Energy absorbed by evaporation

The number is called the latent heat of evaporation. It is about 0.4 MJ/kg, depending on the makeup of the gas.

The heating value is 46 MJ/kg.





Cheers

Greg Locock

thanks greg... from looking at the major oil company sites i've managed to find out that our blends (i'm pretty sure you're an aussie, yeah?) are fairly close to 50/50 propane/butane, so by following that i should be able to obtain most of the physical data i need.

or i could, if i knew exactly what i was looking for!

basically, i figure a 4 litre engine consumes 12000 litres of air/fuel at 6000 rpm.

i seem to recall the ideal air/fuel ratio being approximately 14:1? is that correct? is it measured by weight?

if so, i assume i could figure out what mass of air is consumed per minute, thus calculate the ideal quantity of LPGas to be consumed in the same minute, and therefore figure out how much energy the evaporating gas consumes?

i think as i'm typing it, it's starting to make sense in my head, but i'd still like to check with you if that's cool...

am i on the right track?


thanks,

col.

Here's some calculations for you.

Assuming 100% volumetric efficiency (which of course it won't be, but work with me here), the amount of air pumped by an engine per minute will be equal to the Displacement of the engine multiplied by the RPM divided by 2. For a 4 liter engine running at 6000 rpm, the formula works out as:

= 4 liters * 3000 = 12000 liters

I might point out that this is only true with the throttle wide open: in most normal driving the engine will be markedly throttled, and perhaps only running at 25% of maximum output.

If you recall your college chemistry, you will remember that at STP (standard temperature and pressure, which is about 70 deg F and one atmosphere), 22.4 liters of air will contain a mole of gas molecules. (A mole is 6.023 x 10**23 molecules, if I recall rightly.) The weight of a mole of a gas will be equal to the molecular weight in grams. For instance, if we have 22.4 liters of helium, the weight of the amount of gas will be 4 grams. If we have 22.4 liters of oxygen, the weight of the gas will 32 grams.

For air, a reasonable approximation for molecular weight is 29. If you have 12000 liters of gas a minute, that's equal to (12000/22.4) = 535.7 moles of air. At 29 grams per mole, you have 29*535.7 = 15,536 grams of air per minute. Since there are 453.6 grams per pound, this is equivalent to 15,536/453.6 = 34.2 pounds of air a minute.

Calculations such as this are typically made when trying to determine the appropriate size of a supercharger/turbocharger. Just for fun, let's look at the amount of horsepower it would require to power a turbo/supercharger compressor handling 34.2 pounds of air a minute, with just mild boosting -- seven psi.

To do this requires the use of the ideal gas tables, which I won't bore you with, other than to note that it will require about 16 BTUs per minute to boost one pound of air by 7 psi. If you are running 34.2 pounds, you will need 533.8 BTUs of work a minute to drive the compressor -- if you have a 100% efficient compressor. If you using a turbocharger with a 70% efficient compressor, then 762 BTUs/min will be needed from the exhaust turbine to power the compressor. Translating to horsepower, you are looking at about 18 horsepower for the compressor. If you are using a Roots type supercharger with 50% efficieny, you will need 25 horsepower from the crankshaft to make the compressor work.

If I read your statement sort of between the lines, are you thinking about using some of the vaporization effect as an enhancement for air mass (the poor mans intercooler)?

There is not enough heat tradeoff to provide much benefit. I have tried it in the past with little gain, usually wound up with a ball of ice. The air-exchange medium does not absorb heat quickly enough to recover for rapid LPG demands.

Franz

Doc, interesting reading, thank you.

Franz... yes, you read the invisible lines correctly. Basically, my logic went thus:

in a turbo/super-charged car, you often have an intercooler, to bring down intake temps. In an LPG powered car, you use radiator fluid to add heat to the vaporising gas, to prevent if from freezing. This is two independent systems, and neither of them will be 100% efficient...

So, I thought why can't the LPG cool the intake charge as it vaporises? It won't just be an intercooler, it would be an inter-refridgerator! What I'm trying to figure out is if, in theory, the vaporising gas will absorb enough energy to make an appreciable difference to the hot air coming from the turbo/super-charger.

As to coming up with a system that efficiently delivers the heat transfer, that's just the next hurdle to cross. I need to figure out if the AMOUNT of energy involved in the vaporisation could possibly make this worthwhile...

cheers,

col.

hi

Autocol, you need to compare the amount of heat absorbed by the LPG as it vaporizes with the sensible heat variation of the intake air across the IC. The heat load on the IC can be calculated by the formula

m.Cpair.(Tin - Tout)

where

m: mass flow of air
Cpair: air heat capacity (average between both temps.)
Tin: IC air inlet temperature (same as the temperature after the turbocharger)
Tout: IC air outlet temperature

the heat absorbed by the LPG vaporization is given by the LPG mass flow times its latent heat of vaporization. The stoichiometric AF ration is 14.7 by weight.

combining the two:

m.Cpair(Tin-Tout)=m/14.7*latent heat

so the eq. becomes:

Cpair[J/kg.ºC]*(Tin-Tout)[ºC]=latent heat[J/kg]/14.7

Assuming Tin=120ºC, by solving this eq. you'll have Tout=93ºC, or, an equivalent IC efficiency of 40% assuming ambient air around 25ºC.

hope this answers to your question ans someone please correct me if I'm wrong.

the next problem, as you've corectly mentioned would be, if worthwhile, of course, finding the right equipment to transfer this amount of heat.

regards
Spereira