Oil Analysis

[jl-c]

JFP, this more like it?

http://www.thermal-lube.com/wp-content/uploads/2014/02/In-Line-Fuel-Dilution-Lube-Ex.pdf

[JFP in PA]

JFP, this more like it?

http://www.thermal-lube.com/wp-content/uploads/2014/02/In-Line-Fuel-Dilution-Lube-Ex.pdf

 

Not really, FTIR analysis is useful as a continuous "on-line" analytical tool for diesel engines in environments such as engine test cells where the engine is running and the FTIR is automatically collecting data.  FTIR is not as accurate as either GC or FI techniques, but they are both very expensive and are also "batch" samples techniques (read you take a small sample of the oil and inject it into the instrument), which makes them difficult to automate for long run auto sampling testing.

[JFP in PA]

OK, here is where I'm at as of this juncture:

 

1. Everyone I have spoken to (oil company technical personnel, used oil analytical labs) has basically said the same thing:  When fuel dilutes the oil, a very small (read minute) amount of the oil's ingredients may react with the fuel and form compounds that may be more difficult to separate from the oil, but even that change does not in any way impede a laboratory from separating the fuel from the oil in both a quantitative and qualitative manner.  The most common technique used to determine both the quantity of fuel in the oil, and to analyze this fuel effluent to confirm that it truly is fuel, is head space gas chromatography.  In this technique, the oil sample is gradually heated up to between 300 and 350F so that the instrument can determine both the temperature at which the effluent comes off, and compare the analytical data of the individual components of the effluent with stored known sample data bases to confirm that it actually is fuel.  From this technique, the lab can tell you both how much came off as a percentage of the sample mass, and confirm that it is fuel and not some other diluent.  Using gas chromatography, or flame ionization chromatography  which is able to do the same test with even higher levels of accuracy, this analytical technique is considered to be both accurate and extremely reproducible.
2. No one that I have spoken with concurs with the idea that diluent fuel mixed with common commercial synthetic engine oil will not flash off under normal engine operating conditions found in a Porsche oil sump (coolant around 200-215F, oil around 230-240F, partial vacuum of around 5 inches of water).  And as these conditions are "nominal", meaning that the engine, coolant, and oil are often hotter under warmer ambient conditions, higher speed driving, or aggressive driving situations; the engine should therefore have no problem ridding itself of most, if not all, of the highly volatile components found in normal gasoline.  One individual even noted that water, which boils at a much higher temperature than gasoline, will completely boil off from engine oil at a little over 200F at 5 inches of water vacuum, well within the normal operating conditions of a Porsche engine.
3. Used oil analytical labs I spoke with were fairly adamant in their response to fuel dilution levels continuing to rise in sequential oil samples from the same engine.  The common response was one of two possibilities: Either the engine had a worsening problem that accelerated the fuel dilution problem with time, or the analytical technique used to determine the level of dilution was flawed.  While some noted that as oil aged in use, it was possible for the breakdown by products of the oil itself to flash off at lower temperatures, which could lead to some confusion over the actual amount of fuel in the oil, but noted that either gas chromatography or flame ionization should recognize and discount these by products as not part of normal gasoline, and therefore not count them as being from fuel.

So at this point, I will have to throw the topic back open for further discussion, and welcome input from anyone with other observations or data.

[jl-c]

Well, personally I would say you have nailed it. Thank you for finding all this out. It makes sense of Porsche's comments in the manual and also my own observations. I will keep doing a long run on a regular basis to allow me to get an accurate oil reading as my usual run to and from work on a daily basis is only a 4 mile round trip. Oil changes at every 6,000 miles or less I think. No way 20,000!

[Mijostyn]

 

Well, personally I would say you have nailed it. Thank you for finding all this out. It makes sense of Porsche's comments in the manual and also my own observations. I will keep doing a long run on a regular basis to allow me to get an accurate oil reading as my usual run to and from work on a daily basis is only a 4 mile round trip. Oil changes at every 6,000 miles or less I think. No way 20,000!

Fraid not. But you are probably right jl-c. Change your oil more frequently. If you want to know how frequent do serial UOAs on your car. All of the reliable references I have found recommend to change your oil at a fuel dilution of 2%. A good article on this is at www.machinerylubrication.com/Atricles/Print/1304

 

OK, here is where I'm at as of this juncture:

