2001 Dodge 2500, 370k, 18k miles on RT6

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Mid-OCI oil sample of RT6 CK-4.

From Polaris:
Truck Odometer: 370600 mi
Miles on oil: 18000
5 quarts of makeup oil added.
Fram Ultra filter.

Wear Metals:
(Element (ppm); 235k, 322k, 338k, 352k, 370k)
Iron; 42, 52, 30, 50, 38
Chromium; 3, 3, 1, 2, 1
Al; 4, 5, 3, 4, 3
Cu; 3, 4, 1, 2, 1
Lead; 9, 31, 4, 7, 2

Contaminants:
Silicon; 6, 13, 7, 9, 7
Sodium; 8, 6, 5, 4, 7
Potassium; 62, 0, 1, 1, 4

Multi-source metals:
Moly; 75, 76, 5, 4, 0
Boron; 83, 38, 112, 69, 106

Additive metals:
Mag; 1333, 1198, 179, 163, 94
Calcium; 960, 1111, 2174, 2198, 2053
Phos; 1138, 1104, 1012, 1048, 924
Zinc; 1439, 1471, 1277, 1271, 1143

Fuel Dilution; Soot; Water;
KV100 (cSt); 15.1, 15.1, 15.3, 15.3, 15.4
TBN; 6.54, 6.99, 5.69, 4.27, 4.96
Oxidation; 15, 16, 14, 14, 13
Nitration; 10, 10, 9, 9, 8

Polaris Comments:
No abnormal findings. Resample at normal interval.

A pretty boring UOA, just the way I like it. Only thing is a developing trend with soot, but not a serious amount.
Corrected Iron wear rate is 3.1ppm per kmile. That's OK.

I changed the filter after taking the oil sample. Fram Ultra not available at WalMart, so I bought a Purolator Boss.
Plan to extend the OCI to 36,000 miles.
 
"5 quarts of makeup oil added"

^ Does all the makeup oil skew your sample? What is the sump capacity?
 
Originally Posted By: gman2304
"5 quarts of makeup oil added"

^ Does all the makeup oil skew your sample? What is the sump capacity?


Yes, makeup oil skews the sample. It dilutes the wear metals and contaminants, bolsters the TBN, but doesn't change the additive package if you use the same oil as the original fill. I run an 11 quart fill in the truck, and add oil when it gets 1 qt down, so the average fill is 10.5 quarts. Adding 5 qts over the sample interval has caused a dilution of 5/10.5, or 48%. So to find the actual wear metal concentration, you would have to multiply the UOA readings by 1.48 to get what the concentration would be if there had been no oil consumption. The number on Iron that I quoted was based on the UOA value of 38 multiplied by 1.48.
 
Originally Posted By: A_Harman
Originally Posted By: gman2304
"5 quarts of makeup oil added"

^ Does all the makeup oil skew your sample? What is the sump capacity?


Yes, makeup oil skews the sample. It dilutes the wear metals and contaminants, bolsters the TBN, but doesn't change the additive package if you use the same oil as the original fill. I run an 11 quart fill in the truck, and add oil when it gets 1 qt down, so the average fill is 10.5 quarts. Adding 5 qts over the sample interval has caused a dilution of 5/10.5, or 48%. So to find the actual wear metal concentration, you would have to multiply the UOA readings by 1.48 to get what the concentration would be if there had been no oil consumption. The number on Iron that I quoted was based on the UOA value of 38 multiplied by 1.48.


Not quite true. It is an exponential formula, of the form 1 - exp(-at), where t = hrs use and a = constant which includes rate of consumption.
If half the oil is replaced gradually over time, for example, about 69.3% of the sump is "original" oil, the other 30.7% is replaced oil of various ages.
The full formula is
OSF = (1/R) * (1 - exp(-Rt/V)) OSF = oil stress factor in kWh/gm, R = specific oil consumption in gm/kWh, t = total time in hrs of OCI, V = specific oil capacity in gm/kW
you calculate your kW not from peak hp but using mean fuel consumption, average speed (if vehicle has a nav computer!) and an educated guess at specific fuel consumption: 250gm/kWh for gasoline, 200gm/kWh for diesel. Diesel weighs 3300gm/US gal, gasoline 3000gm/US gal. My Unimog averages 36mph and 9mpg. So 4 gal/hr or 13200gm/hr. That comes out to 66kW average power (peak is 191)
My oil capacity is 29.6L, oil weighs ~860gm/L, so specific oil capacity 29.6*860/66 = 384 gm/kW.
Oil consumption is 1L/2500mi or 860gm/4583kWh = 0.18765 gm/kWh.
I just went an OCI of 703 hrs (25600mi)
1/R = 5.33 kWh/gm of oil
The exponential is 0.71, so multiply 1 - 0.71 = 0.29 x 5.33 = 1.54 to give the oil stress factor. For NO oil consumption the formula reduces to OSF = t/V, in my case 1.83. So I reduced my OSF by about 16% with makeup oil during my recent OCI. Definitely different than one's "linear estimate" of 10/29.6 = 33.8%

