# Does a Hairdryer blow harder with heat

1. Jan 23, 2014

### Jedi_Sawyer

Does a Hairdryer blow harder with heat on than it does with heat off ? and why?

My feeling is the blowing force, or pressure that the air exhaust will exert on a surface that opposes it will stay the same whether the hairdryer is blowing hot or cold air. That the difference will be that the hot blown air will not be as dense as cold blown air. That would be because the heater which is in the air stream will be slowing down the air coming at it thereby making the air stream being dammed up a little, or raising the air pressure. Not as much air will get past the damn but the air that does get past the heater will have more Kinetic Energy. So since the heat radiates in all direction it lowers the mass/sec output of the hairdryer but the mass that comes out is more energetic.

2. Jan 24, 2014

### Staff: Mentor

Hot air does not have more kinetic energy than cold air. Its temperature is simply higher.
Are you thinking that the heating coil isn't in the way of the airstream when the coil is off?

3. Jan 24, 2014

### Staff: Mentor

Probably true. You get about twice as much air out as is sucked in when the drier is switched to HEAT.

Have you tried measuring the force, to see whether the air blast pushes harder with heating?

4. Jan 24, 2014

### Jedi_Sawyer

Hot does mean more kinetic energy

Hot air does mean more kinetic energy than cold air. Heat is energy and the PV = nrT. The only way to have the same blowing force out of a heat gun for cold air as for warm air, is for the cold air to be more dense.

5. Jan 24, 2014

### Staff: Mentor

If there is any difference to the amount of air blown through the dryer I'd guess it's either because of the power requirements for the heating elements or because it's designed to.

I don't see how the ideal gas law applies here. You're talking about the amount of air getting blown out of a hair dryer, not the kinetic energy of the molecules of the gas.

Honestly I can barely understand your original post. You talk about the heating element being in the way, yet I don't see how that makes any difference between the cold and hot air. I also don't see how radiating heat affects the airstream.

6. Jan 24, 2014

### 256bits

Is the output air that hot? I would have thought about 20 to 30% more.

In any case with enough hair dryers and a long enough cord, one should then be able to propel a mini car mimicking propulsion as seen on your regular passenger jet aircraft.

7. Jan 24, 2014

### sophiecentaur

The speed of the air will probably be greater after it leaves the machine if it has been heated by the element by, say 40K. Its volume will have increased by a factor of 340/300 (gas laws) or 13%. The nozzle diameter is the same so this will mean an increase in speed of 13%, or an increase of 28% in Kinetic Energy. This is not a massive difference but would be detectable.
I have noticed a (subjective) air speed increase when I have used a fan heater / hair dryer. It would be easy enough to register an increase by blowing it against a suspended, light object.
Also, the note of the motor always drops when the heater is turned on - whilst this could, in part, be due to a resistive voltage drop in the power lead, I would suggest it's more likely to be because of the extra power being used to supply the extra KE of the blast. (The fan does more work)

8. Jan 24, 2014

### kb7wg

Off. I believe the voltage drop across the blower motor would effect air volume more than air density. I would assume the heating element draws several times the current of the blower motor. The voltage drop across the power cord will increase and voltage to motor will decrease. The blower will slow down.

9. Jan 24, 2014

### jbriggs444

This is almost certainly incorrect. If you increase the pressure load on a fan, the motor does not slow down. It speeds up. This is because it passes a lower volume of air. You can hear this effect very clearly if you block the end of your vacuum cleaner hose.

10. Jan 24, 2014

### sophiecentaur

Density = Mass/Volume
How could you affect one without affecting the other, if the mass is constant?

It would not be difficult to measure any voltage change. I might just do that.
I think my 13% figure for 40K temperature increase must be correct so the motor will certainly have a greater mechanical load on it.

Whenever you block the input flow of air to those shaded pole motors, they always speed up noticeably, so they are very load sensitive (i.e. with the inlet blocked, they can't be shifting as much air. What they actually do is a bit counter intuitive, I know, but you should try it.

11. Jan 24, 2014

### sophiecentaur

See my post above. The air flow through fan is different from a normal 'braking load'. If you take a fan motor and give it an extra mechanical load, it will slow down (as do all motors) and not speed up. But, by starving a fan of air to work on, it cannot do as much work at its normal speed and so it speeds up to a new equilibrium situation. You will notice that you get no rise in pitch of you try to block the output of a fan because there is air available for the fan to work on.
Not intuitive, I know, but an experiment will prove it to be right.

12. Jan 24, 2014

### jbriggs444

I understand that air flow thruogh a fan is different from a normal braking load. That is rather the point. Starving a fan of input air (decreasing the pressure of its input) is identical in this respect to increasing the pressure of its output. The relevant effect is to increase the pressure gradient against which the fan works.

As we both agree, if you increase this pressure gradient, you reduce the air flow and increase the fan blade speed.

Accordingly, I predict that you will get an increase in pitch when you block the output.