• Everyone I have spoken to (oil company technical personnel, used oil analytical labs) has basically said the same thing:  When fuel dilutes the oil, a very small (read minute) amount of the oil's ingredients may react with the fuel and form compounds that may be more difficult to separate from the oil, but even that change does not in any way impede a laboratory from separating the fuel from the oil in both a quantitative and qualitative manner.  The most common technique used to determine both the quantity of fuel in the oil, and to analyze this fuel effluent to confirm that it truly is fuel, is head space gas chromatography.  In this technique, the oil sample is gradually heated up to between 300 and 350F so that the instrument can determine both the temperature at which the effluent comes off, and compare the analytical data of the individual components of the effluent with stored known sample data bases to confirm that it actually is fuel.  From this technique, the lab can tell you both how much came off as a percentage of the sample mass, and confirm that it is fuel and not some other diluent.  Using gas chromatography, or flame ionization chromatography  which is able to do the same test with even higher levels of accuracy, this analytical technique is considered to be both accurate and extremely reproducible.
• No one that I have spoken with concurs with the idea that diluent fuel mixed with common commercial synthetic engine oil will not flash off under normal engine operating conditions found in a Porsche oil sump (coolant around 200-215F, oil around 230-240F, partial vacuum of around 5 inches of water).  And as these conditions are "nominal", meaning that the engine, coolant, and oil are often hotter under warmer ambient conditions, higher speed driving, or aggressive driving situations; the engine should therefore have no problem ridding itself of most, if not all, of the highly volatile components found in normal gasoline.  One individual even noted that water, which boils at a much higher temperature than gasoline, will completely boil off from engine oil at a little over 200F at 5 inches of water vacuum, well within the normal operating conditions of a Porsche engine.
• Used oil analytical labs I spoke with were fairly adamant in their response to fuel dilution levels continuing to rise in sequential oil samples from the same engine.  The common response was one of two possibilities: Either the engine had a worsening problem that accelerated the fuel dilution problem with time, or the analytical technique used to determine the level of dilution was flawed.  While some noted that as oil aged in use, it was possible for the breakdown by products of the oil itself to flash off at lower temperatures, which could lead to some confusion over the actual amount of fuel in the oil, but noted that either gas chromatography or flame ionization should recognize and discount these by products as not part of normal gasoline, and therefore not count them as being from fuel.

So at this point, I will have to throw the topic back open for further discussion, and welcome input from anyone with other observations or data.

Hey JFP, miss me? ;-)

 

Polaris uses Gas Chromatography. The Flame Ionization Detector is the gizmo used in the gas chromatograph to quantify the various fractions. A little hydrogen flame ionizes the molecules increasing the current through the detector. The time and duration identify the substance. What separates the substances is not their vapor pressure but the speed at which they pass through various mediums on their way to the FID. I finally remembered the name of the guy who got me into all this trouble, Francois-Marie Raoult. Raoult's law attempts to sort of quantify the interaction between varying molecules in solution. Don't get too dizzy when you read it. I only make it through the first couple of equations. In one sense I was wrong. There is a relation to vapor pressure but the effect is that it mostly averages out. This is only in solutions of like substances. As soon as you put polar molecules in nonpolar solutions or visa versa they just repel each other. Which is why the water boils off. Good thing too as I can't stand the thought of my engine rusting! Still hate that white crud on the oil filler cap.

Edited August 14, 2014 by Mijostyn

[Loren]

GC-FID I feel my past catching up with me... :huh:

[Mijostyn]

GC-FID I feel my past catching up with me... :huh:

I'm still buried in it Loren. We use all this stuff in medicine. What really is annoying is that my old brain is not near as nimble as it use to be. In my 20s I would see an equation and visualize immediately what it was telling me. Now things are not as immediate or clear. What is it that Einstein said about getting old? Something about you just get dumber?

[Lyn]

^Mijostyn,

^{I always enjoy reading your posts because of your analytical approach to the subject at hand. The subject of which oil brand and viscosity to use in which environment concerns me because of the risk of Intermediate shaft bearing failure and/or cylinder scoring. I can't help but think that, as you surmised, the environment has more of an impact than many of us realize. I live in Phoenix. and for a large part of the year, the oil temperature is never below 100 degrees, even when parked overnight. My 997 3.6 is as much a grocery getter as a long haul car and I put 14-16,000 miles a year on it. I am a bit paranoid about oil dilution and breakdown in the severe temperatures we see in Phoenix and use Motul Excess 5-40 with a 3,500 mile change interval. I realize that  people might think that I'm wasting oil and filters to some extent with that interval, but both are cheap compared to an engine failure. What I'm building up to here, is my concern about oil pressure in high temp environments. With oil temperature in the 225-230 degree range, my oil pressure drops to just over 1 Bar at idle while motoring about town or in heavy traffic. Once on the move, oil pressure goes up to 3 1/2-4 bar. I've considered going up a step to 10-40 or 10-50 during the summer months, but have heard that the vario-cam system is very sensitive to viscosity levels above 5-40 because of the small oil passages. You've done more research than most on the subject of which oil to use and I'd appreciate your thoughts on what direction I might go to bump up my low RPM oil pressure.}

^{Thx, Lyn.}

[JFP in PA]

 

Well, personally I would say you have nailed it. Thank you for finding all this out. It makes sense of Porsche's comments in the manual and also my own observations. I will keep doing a long run on a regular basis to allow me to get an accurate oil reading as my usual run to and from work on a daily basis is only a 4 mile round trip. Oil changes at every 6,000 miles or less I think. No way 20,000!

