# Question about rapidly cooling air, and the resulting density

1. Jul 7, 2008

### JasonWilliam

Hey all. A question has come up regarding the rapid cooling of air, and if that cooled air is "denser" as result. On the surface, this seems trivial to me, but it quickly exceeded my ability to answer (not a physicist, just a hobbyist question asker and answerer at best).

I've tried to draw a picture that summarizes the situation:
http://i193.photobucket.com/albums/z260/JasonWilliamPics/Misc Pics/Junk Drawer/density_question.jpg

(This application is in regards to using water/meth to cool incoming air to an engine, to thus make the engine produce more power on a hot day).

On the right of the intake tube is ambient "air". For simplicity sake, lets say we're at sea level on a typical summer day. Track temps are 120degrees (not uncommon) and humidity is low at around 40%.

On the left of the intake tube is the engine. Its sucking air thru the tube. Some folks have introduced a delivery system into the middle of that tube that sprays a fine mist of 50% water and 50% methanol. They report they see the 120deg air cooled to around 70deg at the exit of the tube. Typically, that's a very good thing.

But... it being a very good thing is based on the assumption that cold air = dense air. So my question is, is that cold air really dense air too? Or is it that the pressure in the tube has changed and the air is no more dense than it was at 120deg?

Keep in mind the air thru the tube is being sucked at a (guessing here) speed of 40mph. So its moving very quickly during this cooling process.

I can make guesses at this all day long. For example:
No, the air is no more dense than it was before. The pressure in the tube dropped.

Yes, the air is denser. But the process of cooling sucked air at an accelerated rate from both down and upstream sources, creating a vacuum.

Ect ect ect...

I just don't have the background or knowledge needed to answer. Do any of you?

Last edited: Jul 7, 2008
2. Jul 7, 2008

### DaveC426913

The relevant question is: nevermind temp. or pressure or density or what-have-you, is more air delivered to the pump?

3. Jul 7, 2008

### JasonWilliam

Yes exactly. That is the overhead question, for sure.

But if we step under that umbrella, we see the method used to try to deliver that "more air" is a cooling based method. Once we establish that a cooling method is being tried, and we accept that its not reeeallly cold air we're after... we're after its property of being more dense in normal situations (again "more air"), my question becomes the heart of the matter I think.

4. Jul 7, 2008

### Andy Resnick

I don't see how "more air" can be delivered; conservation of mass means that the amount of air out is equal to the amount of air in. Is there a blower/compressor that you are not showing? I assume there's other components- adding water would seem to be a bad idea.

5. Jul 7, 2008

### JasonWilliam

No Sir. No blower. This is a debate regarding the validity and benefit of adding a water/meth kit to a naturally aspirated motor, NOT a forced air motor.

You can see my skepticism.

As for the conservation of mass... yes of course you're correct. However, if there is mass to draw from on either end of the tube, would the system tend to change pressure at the point of cooling or would it tend to draw from those sources in order to change density at that same point? And how does the stream moving at 40mph affect things?

Not so clear cut as I might like...

PS: As for the water, used in this way its been shown to be a benefit in FA applications. It actually helps reduce knock since it cannot combust, thus raising your octane rating, some say.

6. Jul 7, 2008

### nucleus

First your diagram is wrong. You cannot have zero speed on one end of a tube and 40 mph at the other end. It is the difference in pressure that creates flow – in this case you have a negative pressure created by the pistons and atmospheric pressure at the other end (ignoring ram effect). Therefore the air would flow into engine.

What you are doing with the water and meth is called environmental cooling. It takes the latent heat of evaporation to cool the charge and it increases the density of the charge. It helps prevent detonation (knock) but not by changing octane rating. Fuel also cools the charge in a carbureted engine.

7. Jul 7, 2008

### JasonWilliam

Hmm... I don't disagree that my diagram is entirely simplistic, but I don't think its wrong. The point was to show that there is no ram effect on the right; its just open to ambient.

Can you please explain how the simple act of cooling air (regardless of the method) increases its density? What if I sealed a container and then cooled the air? No density change there...

Or are you saying something else and I'm not understanding?

8. Jul 7, 2008

### nucleus

All you have to do is google density of air and find that air density changes with temperature.

9. Jul 7, 2008

### JasonWilliam

Unless pressure is allowed to change. Again, I think you need to consider the sealed container example.

10. Jul 7, 2008

### Ignea_unda

Yes, the sealed example creates a new problem.

