JFP in PA

This tool will work on both the M96 and 97 style engines (1997-2008 model years), all of which used the same RMS. The Turbo cars are different as they use a different version engine. This tool has also been evaluated against the nearly $500 OEM tool and was found to correctly install the PTFE RMS seal at the ideal 13MM depth.

It is hard to tell as there are too many un-dimensioned variables here (previous wear on the clutch, driving style, car usage, etc.). I'd suggest driving it and seeing if any problems pop up.

I find that acetone (nail polish remover) works very quickly and leaves no residue. The other solvents should work as well, but I have no reference point to their effectiveness as I have never tried them.

If you are referring to the inner sleeve, leave it in place, the tool should retain it when the seal goes in.

OK, pretty simple actually. If it is not out already, you need to pull the old seal. Normal way is to thread a small screw into the body of the seal and pull it out with pliers, being extremely careful to not let the screw scratch either the crank flange or the seal bore. Using a small lint free cloth and a bit of acetone, carefully wipe off any grease, oil, or residue left by the old seal. Both the bore and flange must be scrupulously clean, no fingerprints or any other materials should remain. I've pulled down this diagram to help with the explanation: Now slide the new seal over the assembly aid, being careful to not get fingerprints on the mating surfaces, clean nitrile gloves are the order of the day for this step. Attach the assembly aid (item #5 above) to the rear of the crank flange. I don't think the old flywheel bolts will do for this, my tool came with two bolts with small Torx style heads that fit inside the two openings. I sure you can find a metric Allan head bolt that would do the same job, but I think the old flywheel bolts have too large a shoulder on them which would get in the way. This assembly aid does not have to be overly tight, I just run it down with a 1/4 drive ratchet until they are snug. Now slide the insertion tool (#2 above) over the two guide pins and up against the seal, holding it there until you can get the large tightening screw started by hand and threaded down until it is holding the seal. Now, using a wrench or ratchet, carefully tighten the screw until the tool just bottoms against the assembly aid, which should set the seal at the prescribed 13MM insertion depth: Now remove the large screw and insertion tool, then the assembly aid, and admire your new correctly installed PTFE seal.

What codes, and what type of diagnostic tool was used to retrieve them? Thank you for your prompt and detailed reply. I just sent the seller an email asking what the codes are. He claims that he uses the Durametric tool. My seller just had surgery, so my reply is slow. Codes are 0130 and 5525. Seller claims that 0130 is mass air flow which he claims to have replaced twice. 5525 he claims is improper data from DME. Any idea as to what is wrong and how much it might cost to repair? OK, I am going to stick my neck out here because I obviously have not seen the car, and say "Walk away". P5525 is troublesome; it is not "improper data from the DME" but rather a data communications fault with the DME, which can cause the car to throw ABS, PSM and other codes. While there have been cases where P5525 was caused by leaking intercooler hoses, in the worst case scenario, a P5525 can result in having to replace the DME, which is anything but cheap. I am concerned about the DME being "reprogrammed" without knowing by whom and to what it was reflashed. I would also like to know if the DME has ever be reflashed to an aftermarket program, some of which cause long term problems with these cars. Porsche's are complicated, Porsche Turbo's are very complicated. You could be buying in to a real can of worms here, and I'm sure there are other Turbo's for sale out there which would pass a PPI with flying colors.

The electrical side would be covered in the wiring diagrams found in either the OEM service manuals or in aftermarket sources such as the Bentley manuals. The vacuum line routing would still be as Loren has already noted.

They are fusible links that go in the current distributor, which is in the upper dash on the passenger's side:

Not without some recoding; the car will see an airbag system fault because the air bag is no longer there.

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.

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.

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.

DOT 4 (or "super" DOT 4) fluids are backwards compatible to vehicles that used DOT 3. Basically, DOT 4 is a dryer, higher boiling point version of DOT 3.

I think the rationale for using the PIWIS stems from the use of the "thermal management system" on the 981/991 cars, which is an overly complicated, electronically controlled thermostat-like setup.

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.

Top of the engine, around and under the intake manifolds.

There are multiple vacuum line diagrams for these cars, which particular area are you looking for?

What codes, and what type of diagnostic tool was used to retrieve them?

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:

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, Po is the vapour pressure of the pure solvent at a particular temperature. xsolv 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.

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 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:

The Turbo oil level sensor has been a bit of a problem child over the years, some have had multiple failures while others have had none. Most likely case is you have either lost the sensor itself , or the wire has become disconnected or even eaten by rodents (not at all uncommon for cars that sit). The sensor is a rod like affair mounted in the oil tank and held in by one bolt. Retails around $125 or so.

Year and model of the car?

+1. It is often very difficult to correctly diagnose these cars without a Porsche specific scan tool (Durametric, PIWIS, PST II).