Author Topic: What is the purpose of the intercooler on a turbo powered car:in terms of PV=nRT  (Read 1852 times)

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Offline Beamin

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I have always been trying to figure this out in terms of PV=nRT.

So Cold air is denser great but why does the engine care? Why can't you just have hot compressed air going into the engine instead of cooler less compressed air? How much does the turbo increase the pressure? The only reason I could see not wanting hot air is to stop pre ignition but the turbo doesn't make that much heat or does it?

All explanation of you tube just say: cold air is denser so its the same reason you put a box around a cold air intake. Yes that makes sense you don't want hot engine air on a NA car but this is different because the amount of air is fixed by ho much the turbo pumps. Not like the engine bay where the NA car  just takes the 1 bar pressure of hot engine air or 1 bar outside engine bay air. Then it does make a difference but in a turbo its a closed system so that doesn't matter so much.

In a turbo set up: Sure cold air is denser but you have the same amount of air regardless of temp from the turbo output to the engine. Why bother to cool it down in between?

Having some real numbers to put into the equation would help.
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Offline Beamin

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I want to keep things on topic. Does this thread belong  in general since its a science topic or is it not electronic so maybe better on a car forum? Don't want to get in the habit of making off topic threads too far off topic.

Sorry
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Offline alanb

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By cooling the air/fuel mixture it becomes more dense, i.e. for a given volume you have more of the mixture. Therefore when you injecting it into the cylinder you are getting more fuel then you would have done without the intercooler.

Sorry I cant give you any figures.
 

Offline sokoloff

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Most turbo installations have a wastegate, which regulates the pressure of air sent to the intake [by adjusting the amount of exhaust that feeds the turbo compressor drive]. Some (few) of them are simple "pop offs", which dump a lot of pressure by opening and venting to the atmosphere in a condition of excess pressure. Others (more) are variable, attempting to target a maximum pressure.

I have a lot of data logs from my airplane installation. Figures vary greatly based on outside air temp and altitude (turbo "works harder" in thinner air), but some typical figures with intercooler are:

CDT : 230°F
IAT : 90°F
OAT : 40°F

where CDT is compressor discharge temp (post-turbo, pre-intercooler), IAT is intake air temp (post-intercooler), and OAT is outside air temp (ambient, and the heat sink temp for the air-to-air intercooler).

In this system, a variable wastegate attempts to keep the manifold pressure at full throttle at 30.0 inHg absolute, meaning the lower air temp is linearly related to more mass air flow into the engine (when expressed in Kelvin/Rankine terms). Here are sample IAT plots comparing intercooled vs non-intercooled applications.

Car turbos are typically designed to boost above ambient and if a car gets to 20K MSL, something has gone terribly wrong, of course...but the physics of induction temp are the same.

 
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Offline sokoloff

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this is different because the amount of air is fixed by ho much the turbo pumps.
The adjustment provided by the wastegate makes for varying amounts of air.

It's not a closed system overall. The turbo pumps a varying amount of air, controlled by the wastegate (which adjusts how quickly the compressor spins, increasing it when the output pressure is below target and decreasing it when output pressure is above target).

So, if you now model the turbo output as a fixed pressure, adjustable volume source of air (like a constant voltage supply), the colder you can make the induction air, the more volume you can use.
 

Offline Beamin

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So they are purposely trying to keep the mass of the air the same but in a smaller volume? Is this because less pressure works better with the fuel injectors or other parts of the engine? So the physics is the same but in practice lower pressures work better for air/fuel mixing or something?
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Offline Kleinstein

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For the same amount (mass) of air, the engine has to work harder to compress the hot air compared to cold air.  The hot air also flows less easy through the valves - so more pressure is lost at the valves. Here it is the higher volume and higher viscosity or hot air. So the inter-cooler is not just allowing more power, but also to get higher efficiency. Cooling the air before the motor also reduces the cooling requirement for the rest of the motor - so the normal cooler can be smaller.

How much pressure the turbo is making depends on RPM and motor construction. AFAIK with normal cars it is something like up to 1 bar over pressure. With F1 racing cars 30 years ago they went to 6 bar and would have gone higher if the regulations would have allowed.