Fraid not. But you are probably right jl-c. Change your oil more frequently. If you want to know how frequent do serial UOAs on your car. All of the reliable references I have found recommend to change your oil at a fuel dilution of 2%. A good article on this is at www.machinerylubrication.com/Atricles/Print/1304

 

OK, here is where I'm at as of this juncture:

• Everyone I have spoken to (oil company technical personnel, used oil analytical labs) has basically said the same thing:  When fuel dilutes the oil, a very small (read minute) amount of the oil's ingredients may react with the fuel and form compounds that may be more difficult to separate from the oil, but even that change does not in any way impede a laboratory from separating the fuel from the oil in both a quantitative and qualitative manner.  The most common technique used to determine both the quantity of fuel in the oil, and to analyze this fuel effluent to confirm that it truly is fuel, is head space gas chromatography.  In this technique, the oil sample is gradually heated up to between 300 and 350F so that the instrument can determine both the temperature at which the effluent comes off, and compare the analytical data of the individual components of the effluent with stored known sample data bases to confirm that it actually is fuel.  From this technique, the lab can tell you both how much came off as a percentage of the sample mass, and confirm that it is fuel and not some other diluent.  Using gas chromatography, or flame ionization chromatography  which is able to do the same test with even higher levels of accuracy, this analytical technique is considered to be both accurate and extremely reproducible.
• No one that I have spoken with concurs with the idea that diluent fuel mixed with common commercial synthetic engine oil will not flash off under normal engine operating conditions found in a Porsche oil sump (coolant around 200-215F, oil around 230-240F, partial vacuum of around 5 inches of water).  And as these conditions are "nominal", meaning that the engine, coolant, and oil are often hotter under warmer ambient conditions, higher speed driving, or aggressive driving situations; the engine should therefore have no problem ridding itself of most, if not all, of the highly volatile components found in normal gasoline.  One individual even noted that water, which boils at a much higher temperature than gasoline, will completely boil off from engine oil at a little over 200F at 5 inches of water vacuum, well within the normal operating conditions of a Porsche engine.
• Used oil analytical labs I spoke with were fairly adamant in their response to fuel dilution levels continuing to rise in sequential oil samples from the same engine.  The common response was one of two possibilities: Either the engine had a worsening problem that accelerated the fuel dilution problem with time, or the analytical technique used to determine the level of dilution was flawed.  While some noted that as oil aged in use, it was possible for the breakdown by products of the oil itself to flash off at lower temperatures, which could lead to some confusion over the actual amount of fuel in the oil, but noted that either gas chromatography or flame ionization should recognize and discount these by products as not part of normal gasoline, and therefore not count them as being from fuel.

So at this point, I will have to throw the topic back open for further discussion, and welcome input from anyone with other observations or data.

 

Hey JFP, miss me? ;-)

 

Polaris uses Gas Chromatography. The Flame Ionization Detector is the gizmo used in the gas chromatograph to quantify the various fractions. A little hydrogen flame ionizes the molecules increasing the current through the detector. The time and duration identify the substance. What separates the substances is not their vapor pressure but the speed at which they pass through various mediums on their way to the FID. I finally remembered the name of the guy who got me into all this trouble, Francois-Marie Raoult. Raoult's law attempts to sort of quantify the interaction between varying molecules in solution. Don't get too dizzy when you read it. I only make it through the first couple of equations. In one sense I was wrong. There is a relation to vapor pressure but the effect is that it mostly averages out. This is only in solutions of like substances. As soon as you put polar molecules in nonpolar solutions or visa versa they just repel each other. Which is why the water boils off. Good thing too as I can't stand the thought of my engine rusting! Still hate that white crud on the oil filler cap.

 

 

I looked into it, according to what I just read, Raoult’s Law applies at limiting low concentrations for non-volatile solutes. At higher concentrations or with volatile solutes, different compounds behave differently. 

 

Are you sure that is the correct reference?

 

And no, I didn't miss you..................... :eek:

[Mijostyn]

Well, personally I would say you have nailed it. Thank you for finding all this out. It makes sense of Porsche's comments in the manual and also my own observations. I will keep doing a long run on a regular basis to allow me to get an accurate oil reading as my usual run to and from work on a daily basis is only a 4 mile round trip. Oil changes at every 6,000 miles or less I think. No way 20,000!

