# Why isn't wind hot?

1. May 19, 2010

I'm a high school science student/aficionado (since it's sort of hard to specialize in anything in high school) and I have some relatively elementary questions:

I learned in chemistry class that temperature is the average kinetic energy of the particles in any given sample of gas, so basically, shouldn't wind be hotter than still air? Is the majority of an atom's or molecule's kinetic energy simply vibration instead of movement along any specific vector, or are their different ways to measure temperature besides average kinetic energy? I am talking more about measured temperature than humanly perceived temperature.

My other question has to do with mass. In class, when you measure something with a balance or scale, the base unit of measurement is the gram. It is my understanding that what we're actually measuring is g*m/s^2, or the force of gravity on that particular mass. It is also my understanding that mass is sort of a measurement of how much inertia effects any particular body. I'm so lost on this one I don't really even know how to phrase a question, other than what's happening with weight measurements? When we deal with outer space e.g a moon landing do we have to measure these things in a different fashion?

If it helps with the explanations, I have an understanding of how ideal gases function and can understand algebra (elementary, of course) and maybe even (a little) bit of calculus. Mathematical explanations not necessarily preferred but appreciated.

2. May 19, 2010

### Char. Limit

I can answer the scale question for you:

Yes, you are correct that a "gram" scale actually measures the millinewtons (g-m/s^2) of force the object exerts by weight. However, since the value of gravitational acceleration on Earth is known, I feel it
likely (I haven't checked this) that the scale then divides the millinewton reading by g to gain the mass. On the moon, a scale would need a different value for g, but the principle would remain the same.

I believe that this is how you get a gram reading, rather than a millinewton reading.

(From a fellow high school aficionado)

3. May 19, 2010

### xcvxcvvc

if you use a scale that has counterweights, you would not need to redesign anything for it to work on different planets.

4. May 19, 2010

### Mute

The problem with the temperature of wind is that wind is not really in equilibrium, so you may not be able to sensibly define a temperature for it. Quantities like temperature and pressure typically require equilibrium (or at least a steady state) to be sensibly defined.

However, consider: When calculating the temperature of a gas you do not take into account the overall centre of mass motion of the gas. That is, temperature is defined with respect to the kinetic energy of the gas, but the velocity of the gas you use to calculate the kinetic energy is the velocity of a particle with respect to the average velocity of the whole gas. So, if you have two identical boxes of gas except that one is moving at a constant velocity, both will have the same temperature.

You might also consider a box of gas which you poke a hole into: the gas that escapes out of the hole will tend to be the particles with the higher velocities, so the escaping gas is in some sense hotter than the gas that remains in the box (even though it is no longer really in equilibrium).

You could weigh something on the moon the same way you weigh something on Earth, except the measurement would give you different weights because g is different on the moon and Earth. If you can calculate/measure g for the moon, then you can calibrate the scale on the moon so that it gives the correct mass when you put something on the moon-scale.

5. May 19, 2010

### Acut

That's a interesting question.
Temperature is associated with the movement of the small molecules of air, and there's no relation with the motion as a whole of air. Wind, on the other hand, represents a movement of air as a whole (air molecules aren't all moving to the same direction, but on average, there's some tendency for them to go to where wind is flowing), and is unrelated with temperature.

You don't get a fever when you're travelling by car, do you?

Also notice that air flows from higher pressure to lower pressure. Pressure reduction cools air.

6. May 19, 2010

thanks for the replies
looks like my mass misunderstanding had less to do with my conceptual understanding of physics and more to do with my lack of knowledge of weighing techniques

Last edited: May 19, 2010
7. May 19, 2010

### sophiecentaur

The average speed of air molecules in air at room temperature is around 400m/s. 80km per hour wind speed corresponds to around 20m/s. So the contribution to KE in even a high wind is only about 5%.
If you measured the temperature of the air in a wind or in flat calm (with a thermometer), the transfer of energy from the air to the thermometer by random collisions of air molecules due to thermal motion would be much more than the effect of the bulk movement of air past it. But I imagine the reading on the thermometer would, in fact, be a finite amount higher for moving air than for stationary air.
Whether or not that means that the temperature is higher . . .

8. May 19, 2010

### D H

Staff Emeritus
Yes and no.

