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An intercooler is a heat exchanger. That means there are
two or more fluids that don't physically touch each other but a transfer
heat or energy takes place between them. Turbo Regals made in 1986/87,
Turbo TAs, GMC Syclones and Typhoons all came with intercoolers to cool
down the hot compressed air coming from the turbocharger. Turbo Regals
and Turbo TAs use outside air as the cooling media; Syclones and Typhoons
use water. Turbo Regals made in 1985 and before did not have intercoolers
as original equipment.
At wide open throttle and full boost the hot compressed air coming from
a turbocharger is probably between 250 and 350 deg F depending on the
particular turbo, boost pressure, outside air temperature, etc.. We want
to cool it down, which reduces its volume so we can pack more air molecules
into the cylinders and reduce the engine's likelihood of detonation.
How does an intercooler work? Hot air from the turbo flows through tubes
inside the intercooler. The turbo air transfers heat to the tubes, warming
the tubes and cooling the turbo air. Outside air (or water) passes over
the tubes and between fins that are attached to the tubes. Heat is transferred
from the hot tubes and fins to the cool outside air. This heats the outside
air while cooling the tubes. This is how the turbo air is cooled down.
Heat goes from the turbo air to the tubes to the outside air.
There are some useful equations which will help us understand the factors
involved in transfering heat. These equations are good for any heat transfer
problem, such as radiators and a/c condensers, not just intercoolers.
After we look at these equations and see what's important and what's not,
we can talk about what all this means.
Equation
1
The first equation describes the overall
heat transfer that occurs.
Q = U x A x DTlm
Q
is the amount of energy that is transferred.
U is called the heat transfer coefficient. It is a measure
of how well the exchanger transfers heat. The bigger the number,
the better the transfer.
A is the heat transfer area, or the surface area of the intercooler
tubes and fins that is exposed to the outside air.
DTlm is called the log mean temperature difference. It is
an indication of the "driving force", or the overall average
difference in temperature between the hot and cold fluids. The equation
for this is:
DTlm = (DT1-DT2) * F
ln(DT1/DT2)
where DT1 = turbo air temperature
in - outside air temperature out
DT2 = turbo air temperature out
- outside air temperature in
F = a correction
factor, see below
Note:
The outside air that passes through the
fins on the passenger side of the intercooler comes out hotter than the
air passing through the fins on the drivers side of the intercooler. If
you captured the air passing through all the fins and mixed it up, the
temperature of this mix is the "outside air temperature out".
F is a correction factor that accounts for the fact that the cooling air
coming out of the back of the intercooler is cooler on one side than the
other.
To calculate this correction factor, calculate
"P" and "R":
P = turbo air temp out - turbo air temp in
outside air temp in - turbo air temp in
R = outside air temp in - outside air temp out
turbo air temp out - turbo air temp
in
Find P and R on "Fchart.jpg" (attached) and read F off the left
hand side.
This overall heat transfer equation shows
us how to get better intercooler performance. To get colder air out of
the intercooler we need to transfer more heat, or make Q bigger in other
words. To make Q bigger we have to make U, A, or DTlm bigger, so that
when you multiply them all together you get a bigger number. More on that
later.
Equation
2
We also have an equation for checking
the amount of heat lost or gained by the fluid on one side of the heat
exchanger (ie, just the turbo air or just the outside air):
Q = m x Cp x DT
Q is the energy transferred. It will have the exact same
value as the Q in the first equation. If 5000 BTU are transferred
from turbo air to outside air, then Q = 5000 for this equation AND the
first equation.
m is the mass flowrate (lbs/minute) of fluid, in this case either
turbo air or outside air depending on which side you're looking at.
Cp is the heat capacity of the air. This is a measure of
the amount of energy that the fluid will absorb for every degree of temperature
that it goes up. It is about 0.25 for air and 1.0 for water. Air
doesn't do a great job of absorbing heat. If you put 10 BTU into
a pound of air the temperature of it goes up about 40 degrees. If
you put 10 BTU into a pound of water, the temperature only goes up about
10 degrees! Water is a great energy absorber. That's why we
use water for radiators instead of some other fluid.
DT is the difference in temperature between the inlet and outlet.
If the air is 200 deg going in and 125 deg coming out, then DT =
200 - 125 = 75. Again, on the cooling air side the outlet temperature
is the average "mix" temperature.