Fraid not. But you are probably right jl-c. Change your oil more frequently. If you want to know how frequent do serial UOAs on your car. All of the reliable references I have found recommend to change your oil at a fuel dilution of 2%. A good article on this is at www.machinerylubrication.com/Atricles/Print/1304

OK, here is where I'm at as of this juncture:

• Everyone I have spoken to (oil company technical personnel, used oil analytical labs) has basically said the same thing: When fuel dilutes the oil, a very small (read minute) amount of the oil's ingredients may react with the fuel and form compounds that may be more difficult to separate from the oil, but even that change does not in any way impede a laboratory from separating the fuel from the oil in both a quantitative and qualitative manner. The most common technique used to determine both the quantity of fuel in the oil, and to analyze this fuel effluent to confirm that it truly is fuel, is head space gas chromatography. In this technique, the oil sample is gradually heated up to between 300 and 350F so that the instrument can determine both the temperature at which the effluent comes off, and compare the analytical data of the individual components of the effluent with stored known sample data bases to confirm that it actually is fuel. From this technique, the lab can tell you both how much came off as a percentage of the sample mass, and confirm that it is fuel and not some other diluent. Using gas chromatography, or flame ionization chromatography which is able to do the same test with even higher levels of accuracy, this analytical technique is considered to be both accurate and extremely reproducible.
• No one that I have spoken with concurs with the idea that diluent fuel mixed with common commercial synthetic engine oil will not flash off under normal engine operating conditions found in a Porsche oil sump (coolant around 200-215F, oil around 230-240F, partial vacuum of around 5 inches of water). And as these conditions are "nominal", meaning that the engine, coolant, and oil are often hotter under warmer ambient conditions, higher speed driving, or aggressive driving situations; the engine should therefore have no problem ridding itself of most, if not all, of the highly volatile components found in normal gasoline. One individual even noted that water, which boils at a much higher temperature than gasoline, will completely boil off from engine oil at a little over 200F at 5 inches of water vacuum, well within the normal operating conditions of a Porsche engine.
• Used oil analytical labs I spoke with were fairly adamant in their response to fuel dilution levels continuing to rise in sequential oil samples from the same engine. The common response was one of two possibilities: Either the engine had a worsening problem that accelerated the fuel dilution problem with time, or the analytical technique used to determine the level of dilution was flawed. While some noted that as oil aged in use, it was possible for the breakdown by products of the oil itself to flash off at lower temperatures, which could lead to some confusion over the actual amount of fuel in the oil, but noted that either gas chromatography or flame ionization should recognize and discount these by products as not part of normal gasoline, and therefore not count them as being from fuel.

So at this point, I will have to throw the topic back open for further discussion, and welcome input from anyone with other observations or data.

Hey JFP, miss me? ;-)

Polaris uses Gas Chromatography. The Flame Ionization Detector is the gizmo used in the gas chromatograph to quantify the various fractions. A little hydrogen flame ionizes the molecules increasing the current through the detector. The time and duration identify the substance. What separates the substances is not their vapor pressure but the speed at which they pass through various mediums on their way to the FID. I finally remembered the name of the guy who got me into all this trouble, Francois-Marie Raoult. Raoult's law attempts to sort of quantify the interaction between varying molecules in solution. Don't get too dizzy when you read it. I only make it through the first couple of equations. In one sense I was wrong. There is a relation to vapor pressure but the effect is that it mostly averages out. This is only in solutions of like substances. As soon as you put polar molecules in nonpolar solutions or visa versa they just repel each other. Which is why the water boils off. Good thing too as I can't stand the thought of my engine rusting! Still hate that white crud on the oil filler cap.

I looked into it, according to what I just read, Raoult’s Law applies at limiting low concentrations for non-volatile solutes. At higher concentrations or with volatile solutes, different compounds behave differently.

Are you sure that is the correct reference?

And no, I didn't miss you..................... :eek:

Absolutely. Let us see if I can explain this correctly.

The total vapor pressure of a solution of non polar molecules such as fuel in oil is, P = p*A XA + p*B XB + ........

P*A is the partial pressure of molecule A and XA is the mole fraction of that substance in the solution and so forth for molecules B, C and onwards. The individual vapor pressure of a given molecule is Pi = Pi* Xi . i is for the sake of the example fuel. Pi is the vapor pressure of fuel in the mixture. Pi* is the vapor pressure of pure fuel and Xi is the mole fraction of the fuel in the mixture. For simplification let's call the mole fraction the simple percentage of fuel in the mixture fuel in oil. This is actually quite close. In our example that would be 2%. Thus Xi is 0.02 and the vapor pressure of fuel is only 2% of the vapor pressure the fuel would normally have. This means that the fuel evaporates at such a slow rate that it is essentially inconsequential and as time goes on the fuel dilution just continues to increase as fuel is being added to the mixture much faster than it can evaporate. This is exactly what the labs are telling us and what my C4S has done, they just don't know why.