The primary effect of a wind is to greatly increase heat flow. If it's hot outside (hotter than body temperature) a wind will feel hot. But if its cold outside a wind makes the air feel much, much colder. That's because the wind mostly blows past you. If you stopped the wind entirely you would see increase the temperature. You are asking about the difference between stagnation and static temperature. Place a sleek, aerodynamically shaped thermometer in a breeze and it will measure static temperature. Some aircraft have specialized temperature sensors that measure stagnation temperature.

A balance scale measures mass. A balance scale will register the same reading for a given quantity of mass whether used on the Earth or on the Moon. A spring scale measures apparent weight. Apparent weight is not the force of gravity. The apparent weight of some object is the sum of all forces acting on the object except for gravity. Take a spring scale to the Moon and it will register about 1/6 of the value that it registered on the Earth. Take a spring scale to the Space Station and it will register nothing, even though the force of gravity is about 90% of that on the surface of the Earth.

9. May 19, 2010

### Jimmy Snyder

I think the reason that wind feels cool is that it evaporates the perspiration on your skin. The state change from water to vapor takes away heat.

10. May 19, 2010

### D H

Staff Emeritus
To an extent. Come down to the Gulf coast in a month or so. There is no perspiration when the relative humidity and the temperature are both in the upper 90s. All a wind does is make one feel even more uncomfortable.

11. May 19, 2010

### Studiot

Firstly the wind is air from someplace else.
So it may be warmer or colder than the natural temperature where you are standing.

The Chinook is an example of warm(er) air wind.
The Mistral is an example of colder air wind.

In addition to this airflow past your body has a cooling effect for two reasons.

Firstly it accelerates the evaporation of moisture from your body so you loose the latent heat of the evaporate. This happens because the air flow clears the evaporating water molecules more quickly, lowering the vapour pressure around you. You may have noticed with hot air hand dryers the ones with the most blow, rather than with the most heat, are the most effective.
Secondly if your body is warmer than the air there will be natural heat transfer on contact. Again the air flow will remove the warmer air molecules more quickly and present a steady supply of new colder ones. You may have heard weather reporters talking about wind chill factor.

Finally there is the question of the body's perception of temperature. Humans actually perceives temperature partly as an actual temperature and partly as a rate of heat gain or loss. This is why plastic at the same temperature as metal 'feels' warmer.

12. May 19, 2010

### LostConjugate

The reason wind feels cool when you turn on a fan is as follows.

You are hot! Lets say the air in your room is 75 degrees. Your skin temp will be about 90 degrees because of the 100watt furnace inside you. So the still air surrounding your skin is much hotter than 75 degrees. When you turn on a fan and put your skin in front of it you are constantly replacing this hot air with the 75 degree room temp air.

Go into a room with 150 degree air and a fan doesn't make you feel any cooler, since your body is not heating up the air around you any longer.

Physically though, you could measure air with a velocity to be slightly cooler because the pressure would be reduced. For the same reason a water hose losses pressure when the water has increased velocity.

13. May 19, 2010

### ehilge

I think explaining the difference between mass and weight might help you out a little, or confuse you more, I dunno we'll see. A gram is not a measurement of weight. Weight is a force and is a vector (has direction). A gram is a measurement of mass, which is a scaler (has no direction). Things can get confusing in English units because the pound is used as both a unit of force and mass.
Something has the same mass in space as it does on earth, but not the same force, or weight. For example, if you're isolated in space, you still have a mass of 70kg or so, but your acceleration is 0 m/s^2, so by F=m*a you weigh 0 Newtons (note that a Newton is a unit of force but a gram is not). But if you're on earth, your acceleration is 9.8 so your weight is technically 9.8*70 newtons.
Also, recall that gravity is not a force, it is an acceleration. You are using the force due to gravity, not the force of gravity. Words like mass and force are many times used interchangeably but with respect to physics, its important to see the difference.

So to answer your questions, if we measure something in kg, we are actually measuring F/g. F would be the force that the scale 'feels'. Check the units and it works out. So if you're on the moon and measured something in kg, both F and g are different and the mass works out to be the same.

I hope that helps, if what I said confuses you more, let me know, I can try to explain everything better.

edit: I thought of something else that might help. Mass can be defined as the amount of inertia an object has as you said. Inertia is the tendency of an object to resist changes in motion. For this reason, it takes more force to move an object that has a large mass than one with a small mass irregardless of gravity.

