[cvhmanchester]

I teach novice students introductory chemistry. One experiment they do is to rub a balloon and attract a stream of water from a tap.  And the explanation of the reason for this behaviour is that the static electricity polarises the stream of water which is attracted and diverted.  But that is literally kitchen science. How is the underlying science quantified?
I have yet to find a worked example of the properties and numbers that I can supply in the "formative" answer:
If you rubbed the balloon 8 times and the stream of water was diverted by 3 cm, the charge generated on the balloon was ?? coulombs.
This would require the student to have measured the diameter of the water stream, length of fall etc. So the worked example would press home the need in every experiment, even in the kitchen, to do the measurements.
I have asked physicists for help, but they, like me, rapidly started waving their hands as an alternative to numbers.
Is there a balloon (or even ink jet) scientist out there?

[Corribus]

The first thing you need to do is articulate exactly what you're trying to calculate. You say "the charge generated". But what does that mean, exactly? Let's consider some of the complexities of the situation, which might highlight why you're not finding much quantitative insight into the problem:

1. Generated charge on the balloon surface is spread over an area. And that spatial distribution likely is not constant in time (it can't be, because the charge dissipates).

2. Water is complex, even when it's not moving. But when water flows over a surface (say, the inside of a pipe), even weirder things happen. In short, flowing liquids generate a static charge due to interactions between the material surface and the flowing liquid. This is why dispensing gasoline in the winter can be dangerous - arcing in the low humidity, low dielectric air can ignite the gasoline. You can read more about charge generation in flowing water here or here.

3. Charge generated on the surface of balloon induces additional charge asymmetry in the water. Those interactions are going to be complex, even assuming a very thin stream of water.

4. Coulombic force is dependent on distance, sure, but because the charge on the balloon surface is not constant in time or space you have to consider that the average force felt between the flowing stream of water and the balloon is the product of the individual force vectors between each unit area element on the balloon and volume element in the water at any time, modulated by the dielectric properties of the air. Even if we assume the charge distribution in the water and the dielectric properties of air are constant (they're not), this is still a complicated problem. And that's not even considering that the (average) force vector you observe visually when a stream of water bends toward a balloon is contaminated by the vector defined by the force of gravity pulling the water down (and possibly the water pressure pushing it down).

All these things considered, I guess you could possibly determine an average Coulombic force experienced by the stream of water. But just from a pure mechanics standpiont it's still a difficult problem when considering the constant flow of water and the geometry of the balloon. If you observe carefully, you'll see that the stream of water curves - this is due to the near constant force of gravity and the spatially variable Coulombic force over the curved balloon surface (again, even assuming the charge distribution is constant, the Coulombic force varies due to the balloon curvature).

Then, even if you manage to calculate an average Coulombic force experienced by the water over a short time element, how do you back translate that into a total amount of charge generated by the initial act of rubbing? My thought process would be that you'd have to somehow plot out how the water trajectory changes in time, then do some spatial and time integrations of the various force vector equations... my head hurts thinking about it.

Anyway, that's a lot of spitballing (hand waving, as you say) but I think it gives some sense of the difficulty of the problem.