The fact of the mater is that fuel does not just "flash off" regardless of temperature. That is just a myth based on a knee jerk comprehension of the problem. Unfortunately, it seems that this myth is believed by many people in important places that should know better. Then those of us that do not have the comprehensive background needed to understand this stuff are deluded into thinking that the myth is true. And away we go.

These discussions are great stuff for all of us. They force use to review and rethink our beliefs in search of the truth not one ups man ship. None of us should feel stupid in the face of these discussions as none of us can learn absolutely everything about this universe.

We all specialize in our own area of expertise.

Come on JFP. I know you miss me.

Edited August 16, 2014 by Mijostyn

[JFP in PA]

A little light reading (from the UC Davis site) on Raoult's Law and how it applies to very dilute ideal solutions (only where the volume change upon mixing is zero, which is not the case of fuel mixing into engine oil) and in which the solute is non-volatile - for example, a solution of salt in water, which obviously does not apply to fuel, which is a volatile.  Be sure to pay particular attention to the "Limitations of Raoult's law" section below:

 

 

RAOULT'S LAW AND NON-VOLATILE SOLUTES

This page deals with Raoult's Law and how it applies to solutions in which the solute is non-volatile - for example, a solution of salt in water. A non-volatile solute (the salt, for example) hasn't got any tendency to form a vapour at the temperature of the solution.

It goes on to explain how the resulting lowering of vapour pressure affects the boiling point and freezing point of the solution.

 

 

 

Raoult's Law

There are several ways of stating Raoult's Law, and you tend to use slightly different versions depending on the situation you are talking about. You can use the simplified definition in the box below in the case of a single volatile liquid (the solvent) and a non-volatile solute.

The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

In equation form, this reads:

In this equation, P^o is the vapour pressure of the pure solvent at a particular temperature.

x_solv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent.

You calculate this using:

Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar. The total number of moles is therefore 10.1

The mole fraction of the water is:

A simple explanation of why Raoult's Law works

There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Because of the level I am aiming at, I'm just going to look at the simple way.

Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again.

Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution.

A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 or 1 in a million, or whatever). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time.

But it won't make any difference to the ability of molecules in the vapour to stick to the surface again. If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place.

The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning.

If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower.

Limitations on Raoult's Law

Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law.

Features of an ideal solution

In practice, there's no such thing! However, very dilute solutions obey Raoult's Law to a reasonable approximation. The solution in the last diagram wouldn't actually obey Raoult's Law - it is far too concentrated. I had to draw it that concentrated to make the point more clearly.

In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent. The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event!

Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion.

Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away.

If there were strong solvent-solute attractions, this proportion may be reduced to 1 in 2000, or 1 in 5000 or whatever.

In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behaviour.

The nature of the solute

There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law (beyond the scope of this site). You may have noticed in the little calculation about mole fraction further up the page, that I used sugar as a solute rather than salt. There was a good reason for that!

What matters isn't actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt.

So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation.

Unless you think carefully about it, Raoult's Law only works for solutes which don't change their nature when they dissolve. For example, they mustn't ionise or associate (in other words, if you put in substance A, it mustn't form A₂ in solution).

If it does either of these things, you have to treat Raoult's law with great care.

 

 

Note:  This isn't a problem you are likely to have to worry about if you are a UK A level student. Just be aware that the problem exists.

 

Raoult's Law and melting and boiling points

The effect of Raoult's Law is that the saturated vapour pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent.

The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points.

 

 

Note:  In common with most phase diagrams, this is drawn highly distorted in order to show more clearly what is going on.

 

If you haven't already read my page about phase diagrams for pure substances, you should follow this link before you go on to make proper sense of what comes next.

Use the BACK button on your browser to return to this page when you are ready.

 

The line separating the liquid and vapour regions is the set of conditions where liquid and vapour are in equilibrium.

It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapour pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams (follow the last link above).

If you draw the saturated vapour pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water.

 

 

Note:  The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong!

 

Suppose you have a solution where the mole fraction of the water is 0.99 and the vapour pressure of the pure water at that temperature is 100 kPa. The vapour pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapour pressure of the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't!

 

If you look closely at the last diagram, you will see that the point at which the liquid-vapour equilibrium curve meets the solid-vapour curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapour all in equilibrium with each other at the same time.

Since the triple point has solid-liquid equilibrium present (amongst other equilibria), it is also a melting point of the system - although not the normal melting point because the pressure isn't 1 atmosphere.

That must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium line) passing through the new triple point. That is shown in the next diagram.

Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines).

Because of the changes to the phase diagram, you can see that:

• the boiling point of the solvent in a solution is higher than that of the pure solvent;

• the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent.

We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions.

The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that doesn't affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed.

You will find it makes no difference whatsoever.