If you know 3 of the 4 main variables on one side of the exchanger (the
amount of heat transferred, the inlet and outlet temperatures, and the
fluidís flow rate) then this equation is used to figure out the 4th. For
example, if you know the amount of heat transferred, the inlet temperature,
and the flow rate you can calculate the outlet temperature. Since
you canít measure everything, this equation is used to figure out what
you donít know.
Caveat:
These equations are all for steady state heat transfer, which we probably
don't really see too much under the conditions that we are most interested
in - drag race! Cruising on the highway you would definitely see steady
state. Perhaps at the big end of the track you may see it too, I don't
know. As various people on the mailing list have pointed out in the past,
the material of the intercooler itself will rise in temperature when you
hit full throttle, absorbing more heat than what these equations would
lead you to believe. For example, at steady state idle the intercooler
body may be at 100 deg F. At steady state full throttle it may be 175
deg F. The energy it takes to heat it up to that temperature comes from
the turbo outlet air, and so the cooling of that air is what is removed
by both the flowing outside air and the absorption of the intercooler
body. How long does it take to get to the new steady state? Beats me,
but the graphs I've seen of intercooler outlet temperatures over the course
of a quarter mile run lead me to believe that it is approached before
you get to the end of the quarter mile, since the intercooler outlet temperatures
reached a steady level.
So, now that we've got these equations, what do they REALLY tell
us?
- Heat transfer goes really well
when there is a large temperature difference, or driving force, between
the two fluids. This is shown in equation 1 as a large DTlm. It doesn't
go as well when there is a small temperature difference between the
two fluids (small DTlm). The closer you get the intercooler outlet temperature
to the outside air temperature the smaller DTlm gets, which makes the
heat transfer tougher.
- The difference between the intercooler
outlet temperature and the outside air temperature is called the approach.
If it is 100 degrees outside and your intercooler cools the air going
into the intake manifold down to 140 degrees, then you have an approach
of 40 degrees (140 - 100 = 40). To get a better (smaller) approach you
have to have more area or a better U, but there is a problem with diminshing
returns. Lets rearrange the first equation to Q/DTlm = U x A. Every
time DTlm goes down (get a better temperature approach) then Q goes
up (transfer more heat, get a colder outlet temperature), and dividing
Q by DTlm gets bigger a lot faster than U x A does. The upshot of that
is we have a situation of diminishing returns; for every degree of a
better approach you need more and more U x A to get there. Start with
a 30 deg approach and go to 20 and you have to improve U x A by some
amount, to go from 20 to 10 you need to increase U x A by an even bigger
amount.
- I would consider an approach
of 20 degrees to be pretty good. In industrial heat exchangers it starts
to get uneconomical to do better somewhere around there, the exchanger
starts to get too big to justify the added expense. The one time I checked
my car (stock turbo, stock IC, ported heads, bigger cam) I had an approach
of about 60 deg. The only practical way of making the DTlm bigger on
an existing intercooler is to only drive on cold days; if you buy a
better intercooler you naturally get a better DTlm.
- You can transfer more heat (and
have cooler outlet temps) with more heat transfer area. That means buying
a new intercooler with more tubes, more fins, longer tubes, or all three.
This is what most aftermarket intercoolers strive for. Big front mounts,
intercooler and a half, etc... are all increasing the area.
A practical consideration is the fin count. The area of the fins is
included in the heat transfer area; more fins means more area. If you
try to pack too many fins into the intercooler the heat transfer area
does go up, which is good, but the cooling air flow over the fins goes
down, which is bad. Looking at the 2nd equation, Q = m *
Cp * DT, when the fin count is too high then the air flow ("m")
drops. For a given Q that you are trying to reach then you have to have
a bigger DT, which means you have to heat up that air more. Then THAT
affects the DTlm in the first equation, making it smaller, and lowering
the overall heat transfer. So there is an optimum to be found. Starting
off with bare tubes you add fins and the heat transfer goes up because
you're increasing the area, and you keep adding fins until the it starts
to choke off the cooling air flow and heat transfer starts going back
down. At that point you have to add more tubes or make them longer to
get more heat transfer out of the increased area.
- Make U go up. You can increase
the U by adding or improving "turbulators" inside the tubes.
These are fins inside the tubes which cause the air to swirl inside
the tube and makes it transfer its heat to the tube more efficiently.
Our intercoolers have these, but I understand that more efficient designs
are now available. One of the best ways to increase the U is to clean
the tubes out! Oil film (from a bad turbo seal or from the stock valve
cover breather) inside the tubes acts as an insulator or thermal barrier.
