn’t make sense as it would produce energy. I think what you mean is a value between 0 and 1. But even then, which value? $\endgroup$
    – divB
    Commented Jan 28 at 22:06
  • $\begingroup$ you are right ... you probably have to multiply the area of the hole by some factor $\endgroup$
    – jsotola
    Commented Jan 28 at 23:39

2 Answers 2

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If this hole is the only opening in the room, there will be a buoyancy-driven exchange flow through it (if the interior is hotter than the exterior, in through the bottom half of the hole and out through the top half of the hole, and vice versa). This exchange flow is known as "mixing mode ventilation". Since the outgoing air is at a different temperature from the incoming air, this means a net energy flow. I'd expect the pressure differences driving the flow to be proportional to $\mathrm{\Delta}T$, and therefore the flow rate (through the usual formula for an orifice flow) to be proportional to $\sqrt{\mathrm{\Delta}}T$. The rate of energy transfer $P$ will be proportional to the product of flow rate and $\mathrm{\Delta}T$, i.e. $P$ will be proportional to $\mathrm{\Delta}T^{3/2}$ and your $R$ value proportional to $1/\sqrt{\mathrm{\Delta}}T$, with a constant of proportionality that depends on the top-to-bottom height of the opening. I'd guess a good place to look for a detailed formula for that constant of proportionality would be volume A of the CIBSE guide (or ASHRAE's equivalent if you're of the American persuasion), but I don't have access to a copy right now to check.

If there's another opening somewhere at a different height, there's more likely to be a one-way flow through each hole ("displacement mode ventilation"), and the overall exchange flow (and therefore the $R$-value) will be a combined property of the pair of openings, so there's not a well-defined $R$ value for one of the openings in isolation. Nevertheless, I'd still expect the proportionality of the combined $R$ value to $1/\sqrt{\mathrm{\Delta}}T$ to hold up, just with a different constant of proportionality that depends on the height difference between the two openings and the ratio between their areas. Again, volume A of the CIBSE guide is probably the place to look for a detailed formula.

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The issue is that the formula you use for the window R value does not account for mass movement, whereas an open window permits both mass movement and draft.

The dynamics of the question vary depending on a number of factors, including the window's location and the air circulation it creates inside the room through its vent action.

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  • $\begingroup$ How can I take mass movement or draft into account? Background for my question is that I have a wall adjacent to my laundry room which is unconditioned and I want to calculate the R value of that wall. sustainableengineering.co.nz/… provides an equation how to do this. Just one problem: There are stairs into the crawl space without door. So I am not sure how I should account for this "hole". $\endgroup$
    – divB
    Commented Jan 27 at 8:37
  • $\begingroup$ I would imagine one can use Ansys, Simscale, or some other symulation program! $\endgroup$
    – kamran
    Commented Jan 27 at 19:04

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