[jl-c]

Wibble , Wibble , Wibble, (gently rocking to and fro!!) :-)

OK enough........ Brain is in overload!

[JFP in PA]

Wibble , Wibble , Wibble, (gently rocking to and fro!!) :-)

OK enough........ Brain is in overload!

 

A little bit of science often goes a very long way; it is a lot like nuclear waste, interesting in small doses and from a distance, but most people do not want to get any on themselves.................. :thumbup:

[Ahsai]

My 2cents below just because it's fun :)

1. Raoult's law applies to volatile solution mix too. There are some examples about benzene and toluene mixture on the same page John quoted above

2. Raoult's law applies only to an enclosed container environment. Once we introduce vacuum (~5mm H2O), the law may not apply. With vacuum, there's no concept of vapor pressure anyway.

3. Raoult's law applies only if the solute does not chemically form a different substance with the solvent (also from the page John quoted above)

I tend to agree with John that part of the fuel reacts chemically with the engine oil, which may not be boiled off at engine temp. However, for the part that doesn't react with the engine oil, it should get burned off easily due to its high vapor pressure (much higher than water) and the slight crankcase vacuum and engine temp. Just think about how well pure fuel evaporates even within the gas tank even without vacuum and engine temp applied.

So I think it boils down to how much of gas chemically reacts to the engine oil.

[JFP in PA]

So I think it boils down to how much of gas chemically reacts to the engine oil.

 

Which I've been told by multiple sources is very small.

 

And just for the record, I  am not say that Raoult's law is wrong, but I am questioning its applicability to the situation.

[Ahsai]

So I think it boils down to how much of gas chemically reacts to the engine oil.

 

Which I've been told by multiple sources is very small.

 

And just for the record, I  am not say that Raoult's law is wrong, but I am questioning its applicability to the situation.

Understood and we are saying the same e.g., Raoult's law assumes the presence of vapor pressure that exists during equilibrium where the number of molecules escaping the liquid and those entering the liquid are the same for the same solute. That can't be true in the slightly vacuum crankcase.

[JFP in PA]

 

 

So I think it boils down to how much of gas chemically reacts to the engine oil.

 

Which I've been told by multiple sources is very small.

 

And just for the record, I  am not say that Raoult's law is wrong, but I am questioning its applicability to the situation.

Understood and we are saying the same e.g., Raoult's law assumes the presence of vapor pressure that exists during equilibrium where the number of molecules escaping the liquid and those entering the liquid are the same for the same solute. That can't be true in the slightly vacuum crankcase.

 

 

One of the things that bothers me most about Raoult's law is that  it only applies to "ideal solutions", which tend to be rare in the real world.  One characteristic of an ideal solution is that when the solvent and solute are mixed, there is no change in total volume.  Anyone can pour 95 cc of fresh engine oil into a graduated measuring cylinder, then add 5 cc of fuel and mix, and they will end up with 100 cc of total mix and a 5% fuel dilution level.  Oil and fuel do not form an "ideal solution", so Raoult's law does not apply.

[Ahsai]

So I think it boils down to how much of gas chemically reacts to the engine oil.

 

Which I've been told by multiple sources is very small.

 

And just for the record, I  am not say that Raoult's law is wrong, but I am questioning its applicability to the situation.

Understood and we are saying the same e.g., Raoult's law assumes the presence of vapor pressure that exists during equilibrium where the number of molecules escaping the liquid and those entering the liquid are the same for the same solute. That can't be true in the slightly vacuum crankcase.

 

One of the things that bothers me most about Raoult's law is that  it only applies to "ideal solutions", which tend to be rare in the real world.  One characteristic of an ideal solution is that when the solvent and solute are mixed, there is no change in total volume.  Anyone can pour 95 cc of fresh engine oil into a graduated measuring cylinder, then add 5 cc of fuel and mix, and they will end up with 100 cc of total mix and a 5% fuel dilution level.  Oil and fuel do not form an "ideal solution", so Raoult's law does not apply.

That too but at least at low concentration (which is the case here with 9L of engine oil with small amount of fuel dilution), the law can still act as a good approximation (per the we page you referred to). To me, the whole premise of vapor pressure breaks down because vapor pressure only exists in a closed system http://en.m.wikipedia.org/wiki/Vapor_pressure where molecules are constantly going back and forth between the gas and liquid to attain equilibrium, whereas in the crankcase, molecules in the gas are drawn out constantly to the intake manifold for burning and never returned to the oil in the crankcase.

