Originally Posted by RDY4WAR
Related to the topic, is there a way to work this backwards to get a ballpark of the HTHS at lower temperatures at KV100 and KV40? (which I guess would be "high temp" at that point, but you know what I mean) I'm curious for a drag racing application. Easing into the beams at the start, the oil temp is ~140*F. At the top end of the track, crossing the 1/4 mile, the oil temp is usually 160-165*F, reaching a max temp of 175-180*F on the idle/coast back to the pits.
Assuming that oil temperature in the bearings is ~60*F higher than in the pan (I'm not sure how accurate that is), this would put the peak bearing oil temperature at wide open throttle at about 220-230*F. It would seem an HTHS100 value would be more relevant here than an HTHS150. Is there anyway, with a known KV150, HTHS150, and KV100, that the HTHS100 could be calculated within a small margin of error and assuming the oil contains no VIIs?
I didn't study the temperature dependence of the VII shear but the paper did and they verified a nice formula that gives the temperature dependence.
In your case, you have a monograde oil; so, there is no VII. This makes the calculation of HTHS viscosity as a function of temperature straightforward if you neglect the shear of the detergent inhibitor (DI) pack.
Since there is no VII, you have:
HTHS(T) = density(T) * KV(T)
Again, this is neglecting the shear of the DI pack, which could be significant.
You can easily obtain KV(T) from KV40 and KV100 using the Widman operational-viscosity calculator based on ASTM D341:
https://www.widman.biz/English/Calculators/Operational.html
To calculate density(T), which decreases with the increasing temperature T, simply multiply density(15.6 °C) with the corresponding number in this table:
Code
T (° C) density(T) / density(15.6)
15.6 1.000
20 0.997
30 0.990
40 0.983
50 0.976
60 0.969
70 0.962
80 0.955
90 0.948
100 0.941
110 0.934
120 0.926
130 0.919
140 0.912
150 0.905
160 0.898
170 0.891
180 0.883
190 0.876
200 0.869
The values in the table are based on the formula:
density(T) = density(15.6 °C) * exp{-d * (T - 15.6) * [1 + 0.8 * d * (T - 15.6)]}
The density correction factor d varies from oil to oil, but the typical value of d = 0.000691325 I took usually gives 1% accuracy.
Note that we neglected the shear of the DI pack. However, the DI pack shears as well and you need to adjust your HTHS(T) values accordingly. To a first approximation, you can decrease the HTHS(T) by the same percentage for all temperatures as you see for the shear of the DI pack experienced at 150 °C since you know HTHS(150 °C) by measurement. In other words, if HTHS measured / HTHS calculated is 0.98 at 150 °C, decrease HTHS calculated by 2% for all temperatures -- not accounting for the variation of the shear with temperature but as a first approximation. Since we are talking about a few percent here, it shouldn't be a concern for practical purposes.
This also answers your previous question regarding why your monograde boutique oils appear to have VII in them according to my calculator: The DI pack shears as well. This was discovered in the paper about shear-thinning I linked above.