Quote:
Originally Posted by rotrex
Mass air flow will increase and hence total oxygen.
Someone once did the numbers (enthalphy of evaporation, head capacity of air, vapor density of methanol etc.) and found that for every 1% of methanol (1% of total air flow), mass air flow increases by 2%.
Now evaporating an extra of 4 gram/sec (5 ml/sec or 300ml/min) of methanol wil increase air mass flow by 8 gram/second which would result in an other 10HP from the extra air alone.
Let's try an approximation outselfs:
Methanol has roughly 1.15 kJ/g heat of evaporation and a vapor density of 1.1 g/l. Let's say 1 kJ/g. Makes life easier. The vapor density is very similar to air. The heat capacity of air is roughly 1 kJ/kgK. So evaporating 4 g of methanol per second (300ml/min) consumes about 4kJ of heat per second (4kW). That 4kJ can cool 250g (good for 300HP) of air down by 16K. At the same time it releases 4 g of vapor. The 16 K temperature drop increases air density by about 4%. The vapor replaces about 4/250g or 1.6% of the air. So you gain 4% and lose about 2%. :-)
Water has twice ther heat of evaporation. But it evaporates very slowly compared to methanol. It therefore contributes much less to intake air cooling unless things are very hot.
At 20PSI air is very hot after the turbocharger. Say 150C. Here the water contributes a lot to cooling. Once things get below the boiling point, evaporation speeds drops a lot.
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The theory of methanol is good since meth itself is an oxygenated fuel.
Although water does not create oxygen while evaporation but more works as a potent air charge coolant vs meth.
Here comes the thing, water as the best air charge coolant dissipates more air than meth through evaporation.
So would't the theory of denser air work here?