Source CIMAC paper #21 2004 "Oil stress Investigations in Shell's medium speed laboratory engine"

Charlie
 
I think you are over-complicating the correction to a relatively simple problem in the calculation of concentration of a solution, and are introducing factors which you have no way of accurately calculating. You cannot use 200 g/kW*hr as an average for fuel consumption for a diesel engine. Yes, that is a good estimate for a diesel running at peak torque, but is not accurate for engines running at less than full load.

But you may be right that I am oversimplifying the correction factor for my contaminant concentration. I am looking for a correction factor to account for the makeup oil added prior to an oil sample so I can accurately assess the concentration of contaminants measured in the sample. If I were to keep running on an OCI until I added 11 quarts, there would still be 50% original oil in there, and 50% oil of varying ages. Maybe my linear calculation doesn't account for this.

How about this:

Concentration of contaminants = 1-(number of makeup quarts/(base oil fill volume + number of makeup quarts))
The series would go like this:
1 quart added: 1-(1/12)=.916
2 qts added: 1-(2/13)=.846
3 qts added: 1-(3/14)=.785
4 qts added: 1-(4/15)=.73
5 qts added: 1-(5/16)=.688
.
.
.
11 qts added: 1-(11/22)=.50

So at 5 quarts added, the reported concentration of Iron in my oil sample is diluted to 68.8% of what it would have been. Translating that sentence into an algebraic equation:
Reported concentration = .688 x Corrected concentration

So to get the Corrected concentration, the reported concentration has to be divided by the correction factor:
Corrected concentration = Reported concentration / .688
Corrected concentration = 38 ppm Iron/.688 = 55 ppm Iron

Have I gone wrong somewhere in my train of logic?
 
Originally Posted By: A_Harman
I think you are over-complicating the correction to a relatively simple problem in the calculation of concentration of a solution, and are introducing factors which you have no way of accurately calculating. You cannot use 200 g/kW*hr as an average for fuel consumption for a diesel engine. Yes, that is a good estimate for a diesel running at peak torque, but is not accurate for engines running at less than full load.

But you may be right that I am oversimplifying the correction factor for my contaminant concentration. I am looking for a correction factor to account for the makeup oil added prior to an oil sample so I can accurately assess the concentration of contaminants measured in the sample. If I were to keep running on an OCI until I added 11 quarts, there would still be 50% original oil in there, and 50% oil of varying ages. Maybe my linear calculation doesn't account for this.

How about this:

Concentration of contaminants = 1-(number of makeup quarts/(base oil fill volume + number of makeup quarts))
The series would go like this:
1 quart added: 1-(1/12)=.916
2 qts added: 1-(2/13)=.846
3 qts added: 1-(3/14)=.785
4 qts added: 1-(4/15)=.73
5 qts added: 1-(5/16)=.688
.
.
.
11 qts added: 1-(11/22)=.50

So at 5 quarts added, the reported concentration of Iron in my oil sample is diluted to 68.8% of what it would have been. Translating that sentence into an algebraic equation:
Reported concentration = .688 x Corrected concentration

So to get the Corrected concentration, the reported concentration has to be divided by the correction factor:
Corrected concentration = Reported concentration / .688
Corrected concentration = 38 ppm Iron/.688 = 55 ppm Iron

Have I gone wrong somewhere in my train of logic?


Yes, you are still oversimplyfying.
Because as oil gets consumed, both "original" oil and and added oil get consumed. If oil is consumed continuously and added continuously, the solution involves an exponential. If it is done in relatively small steps it still approximates an exponential solution. Applying the exponential approximation to the problem of an 11 qt pan and 11 qts of makeup oil, the "corrected ppm" goes up by 1/(ln 2) = 1/.693 ~ 1.4.
And there is another consideration. There are 2 reasons for UOA: Figure out what's happening in the motor, and seeing if oil is still OK for longer OCI next time. "Correcting" numbers is reasonable for the 1st reason but not for the second.
Changing the fuel consumption coefficient will obviously change results somewhat. But continuous mixing problems are always differential equations and the solution involves integration, so an exponential solution is very likely in any sort of continuous mixing problem.
Charlie
 
The accurate "correction factor" is (Rt/V)/(1 - exp (-Rt/V)). In other words "no oil consumption OSF" divided by "real OSF".

Charlie
 
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