I used to have a re-mapped Mercedes S600 V12TT that caused all sorts of trouble, but was great fun on a good day. I spent a lot of time trying to figure out the charge cooling system, and modifying it to improve performance and reliability. I wrote this thread six years ago, and some of it's applicable to the RS6, especially as I'm planning to buy one and fit a charge cooling system.
https://mbworld.org/forums/m275-v12-bi- ... pumps.html
How much heat does a charge cooler have to remove? Engine cooling systems are mature and well-proven, so we know they work. The charge cooler is a bit more difficult. How hot would the intake air be if it wasn’t cooled? When you compress air, it gets hotter, and the pressure increases by more than the volume decreases. If its compressed quickly and doesn’t have time to cool, then it’s an adiabatic compression, for which the equation is:
P1*V1exp1.4 = P2*V2 exp1.4
(where 1.4 is the ratio of constant pressure / constant volume specific heat capacity for air)
This shows that if the intake air is compressed to 2 bar at some temperature, then the volume of the air becomes 61% of the original volume. Knowing the compressed pressure and volume, we now use the general gas equation to get the compressed air temperature:
P1*V1/T1 = P2*V2/T2
If the ambient temp is 300K/27C, the compressed temp becomes 366K/93C. So one bar of boost increases the charge temp by typically 66C just because of the increase in pressure, and that’s an intrinsic change – it doesn’t depend on engine size, speed or power. Having said that, it will probably be higher, as hot turbos tend to heat the intake air even when it’s not being compressed. And if the boost is more than one bar, both the temperature and the mass flow will be higher, so there’s a double whammy for tuned engines.
The air consumption is roughly 5000 rpm times 5 litres equals about 200 litres/sec. The density of 2 bar air is about 2g/litre at those temps, so that’s 0.4kg/sec of air flow at full power. The heat capacity of air is 1.0 kJ/kg/C, so assuming the charge cooler needs to drop the temp at least 50C, that means a cooling dissipation of 0.4kJ/s/C x 50C which equals 20kW.
Now, what about the IC coolant? Looking at the pump characteristics chart, it seems the normal operating point is around 20lpm at 30kPa. So the coolant mass flow rate happens to be very similar to the air mass flow rate – though we do need the reduction in air temp to be greater than the increase in water temp.
The heat capacity of water is 4.18kJ/K/kg, so the coolant has a capacity of about 1.4kW/K. At full power, that means a temp increase of 20/1.4 or 14C, or around a quarter of the air temp reduction, as you’d expect from the flow rates and heat capacities. I’m not certain, but from a few articles I’ve read I think that’s pretty similar to the temp drop across the engine radiator. Most people expect to see around 10C or 20F across any radiator, whether it’s on a go-cart or a nuclear power station.
My first impression is that the stock IC system was designed against the requirements of the stock engine, at best.
Second, I hadn’t considered that tuning would increase the cooling requirements exponentially – due to the collective increases in intake air pressure, temperature, density and flow rate.
Third, I’m also surprised that the heat carrying capacity of the water isn’t higher – if it was comfortably high enough, the temp delta might be a few degrees, but 14C sounds kind of marginal for a low temp system.
Fourth, in my earlier reasoning I hadn’t considered how high the intake air mass flow rate - and hence the cooling requirements - would be. If the coolant rises 14C in one cycle, that doesn’t give it much margin or storage capacity – just a few seconds at WOT and it’s all heated up.
Finally, none of this considers the heating contribution from the turbos themselves – aside from compressing the air.
By way of getting a feel for what charge coolers do, and what makes them tick, I've been expanding on some of my earlier sums. I used some simple assumptions and thermodynamics equations to estimate how much thermal energy and power was involved, but that was only a baseline. Upgraded intercooling is often wanted for re-mapped engines, but I said before that re-mapping makes things exponentially harder for the IC. But how much harder? I ran the sums for a range of absolute MAP starting with ambient.
P,V,T,D are absolute pressure, volume, temperature, density & 1 = ambient, 2 = compressed
P1 . . . . . V1 . . . . . T2/T1 . . . T1 . . . . T1 . . . D1
1.0 bar . . 1.0000 . . 1.0000 . . 300K. . . 27C . . . 1.000
P2 . . . . . V2 . . . . . T2/T1 . . T2 . . . . T2 . . . . D2
1.5 bar . . 0.7485 . . 1.1228 . . 337K. . . 64C . . . 1.336
2.0 bar . . 0.6095 . . 1.2190 . . 366K. . . 93C . . . 1.640
2.5 bar . . 0.5197 . . 1.2993 . . 390K. . . 117C . . 1.924
3.0 bar . . 0.4562 . . 1.3687 . . 410K. . . 137C . . 2.192
V2 = V1*(P2/P1)exp-0.71428 (negative reciprocal gamma)
T2/T1 = (P2/P1)*(V2/V1)
D2 = D1*(V1/V2)
These figures show what happens to compressed intake air if you don't cool it down. The density column shows how much extra air mass flow you get for the extra boost pressure. If you could keep the IAT down to ambient, D2 would be the same as P2, but it illustrates the loss of charge density and presumably torque caused by the IAT increase.
Consider the case of a tuned turbo with the boost turned up from 1.0 bar to 1.5 bar. Relative to stock, the mass flow goes up by 1.924/1.640 or 17%, but the air temp goes up 90/66 or 36%. Therefore the total extra thermal load power on the charge cooler goes up by 1.17 x 1.36 or 60% over stock. That 17% extra air mass flow probably translates quite directly into a 17% increase in torque. Turbo tuning doesn't increase power as much as torque due to turbo throttling and volumetric efficiency, so real world peak power probably goes up 10%.
Therefore a 10% power increase costs you a 60% increase in charge cooler thermal loading. That's what I meant by exponential increase. So for a typical tuned car, the IC needs to remove 20kW x 1.6 = 32kW heat power. Divide that by 1.4kW/C stock cooling capacity gives 23C, which is the coolant temp increase in one pass through the IC. This suggests the charge cooler needs to be upgraded.
As I see it, the turbos put energy into the intake charge in three ways:
1. Heating the air by compressing it (as above)
2. Heating the air because the turbos are hot (see below)
3. Increasing the pressure of the air
I understand (1) through the posts above. It’s a necessary by-product of forced induction, and it also happens with natural aspiration during the compression stroke. The heat energy put into the air needs to be removed by the IC. The figures above don't quite match up - the IAT is higher in practice than (1) predicts, so I assume that (2) is a significant effect.
The useful part of what turbos do is (3), and (1) and (2) are just throwing away waste. However, I didn't have a feel what the useful power was, so here it is for interest.
•Air Power = 100 kPa x 200 l/sec = 20 kW
Turbos are at best three quarters efficient, but max power sits to the right on the turbo map, and max torque towards the top, so lets call it two-thirds. Therefore charge air heating due to compressor inefficiency is about half the compression power, or 10kW @ 1 bar, so the total heating is 30kW in this example. This gives a temperature rise of 99degC, which is consistent with what car manufacturers say (charge air temp around 130degC).
Therefore the heat power is greater than the pumping power, so most of the work done by the turbocharger just goes into heating the air, and needs to be thrown away.
Furthermore, when you increase boost pressure you increase the mass flow rate AND you increase the temperature two fold, so you get an exponential increase in heat power to dissipate.
This is why I've been saying that the heat dissipation load on intercoolers increase exponentially rather than linearly with increasing power.
Nick