Intake Manifold Temperature 1

[SBBlue]

Here's a little bit of a strange question.

What is the typical temperature in the intake manifold for a naturally aspirated spark ignition engine?

Here's why I was wondering. . . .

The intake manifold is, of course, below atmospheric pressure for a NA engine except for full throttle conditions. Since the engine has to perform work on the air to get it below atmospheric pressure, the temperature of the air will drop.

According to my calculations, dropping the pressure of a container of air from one atmosphere of pressure to one-fourth atmosphere will drop the temperature from 80 deg F (assumed ambient) to something like 80 degrees below zero.

This phenomena does, of course, account for icing that is frequently encountered with piston airplane engines.

The intake manifold air will be heated by the surrounding engine, but to how great an extent?

Since I started thinking about this I haven't had a chance to put a temperature probe in an intake manifold.

Has anybody ever done that?

 

[strokersix]

http://www.qrg.northwestern.edu/projects/NSF/Cyclepad/cpadw047.htm

I am interested in comments from those who can connect thermodynamics to practical application. Temperature across a throttle is usually considered constant. Therefore the cooling observed must be due to fuel evaporation. Correct or no?

 

[sreid]

As I previously posted, I believe that there is a temperature drop across the throttle body due to the adiabatic gas law. See the link below

http://bednorzmuller87.phys.cmu.edu/demonstrations/thermodynamics/manyparticlesystems/demo524.html

If a gass is compressed isothermally, there will be a temperature increase. If a gas is expanded isothermally there will be a temperature decrease (this is not always true-see the Joule-Thompson coefficient from a previous post but it is true for air at standard temperature and pressure).

Having said that, the gas temperature decrease is probably much greater due to the vaporization of the gasoline and there is most certainly a temperature rise in the intake manifold due to the temperature of the intake manifold.

 

[SBBlue]

Sreid wrote:

"Having said that, the gas temperature decrease is probably much greater due to the vaporization of the gasoline and there is most certainly a temperature rise in the intake manifold due to the temperature of the intake manifold."

=============================

Actually, we can do some calculations relatively easy to determine the temperature drop due to vaporization of fuel.

First, an assumption; all of the fuel will be assumed to vaporize completely in the intake manifold. That, of course, is not correct, but it will give us the "worse case" scenario.

Second: Heat of vaporization for gas. Actually, I'm not sure -- but I do know the heat of vaporization for Butane (~140-170 BTU/lb, depending upon temperature), Isobutane (same), propane (about the same), and turpetine (again, about the same). I would suspect Octane (the gasoline equivalent) would be about the same.

Third, we will assume stoichometric combustion, so the air/fuel ratio will be about 14.5:1.

So what do we get? One pound of hydrocarbon will take, say 150 BTUs of heat for vaporization. This 150 BTUs will be supplied by 14.5 lbs of air, which means that each lb of air will lose about 10.3 BTUs of heat. Air at a temperature of 80 deg F will drop to about 36 deg F.

Of course, you must remember that not all of the gasoline will be vaporized in the intake manifold. I'm not completely sure, but I suspect that the bulk of it would be vaporized in the cylinder.

 

[Turbinator]

Chumley,
Let's get some context here:

My post was in reply to SBBlue's opening post:

"According to my calculations, dropping the pressure of a container of air from one atmosphere of pressure to one-fourth atmosphere will drop the temperature from 80 deg F (assumed ambient) to something like 80 degrees below zero."

I believe the formula he used in this calculation was that for isentropic expansion (Efficiency=100%)

T2=T1*(P2/P1)^0.286

(SBBlue, please correct me if I'm
wrong). That would explain his calculated temperature drop of 160 deg. F or so.

As I said earlier, the formula does not work for a throttling proccess, which simply lowers the pressure without taking energy (heat) out of the air stream. Here's the extended version of the formula:

T2=T1*[1-efficiency*(1-[p2/p1]^0.286)].

If the isentropic efficiency is 100%, you get the simpler formula. 100% isentropic efficiency means the maximum amount of energy is being extracted from the fluid stream. For a throttling process, no energy is being extracted. Therefore, the isentropic efficiency is 0%. Plug that into the formula, and you get T2=T1. So, !!as far as the expansion process alone!!, their is no temperature difference between the inlet and outlet of a throttle. Now, as explained by other members their which does cause the temperature to lower.

 

[Turbinator]

Edit to last line above:

Now, as explained by other members there is evaportaion which does cause the temperature to lower.

 

[Turbinator]

Here are some further clarifications:

For a steady flowing gas, if there is a temperature change, energy has either been added to it or removed. For relatively low velocity change between points 1 and 2:

Change in Energy=Cp x (T2-T1).
where Cp is Specific heat at constant pressure, assumed constant for small temperature differences.

Energy is either heat being transfered or shaft work, neither which occurs to a significant amount across a throttle. So, if there is no change in the energy, their is no change in the temperature.

Again, this is related only to the throttling process and not to evaporative cooling effects of the fuel or moisture in the air.

