Here I am

Pre & Post Intercooler temps

Attention: TDR Forum Junkies
To the point: Click this link and check out the Front Page News story(ies) where we are tracking the introduction of the 2025 Ram HD trucks.

Thanks, TDR Staff

add propane?

Getrag oil opinions/reccomendations

Status
Not open for further replies.
FWIW- I took some temperature readings of the pre and post intercooler intake temps. I’ve been curious how much heat the turbo made and how much the intercooler cooled.



Pre intercooler temp readings were taken at the boot just before the intercooler inlet and the post temps were taken at the boot connecting to that air horn. Temp readings were taken using a thermocouple and a Fahrenheit temp adapter for a digital multi-meter.



Outside temp 65



Intake temps



Pre Intercooler



Idle= 85-90

In town= 85-100

Highway 70mph 2-5psi boost= 140-150

Highway accelerating steady 10psi boost= 170-180

WOT 28psi boost up to 90mph= 240-260



Post Intercooler



Idle= 85-90

In town= 70-75

Highway 70mph 2-5psi boost= 70-80

Highway accelerating steady 10psi boost= 115-120

WOT 28psi boost up to 90mph= 135-140



With AC on high add 5-10 deg to all post temps
 
Great info

I have been toying with a ramair setup for over a year now and have discovered that empty I can now maintain ambient temp at my airhorn at all times, and if I close off the ramair inlet the temp at the horn goes up 20* above ambient . havent towed with it yet but will soon, I have several pics on this in readers rigs if interested.



cheers, Kevin



#ad
 
Last edited by a moderator:
Were your temps taken with the probe in the air stream, or just the surface temp of the boot? The reason I am asking this is because your readings tell me that you have a 95% efficient turbo at 28 PSI of boost :eek: :eek: ..... and I would trade my PDR40 with you straight across for the turbo you are currently running (note the hint of sarcasim) for numbers like that.



The figures also tell me that your CAC is only 63% efficient.



I am assuming you took the temps off the external surface of the boots.
 
Just for comparison, I have an exposed junction TC in my intake horn (in the actual air stream).



On a load type dyno, I had an aftercooled air temp of only 91F at 38 psi of boost with the stock turbo on a 65F day... . no external fans either. This was a 15 seccond run, and the CAC had no time to heat soak.
 
Geez tough crowd :-laf



Diesel Freak- explain the 95% effiiency? I'm not sure what your refering too. Boost readings are taken from the head only. Now if you still want to trade turbos..... (no sarcasim here :D ) What are you using for a temp gauge?



Here is a pic of the thermocouple.
 
An adiabatic process is one in which heat does not enter or leave the system. Though this is not strictly true in a turbocharger, it is a close approximation over a very short time period. An adiabatic temperature or pressure change can be calculated from the volume change using the following two equations (which usually are found in your physics texts):



(T2/T1) = (V1/V2)^g-1 and (P2/P1) = (V1/V2)^g,



where g is the ratio of air's heat capacity at constant pressure to air's heat capacity at constant temperature; T1, P1, and V1 are the initial temperature, pressure, and volume; and T2, P2, and V2 are the final temperature, pressure, and volume. Taking the log of both of these equations, and knowing that log ab = b log a, the following equation relating adiabatic temperature change to pressure change can be derived:



(T2/T1) = (P2/P1)^[(g-1)/g].







Physics texts will tell you that for the diatomic gasses like N2 and O2 in air, g is assigned a value of 1. 4. By substituting and rearranging we get



T2 = T1 x (P2/P1)^0. 286.



Actually, the exponent in the above equation that you usually see is 0. 283, meaning that g is actually closer to 1. 395. You can use either number, the difference it makes in the calculations is about 0. 25%. We need to note that temperature in all of these equations must be absolute temperature in either Kelvin (centigrade scale) or Rankine (Fahrenheit scale). If you want to work in C then K = C + 273. 15. If you are more comfortable with F then R = F + 459. 69. Also note that 1 degree on the Kelvin scale equals 1 degree on the Centigrade scale, and 1 one Rankine degree equals 1 Fahrenheit degree. I prefer the Fahrenheit scale and will round off 459. 69 to just 460 for the examples here. I'll assume that we are at sea level at a standard pressure of 14. 7 psi with an ambient air temperature of 70F.



The expression (P2/P1) represents the turbocharger compressor pressure ratio (PR), where P1 is atmospheric pressure on the inlet side and P2 is the absolute pressure on the outlet side. T1 is the air temperature entering the compressor and T2 is the air temperature leaving the compressor.



Many people think that 12 psi in the manifold is a moderate boost level so I'll use that as an example. If we can assume there is a 2 psi pressure loss from the turbo to the manifold, then the PR at the compressor is about 1. 95 ( (14 +14. 7)/14. 7). Let's also assume we manage to keep the intake air temperature down to 800F (or 5400R). The air temperature leaving the compressor, due to adiabatic heating only, would be 194F which is 654R = 540R x (1. 95)^0. 286. This is an increase of 114F! But the air is actually hotter than that.



