Temperature change.

[Gareth56]

I read somewhere that water has PE at the top of a hill (waterfall) which is mostly converted into heat by the time it reaches the bottom of the hill (waterfall). So is it possible to measure the change in temperature (if there is any) for water than falls over the top of Niagra Falls and finishes at the bottom of the Falls?

[insane_alien]

its possible but in the case of the niagra falls it is not a closed system so the water will lose the heat energy it generates.

 

you could do it relatively easily in a lab though.

[Klaynos]

As I understand it.

 

Most of the PE is converted into KE (the water at the bottom of the waterfall has speeded up), but at the bottom it is quickly slowed down vertically, some of the PE then will be converted into thermal energy, but it'd be such a mess that it'd be hard to measure as there are so many varying frictions acting on it...

[Gareth56]

Out of interest, how could you do this in the lab?

[insane_alien]

well, firing a jet of water(to simulate it falling a few hundred feet) into a collection vessel with temperature sensors in the nozzle and vessel will do the trick.

 

the more insulation you wrap round it the better.

[Klaynos]

Best to do it in a vacuum as well

[Gareth56]

Wouldn't that temperature reading include some other form of energy (friction?) from the water hitting the sensor at a certain velocity rather than just relying on gravity as what occurs in a waterfall?

[insane_alien]

well the water is falling at a certain velocity if it is falling off a waterfall.

 

water travelling at 20m/s is water travelling at 20/ms whether it is off a waterfall, water jet or whatever. it still has the same energy and it'll still be converted to heat when it gets slowed down by friction.

[DJBruce]

I would think the amount of energy lost to heat would be minimal in comparison with the amount lost to sound and to Kinetic Energy.

[Klaynos]

I would think the amount of energy lost to heat would be minimal in comparison with the amount lost to sound and to Kinetic Energy.

 

I would too...

[Edtharan]

Wouldn't that temperature reading include some other form of energy (friction?)

Friction is not a form of energy. It is a process that allows kinetic energy to be turned into some other kind of energy (heat or sound).

 

But yes. The kinetic energy will not be all turned into heat energy, most of it would be turned into sound energy (waterfalls are quite loud), and it will also be used to do work (wear away the rocks, push water further down stream, displace the water near the waterfall as spray, and so forth).

[tvp45]

You might look at the attempt by Joule to do this. If I recall correctly, someone (probably Kelvin) later performed a lab demo of Joule's calculation.

[DrP]

As the water falls I reckon it would break up into droplets - thus I think that it would loose more thermal energy to the surounding air as it falls (like blowing on it to cool it - the surface area of the droplets would be alot greater than the flowing water). Also I think more would evapourate as it falls in drops (surface area and air movement again) as opposed to when it flows normally - thus more cooling. I do not know what the net change in temperature would be though.

 

 

PS - did you post this question on the chemforums site as well?

[swansont]

Heat capacity of water is a calorie per degree K cc, and a gram of water is going to gain 9.8e-3 J of kinetic energy per meter. Since it's 4.18 J (1 calorie) to raise the temperature 1 K, that's a paltry 2.3 milliKelvin per meter

 

 

If you could build it properly, just drop a bomb calorimeter. Falling water is falling water — it doesn't have to be a "waterfall" You'll want the mass of the water to be the dominant mass term, and a low-mass material that doesn't deform, so that the water's energy is what you're actually measuring.

[SkepticLance]

I appreciate that a lot of the energy will be lost to other forms, such as sound, accelerating water, accelerating air etc. However, that portion of the energy that is converted simply to heat should be measurable.

 

Seems to me that if you simply took an accurate temperature of the water at the top of the Niagara Falls, and another at the bottom of the falls, you should get a temperature increase. Sure, it is not the whole story, but it would be enough to demonstrate the principle.

[Pete]

I read somewhere that water has PE at the top of a hill (waterfall) which is mostly converted into heat by the time it reaches the bottom of the hill (waterfall).

It may seem trivial to mention this but it is imporant to understand if you want to understand thermal physics (so don't get the impression that I'm trying to be nitpicky).

