by sergei » Sat Mar 20, 2010 11:49 pm
Has any one considered amount of energy needed to dissipate vs surface area and water volume?
There is a reason why intercoolers are bulky, even W/A ones that are generally smaller still, have huge surface area where water indirectly contacts the air flow.
That ring, even if you put 10 of them, hasn't got enough surface area to allow efficient heat transfer. The fins do not have capacity to conduct heat either. They are long and thin.
I can bring calculations into this... but I am too lazy.
Here is the basic break down:
Water heat capacity is ~4.2 kj/ kg K
In other words it means that 4.2kJ needed to raise temperature of 1kg (1L) of water by 1K (1'C).
Now lets assume following things: guestimate % of heat lost through turbo is ~10% of the power. So if 200kW engine 20kW is heat created by turbo.
20kW is 20kJ per second.
I assume the following: temperature drop is 30'C (before intercooler is 80'C, after is 50'C).
so we have the following numbers:
4.2 kJ / kg K
30K
20kJ / s
We can calculate water flow needed to soak that heat.
20kW / (30K * 4.2 kJ / kg K) = 0.159 kg / s (or ~9L/min).
That is all good and dandy, until you start calculating the capacity of certain area to conduct heat. What people expect is all of those 20kW go through that small area (~10sq cm). The smaller the surface the greater thermal resistance. Thermal resistance of that desing is huge.
Thermal conductivity of aluminium is: 237 W/ mK
The thickness of that unit (eyeballing) is ~ 5mm. The area ~ 10sq cm
So:
237 W/ mK * 30K * 0.001sq m / 0.005m =1420 Watts.
This unit with those parameters (temp difference is 30K and dimensions eyeballed from picture, you are welcome to supply real dimensions), the unit can only dissipate 1.42 kWatts!. That number is irrelevant as the fins are not able to conduct that heat to the rings anyway (same reason - low thermal conductivity).
To improve my calculations one can measure real dimensions (I need surface area of the area covered by ring and thickness of the tube wall).