In addition to heating by the compression, there is also some heat transfer from the hot side of the turbo.
 

Offline mbless

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So they are purposely trying to keep the mass of the air the same but in a smaller volume? Is this because less pressure works better with the fuel injectors or other parts of the engine? So the physics is the same but in practice lower pressures work better for air/fuel mixing or something?

No, the volume is the same (cylinder volume) but with more air mass. The air-fuel ratio is (fairly) constant, so by increasing the air mass, the fuel mass must be increased. More fuel = more power.
 

Offline sokoloff

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So they are purposely trying to keep the mass of the air the same but in a smaller volume? Is this because less pressure works better with the fuel injectors or other parts of the engine? So the physics is the same but in practice lower pressures work better for air/fuel mixing or something?
Mass airflow is what matters. Mass is what is chemically reacted.

The reason to limit or control pressure is that that's easy to sense and have it feedback into the wastegate, makes for a more responsive engine, but it also changes how much pressure the clamps and hoses see, as well as limiting the stress on the engine overall. On the more responsive engine point, a turbo without a variable control is mismatched to the engine at some RPM. At low RPM, it struggles to "make enough air". At high RPM, it "makes too much air". At some sweet spot in the middle, it's "just right". Street engines need to run from about 600 RPM to over 6000 RPM and need to work well from perhaps 1200 to 5500 RPM. That's a really wide range of exhaust flow rates and so fixed turbo installs are rarely done. (The relationship between exhaust flow and intake mass flow is not linear. A belt driven mechanical supercharger is close[r] to linear and so often does not need a variable wastegate [on the output].)

The variability is a compensation mechanism for the non-linearity of the turbo.
 

Offline Myrv

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I have always been trying to figure this out in terms of PV=nRT.

For a any particular engine you are generally limited by a maximum pressure.  So for fixed (max) pressure (P) and fixed volume (V) if you lower the temperature (T) then you can increase 'n' which gives you more air (oxygen) in the cylinder to burn fuel with giving extra power.   

Put another way, to get any (significant) benefit from an intercooler the assumption is you can pump enough air to maintain the pressure so a lower temperature results in more air mass.

How much does the turbo increase the pressure?

This is highly variable depending on the car and the desired use of the turbo.  On a small econobox engine the boost may be quite low 5-10psi just to improve power slightly while maintaining fuel mileage.   On a more performance oriented car the boost can reach mid 20's or higher (a stock Evo X apparently runs as high as 24psi, a M3 sits around 19 psi).
 

Online joeqsmith

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I have always been trying to figure this out in terms of PV=nRT.

So Cold air is denser great but why does the engine care? Why can't you just have hot compressed air going into the engine instead of cooler less compressed air? How much does the turbo increase the pressure? The only reason I could see not wanting hot air is to stop pre ignition but the turbo doesn't make that much heat or does it?

All explanation of you tube just say: cold air is denser so its the same reason you put a box around a cold air intake. Yes that makes sense you don't want hot engine air on a NA car but this is different because the amount of air is fixed by ho much the turbo pumps. Not like the engine bay where the NA car  just takes the 1 bar pressure of hot engine air or 1 bar outside engine bay air. Then it does make a difference but in a turbo its a closed system so that doesn't matter so much.

In a turbo set up: Sure cold air is denser but you have the same amount of air regardless of temp from the turbo output to the engine. Why bother to cool it down in between?

Having some real numbers to put into the equation would help.

Odd question for an electronics site.  There are some decent forums dedicated to turbocharging that you may find helpful.  I have a few books on the subject that are fairly easy to read.  If you are interested, I can provide some ISBNs.

As you compress the air, with say a tire pump, it gets hot.  Imagine taking that hot air and compressing it even further.   One problem with the heated charge is that it can auto ignite from the heat.  If you wanted to use gasoline, there are some blends that will help with this.   I use some of it to test hand held multimeters.    There is a limit what you can do with even the best gasoline.  Methanol may be a good choice.    Assuming the fuel is uses some sort of ignition source rather than heat (diesel) you may be able to change the angle where the ignition takes place to help out as well.    I imagine you would be pretty limited on what you could do with pump gas, especially without cooling the intake charge.  There used to be water/methanol injection systems that would help out.  There are other ways to cool the intake charge, for example N2O. 