14. May 19, 2010

### CanisMajor

well if you make the assumption that when light travels through air no energy is absorbed because its a transparent gas, and taking into account the ideal gas law; PV = mRT then when a gas is heated up in a specific location due to convection, miles away, the gas is hotter and expands but the expansion is reprocussed as a wave, much like electrons in a wire, then the gust of wind you experience is from an expansion outside of your local control volume and so it doesnt appear to be of any specific heat difference.

15. May 21, 2010

### Gerenuk

The most general definition of temperature is not mean kinetic energy! It has to do with macrostates and microstates and it's not that easy to define that for a moving gas. In fact you could define temperature for just anything like a set of marbles, but that's another story.
Anyway, for stationary gases the mean kinetic energy happens to be proportional to the temperature.
While it's not that easy to define temperature for a moving gas (but it can be done!), you can still think about what happens when a moving gas is finally caught in a container and it's temperature is measured. We assume that the total energy of the gas is constant and merely the center of mass kinetic energy transforms into random motion. In that gas the change of temperature for nitrogen molecules (main part of air) is
$$\Delta T/K=\frac{(v/ms^{-1})^2}{926}$$
So indeed wind is hotter (when caught), but not much.

Yes, the scale should rather say Newtons not Kilogrammes. The conversion is done by knowledge about the planets you are on.

There could be two different masses since inertia and gravity are different effects. I actually have to think for myself what it means. By doing experiments you can determine force ratios. But this can be done separately for F=ma (maybe use electrostatics?) and also F=Gm1 m2/r^2. I'll get back when I have an idea.
Somehow, it's an important inertia mass and gravitational mass are the same.

Last edited: May 21, 2010
16. May 21, 2010

### sophiecentaur

A true 'balance' compares (the weight of) one mass with (the weight of) a standard mass. It is independent of g.
A 'spring balance' is really a direct force measurer so it scale should, strictly, be in Newtons and leave you to do the F = mg sums appropriate to where you actually do the measurement (Equator / Poles / on the Moon).

17. May 21, 2010

### Staff: Mentor

In my line of work, the heat from wind (fans) is non-trivial.

 When wind washes over you, it removes heat from you via convection, but drag (both pressure and friction) from your body slows the wind and converts the uniform motion energy into heat energy. For your body's interaction with the air, this amount is trivial, for industrial HVAC it is not.

For every 1" w.g. of static pressure drop in a moving air mass, there is a 0.11 F temperature increase (this doesn't include lost fan energy due to inefficiency). For your home's HVAC system, there may only be 0.5-1" of static pressure, but commercial/industrial HVAC systems start at about 4" of static pressure and for specialized systems can be as high as 8". With a 20F cooling delta-T being common, you could end up with 1F of a temperature rise, or 5% (not quite that simple...). Not a huge fraction, but it is necessary to take it into consideration.

Last edited: May 21, 2010
18. May 21, 2010

### fawk3s

I've been thinking about this in the back of my mind for some time now and just cant reach the solution. I even checked the wikipedia which says that if an object has a centrifugal acceleration, the apparent weight isnt equal to the actual weight.
But I dont understand why?
This is the way I understand it:
[PLAIN]http://img405.imageshack.us/img405/1631/centrifugal.png [Broken]

There is a centrifugal acceleration which changes the speed vectors direction. But as Newtons Third law states, the forces which are created are equal and opposite. So there must be a force pulling the object outward. I dont exactly know what it is but I'd guess inertia.
So say you measure an objects weight with a spring scale. What happens? Why does it show 0? I just cant wrap my mind around it.
Say you are holding the spring scale in your hand, standing at that point. Shouldnt it act like in a normal gravity field?

Thanks if anyone bothers to explain

Last edited by a moderator: May 4, 2017
19. May 21, 2010

### sophiecentaur

I don't know what you're implying here. Are you suggesting that a spring balance wouldn't work in a centrifuge? Having ridden on 'The Rotor' in a fairground, I can confirm that all the familiar forces act - but pulling / pushing you against the wall.

Even in an orbiting (rigid) spacecraft, there is a difference in net gravity / acceleration forces when you are not at the same distance of the CM of the craft. This is because the CM experiences perfectly balanced forces but, in parts of the ship which are further out, the G is a bit less further in the G is a bit more - the radii of rotation around the Earth are also different. Hence the expression 'microgravity' when in orbit.