However look at this with PV=nRT. If you you lower the temperature of a gas (air in this case) you can see that either pressure (and/or) volume decrease for the same number of moles of air you have. If the pressure changes then you are creating a difference in pressure that will equal out by the greater pressure at the intake versus the pressure after the cooling system--this will decrease the volume. As the volume decreases for the same ammount of matter then you have increased density. With increased density you would be delivering "more air" to the engine than before.

This system also allows for more expansion in the combustion chamber as the cooler air is more dense and expands more with the combustion and change of temperature lending more power per stroke.

11. Jul 7, 2008

### JasonWilliam

So you're asserting that cooling air with the introduction of no more matter (moles) changes the volume of that air? You must be allowing pressure to stay a constant then? And thus so changes the density? Hrm...if you have time, can you show some quick calculations that prove this change in density, given the variables in my chart? I'd like to see that.

As for your second point, you're referring to forced air. Unfortunately in naturally aspirated applications, the water and meth have displaced available o2, so there's actually less to burn. In other words, you start running pig rich.

This can get really deep. I'm trying to keep it simple and focus on the "what happens when" part, since there's a great need to understand FA vs NA and how a motor responds to either with and without the introduction of this mix, when trying to dig deeper.

12. Jul 7, 2008

### brewnog

Interesting. The key bit is that more matter is introduced.

In truth, what happens after the point of water injection is a combination of increased density and lower pressure. The increased density part is what the engine sees; a greater amount of air can be admitted per stroke, so more fuel can be burnt. The lower pressure sets up a greater differential between the atmosphere and the port; thus allowing more air to be aspirated.

In a sense, a naturally aspirated engine is "forced aspiration"; it's just atmospheric pressure forcing air into the inlet tract.

Although it's an interesting thought experiment, the "sealed container" example isn't appropriate. Since we are dealing with an (essentially) continuous process, the mass of air in the system is indeed subject to change depending on conditions within the system.

Having said that, I believe that in a naturally aspirated engine, the only tangible benefit from water injection is a decrease in charge temperature, thus lowering NOx emissions and widening the knock margin.

13. Jul 7, 2008

### nucleus

The formula for horsepower for an engine is PLANK/33000
Where P = break mean effective pressure
L = length of stroke
A = area
N = speed (rpm)
K = number of cylinders
This gives volume equal to length times area, which you cannot change easily. In any given engine at fixed RPM LANK will be constant. So that gives the only thing you can change at any one moment is P.
By definition: Density is a measure of how much mass is contained in a given unit volume (density = mass/volume) and pressure = Force/Area, therefore if you have a fixed area and add more mass you will have more pressure and hence more HP, everything else being constant..

14. Jul 7, 2008

### DaveC426913

It seems to me though, that a factor being overlooked is that, if the volume drops, lowering pressure, that will cause a pressure differential from BOTH ends of the tube, i.e. including the end of the chamber where the air pump is - it's not like the parital vacuum only works one way. So the net effect is a drop in flow into the air pump.

15. Jul 7, 2008

### JasonWilliam

brew: I follow that thinking perfectly. If we grant that there is a density change, there almost certainly is a larger pressure change. And we really ought to see what that pressure change means to the system as a whole. Great point!

Dave: you're spot on. I brought this up elsewhere and it wasn't addressed. The pressure drop just cant be ignored on either end.

I'm pretty well satisfied at this point guys. I'm willing to grant that there might be a density change. But if I do, the real question becomes what is the pressure change that is sure to be more dramatic to the system, doing to the system. THAT I think is a matter for experimentation.

Thank you for all your help. Of course if anyone else has anything to add, please do!

16. Jul 7, 2008

### Ignea_unda

I agree experimentation is key at this stage. Try a setup with an air pump as a car intake would have and measure the airflow after the air pump, then set up a similar setup with the air cooling system. This will determine if the net flow of air is greater or diminished or the same. If you are just concerned on the effects on a car, try to find an analysis of an engine with and without the cooling system (with no other adjustments to the engine).

Dave: I see where you are coming from, but would the pressure drop act as a ram to draw more air from the outside equalizing the pressure? I don't know. I'm wondering if the one side being sealed by the pump would not have the pressure loss that the intake side would. If you could explain this a little further maybe I would understand more. Maybe I am missing something in your explanation. Thanks!

Last edited: Jul 7, 2008
17. Jul 7, 2008

### nucleus

What’s he talking is not new. Engineers have known about it since the 1920’s. It’s call water injection and anti-detonation injection and was used during the Second World War in piston engine aircraft.

A goggle search comes up 41000000+ hits. It has been argued to death on several forums for years.

If the OP wants to find out how it works I suggest reading the NACA reports and SAE Transcripts from the 1930’s and 40’s and reading work by engine experts like Taylor, Stone etc.