Obviously, cheap fuel with low pressures could keep things safe but whats the fun in that??   

To get an idea about the more complex control systems, maybe have a look at HCCI.   
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Offline JohnnyMalaria

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As a physical chemist, this question caught my eye.

In short, it is to increase the efficiency of the combustion process.

There's a great explanation here, part of which reads:

Quote
In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors.
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Online joeqsmith

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As a physical chemist, this question caught my eye.

In short, it is to increase the efficiency of the combustion process.

There's a great explanation here, part of which reads:

Quote
In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors.

Of course, once the engine has been damaged all that combustion efficiency becomes a moot point. 

http://www.superchargersonline.com/index.php?main_page=page&id=46
https://www.procharger.com/what-detonation-and-how-can-it-be-controlled
https://mechanics.stackexchange.com/questions/17490/why-do-we-need-to-cool-air-after-it-leaves-a-turbocharger

You can make a fair amount of pressure just before things melt.  :-DD

It does take a bit of steel to hold things in place if it stays together.   Turbos are fun!
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Offline JohnnyMalaria

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I get the first picture but what's the second? It looks like a recumbent bike frame or maybe some funky dog sled.
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Offline IanB

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As a physical chemist, this question caught my eye.

In short, it is to increase the efficiency of the combustion process.
This is a strange answer, which I have trouble agreeing with. Thermodynamics says the efficiency of a heat engine increases as the operating temperature increases (strictly the hot source temperature). So feeding hot air into an engine will increase the combustion temperature which should increase the efficiency. (Combustion engines are typically designed to run as hot as possible subject to material constraints and NOx emissions among other things.)

However, what will not increase will be the total power output. Total power is increased by feeding as much air and fuel into the engine as possible. Since cold air is more dense than hot air the mass flow is increased, which means the corresponding fuel flow can be increased, which means more power. This is what the quote below says (more power). The comment about efficiency is interesting, but there are many variables at play, and I don't know to evaluate that statement.

Quote
There's a great explanation here, part of which reads:

Quote
In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors.
I'm not an EE--what am I doing here?
 
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Offline IanB

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The only reason I could see not wanting hot air is to stop pre ignition but the turbo doesn't make that much heat or does it?

Yes, it does:

https://youtu.be/4qe1Ueifekg
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Online joeqsmith

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I get the first picture but what's the second? It looks like a recumbent bike frame or maybe some funky dog sled.

You are not too far off.   :-DD
https://youtu.be/h9pixcy3GOU?list=PLZSS2ajxhiQA9yWh-hvVtv2ib-3ObOHGV&t=747

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Offline JohnnyMalaria

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As a physical chemist, this question caught my eye.

In short, it is to increase the efficiency of the combustion process.
This is a strange answer, which I have trouble agreeing with. Thermodynamics says the efficiency of a heat engine increases as the operating temperature increases (strictly the hot source temperature). So feeding hot air into an engine will increase the combustion temperature which should increase the efficiency. (Combustion engines are typically designed to run as hot as possible subject to material constraints and NOx emissions among other things.)

However, what will not increase will be the total power output. Total power is increased by feeding as much air and fuel into the engine as possible. Since cold air is more dense than hot air the mass flow is increased, which means the corresponding fuel flow can be increased, which means more power. This is what the quote below says (more power). The comment about efficiency is interesting, but there are many variables at play, and I don't know to evaluate that statement.

Quote
There's a great explanation here, part of which reads:

Quote
In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors.

I think the term efficiency isn't used thermodynamically but more in a miles (kilometers) per gallon (liter) sense.

Nevertheless, the greater the temperature difference, the greater the amount of energy released (per the Gibbs function - WARNING! Contains potential triggers of painful memories of p-chem class).
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Offline IanB

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I think the term efficiency isn't used thermodynamically but more in a miles (kilometers) per gallon (liter) sense.

