Lets start with some counter intuitive but nonetheless correct stuff.

there are four forces in physics, not three, not five, not nine, not any number other than 4. These four are

1/ Electromagnetism

2/ Gravity

3/ Weak nuclear force

4/ Strong nuclear force

Everything that happens inside an internal combustion engine, of any description, is defined by the electromagnetic force.

If you think this is hard to reconcile, for example what happens when you run out of lube and pick up a big end shell, it has nothing to do with gravity, it has nothing to do with the weak or strong magnetic forces, therefore it must be electromagnetism at work, even though there is no apparent electricity of magnetism evident, the failure of the oil shear etc happens at the molecular level because of electromagnetic forces and laws.

Similarly efficiency can be counter intuitive, just because there are 140,000 btu locked up in a gallon of diesel doesn't mean you can get 140,000 btu out of a gallon of diesel, what it does mean is that whatever part of the 140,000 you did not get out as obvious work done or obvious waste heat is still locked up somewhere in the system.

It takes about 1000 btu per pound of water to boil said water, this is an approximation of course, winter tap water is colder than summer tap water, not all boilers are equal, the council of industrial boiler operators will tell you about two identical stoker fired boilers in indiana, identical design and constructed side by side, both burning exactly the same fuel, but with very different performance characteristics....

so anyway, lets say it takes 1000 btu to boil a pound of water, and lets say we have a very efficient diesel boiler that has an efficiency of 70%.

1000 btu per pound of water.

1000 / 0.7 (efficiency) = 1428 btu applied to boiler heat exchanger.

of course the diesel flame is not homogenous or perfect, different parts of a flame are at different temperatures and different combustion efficiencies, so let say again we have a highly optimised flame that is 70% efficient.

1428 btu / 0.7 = 2040 btu potential energy in diesel burnt.

140,000 btu in a gallon of diesel, so 140,000 / 2040 = 68.6 pounds of water boiled per gallon of diesel.

so, is this boiler 70% efficient, because 70% of the heat energy generated by combustion of the fuel is transferred to the water, or is it 0.7 x 0.7 = 0.49 or 49% efficient, because only 49% of the heat energy contained in the fuel is transferred to the water?

If we say it is 70% efficient then we are saying 1000 btu is going into the water and 428 btu is going up the chimney

if we say it is 49% efficient then we are saying 1000 btu is going into the water, 428 btu of HEAT energy is going up the chimney, and 612 btu of wasted fuel potential energy is going up the chimney.

Nobody will buy a 49% efficient boiler when there is an identical boiler sat next to it that is 70% efficient, but the 49% efficient boiler is the only one telling the truth.

it gets worse.

we have boiled our water, now we wish to keep it at boiling point, ready for use, ok lets make is 99 degrees centigrade at sea level so we avoid issues about phase change and latent heat.

obviously maintaining our boiling water at a steady temperature is going to take far less energy that boiling it in the first place.

So lets say our boiler is badly insulated, and it takes 1000 btu to boil a pound of water, but 100 btu to keep it at that temperature for an hour.

so our burners, which we were told worked at a peak of 70% efficiency, and now turned right down, just enough to maintain the water temperature.

efficiency at this setting is no longer 70%, it is 50%, the water is 80 or 90 degrees closer to the flame temperature, less slope, less ease of energy transfer.

we are losing 100 btu to radiation, convection and evaporation from our WATER, so at 50% efficiency we need to apply 150 btu of heat to the boiler heat exchanger.

100 btu will go through the heat exchanger to the water, 50 btu will go up the chimney as waste heat.

but wait, at lower flame settings the combustion process is also less efficient, it is no longer 70% efficient, it is lets say 40% efficient, so 150 btu required at the heat exchanger / 0.4 = 375 btu of fuel.

140,000 btu per gallon of diesel / 375 = 373, since up there we boiled 68.6 pounds of water with our first gallon of diesel fuel, 373 / 68.6 = 5.43 hours that we can maintain our 68 pounds of fuel at boiling point with our second gallon of fuel.

OK, time to recap, bring back in the common sense.

We just boiled from cold 68 pounds of water, just over 6 gallons, about a cubic foot, quite rapidly with a pretty efficient diesel boiler in a few minutes, we did this for a gallon of diesel, so we are quite happy.

We then keep this just over 6 gallons of water "on the boil" for nearly six hours with our second gallon of diesel, which seems pretty reasonable, quite efficient no?