 

Heat is defined as the any spontaneous slow of energy from one object to another, caused by a difference in temperature between two objects.

 

The temperature of water increases in this case because work is done on it. The first law of thermodynamics states

 

[math]\Delta U = Q + W[/math]

 

where [math]\Delta U[/math] is the increase in energy of a system, Q is the heat which flows into the system and W is the work done on the system. In the present case Q = 0 and W = work on water due to impact. This gives [math]\Delta U = W[/math] so the increase in thermal energy is due entirely to work done on the system (in this case the system is water).

 

Pete

[swansont]

But yes. The kinetic energy will not be all turned into heat energy, most of it would be turned into sound energy (waterfalls are quite loud), and it will also be used to do work (wear away the rocks, push water further down stream, displace the water near the waterfall as spray, and so forth).

 

Sound doesn't really carry off much energy. 90 dB — which is pretty loud — represents a milliWatt. There's a trivia bit about how yelling for ~8.5 years has the same energy as it takes to heat a cup of coffee.

 

So let's do a quick calculation: The splash of a liter of water over the course of a second, from a height of 1 meter, is almost 10 watts. This is probably less than 60 dB (a microWatt) if you're a meter away. Assume a point source. The sphere is 10^5 cm^2, so what fraction of that area do you detect with your ear? 1 cm^2? 10 cm^2 That means that the sound is less than 1% of the energy, and probably much less.

 

I appreciate that a lot of the energy will be lost to other forms, such as sound, accelerating water, accelerating air etc. However, that portion of the energy that is converted simply to heat should be measurable.

 

Seems to me that if you simply took an accurate temperature of the water at the top of the Niagara Falls, and another at the bottom of the falls, you should get a temperature increase. Sure, it is not the whole story, but it would be enough to demonstrate the principle.

 

The Horseshoe falls are about 50 m tall. That's a tenth of a degree or so, assuming all potential energy is converted into thermal energy.

[DrP]

So Swansont, does more get lost due to surroundings on a large fall if the water droplets breack up to increase SA? Like blowing on tea?

[swansont]

So Swansont, does more get lost due to surroundings on a large fall if the water droplets breack up to increase SA? Like blowing on tea?

 

That breakup probably comes at the expense of the water's kinetic energy, and you'll recover it when the water coalesces at the bottom.

[CaptainPanic]

It took me several days to figure out that we're talking about Potential Energy ("PE") and Kinertic Energy ("KE"). I re-read the 1st post a couple of times, and every time I just clicked the "back" button after 3 seconds, because I cannot be bothered to think about abbreviations. Is it really such an effort to write out a few words??? Comon, lazy bums, press those keys on your keyboard... we're just typing, it's not calligraphy.

 

People think it looks "scientific" to use abbreviations, I think it looks stupid, especially if you fail to explain them.

 

On topic:

Evaporation will cool the water. To my knowledge the falling doesn't heat up the body, it accelerates it. It's the friction that follows that heats the water. But since it comes to an almost complete stop at the base, we can say that it is converted to heat by friction almost completely. Sound is so marginal that I propose we neglect it. Evaporation is the only other parameter in the energy balance.

 

Back-of-the-envelope calculation:

Potential energy turns to kinetic energy, which turns to heat by friction:

Height of Niagra Falls: 52 m.

1 kg of water has m*g*h = 1*9.81*52 = 510.1 J of potential energy.

This is able to heat up the water by dT = Q / ( m * Cp ) = 510.1 / ( 1 * 4.2E3) = 0.12 degrees C... which is most certainly measurable.

 

Evaporation:

But, evaporation occurs. Let's look how much we need to completely negate this temperature increase (we'll need a couple of assumptions, because this is weather dependent):

 

Assuming a temperature of 25 deg C. The vapor pressure of water is then: 23.76 mmHg, or (in real SI units) 3126 Pa.