And there's a small matter of the Principle of Equivalence which says that you can't distinguish a gravitational force from an acceleration force. Of course, if you try to move around in a rotating frame, you will experience Coriolis force so you you could detect the difference in circumstances.

Last edited: May 21, 2010
20. May 21, 2010

### D H

Staff Emeritus
Let's look at this from two perspectives: An inertial frame and a frame rotating with the planet. To make things a bit simpler, I'll consider a person standing on a scale on the equator and ignore things like buoyancy from the atmosphere.

Inertial explanation
There is no centrifugal acceleration in an inertial frame. The centrifugal force is a fictitious force. From the perspective of an inertial observer, the person standing on the scale is undergoing uniform circular motion about the planet's rotation axis. With a rotation rate of ω=2π/day, the centripetal acceleration is anet2Re. From a purely kinetics point of view, the net force on the person of mass m must be Fnet=mω2Re.

From a dynamics point of view, there are two forces acting on the person (I'm ignoring buoyancy here): The downward force due to gravitation, and the upward force exerted by the scale. The net force on the person is the vector sum of these two forces. We already know the net force is not zero. Thus the upward force exerted by the scale is not equal but opposite to the downward force of gravity. The magnitude of the force exerted by the scale is instead

$$F_{\text{scale}} = mg - m\omega^2 R_e$$

Planet-fixed explanation
From the perspective of an observer fixed with respect to the Earth, the person standing on the scale is not moving. The net apparent acceleration is zero. In this frame, there are three apparent forces acting on the person: The downward force due to gravitation, the upward force exerted by the scale, and the upward force due to the fictitious centrifugal force. These three forces must sum to zero, once again leading to

$$F_{\text{scale}} = mg - m\omega^2 R_e$$

21. May 21, 2010

### cstoos

The wind problem boils down to velocity. As velocity changes the convection heat transfer coefficient "h" changes. If the wind temperature is lower than your body temperature it will cool you more as velocity increases. If the temperature is higher than your body temperature it will heat you more as velocity increases.

That is why in heat exchangers flow is a major factor.

22. May 21, 2010

### Studiot

23. May 21, 2010

### fawk3s

Ok, when I read your post, it seemed to make sense. But I guess Im that kind of a person who needs real simple explanation to understand the situation fully. So Im going to post my perspective on the thing, which is probably probably wrong, but I hope you guide me to the right direction.
Sorry if Im just plain dull and didnt understand your explanation.

[PLAIN]http://img691.imageshack.us/img691/1631/centrifugal.png [Broken]

Ok, so there's me in an awesome circular space station, holding my spring scale with a random mass attached to it and the whole thing is happening from my perspective.
The black arrow represents the circular velocity of the space station, the red one the centripetal acceleration pointed to the center of the circular motion, and the green arrow an opposite force to it. (Is this logic faulty?)

So there's an artificial gravity pulling me, the scale and the mass to the outer wall. As im holding the scale in my hand, there's a constant force on it, since the radius from the centerpoint to it doesnt change.
But now, shouldnt the mass and a spring reach an equilibrium since the mass is pulled towards the outer wall?

fawk3s

Last edited by a moderator: May 4, 2017
24. May 21, 2010

### cstoos

I don't have my heat transfer or thermo books in my office, but essentially the heat transfer coefficient is based on mass flow. The higher the flow mass flow the higher the velocity (given no pressure difference).

So, the higher the heat transfer coefficient the faster the substance will gain or lose heat.

The flow thing is really an engineering problem. Ask it in the Mech Engr forum and you will get better answers than my vague recollections.

25. May 21, 2010

### sophiecentaur

The point here is that the wind is moving a lot slower than the mean velocity of the air molecules (I discussed the relative values earlier). The rate of energy transfer between the air and the object is proportional to the difference in temperature (difference in average KE). If new air is constantly brought next to the object the rate of transfer will be higher than if air stays stationary around the object. Whether the air is hotter or colder, the rate is increased by movement of the air - this shows that the effect of the bulk movement / KE is very much smaller than the effect of the temperature difference (the internal random KE situation) - so the air speed doesn't contribute to the temperature difference.
If this weren't the case then things would heat up quicker than they cool down for a given magnitude of temperature difference.