vsv123

Ok, 5% of HiGear diesel cetane booster increase some power (comparable to 3% of poppers but still less), does not affect compression. Overall incomparable with 3% of poppers

Recycled Flyer

I am under the impression that a higher cetane rated fuel shortens the ignition delay thus allowing less compression to be used for a comparable setting.

How does lowering the comp to achieve a comparable setting relate to more power?

Does it add calorific value to the base fuel?

Recycled Flyer

No one know?

Its just that the original post claims a 50 to 60% power increase with less compression from the 3% addition of an ignition improver so I am calling shenanigans.

vsv123

Well this is only for Cox049 surestart an increase from 12000 RPM to 14000 RPM from an addition of ignition improver, and this is (14000/12000)^3 times more power output (1.17^3 = 1.59)

qazimoto

My Feedback: (1)

Well answered there vsv123

:-)

Recycled Flyer

Ok, I am learning here but there seems to be the power absorption coefficient of the propeller missing, as in BHP=P X RPM3.

Where P is that coefficient and running on the assumption that the recommended prop for a Cox 049 diesel is indeed a 6x3 the power increase between 14000 rpm and 16000 could be as low as 1 ounce/inch (going from 0.45 to 0.55) bearing in mind that small diameter props have a far flatter power increase curve than larger ones.

Or as high as 0.75 to 1.5 when using a 7X3 APC prop.

Based on -
http://adriansmodelaeroengines.com/c....php?cat_id=57

But either way 50% is a massive power increase from just a 3% fuel additive.

Thanks.

vsv123

Well, of course this is a mathematical estimate, which does not take into account propeller's efficiency. To calculate the exact amount of power increase we need to measure actual thrust as well as RPM

qazimoto

My Feedback: (1)

Gordon Connel's book has a good section on this. It's the traditional Mathematical Modelling Technique for expressing model engine performance used in engine tests since the early 1950's. For various reasons It's only a good estimation.

This is how I understand it, but it's off the top of my head without consulting my notes which are somewhere else..

Your equation is that for a Cubic Polynomial, which you may remember from Algebra at school.

It's the same as the general equation Y = M.X^3

where: Y = Power, M = the P Coefficient, a Constant and a number characteristic of a particular prop, determined (rather crudely) from it's ability to absorb engine power, and X = engine RPM on that prop.

Model engine power curves are assumed to be that close to a cubic polynomial curve that any difference doesn't matter.,

The same prop is used in both cases, with and without the DII in the fuel, so the K factor in both cases will be the same..

Guess a number for the K Factor, and use it in both places in the calculation. Since the general form of the equation has the same Constant in both Numerator and Denominator it cancels out.

So change in Power (really ratio of power) = (K x (rpm2)^3) / (K x(rpm1)^3)

Let's call K =1, rpm1 = 12000, rpm2 = 14000

Substituting, Ratio of Power = (1 x 14000^3) / (1 x 12000^3) = 1.587 or 1 to 1.587

So change in power is 59% which is what our Russian friend got.

The power increase is simply the difference between the Cox running very poorly without any DII and very well on sufficient DII.

Sounds fair enough to me.