[Mijostyn]

^Mijostyn,

^{I always enjoy reading your posts because of your analytical approach to the subject at hand. The subject of which oil brand and viscosity to use in which environment concerns me because of the risk of Intermediate shaft bearing failure and/or cylinder scoring. I can't help but think that, as you surmised, the environment has more of an impact than many of us realize. I live in Phoenix. and for a large part of the year, the oil temperature is never below 100 degrees, even when parked overnight. My 997 3.6 is as much a grocery getter as a long haul car and I put 14-16,000 miles a year on it. I am a bit paranoid about oil dilution and breakdown in the severe temperatures we see in Phoenix and use Motul Excess 5-40 with a 3,500 mile change interval. I realize that  people might think that I'm wasting oil and filters to some extent with that interval, but both are cheap compared to an engine failure. What I'm building up to here, is my concern about oil pressure in high temp environments. With oil temperature in the 225-230 degree range, my oil pressure drops to just over 1 Bar at idle while motoring about town or in heavy traffic. Once on the move, oil pressure goes up to 3 1/2-4 bar. I've considered going up a step to 10-40 or 10-50 during the summer months, but have heard that the vario-cam system is very sensitive to viscosity levels above 5-40 because of the small oil passages. You've done more research than most on the subject of which oil to use and I'd appreciate your thoughts on what direction I might go to bump up my low RPM oil pressure.}

^{Thx, Lyn.}

Hey Lyn, I almost missed your post! You live in a very warm environment. I think the Motul Excess 5W 40 you are using is perfect. Your oil pressure is absolutely fine. Just because your oil pressure looks higher with thicker oil does not mean things are squirting any better.

Your engine has oil passage ways and jets with holes of a specified diameter to work well with oils of a specific viscosity. The thicker the oil gets the harder it is to squirt which is why your pressure goes up with cold oil and oil of a higher viscosity. If the oil is too thick you may not spray stuff like your pistons as well! If you were just racing a car a 10W 50 oil would be OK as the oil temps are going to be much higher for a sustained period of time but you can't drive your car on the street like that.

You can't change your oil too early, only too late. But, I think you could easily and safely go longer than 3500 miles. Our modern synthetic oils are very tough. The limiting factor is the fuel dilution. Just do serial oil analysis on your car every 3000 miles. It only costs about $25.00. Once your oil is 2% diluted you are done. In my car that happens by 6000 miles. The oil otherwise is in great shape.

Just change your oil filter and send the oil in the filter housing to the lab, replace the filter and top your oil off, about 1/2 liter.

[Mijostyn]

A little light reading (from the UC Davis site) on Raoult's Law and how it applies to very dilute ideal solutions (only where the volume change upon mixing is zero, which is not the case of fuel mixing into engine oil) and in which the solute is non-volatile - for example, a solution of salt in water, which obviously does not apply to fuel, which is a volatile. Be sure to pay particular attention to the "Limitations of Raoult's law" section below:

RAOULT'S LAW AND NON-VOLATILE SOLUTES

This page deals with Raoult's Law and how it applies to solutions in which the solute is non-volatile - for example, a solution of salt in water. A non-volatile solute (the salt, for example) hasn't got any tendency to form a vapour at the temperature of the solution.

It goes on to explain how the resulting lowering of vapour pressure affects the boiling point and freezing point of the solution.

Raoult's Law

There are several ways of stating Raoult's Law, and you tend to use slightly different versions depending on the situation you are talking about. You can use the simplified definition in the box below in the case of a single volatile liquid (the solvent) and a non-volatile solute.

The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

In equation form, this reads:

In this equation, P^o is the vapour pressure of the pure solvent at a particular temperature.

x_solv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent.

You calculate this using:

Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar. The total number of moles is therefore 10.1

The mole fraction of the water is:

A simple explanation of why Raoult's Law works

There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Because of the level I am aiming at, I'm just going to look at the simple way.

Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again.

Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution.

A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 or 1 in a million, or whatever). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time.

But it won't make any difference to the ability of molecules in the vapour to stick to the surface again. If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place.

The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning.

If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower.

Limitations on Raoult's Law

Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law.

Features of an ideal solution

In practice, there's no such thing! However, very dilute solutions obey Raoult's Law to a reasonable approximation. The solution in the last diagram wouldn't actually obey Raoult's Law - it is far too concentrated. I had to draw it that concentrated to make the point more clearly.

In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent. The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event!

Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion.

Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away.

If there were strong solvent-solute attractions, this proportion may be reduced to 1 in 2000, or 1 in 5000 or whatever.

In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behaviour.

The nature of the solute

There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law (beyond the scope of this site). You may have noticed in the little calculation about mole fraction further up the page, that I used sugar as a solute rather than salt. There was a good reason for that!

What matters isn't actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt.

So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation.

Unless you think carefully about it, Raoult's Law only works for solutes which don't change their nature when they dissolve. For example, they mustn't ionise or associate (in other words, if you put in substance A, it mustn't form A₂ in solution).

If it does either of these things, you have to treat Raoult's law with great care.

Note: This isn't a problem you are likely to have to worry about if you are a UK A level student. Just be aware that the problem exists.