 

This means that 1 liter of air can contain n = PV/RT = 3126 (Pa) * 1E-3 (m3) / (8.3145 (J/kgK)* 298 (K) ) = 1.26E-3 mol of water, or 0.023 g of water (22.3E-6 kg). To evaporate this, it costs 51 J. So, to compensate the 510.1 J of kinetic energy, we need to saturate 9.93 liter of air.

 

This means that if the air is totally dry (0% water), and the dispersion of water is 1 liter of water in 10.93 liters total (9.93 liter of air + 1 liter water) and the air is totally saturated at the base of the waterfall, then the temperature effect it 0. This also assumes that the air is refreshed all the time. Each liter of water falling needs 9.93 liters of fresh air.

 

The volumetric flowrate of the waterfall is 5,720 m³/s (season dependent, I think). This means that the volumetric flowrate of (perfectly dry) air needs to be 56800 m3/s.

 

The surface area of the waterfall is harder to estimate, because there is more than 1 actual water fall. The "horse shoe falls" are 792 m wide and 53 m high (why is wikipedia contradicting itself here? Earlier it was 52 m high). Anyway, this means that the frontal surface area is 41976 m2. Assuming a frontal wind, the wind speed must me 1.4 m/s, which is totally realistic.

 

p.s. I was also assuming that the (cold) water doesn't cool down our nice and comfortable 25 deg C air.

 

So, it is possible that you measure actual cooling in stead of heating despite the drop in potential energy.

 

You can now grab your own calculators, and change some of my parameters to see the effect of higher AH, LT, LDoW on the FT.

[DrP]

That breakup probably comes at the expense of the water's kinetic energy, and you'll recover it when the water coalesces at the bottom.

 

Thanks. I'm still not 100% sure though swan, I reckon you'ed get more evapouration as it falls in droplets which would cause some cooling. CP's post above seems to suggest the same. (?)

[swansont]

Thanks. I'm still not 100% sure though swan, I reckon you'ed get more evapouration as it falls in droplets which would cause some cooling. CP's post above seems to suggest the same. (?)

 

I was not considering evaporation at all, just the issue you presented, which was surface area.

[DrP]

I was not considering evaporation at all, just the issue you presented, which was surface area.

 

A-ha, that's what I mean. If the SA is greater then more air is passing over the droplets as they fall (and faster) and, I figure at least, that there will be alot more evapouration than when the river is just flowing along normally. Hey, it may not be the case that it reduces the temperature by alot, but it must have some effect even if it is small (as CP showed). Again, I'm not sure which effect will be greater - the heating from friction or the cooling from falling. I would have thought both would be quite small?

[CaptainPanic]

As I showed, both are quite small. We're talking tenths of a degree temperature differences here. I can adapt parameters such that droplets cool, and also so that they warm.

 

I think I also showed that there are so many parameters which are not fixed (such as the weather: temperature and air humidity) that it's really difficult to make a prediction. Additionally, a lab experiment is already difficult to model:

 

A decent model would take into account:

• Drops heat from conversion of kinetic energy into heat

• Water evaporates from surface (droplet velocity (not constant) and droplet shape (function of velocity))

• Other droplets influence the speed of each other, and therefore the shape of the droplets which in turn influences the mass transfer

• Air and water are not of the same temperature, and even if they are at the top of the waterfall, the previously mentioned effects will change the temperatures of the particles, so that there will be a temperature gradient. This means there is also heat transfer.

• composition of surrounding air is not constant, which influences air density, and heat and especially mass transfer

• Evaporating water influences the droplet size, and that influences everything mentioned before

 

Input parameters:

• Temperature (both water at the top and air)

• Air pressure and humidity

• droplet size at the top (distribution, if you want to make life even harder)

 

Then you integrate the whole thing with respect to height, and pray.

 

I don't see how you can take things as landscape (rocks and such) into the model without spending weeks or even months or years on it. Therefore macroscopic air turbulence is not put into a model easily. If rocks influence wind, and create an updraft (countercurrent system) or a downdraft (co-current), that's a massive difference.

 

Therefore I conclude that I think it's easier to take an airplane and go there to measure it, rather than calculate it with a really decent calculation. Might save you a lot of time anyway.