It keeps heat from moving from the air to the tube wall. This is expressed
in our equation as a lower U. Lower U means lower Qs which mean hotter
turbo air temperatures coming out of the intercooler.
- Air-to-water. If we use water
as the cooling medium instead of outside air, we can see a big improvement
for several reasons: Water can absorb more energy with a lower temperature
rise. This improves our DTlm, makes it bigger, which makes Q go up and
outlet temps go down. A well designed water cooled exchanger also has
a much bigger U, which also helps Q go up. And since both DTlm and U
went up, you can make the area A smaller which makes it easier to fit
the intercooler in the engine compartment. Of course, there are some
practical drawbacks. The need for a water circulation system is one.
A big one is cooling the water down after it is heated (which means
another radiator). This leads to another problem: You heat the water,
and cool it down with outside air like the Syclone/Typhoon. You can't
get it as cool as the outside air, but maybe you can get it within 20
degrees of it. Now you are cooling the turbo air with water that is
20 hotter than the outside air, and you can only get within 15 degrees
of that temperature so coming out of the intercooler you have turbo
air that is 35 degrees hotter than outside! (turbo air is 15 deg over
water temp which is 20 deg over outside temp). You could have easily
done that with an air to air intercooler! But... if you put ice water
in your holding tank and circulate that... Then maybe the air temp coming
out of the intercooler is 15 deg above that or 45 to 50 deg. Hang on!
But after the water warms up, you're back to the hot air again. So,
great for racing, not as good for the street.
- Lower the inlet temperature.
The less hard the turbo has to work to compress the air then the lower
the temperature the air coming out of the turbo is. This actually hurts
the DTlm, but still if it's cooler going in it will be cooler coming
out. You can work the turbo less hard by running less boost, by improving
the pressure drop between the air filter and the turbo, or by having
a more efficient compressor wheel. You can also reduce the pressure
drop in the intercooler, which allows you to run the same amount of
boost in the intake manifold while having a lower turbo discharge pressure.
More on this later. If you can drop the turbo outlet pressure by 2 psi,
or raise the turbo inlet pressure by 1 psi, that will drop the turbo
discharge temperature about 16 degrees (depending on the compression
efficiency and boost level). If the turbo air is going into the intercooler
16 degrees colder then it may come out only 10 degrees colder than before,
but that is still better than what it was.
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Another aspect of intercoolers to be considered is pressure drop.
The pressure read by a boost gauge is the pressure in the intake manifold.
It is not the same as the pressure that the turbocharger itself puts out.
To get a fluid, such as air, to flow there must be a difference in pressure
from one end to the other. Consider a straw that is sitting on the table.
It doesn't having anything moving through it until you pick it up, stick
it in your mouth, and change the pressure at one end (either by blowing
or sucking). In the same way the turbo outlet pressure is higher than
the intake manifold pressure, and will always be higher than the intake
pressure, because there must be a pressure difference for the air to move.
The difference in pressure required for a given amount of air to move
from turbo to intake manifold is an indication of the hydraulic restriction
of the intercooler, the up pipe, and the throttle body. Let's say you
are trying to move 255 gram/sec of air through a stock intercooler, up
pipe, and throttle body and there is a 4 psi difference that is pushing
it along (I'm just making up numbers here). If your boost gauge reads
15 psi, that means the turbo is actually putting up 19 psi. Now you buy
a PT-70 and slap on some Champion heads. Now you are moving 450 gm/sec
of air. At 15 psi boost in the intake manifold the turbo now has to put
up 23 psi, because the pressure drop required to get the higher air flow
is now 8 psi instead of the 4 that we had before. More flow with the same
equipment means higher pressure drop. So we put on a new front mount intercooler.
It has a lower pressure drop, pressure drop is now 4 psi, so the turbo
is putting up 19 psi again. Now we add the 65 mm throttle body and the
pressure drop is now 3 psi. Then we add the 2.5" up pipe, and it
drops to 2.5 psi. Now to make 15 psi boost the turbo only has to put up
17.5 psi. The difference in turbo outlet temperature between 23 psi and
17.5 psi is about 40 deg (assuming a constant efficiency)! So you can
see how just by reducing the pressure drop we can lower the temperatures
while still running the same amount of boost.