 

[Rick360]

With a drop in pressure there will be no change in the amount of heat contained in a given mass quantity of air, but since this constant amount of heat is expanded into a larger area, due to the drop in pressure, the temperature will drop.

Rick

 

[sreid]

I also still believe that there is a temperature drop when air is throttled. See the following link and go to "Joule-Thompson Coefficient." The definition is dT/dP.

http://www.public.asu.edu/~dmatyus/teaching/chm341/lecture_notes/chapter3/chap3.pdf

One can also search the Web for Isenthalpic and Throttle and find similar results.

 

[SBBlue]

I think what may be causing some of the confusion is that there is more than just a throttling process going on -- there is also an active process pumping gas out of the manifold.

If you have a sealed container -- no throttle at all -- and pump half of the gas out, you will see a significant pressure drop in the gas that remains.

The pumping process is what is causing the temperature drop -- not the throttling process. And we can regard the pumping process as isoentropic for calculation purposes.

 

[sreid]

I once had the seamingly bright idea that I could increase the HP of a racing engine by cooling the gasoline. My theory was that the gasoline would cool the air making it more dense. So on a dyno we ran the gas through a copper tube imersed in a dry ice alcohol mix to get some really cold gas (and it was cold, frost build up on the carburetor bodies). But no HP increase even using bigger primary jets.

I think the reason it didn't work is that the air needs to be cooled before it gets to the carburetor (or today, FI). Cold air definately makes more HP, you correct dyno HP for air temperature and pressure. Cooling the air would also precipitate out water vapor which also hurts HP.

A side note. The F1 regs have three separate rules against cooling the gasoline.

 

[schmidtj86]

Throttles are isothermal and IRREVERSIBLE, unless you get some work out of the (dry) working fluid. Iso = same, therm = temperature. Those who are saying "think about pumping the air out of a volume, that gets cold" nope, not the same situation, except right at start up, or in a negative throttle movement. Think about this, poke a hole in a balloon, it's not freezing is it? The only reason it may feel cooler at all, is if it's blowing on a moist skin surface. Also, at a throttle, if it gets past the throttle without changing temperatures, then it won't get cooler due to some magical "now the pistons are emptying the manifold causing a pressure drop causing it to cool." The throttle creates a restriction to the pistons moving air through other restrictions called valves. That restriction drops the pressure.
When the piston starts its downstroke, and lowers the pressure, that process of lowering the pressure may lower the temperature, but once the valve opening throttles the air flow (which it may not do). But that is totally not a steady state process. A throttle is very close to steady state (except during transients, i.e. angle changing).
So, when the 70 degree air sees a throttle and a lower pressure, it doesn't go in and get colder, it just looses energy across the throttle, like a hole in a balloon. On one side of the throttle is 1 bar 70 degree air, on the other, there is .5 bar 70 degree air. That pressure differential keeps the air flowing, the pistons are filled with that pressure and temperature of air.
If this were a single cylinder engine, with the manifold volume at 1bar 70 F momentarily, and then the piston is dropping the pressure, the temperature may drop momentarily, but the inflow of 70 F air is still fighting it.
But, since the situation is a manifold with several exits, creating a constant flow (whether it's pistons or a fan sucking, same difference to the manifold), and this manifold is staying the same size, and there is a leak (throttle) into it, and equal flow out of it, it's just an intermediate area, the temperature stays the same. It's fluid flow, not thermodynamic expansion.
If it were a container of air, with no inlet, and either a piston pulling down, or some pressure drop through some hole, then its contents would cool, but engines can't cool ambient air, since it's not in a container of constant volume. If it were the piston pulled down, and the container volume were still the same (and insulated), and there was the original pressure on the opposite side of the piston, then the work done on that piston could have been recouped, as it is sucked back in. In that case, it would have cooled from the expansion, and warmed back to original during the contraction (assuming 100% efficiency).
blah blah blah, yada yada yada.
I'm just trying to paint the right pictures in your heads here.
Vaporization (of fuel) on the other hand, that'll suck the heat outta the air and manifold quickly, especially at reduced pressures (such as in a manifold).
A dry manifold, such as a MPFI manifold, will not drop the air temperature below ambient.
I'm tired so I'll stop now, but the REASON for the temperature change (if any) is not the thermodynamics of a throttle, manifold, and dry air.

 

[strokersix]

Thank you schmidtj86 for reiterating the theory!

It may come as a surprise to some of the previous posters to know that in some cases temperature of dry gas actually will rise across a throttle, depending on the sign of the Joule-Thompson coefficient!

 

[SBBlue]

Not entirely correct.

It is true for an ideal gas, but nitrogen and oxygen aren't ideal gases.

If what TJ Schmidt said was true for nitrogen, then liquid nitrogen and oxygen couldn't be produced. It is produced by pressuring gas, cooling the air down, and then letting it cool itself as it expands. This cycle is repeated several times until the temperature gets several hundred degrees below zero and and the remaining air liquifies.