Raoult's Law and melting and boiling points

The effect of Raoult's Law is that the saturated vapour pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent.

The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points.

Note: In common with most phase diagrams, this is drawn highly distorted in order to show more clearly what is going on.

If you haven't already read my page about phase diagrams for pure substances, you should follow this link before you go on to make proper sense of what comes next.

Use the BACK button on your browser to return to this page when you are ready.

The line separating the liquid and vapour regions is the set of conditions where liquid and vapour are in equilibrium.

It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapour pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams (follow the last link above).

If you draw the saturated vapour pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water.

Note: The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong!

Suppose you have a solution where the mole fraction of the water is 0.99 and the vapour pressure of the pure water at that temperature is 100 kPa. The vapour pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapour pressure of the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't!

If you look closely at the last diagram, you will see that the point at which the liquid-vapour equilibrium curve meets the solid-vapour curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapour all in equilibrium with each other at the same time.

Since the triple point has solid-liquid equilibrium present (amongst other equilibria), it is also a melting point of the system - although not the normal melting point because the pressure isn't 1 atmosphere.

That must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium line) passing through the new triple point. That is shown in the next diagram.

Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines).

Because of the changes to the phase diagram, you can see that:

• the boiling point of the solvent in a solution is higher than that of the pure solvent;
• the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent.

We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions.

The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that doesn't affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed.

You will find it makes no difference whatsoever.

Thanks JFP. He is a little weird about they way he explains some things but, what the hey. EXACTLY as I said. Raoult's law explains the very general but observed behavior of molecules as they interact with each other in solution. Ahsai is absolutely right. This author chose a solution of polar molecules. Our example is a solution of non polar molecules. Remember, I said the law only works with molecules of like behavior. He does mention that the law is most accurate when the concentration of the solute is very low which is exactly what we have in our example. The solute, fuel is max 2% in our solvent , the oil. There is no line between solvent and solute by the way. It is any substance in solution. The vapor pressure of the oil essentially remains exactly the same as it composes 98% of the solution. The equation at the top, the one I mentioned in my previous post works just fine. Yes, the law works best under ideal circumstances which the crank case of an engine is not as Ahsai mentioned but as a generality it explains the situation quite well. So, when you argue about a hypothesis it is time to do an experiment.

My C4S has serially increasing fuel dilution. As time goes on more fuel is being added to the oil than boils off. (Measured by gas chromatography) so, everyone who is doing UOAs look back on your reports, plot your fuel dilution and tell us what you have. I will do the same experiment on my Turbo S. I should hit 3000 miles by the end of the Summer. (If I don't get arrested)

Oh, JFP, I noticed something interesting with the oil pressure in the Turbo. Say I'm going 75 mph in 7th gear, revs about 2000 rpm. If I step hard on the gas, boost goes up and the oil pressure shoots right up before the revs go anywhere. How does it do that?? Is the oil pump being operated by a separate electric motor?.

Edited August 18, 2014 by Mijostyn

[JFP in PA]

I've only seen a couple of the new non-Mezger Turbo engines to date, so I'm not really sure why the oil pressure would behave that way.  I wouldn't bet against the oil pump being electrically operated, but lean more towards a valve opening or closing somewhere in the oil system.  I'll let you know if I find an answer.

[JFP in PA]

Looks like you and I may both be right, from a Turbo press release: "To reduce drive losses and increase efficiency, an electronic on-demand oil pump is used. This means that the oil pump is operated at high power when there is high demand and at low power when there is low demand."

 

From what I can decipher from the press releases, the pump sounds mechanically driven, but with an electronic control system that can quickly vary oil pressure to engine demand.  This sounds like a bypass system which unloads the pump when the engine itself is not under load, which would also impact both HP and long term fuel consumption as well.  Apparently, Porsche has been using this concept for some time.

[Mijostyn]

Looks like you and I may both be right, from a Turbo press release: "To reduce drive losses and increase efficiency, an electronic on-demand oil pump is used. This means that the oil pump is operated at high power when there is high demand and at low power when there is low demand."

 

From what I can decipher from the press releases, the pump sounds mechanically driven, but with an electronic control system that can quickly vary oil pressure to engine demand.  This sounds like a bypass system which unloads the pump when the engine itself is not under load, which would also impact both HP and long term fuel consumption as well.  Apparently, Porsche has been using this concept for some time.

 Thanx JFP. It had to be something like that. The pressure jumps from 2 bar right up to 4 bar before the engine has gained 200 rpm.

 

The car is so complicated it is scary. I have a 7 year 100,000 mile warranty on it.

 

Finally did launch control last week. I did it twice and started to get sick. It pulls more Gs longer than any amusement park ride I can remember. I have no idea how that guy did it 50 times in a row.