13. Jan 24, 2014

### sophiecentaur

If it were as simple as that, the greater the pressure difference, the more work the fan would do and it would slow down - as it will with an applied braking load. So it cannot be as simple as that. There is some sophisticated fluid flow going on and it's not like a simple piston pump, working between two reservoirs at different pressures.
My point is that the air which gets to the fan needs to be flowing past it at an appropriate speed and direction for the fan to deliver the appropriate amount of energy to it. When you block the inlet, air circulates round, from exit side to inlet side because new air cannot get there. I guess this will mean that the air will follow a helical / circular path, going faster than it would usually. Under normal circumstances, the fresh air would be arriving parallel to the axis and would need to be accelerated by the fan, taking more power.

In any case, I would need a lot of convincing that the motor actually increases its speed under a load.

And my point about the greater speed for hot air exiting has to be correct, I think. There's just more volume passing through per second.

I blocked both input and output of our vacuum cleaner and the inlet block made the predicted speed increase. Not so, the outlet. This would be because the work situation is different and the outlet air flow is actually directed by the blades but the flow to the back of the blades is just due to a general pressure difference. There's no reason why the situation would be symmetrical.

14. Jan 24, 2014

### jbriggs444

You are arguing with a straw man.

The minimum power requirements are given by pressure gradient multiplied by flow rate. If you reduce the flow rate by increasing the pressure gradient, the effect on the flow rate may result in either an increase or a decrease in power.

We agree on this.

Again, you are arguing with a straw man. We are dealing with a shrouded fan. A shrouded fan, as you say, does not behave like a piston pump.

We agree on this.

Yes. If the shroud on the fan is not tight, you get ambient pressure outlet air leaking back to the inlet.

For a given pressure gradient, this bypass path will work to increase fan load and reduce motor speed. The tighter the shroud, the less bypass, the less resulting load and the higher the motor speed until, in the limit, the fan is simply free-wheeling, spinning some air in place with essentially zero power requirement.

This fact works against the point that I thought you were trying to make. So perhaps I have misunderstood what you were trying to explain.

Another straw man. The motor increases its speed as its load decreases. The load presented to the motor by the fan decreases as the air flow rate decreases. The air flow rate decreases when you put your hand over the inlet.

We agree on this.

I claim that the flow rate decreases because of the increased pressure gradient that results from your hand. Possibly you disagree with that.

For an incompressible fluid (and air is largely incompressible at the pressure ranges we are considering) the situation is indeed symmetrical. Can you explain better what asymmetries you see in light of this apparent symmetry?

15. Jan 24, 2014

### kb7wg

Air temperature and air density have nothing to do with this question. The heating element is after the fan blower. The air that the fan is moving is at constant temp. and pressure. The only thing that can effect fan speed is motor voltage. Anyone whose has worked around electric motors very long can tell. When you turn high heat on.....you can hear the speed go down.

16. Jan 24, 2014

### Jedi_Sawyer

I agree that the hot air is traveling faster from a hair dryer than cold air for any given fan setting. The part about more volume/sec passing through is mysterious as to what that means. I think he is saying that the hair dryer should blow harder when it is pushing hot air than cold air and I don't think it does. The experimental evidence is that it does not blow harder see You Tube video

I believe that this is a correct analysis for the vacuum situation. However the hairdryer changing pitch is different. I think that the change in pitch adding heat is due to two effects. First there is line loss and it is not much. I measured 119.5VAC no heat to 117.9VAC with heat. The heater fan itself is DC and is 12 or 24 V depending on speed setting and it would also be subject to the AC changing unless they do expensive regulation I am sure they don't. The other more important effect is that the load increased due to the air damming effect of heat being added I suspetct

Last edited by a moderator: Sep 25, 2014
17. Jan 24, 2014

### sophiecentaur

Yes. The position of the fan vs the element is not too relevant. The air expands as it goes over the element and that will increase the pressure that the fan has to work into, one way or another - hence the load.

Perhaps we should be talking in terms of momentum when considering the effect in the movie. The same mass of air is being ejected per second and the fan could be imparting the same amount of momentum per second (well - I think so) so the same amount of momentum is expended on the sheet in the movie per second. That sort of explains that result.
However, when you blow a hairdryer into your face and turn the heating on, it 'feels' stronger. (Whatever that means in objective terms.)

Last edited by a moderator: Sep 25, 2014
18. Jan 24, 2014

### nsaspook

Yes, the device is usually designed to increase the fan speed with increasing heat to prevent over-temperature.

A typical example:
http://img.docstoccdn.com/thumb/orig/41771342.png

19. Jan 25, 2014

### sophiecentaur

For many decades, blown heaters have had independent switching. The only design feature has been to ensure the fan is on whilst an element is in operation. That attachment does not describe common practice.

20. Jan 25, 2014

### 256bits

Not the usual design, maybe for a higher priced model
Many and most models have temperature control separate from speed control.

( I see Sophiecentaur has already explained common practice )