18. Jul 7, 2008

### JasonWilliam

Nucleus... you're referencing studies and transcripts that discuss the benefits in a FORCED AIR system. My question is regarding a NATURALLY ASPIRATED system. So they simply do not apply.

I thank you for your attempted help though. You're right, there's a TON of stuff on this method. Unfortunately for me, that 'stuff' usually ends with a line saying something like "The usefulness of this in naturally aspirated applications is the subject of great debate". and the document ends.

19. Jul 7, 2008

### brewnog

Nucleus, the purpose of water injection in early aero engines was to cool the charge and (as you say) inhibit detonation. Charge density was not a key objective (unlike the use of supercharging).

20. Jul 9, 2008

### GT1

This is what I remember:
The mass inside the cylinder is m=$$\rho$$*V, the mass is proportional to the power of the engine. You can increase rho by lowering the temperature or by increasing the pressure. There are few methods for increasing the pressure: Supercharging, Inlet resonance, Comprex, Sendyka etc.
The method I know for lowering the temperature is Water injection, which increases the density. by spraying 2%-3% of water you can increase the power of the engine by 10%-15%.

21. Jul 9, 2008

### xArcherx

Looks interesting, there was a lot to read so sorry if this repeats with anyone. I'm gonna have a go....

So you have the tube with one end open to atmosphere and the other connected to the inlet of a pump. The pump turns on and displaces, say 100cc (1m^3) of air per second. The 100cc of air has a mass of say 1.2kg (dry air) at 20 deg C. When you cool the air as it comes in, the pump will still only displace 100cc of air per second (positive displacement pump). However the density of the air will increase. The only way to get an increase in density and displace the same volume of air is if more air (molecules) occupied the volume being displaced. This will cause an increase in mass of the same volume of air. If the density of air is directly proportional to temperature keeping everything else the same (which I think it is) then cutting the temp in half should result in doubling the mass, keeping the volume the same.

So in order for the pump to displace a greater mass of air (but same volume) is if the opening to atmosphere draws in more air. So if the mass is doubled after the cooling, then before the cooling at the opening to atmosphere, the displacement of air will need to double to 200cc to meet the demand of the conditions after the cooling.

22. Jul 10, 2008

### Ignea_unda

That sounds right to me. It also makes sense that the velocity of the air at the inlet could increase therefore increasing the displacement. Everything sounds plausible. The only question now would be the impact of the condensing air on the pump itself. Would it create resistance on the ammount that the pump could handle?

23. Jul 10, 2008

### xArcherx

Good question. The only thing that I can see causing a greater resistance on the displacement of the pump is if there exists a positive pressure at the outlet and/or a negative pressure at the inlet. Now with the opening to atmosphere, there shouldn't be a negative pressure at the inlet side (except for a decrease in pressure that accompanies an increase in velocity which shouldn't cause any significant resistance). As for the outlet side, that depends on what is there.

24. Aug 17, 2008

### NightSwimmer

Yes. The cooler air is more dense than warmer air. Conservation of mass is not an issue. X number of molecules occupy a different volume at a given pressure when at different temperatures.

The volume of airflow in a closed conduit will be different at different temperatures, but the mass of air entering your tube will equal the mass of air exiting your tube. While your numbers aren't accurate, you imply this in your drawing by noting the different velocities of the air stream prior to and after the cooling mechanism.

Cooling your intake air will provide more oxygen for combustion on a normally aspirated engine without an external pressure pump. The bigger question is whether the coolant being added to the air stream is beneficial to combustion. As stated in a previous post, the addition of water and methanol could decrease pre-ignition, but it will also decrease the BTU value of the fuel mixture introduced into your cylinders. If pre-ignition is not a problem, then this should result in a net loss of power generated during the combustion cycle.

Another consideration is whether your fuel mixture is being dynamically adjusted to optimize the fuel to air mixture in the cylinders. Fuel injection systems typically measure the air mass entering the intake manifold in order to calculate the appropriate amount of gasoline to inject. If the amount of fuel added to the mixture is not dynamically controlled, (as with a standard carburetor) then you could easily exceed the optimum air to fuel ratio in the combustion chamber by introducing super-cooled air.

A better air cooling system would cool the incoming air via a heat exchange system that does not introduce foreign matter into the air stream.

25. Aug 17, 2008

### Staff: Mentor

Assuming I understand the question then this is easy to resolve experimentally if you can measure the airflow a little bit before the water/methanol injector. Run it once with the injector and once without, if the flow with is > the flow without then more air is going in. Otherwise any temperature effects are being cancelled out by a pressure change.