I have seen some misunderstandings regarding intercooler pressure drop
and how it relates to heat transfer. For example, one vendor's catalog
implies that if you had little or no pressure drop then you would have
no heat transfer. This is incorrect. Pressure drop and heat transfer are
relatively independent, you can have good heat transfer in an intercooler
that has a small pressure drop if it is designed correctly. It is easier
to have good heat transfer when there is a larger pressure drop because
the fluid's turbulence helps the heat transfer coefficient (U), but I
have seen industrial coolers that are designed to have less than 0.2 psi
of drop while flowing a heck of a lot more air, so it is certainly feasible.
Pressure drop is important because the higher the turbo discharge pressure
is the higher the temperature of the turbo air. When we drop the turbo
discharge pressure we also drop the temperature of the air coming out
of the turbo. When we do that we also drop the intercooler outlet temperature,
although not as much, but hey, every little bit helps. This lower pressure
drop is part of the benefit offered by new, bigger front mount intercoolers;
by the Duttweiler neck modification to stock location intercoolers; by
bigger up pipes; and by bigger throttle bodies. You can also make the
turbo work less hard by improving the inlet side to it. K&N air filters,
free flowing MAF pipes, removing a screen from the MAF, removing the MAF
itself when switching to an aftermarket fuel injection system, the upcoming
3" and 3.5" MAFs from Modern Muscle, these all reduce the pressure
drop in the turbo inlet system which makes the compressor work less to
produce the same boost which will reduce the turbo discharge temperature
(among other, and probably greater, benefits).
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Wondering if your intercooler is up to snuff? The big test: measure
your intercooler outlet temperature! When I did this I got a K type thermocouple,
the thin wire kind, slid it under the throttle body/up pipe hose and down
into the center of the up pipe, and went for a drive. On an 80 to 85 deg
day I got a WOT temperature of 140 deg, for a 55 to 60 deg approach. That
tells me that I need more intercooler. If I can get the temperature down
to 100 deg, the air density in the intake manifold goes up by 7%, so I
should flow 7% more air and presumably make 7% more hp. On a 350 hp engine
that is 25 hp increase. On a 450 hp engine that's a 30 hp increase. Damn,
where's my check book…
Another check is pressure drop. Best way to check it is to find a pressure
differential gauge, which has 2 lines instead of the single line a normal
pressure gauge has. It checks the difference between the 2 spots it is
hooked up to, as opposed to checking the difference in pressure between
the spot it is hooked up to and atmospheric pressure, which is how a normal
pressure gauge works.
Hook one line of the gauge to the turbo outlet and one to (preferably)
the intercooler outlet. The turbo outlet/intercooler inlet pressure is
easy, just tee into the wastegate supply line off the compressor housing.
It would be nice to get the intercooler outlet pressure directly, but
there's no convenient spot to hook up to. Hooking into the intake manifold
(such as via the line to the boost gauge) is quite convenient, but gives
the total pressure drop: intercooler + up pipe + throttle body. That'll
give you a pretty good idea though.
Instead of the differential pressure gauge you could use 2 boost gauges,
one in each spot, but then you have to worry about whether both gauges
are calibrated the same, try to read both at the same time while driving
fast, etc AND you may spring (ie, ruin) the gauge on the turbo outlet
since when you close the throttle you get a big pressure spike that your
normal boost gauge never sees.
If you find more than 4 or 5 psi difference between the intercooler inlet
and intake manifold (and I'm just giving an educated guess here, you'd
probably want to refer to one of the intercooler manufacturers for a better
number) then I would suspect that a larger, lower pressure drop intercooler
would offer you some gains.
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Well, you've made it this far. If you'd like to see some examples
using the formulas outlined in the beginning, read on. If not, well, I'm
done. It's pretty easy to make a spreadsheet up to do all these calculations.
Please remember that all these numbers have been made up! Any resemblance
to real life is a happy coincidence.
Stock intercooler, stock turbo.
Given 40 lb/min air flow @ 300 deg F and 19 psig from the turbo to make
15 psig boost in the intake manifold; 85 deg F outside temperature; an
intercooler outlet temperature of 140 deg F has been measured, as has
the cooling air temperature of 160 deg. What is the UA of the stock intercooler?
First, calculate Q
Q = m * Cp * DT
Q = 40 lb/min * 0.25 BTU/lb-F * (300-140 F) = 1600 BTU/min
Calculate DTlm
DT1 = turbo air temperature in - outside air temperature out = 300 - 160
= 140
DT2 = turbo air temperature out - outside air temperature in = 140 - 85
= 55
P=0.74, R=0.47, F=0.875
DTlm = F*(DT1-DT2)/ln(DT1/DT2) = 0.875*(140-55)/ln(140/55) = 74.4/0.934=79.6
F
Calculate UA
UA = Q/DTlm = (1600 BTU/min)/79.6 F = 20.1 BTU/min-F
What is the cooling air flow?
