Effect of humidity on length of the day Do we have any data on how humidity in the air or storms and tornados change Earth's angular momentum by moving masses of water at higher elevations. By doing so they change earth's angular momentum and hence its rotational velocity?  
Has there been any study on this? In some big storms millions of tons of water are sucked from the ocean and lifted to heights of several miles?
Could this process slow down the rotation of earth and make days longer slowly?
 A: Let's assume that an excess of 2 cm of water evaporates across the entire surface of the Earth, and settles at an altitude of 3000 m. It is possible to compute the increased moment of inertia of the Earth, and thus the theoretical slowdown.
Moment of inertia of the earth can be found from data on the NASA website http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
moment of inertia I/MR^2 = 0.3308
mass                     = 5.97E24 kg
radius                   = 6.371E6 m

from which we get $I = 8.03\cdot 10^{37} kg m^2$ (confirmed by Wolfram Alpha).
A shell of water radius $R$ and thickness $dr$, density $\rho$ has a moment of inertia $I=\frac23 m r^2 = \frac23 (4\pi R^2 \rho dr) R^2$
For the given dimensions, this becomes 2.8E29 kg m$^2$.
When it rises to an altitude of 3000 m, the moment of inertia increases to
$$I = \frac23 (4\pi R^2\rho dr)(R+h)^2$$
an increase of about $\frac{2h}{R}$ or roughly 0.1%
But that's 0.1% increase for the moment of inertia represented by the water, which itself is only 3.5E-9 of the moment of inertia of the earth.
This will therefore change the rate of rotation by 3.5E-12 , meaning that a day would get longer by 0.28 µs
By comparison, the big earthquake in Japan in 2011 increased the rate of rotation of the earth by 1.8 µs (see http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html), and the tidal drag of the Moon on the Earth is causing the day to lengthen by 2.3 ms per century (much less on a per-day basis, but it keeps going day in and day out...)
Finally - the mass of water I estimated (2 cm covering the entire surface of the Earth including both sea and land) would be approximately 10$^{16}$ kg ; now the density of a cumulonimbus ('fluffy') cloud is about 0.3 gram / m$^3$ (source); if you need clouds that have a mass of 20 kg per square meter of Earth's surface, this implies clouds that would be 60 km high. Or a lot more dense. Either way, my estimate of the amount of water that might evaporate is pretty optimistic, meaning that the calculated (tiny) shift in the length of the day is an upper estimate.
On the other hand - if it's that muggy, the day might seem a lot longer....
