Today, the question that is on my mind is: How much does a water droplet heat up when it goes over niagara falls?

(image from http://grandcanyon.free.fr/)

More quantitatively, assume the water is going to fall from some height h to the ground. The gravitational potential energy of our water droplet is given by

\[ PE=mgh \]

where m is the mass of the droplet and g is acceleration due to gravity. We assume that the particle starts with no initial velocity. It is very easy to place an upper bound on how much the droplet will warm. The maximum heating will happen if all of the potential energy were converted to thermal energy. The temperature change wound be

\[ mc\Delta T = mgh \]

so

\[ \Delta T = \frac{gh}{c} \]

Where c is the specific heat of water.

Niagara falls is 51m tall, g is 9.8m/s*s and c is 4.1kJ/kg*K so this gives a maximum temperature change of .12C=.22F, fractions of a degree.

However, we can do better than this.

A small droplet will experience a linear drag force, and a bouyant force from the air. The combination of these will result in a terminal velocity of

\[ v_t=\frac{2}{9}\frac{(\rho_p-\rho_f)}{\mu} g R^2 \]

Where $\rho_p \text{ and } \rho_f$ are the droplet and air density respectively, $\mu$ is the dynamic viscosity and R is the radius. This gives the kinetic energy the particle gains and we assume the rest of the potential energy goes to thermal energy. So,

\[mc\Delta T = mgh-\tfrac{1}{2}mv_t^2\]

\[\Delta T = \frac{gh-v_t^2}{c} \]

Now we can plug in some numbers. A little bit of searching will turn up appropriate values for the density of water (~1000kg/m^3), the density of air (~1.2kg/m^3) and the dynamic viscosity of air (~.8mPa*s). We estimate the radius of a small droplet of water is 1.5mm (~1/16"), which seems about a medium sized raindrop. This gives a terminal velocity of 6m/s. Using the above equation this gives a temperature change of

\[ \Delta T = .0024C/m*h-.009C \]

It looks like our calculation only makes sense after the after droplet has fallen more than 4.5m (this is probably about the distance it takes to reach terminal velocity)! Niagara falls is 51m tall, so this gives a change in temperature of .11C=.2F. The water only heats up by a fraction of a degree. This more realistic estimate is still rather close to our upper bound.

I leave it as a question to you, the reader, to estimate the temperature change of a typical raindrop.

What about evaporation? Unless the humidity is 100% the water droplet will lose latent heat of evaporation as it falls.

ReplyDeleteAlso, the drag forces might to some extent heat the air rather than the falling drop.

Finally, you might look into the results that Joule obtained in 1818 measuring the temperature of water at the tops and bottoms of waterfalls.

You're right. There are some assumptions here. I've ignored any kind of evaporative cooling (though one might argue that if water droplets are moving en masse through the air the humidity is around 100%), and I also assumed that all heat transfer is into the droplet. At least for the second piece, multiply by x where x is the fraction of heat that goes into the droplet, and the answer scales exactly.

ReplyDeleteI had actually almost forgotten about Joule's experiments, but as I recall that is what suggested conservation of energy to him, if memory serves. Which supports our approach here.

although i`m not a science student , but when ever water or any liquid falls from height it cools down because of evaporation. most cooling towers run on this principle.

ReplyDelete