Turbochargers compress the air by increasing the velocity of the molecules. When the air molecules don't move in a direction toward the intercooler (IC), they serve only to heat the air. The more the air is heated above that predicted by adiabatic compression, then the less efficient the turbocharger is. Compressor flow maps usually show at various flow amounts and pressure ratios the measured efficiency of a turbocharger. Compressors usually operate in the 55-75% efficiency range depending on engine load, pressure ratio, turbine speed, and compressor size (65-70% is average, 75% or more is where we would like to be).



When we factor in compressor efficiency (CE) the final temperature of the air leaving the compressor is



T3 = T1 + (T2-T1)/CE.



If compressor efficiency is 70% in the above example, then the final temperature is 242F, more than hot enough to boil water. If we increase the PR to 2. 36 (18 psi manifold and 2 psi IC loss), the temperature would increase to 295F. These high temperatures explain why intercoolers are so important when boost pressure is above a few psi.
 
:--) :--)



You know the really depressing part of that... ..... it's the simplified version too... ... .

Have I mentioned I suck at formula's... ... :D;)



bob.
 
Good info. I've been wanting to try this for along time but just have'nt got around to it. I've seen quite a few posts over the years about 2nd gen guys (cac) post intercooler temps. running just 20-40* above ambient temperature. This is just proof to me that a 1st gen cac is not very efficient ( including Banks ) and our trucks could benefit from a 2nd gen intercooler. So, who will be the first to try to squeeze one in ( and succeed ) a 1st gen truck with a/c? I've seen this topic come up before but not much information. Where there's a will there's a way. :)

Kurt
 
Last edited:
As for Jay's 95% efficient compressor at 28psi boost... ... . For the given temperatures at 28psi, the CE does come out to about 95%. But at the lower pressure of 10psi, the CE is about 65% I think the problem is that the thermocouple does not have enough time to reach equilibrium with the surrounding air. This is due to the heat capacity of the probe sucking up heat and creating a lag in the temporal accuracy of the reading. (also, as the delta T decreases the rate of heat transfer also decreases) Using a probe with very low mass would be preferable.



Note: when performing these calculations you MUST use absolute temperature. Oh, and just to be a PIA, there are no "degrees" on the Kelvin scale,

log ab DOES NOT = b log a

log ab = (log a) + (log b)

log a^b = b log a :D



Cheers,

Sean
 
Last edited:
I think I know where the problem is then. The hose I used is of the cheap rubber variety. On the post CAC side I could hold 28psi for a few seconds. About the time lag of of the thermocouple. On the Pre CAC side the thermocouple would blow out of the hose almost as soon as I hit 28psi. I'll try it again with a better hose and see what I get.



Thanks for explaining the math side of it.



Jay
 
Originally posted by formula

This is due to the heat capacity of the probe sucking up heat and creating a lag in the temporal accuracy of the reading. (also, as the delta T decreases the rate of heat transfer also decreases) Using a probe with very low mass would be preferable.



Sean



My thoughts exactly... .



Note: when performing these calculations you MUST use absolute temperature. Oh, and just to be a PIA, there are no "degrees" on the Kelvin scale,

log ab DOES NOT = b log a

log ab = (log a) + (log b)

log a^b = b log a :D




sorry, that was a mistake in the original write up.



-Cliff
 
:rolleyes: I'll spend spring break working on this one...

I was beginning to think the 1st-Generation CTD was such a simple machine:eek:

Patrick.
 
Lets see if I have this straight. Taking my data using the Rankine scale



Ambient air temp = 65*

Std air psi = 14. 7

Turbo boost = 28 psi

CE = 70% or 0. 7



P1= 14. 7

P2= 28

Pressure ratio (PR) or (P2/P1) = 2. 9



T1= 65 + 460 or 525

T2= T1 x (P2/P1)^0. 286 or 525 x (2. 9)^0. 286 = 712



T3 = T1 + (T2-T1)/CE



525 + 187/ 0. 7 = 792



792 – 460 = 332*



100% CE would = 252*



Does this look right? Took me a while to figure out what "^" meant.



Now would someone be so kind as to explain this.



"log ab DOES NOT = b log a

log ab = (log a) + (log b)

log a^b = b log a"



Jay
 
Originally posted by jbolt

Lets see if I have this straight. Taking my data using the Rankine scale... ... ... ... ... ... ... ... ... ... ...

Now would someone be so kind as to explain this.



"log ab DOES NOT = b log a

log ab = (log a) + (log b)

log a^b = b log a"



Jay



At a glance, the calculations look ok (not enough time to check though)



As for the explaination of the properties of logarithms... ... in short I'll give the definition of "log" which is actually "log base 10" this means that for "log x = y" that "10^y = x" If you do some calculations, you will find that the above statements are true. I posted the correction to the original equations to prevent any errors when rearranging the equations. (T2/T1) = (V1/V2)^g-1 and (P2/P1) = (V1/V2)^g So, in order to derive the equaiton (T2/T1) = (P2/P1)^[(g-1)/g]. you would have to first go through the steps "log (T2/T1) = g-1 log (V1/V2)" and "log (P2/P1) = g log (V1/V2)"



Sean
 
Status
Not open for further replies.
Back
Top