Another example where it isn't true can be seen with explosive depressurization that can be seen with high flying aircraft. Anybody who's ever taken an altitude chamber "ride" is very familiar with this. What happens? There is immediate condensation of water vapor in the air, creating a cloud. The temperature drops so much that the vapor causes frost on the inside of the windows.

This effect can also be seen if you look at air just before it enters a gas turbine at full power. In some instances a water vapor cloud is produced. What is the mechanism for that other than the fact that the air temperature suddenly decreases as the air pressure in front of the turbine decreases from air being sucked into the turbine.

 

[Turbinator]

After a quick search, I couldn't find a chart on the internet, but there is the "Generalized Compressibility Chart" which gives the gas law for a real gas, Z=Pv/RT. For an Ideal gas, the Compressibility Factor, Z, is 1, thus resulting in Pv=RT.

Z is a function of:
Reduced Pressure, P/P-critical, where for P-critical is 547 psia for air
and Reduced Temperature, T/T-critical, where T-critical is 238.5 R for air.

Looking at the chart, the compressibility factor Z gets farther from 1 as P-reduced increases and as T-reduced decreases.

So, using an extreme example:
Say the air before the throttle has been compressed to 65 psia, giving a Reduced Pressure of 0.12, and the Temperature is -100 F (360 R) giving a Reduced Temperature of 1.5.

According to the chart in front of me, the Compressibility factor would be Z=0.99, which is about the largest deviation from the ideal gas law (Z=1) as would be encountered at the throttle of a common air breathing engine.

 

[Tmoose]

"snip........ the formula does not work for a throttling proccess, which simply lowers the pressure without taking energy (heat) out of the air stream. "

Isn't some of the kinetic energy being taken away to push the gas into a bigger volume?

Wouldn't the expansion be similar to warm air rising? Cooling due to expansion is often cited as the reason for clouds forming.

 

[strokersix]

I'll quote from Marks' Handbook 6th ed:

"The investigations of Joule and Lord Kelvin showed that a gas drops in temperature when throttled. This is not universally true. For some gases, notably hydrogen, the temperature rises for throttling processes over ordinary ranges of temperature and pressure. ..... The ratio of observed drop in temperature to the drop in pressure, i.e. dT/dp, is the Joule-Thomson coefficient."

 

[schmidtj86]

I also have been told, but am not sure, that as the air goes past the throttle, the temperature decreases (allowing icing on throttles), but recovers to whatever the Joule-Thomson calculation says, and for ideal gases, which air is close to, it recovers fully.
I don't have time or energy to research this, but if anyone else does, please post it here.

 

[Tmoose]

So, it seems as long as we are talking about air, or mixtures resembling air, whose molecules find each other somewhat attractive, there will be >some< cooling, just as my frosty tire valve suggests.

EXAIR, who make all kinds of air ejectors, does not seem to claim any cooling benefit, aside from the air flow itself.

http://www.newton.dep.anl.gov/askasci/eng99/eng99082.htm

http://www.mindspring.com/~divegeek/lift.htm

 

[MaxRaceSoftware]

4.500 x 4.500 = 572.6 cid BBC Chevy Marine engine

Chevron 92 premium pump gas

Part Throttle, very light Dyno load

Engine----CAT----Vacuum----Manifold
RPM-------degF----in Hg---- deg. F.
1435-------105----16.1--------111
1754-------106----16.3--------111
2064-------105----17.6--------110
2305-------104----17.3--------108
2632-------104----15.8--------107
3141-------104----13.7--------104
3712-------105----12.3-------- 99

Full Load at 600 Rpm/Sec Acceleration Rate SF-901 Dyno
Engine----CAT----Vacuum----Manifold
RPM-------degF----in Hg---- deg. F.
3200------102------0.0------95
3300------102------0.0------95
3400------102------0.0------95
3500------103------0.0------95
3600------103------0.0------94
3700------103------0.0------94
3800------103------0.0------94
3900------103------0.0------94
4000------103------0.0------94
4100------103------0.0------94
4200------103------0.0------94
4300------103------0.0------94
4400------103------0.5------94
4500------103------0.5------94
4600------103------0.5------93
4700------103------0.0------93
4800------103------0.5------93
4900------103------0.5------93
5000------103------0.5------93
5100------103------0.6------93
5200------103------0.6------93
5300------103------0.7------92
5400------102------0.7------92
5500------102------0.7------91
5600------102------0.8------91
5700------102------0.8------91
5800------102------0.8------91
5900------102------0.7------91
6000------103------0.9------90
6100------103------0.8------90
6200------104------0.8------90
6300------104------0.9------90
6400------104------0.9------91
6500------104------0.9------90


Larry Meaux (maxracesoftware@yahoo.com)
Meaux Racing Heads - MaxRace Software
ET_Analyst for DragRacers
Support Israel - Genesis 12:3

 

[schmidtj86]

Larry,
Is that with a carburetted engine?
The fuel vaporization will pull heat energy from the air if it is, explaining the temperature drop for the most part. I still think the argument is whether the throttle itself is causing the temp drop which I'm pretty sure it isn't.