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Engine/Transmission (1998.5 - 2002) Anyone monitoring intake air temp?

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Engine/Transmission (1994 - 1998) lift pump

Engine/Transmission (1998.5 - 2002) Cooling ?'s

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Wow, alot of good posts on the subject. I originally started the thread just out of curiousity to find out what range of IAT my engine was subjected to. It sounds like it starts at slightly above ambient (no load no boost condition) to about 20 to 30 degrees > OAT under a normal load scenario.



How hot can it get, full load, full boost, all out? 300 degrees? too high or too low?



NoSeeUm, Using Case 1 above, what does T3 work out to using 75psia (simulating 60 psi boost) for P2?



Thanks for all the info...
 
NoSeeUm, Using Case 1 above, what does T3 work out to using 75psia (simulating 60 psi boost) for P2?



Thanks for all the info...



Case 4

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 75%

IE = 73%



T2 = 489F

T3 = 194F



Case 5

T1 = 85F

P1 = 12. 6psia (4000')

P2 = 73psia

CE = 75%

IE = 73%



T2 = 530F

T3 = 205F



Keep in mind that IE of 73% is basically a best guess as it is based on nominal standard conditions. Heat transfer rate is porportional to temperature difference and flow rate. So by raising T2 with T1 constant or or by increasing air flow the IE would likely increase.



Case 6

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 85%

IE = 73%



T2 = 441F

T3 = 181F



A better turbo.



Case 7

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 65%

IE = 73%



T2 = 551F

T3 = 211F



Not so good turbo.



Jim
 
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So, is there not a wind chill factor involved here?

Edited: I guess thats the flow rate variable in the previous post.

Edited again: Thinking out loud, I guess with wind chill, the air doesn't actually get colder, it just pulls the heat out of everything it touches faster which in our case cools the motor, but the air never got colder. Right?
 
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So, is there not a wind chill factor involved here?



Edited: I guess thats the flow rate variable in the previous post.



Edited again: Thinking out loud, I guess with wind chill, the air doesn't actually get colder, it just pulls the heat out of everything it touches faster which in our case cools the motor, but the air never got colder. Right?



I would suspect that situations such as high humidity, rain or fog would create a "natural" sort of external cooling effect that might contribute to reducing the internal temps within the intercooler tract to a value somewhat lower in relation to ambient than would otherwise occur. And yes, as I understand it, the effect of ambient airflow and humidity are what determines "chill factor"...
 
So it comes down to. .

How cold does the intercooler get while cruising at speed? If "wind chill" cools the intercooler enough to pull significant heat out of the charge air, then could it get below ambient?

We're talking about charge air that has been compressed and sliding across warm components all the way to the intake valves. The intercooler would have to be VERY efficient to outcool all the warming factors. Its not happening. .

Added after reading Gary's comment above: Agreed, and Im with you on this one. Dont see how the charge air can be lower than ambient without some kind of cooling mist or added factor. It would be helpful to know what the air temps are at say 30psi. .
 
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So it comes down to. .



How cold does the intercooler get while cruising at speed? If "wind chill" cools the intercooler enough to pull significant heat out of the charge air, then could it get below ambient?



We're talking about charge air that has been compressed and sliding across warm components all the way to the intake valves. The intercooler would have to be VERY efficient to outcool all the warming factors. Its not happening. .



Added after reading Gary's comment above: Agreed, and Im with you on this one. Dont see how the charge air can be lower than ambient without some kind of cooling mist or added factor. It would be helpful to know what the air temps are at say 30psi. .



You are correct in your thinking, almost. In your case, with higher humidity it is because the water in the air improves the heat transfer ability of the air and therefore the heat transfer ability of the IC. Remember, the higher air flow as well as the higher the density of the air and the greater the temperature difference the more heat will be transfered. Air flow is on both sides of the IC, just as it is for the temperatures. So as the truck goes faster the IC effecientcy will increase, because air flow on both sides of the IC is greater. Greater IE means effectively that the IC outlet temperature will decrease.



Basically the Heat Transfer rate formula goes something like:



(air flow outside * density factor / air flow inside * density factor) * ((temperature outside - temperature inside)*funny factor) * ((Heat Transfer co-efficient of the metal of the IC)*(contact area of the the IC)*(another funny factor))



It can get really complicated, with so many variables in the real world to make my head spin. I only have an inkling of calculus background, so the actual calculation is way beyond me. Thats why it is easiest to use a nominal value for IE of for example 73%.



Where:

T3 = T2 * IE



Case 4

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 75%

IE = 73%



T2 = 489F

T3 = 194F



Case 7

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 75%

IE = 85%



T2 = 489F

T3 = 145F



The concept of Wind Chill exists only for warm blooded creatures and weather forcasts. Where you end up with a "feels like" temperature, as opposed to an actual temperature. The laws of heat transfer i have described above hold true. With higher winds heat is being removed from your body faster, your body tries to hold the temperature inside constant so it feels colder. For example, water will not freeze at 34F (ASL) wether the wind is blowing 10mph or 50mph. However, if the temperature is 30F then the water will freeze at a much faster rate if the wind is blowing 50mph. An IC has no "feels like" temperature, it is only the actual temperatures that matter.



Jim
 
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Basically the Heat Transfer rate formula goes something like:



(air flow outside * density factor / air flow inside * density factor) * ((temperature outside - temperature inside)*funny factor) * ((Heat Transfer co-efficient of the metal of the IC)*(contact area of the the IC)*(another funny factor))



It can get really complicated, with so many variables in the real world to make my head spin. I only have an inkling of calculus background, so the actual calculation is way beyond me. Thats why it is easiest to use a nominal value for IE of for example 73%.



Where:

T3 = T2 * IE



Case 4

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 75%

IE = 73%



T2 = 489F

T3 = 194F



Case 7

T1 = 85F

P1 = 14. 69psia (sea level)

P2 = 75psia

CE = 75%

IE = 85%



T2 = 489F

T3 = 145F Jim





Uh, what? :-laf
 
BHAF unshielded

BHAF unshielded

Cruising near 60mph with



AIT (according to Juice) OutsideTemp

78 34

100 61

134 101

avg avg difference

116 38. 6 degrees
 
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