## Wednesday, April 7, 2010

### Falling water - hot or cold?

Hello everyone! Since this is my first post as one of the virtuosi, I should probably introduce myself a little. I'm a first year graduate student in physics at Cornell university. I did my undergraduate work at Oberlin college (I know, you've never heard of it), and I'm currently just trying to keep my head above water and take in as much physics as I can. Additionally, I'm trying to find work for the summer, I might post more on that later.

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/)

Let's begin with a little motivation. Why would a water droplet heat up when it falls? Well, the physical mechanism is that as it falls through air, air resistance dissipates energy. This energy is dissipated mostly as heat, so we expect to put some additional heat into our water droplet as it falls.

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.