# How big must a building/spaceship be to create an indoor weather system?

The NASA Vehicle Assembly Building and the proposed Nazi Volkshalle are cited to be "big enough for weather systems to form inside them".

The building has at least 10,000 tons of air conditioning equipment, including 125 ventilators on the roof supported by four large air handlers (four cylindrical structures west of the building) to keep moisture under control. Air in the building can be completely replaced every hour. The interior volume of the building is so vast that it has its own weather, including "rain clouds form[ing] below the ceiling on very humid days",[8] which the moisture reduction systems are designed to minimize.

The same problem would undoubtedly occur to spaceships, which would also suffer from the problem of interior cloud formation.

How large can an indoor chamber (i.e. a bubble habitat, spaceship) become before it will start to rain indoors due to condensation? Assume 50% RH in the air at ground level, for the purposes of this question, as such habitats are almost certainly going to be humidity controlled.

## 1 Answer

A 250m high building will generate clouds at 50% R.H.

Relative humidity fluctuates with temperature. Thanks to the way it is measured (partial pressure of water vapor divided by the equilibrium partial pressure of water vapor), if any section of the atmosphere hits 100%, it begins condensing.

The key to creating condensation and rain is to have areas of warmer temperature and areas of colder temperature. The warmer temperatures can have a low relative humidity, low enough to evaporate water off of surfaces and people. The colder temperatures can have a high relative humidity at the same partial pressures, allowing it to condense.

For a moment, let's ignore condensation onto surfaces, which can easily be cold enough to condense water in any building, not just big ones. Lets concentrate on cloud formation, which requires a 100% relative humidity mid air. We'll ignore the requirement for nucleation sites like dust; those will be easy to come by.

For your question, we need to determine how large a building needs to be to generate a temperature gradient sufficient to have 50% RH at the bottom and 100% RH at the top (bottom-to-top because natural convection will create the temperature gradient in that direction).

Step 1: $\Delta T$ required to create the required equilibrium ratio

I'm going to use the Arden-Buck equation. Why? Its simple enough in form, and Wikipedia recommends it (and who wouldn't want to use their recommendation ;) )

$$e^*_w = (1.0007 + 3.46 \times 10^{-6} P) \times (6.1121) e^{\left(\frac {17.502 T} {240.97 + T}\right)}$$

(Like all empirical equations, units are specified: Pressure is in hectopascals (absolute), temperature is in Celsius)

We are interested in a case where pressures are [roughly] constant and the equilibrium pressure of water vapor is twice as high where it is warm as where it is cold. Using these two data points, we get

$$2 = \frac{e^{\left(\frac {17.502 T_{warm}} {240.97 + T_{warm}}\right)}}{e^{\left(\frac {17.502 T_{cold}} {240.97 + T_{cold}}\right)}}$$

Let $f(T) = \frac {17.502 T} {240.97 + T}$ for readability

$$2 = \frac{e^{f(T_{warm})}}{e^{f(T_{warm}-\Delta T)}}$$

Bugger: two variables. I'm being quick here, so lets assume $T_{warm} = 23 ^\circ C$. Now I get even lazier and turn to Wolfram Alpha to solve this for me

$$\Delta T = 10.9759 \approx 11 ^\circ C$$

These were empirical equations anyway. Now we know we need a 11 degree difference in temperature from top to bottom.

Step 2: Thermal Transfer

We're going to assume there is no air conditioning circulating air. The example has air conditioning, but they also point out that clouds are forming on "very humid days," and Florida averages for R.H easily clear 80%, so "very humid" is a high bar. Let's assume this higher R.H. and the presence of air conditioning cancel out, so that we can tackle the simpler problem of a building with clouds at 50% R.H. with no A/C

For this we will need one more number: how much heat is generated at the warm side of the room. Kennedy space center consumes an average of 725TJ per year, or 23MW continuously. Arbitrarily assuming 1/50th of the center's power goes to the building in question (based on me squinting my eyes at their map and a lot of fudge factors), there is a .46MW rate of power consumption in the building. Because I've got so many fudge factors already, I'm just going to round that to .5MW or 500kW.

The VAB is 34453m2 in surface area, meaning each square meter averages 14W of heat output. Now we have every number we need to determine the minimum height of a building to dissipate 14W/m2 with a 11 degree temperature gradient.

As long as the building is large enough to support convection (which any building of this scale should easily accomplish), we can measure the heat transfer by mass-flow rates.

We have an equation for mass flow in a convection scenario: $$Q = c_2 P^{1/3} l^{5/3}$$

Where $Q$ is the volumetric flow rate in $m^3/s$, $c_2$ is an application specific coefficient, usually 0.06, $P$ is the heating power from the source in kW, and $l$ is the distance available above the heat source in m. Treating our per-unit-area terms as though they were individual heaters.

The heat transfer from the heater source on the floor must be the same as the heat transfer done by the air to the roof, which is volumetric flow with a known temperature

$$P = Q * s_{air} * 11 ^\circ C \approx 0.0091\frac{kW-s}{m^3} * Q$$

$$Q = 109\frac{m^3}{kW-s} * P$$

Combining these equations around Q (the only remaining unknown), we get

$$109\frac{m^3}{kW-s} * P = c_2 P^{1/3} l^{5/3}$$

from here it is simply a solution $l = 249.248 m \approx 250m$

Thus, after way too much work and far too many variables, a 250m high building will be sufficient to generate clouds at 50% R.H. at the ground level.

Interestingly enough, the VAB is only 160m tall. However, remember that all of these numbers were based around a notional R.H. of 50%. Given their higher R.H, it would be easy for this minimum height to change. There are also perhaps a half-dozen number I had to pull from thin air along the way. These are also potential sources of clouds at smaller heights.

• Pulling from the air seems appropriate concerning the subject matter of condensation. Commented Dec 26, 2014 at 3:43
• @jdlugosz: Yes, but it makes the calculations far more difficult. This seemed like a reasonable enough approximation given the information in the original question. It would be very difficult to approach this question accounting for pulling water from the air because that also changes how the air circulation works, which affects the temperature gradient in a hard to define manner. If one had a building in mind, along with a layout of where the air handlers are, one could start the more difficult question using CFD. Commented Dec 26, 2014 at 5:55
• If desired, rephrase the question as "Florida can hit 99% humidity, but assume the dehumidifiers reduce humidity to 50% R.H. at ground level" without loss of generality. Commented Dec 26, 2014 at 5:56
• I was making a pun on your closing line, pulling numbers fron thin air. Commented Dec 26, 2014 at 7:41
• Ahh, thank you for the clarification. Leave it to me to use a wise-crack wording and forget I did so! Commented Dec 26, 2014 at 17:39