Q = m * Cp * DT, or Q/(Cp * DT) = m,
m = (1600 BTU/min)/[0.25 BTU/lb-F * (160-85 F)] = 85.33 lb/min of outside
cooling air
Stock intercooler, big turbo
How will the same stock intercooler perform with a bigger turbo and more
boost?
Given 53 lb/min @ 350 deg F and 27 psig from the turbo to make 22 psig
in the intake; 85 deg F outside temperature. Cooling air flow is still
85.33 lb/min.
This requires some trial and error to solve since we don't know the intercooler
outlet temperature. There IS a way to calculate it directly, but that
involves some more equations and is a little tedious so I'll skip it and
do it the hard way, by assuming an intercooler outlet temperature and
then checking to see if it is right. I'll do that by calculating Q for
the overall exchanger and then Q for just the turbo air; if they come
out the same then my guess was correct.
m=53 lb/min, Cp=0.25, U*A=20.1
lets start by assuming that the intercooler outlet temp = 140
Q = m * Cp * DT
Then DT = (350 - 140 ) = 210 and Q = 2782.5 BTU/min
Cooling air flow = 85.33 lb/min
DT for the cooling air = Q/(m*Cp)
DT = 2782.5 BTU/min / (85.33 lb/min * 0.25 BTU/lb-f) = 130.4 F
since DT = T out - T in, then 130.4 = T out - 85 and T out = 215.4 F
So the cooling air inlet is 85 F and the outlet is 215.4 F, and the turbo
air inlet is 350 F and the outlet is assumed to be 140 F. Now calculate
DTlm:
P=0.792, R=0.62, and F=0.75
DT1=134.6, DT2=55
DTlm=(134.6-55)/ln(134.6/55) * 0.75 = 66.7
Now calculate a new Q, Q= UA * DTlm
Q=20.1*66.7=1340.7
Since this isn't the same Q we got when we assumed an outlet temp of 140
deg, we have to get a new outlet temp and run through all this again.
I'll assume a new intercooler outlet temp of 170.
Q=(m*Cp*DT)=2385
cooling air DT = 2385/(85.33*0.25) = 111.8
Cooling air outlet = 85 + 111.8 = 196.8
P=0.68, R=0.62, F=0.84
DTlm=97.3
Q=1954.7 still not close enough
Last try!
T IC out = 182
Q=(m*Cp*DT)=2226
cooling air DT = 2226/(85.33*0.25) = 104.4
Cooling air outlet = 85 + 104.4 = 189.4
P=0.63, R=0.62, F=0.88
DTlm=111.0
Q=2232 close enough
Well, this time the Q we guessed at (by guessing the IC outlet temp) and
the Q we calculated from the overall equation are pretty close, so we
can say we've found the answer. It appears that this intercooler, which
worked fine in a basically stock application (cooling the air to the intake
manifold to 140 deg F) isn't working as well in this high HP application,
being able to cool the air down to only 182 deg!
Last example: same turbo and air flow as before, but we have a new intercooler
with the same heat transfer coefficient but 50% more area (intercooler
and a half). We'll assume that it also flows 1.5 times the cooling air
flow.
U * A old IC = 20.1
U * 1.5 * A = 1.5 * 20.1 = 30.15 = UA for new intercooler
m turbo air = 53 lb/min, Cp = 0.25 BTU/lb-F, T in = 350 deg
m cooling air = 1.5 * 85.33 = 128 lb/min, Cp = 0.25 BTU/lb-F, T in = 85
F
Assume intercooler outlet temp = 140 F
Q = m*Cp*DT = 53 * 0.25 * (350-140) = 2782.5
cooling air DT = 2782.5/(128*0.25) = 87
Cooling air outlet = 85 + 87 = 172
P=0.79, R=0.41, F=0.85
DTlm=89.0
Q=2684 not too bad, we'll try it once more
Assume intercooler outlet temp = 142 F
Q = m*Cp*DT = 53 * 0.25 * (350-142) = 2756
cooling air DT = 2782.5/(128*0.25) = 86.1
Cooling air outlet = 85 + 86.1 = 171.1
P=0.78, R=0.41, F=0.86
DTlm=91.7
Q=2763 close enough
So this tells us that in this high performance car the intercooler-and-a-half
outlet temperature is about the same as the outlet temperature of the
stock turbo/stock intercooler car.
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