Nevertheless, the greater the temperature difference, the greater the amount of energy released (per the Gibbs function - WARNING! Contains potential triggers of painful memories of p-chem class).

I was thinking more in terms of the Carnot efficiency, \$ \eta = (T_H - T_C) / T_H \$, where \$\eta \rightarrow 100\%\$ as \$T_H \rightarrow \infty \$

Of course real engines are not Carnot engines, but similar principles apply.
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Online joeqsmith

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I think the term efficiency isn't used thermodynamically but more in a miles (kilometers) per gallon (liter) sense.

Maybe meters or feet per gallon.  :-DD 






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Online emece67

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Hi all,

Maybe I arrived late to this discussion but... the torque delivered by the engine is, in a 1st order approximation, proportional to (or, at least, highly dependant on) the energy released by the chemical reaction between the O2 in the air and the fuel. The more O2+fuel you burn on each stroke, the greater the torque. From PV=nRT you get n = PV/RT, so if you want, in order to get big torque, to burn big moles of O2+fuel, you need high pressure, high volume and low temperature. Being the engine displacement fixed (V), you rise pressure using a compressor. Although the compression in the turbine is not exactly isentropic, it is in a first approximation, so the compression also rises T, partially masking the effect of the increased pressure. Thus, you need to lower T after the compression to fully benefit from this approach.

The simplest EFI systems at least monitor the values of: the Manifold Absolute Pressure (MAP, this is the intake air pressure); the Intake Air Temperature (IAT); and, by means of a flowmeter or, in simpler systems, by means of the known engine displacement and the concept of "volumetric efficiency", the volume of air entering the engine. The injectors remaining opened a time proportional to MAP and aspirated air volume and inversely proportional to IAT (all this in a first order approximation). The throttle pedal, by means of the butterfly, directly controls MAP, and thus, torque.

Of course P & T also take a role in the thermodynamic efficiency of the engine, but their effect, via this mechanism, on final engine performance is much lower than the previous, direct, effect.

Hope this helps, regards.
« Last Edit: May 18, 2018, 06:44:10 pm by emece67 »
Information must flow.
 
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Online james_s

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The only reason I could see not wanting hot air is to stop pre ignition but the turbo doesn't make that much heat or does it?

I have measured the air temperature before and after the intercooler in one of my cars, 1984 Volvo Turbo so not exactly modern tech but physics have not changed much in the last 40 years. At 15 PSI boost I was seeing temperatures of around 300F at the inlet to the intercooler and 90-120F at the exit, ambient temps in the low to mid 60s. Preignition is a significant limiting factor to the maximum boost pressure you can run, and thus the amount of fuel you can burn. Non-intercooled models had the boost set to around 6 psi, intercooled models raised that to 10.5 psi stock. If you try to run that much boost without an intercooler it will usually ping, even with premium fuel.
 

Offline amyk

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...and also why diesels can use much higher boost --- there's no fuel in the air going into the cylinder, it gets injected exactly when it's needed. It's common to see 20PSI+ of boost in a diesel.
 

Offline IanMacdonald

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Basically a manifestation of Charles's Law, that when a gas is heated it tries to expand.  If it cannot, its pressure increases. The converse must be true, that if a gas is compressed its temperature will rise. Hence when you compress a gas it becomes hot, and that makes it harder to compress any further. Cooling the output thus means the turbo doesn't have to work so hard.

Also, the lower charge temperature means a greater mass of gas can be put into the cylinder at the same pressure, hence more engine power.

Filling of oxygen bottles has to be done slowly, for the same reason. 
 

Offline gildasd

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On marine engines, let’s say the exhaust side of the turbo is in the 480C before and 370 after range... that is a lot of energy to dump in a short distance. A lot of it goes to compressing the inlet air heating it.
The compressed air then goes through a two stage water intercooler, the second stage is pretty normal. Howether, the heat from the first intercooler stage can be used as energy to run the fresh water generator, so is not 100% lost energy.
The exhaust after the turbo is still hot enough to heat coils of thermal oil that is used to heat up fuel tanks etc.

 Just to say that calculations like you asked for in the beginning are hard to do in real life due to the cooling of the compressed air not being the end of the system.
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