No?

look at those sums again.

Takes 100 btu per pound of water to MAINTAIN temperature, now it doesn't matter if it is 100, or 10, or 1000, but for our example I chose 100 btu.

it takes 150 btu of heat to transfer 100 btu into the water to MAINTAIN its temperature at boiling point.

Is efficiency 50%? or is it 0%, because there was no net energy increase in the water.

But we consumed fuel with a POTENTIAL 375 btu to get our 150 btu of heat.

375 btu of fuel energy = 100 btu transferred into the water + 50 up the flue as waste heat + 225 btu wasted fuel potential energy up the fuel.

if efficiency was 0% when we used the marketing droids 150 btu figure, what is efficiency when we use the true 375 btu figure?

www.cibo.org has some interesting whitepapers for those who want to look at this.

so back to listers.

140,000 potential btu in a gallon of diesel.

3400 btu in a kWh

so 140,000 / 3,400 = 41.17 potential kWh if we have 100% efficiency.

my 6/1 2.5 kW s-0-m burns a gallon every 4 hours so 41.17 / 4 = I should be getting about 10 kW out of it, I can get 2.5, so what is efficiency? 25% OK

If efficiency is 25%, and a quarter of the 35,000 btu worth of diesel burned every hour is coming out as electricity, where is the rest going? Waste heat? IS it fuck.

35000 btu hour fuel use, eg quarter gallon of diesel an hour.

2.5 x 3400 = 8500 btu coming out as electric

35000 - 8500 = 26500 btu left per hour.

26500 / 3400 (kWh) = 7.794 Kwh

does anyone here know of a lister that rejects heat like 8 x bar electric fires? I don't. There is about a kilowatt (3400 btu) to be had in radiant heat and and kilowatt (3400 btu) in the coolant circuit and a kilowatt and a half (4900 btu) in exhaust heat.

3400 + 3400 + 4900 = 11700

26500 - 11700 = 14800 btu disappearing into the ether, except it isn't, it is potential fuel energy, not heat, disappearing up the exhaust pipe.

if you install it in your basement you can trap the radiant and coolant heat, 6800 btu, and mebbe a quarter of the exhaust heat, 4900 / 4 = 1200ish btu, 8000 btu

so

35000 btu an hour, 8500 is coming out as electric. as an electricity generator it is 25% efficient.

35000 btu an hour, 8000 is coming out as useful heat if you cogen with the lister in the basement, 23% efficient

add em both together, 35000 btu, 16500 coming out as useful energy, 48% efficient.

35000 btu an hour as a space heater running at optimum efficiency = 70% efficient.

35000 x 70% = 24500 useful btu

35000 x 48% = 16800 useful btu

24500 - 16800 = 7700 btu, or about 2 kWh

and I say it that way because if you could find some theoretical stirling engine or heat pump that was 100% efficient, you could get that 2kWh back and equal the lister, near as dammit, on electrical generation while beating it hands down in cogen mode.

of course, the entire 24500 useful btu is THEORETICALLY available to power your stirling engine or peltiers, mega insulate your home and build in a dedicated heatsink area where all the heat energy is lost to the outside enviornment, if you could make your house 50% efficient as an insulator you would have up to 12000 btu of heat energy at your ideally located heatsink site up in the roof space, and since you only need 8500 btu of that 12000 converted to electric to equal the electrical output of the startomatic, you only need 65% odd efficiency.

could be done, should only cost about 300,000 bucks per household.

so efficiency ain't what you think it is, a 70% efficient space heater will never heat up a building with no doors, it will eventually chew through all the fossil fuel on the planet and still not warm up the building, except perhaps by global warming.

you cannot create or destroy energy, you can only transform it, and for mankind we can only get any kind of useful work out of that if we make that transformation happen in such a way that there are steep slopes, high differentials and such like, there are gigawatt hours of energy tied up in the temperature of the atmosphere, but you'll never run a truck off it. the slopes are too shallow and the differentials are too low.

efficiency is about making those slopes steeper, not creating or destroying energy, but making the slope steeper so it is easier to catch, the harder it is to catch, the less efficient it is.

if you measure your efficiency by the steepness of the slope, as most of you seem wont to do, you by definition exclude the energy required to make that slope steep enough in the first place.

here is an analogy for you.

Reality TV competition #1, the one you all want to play at.

Make a vehicle that will carry a man the maximum distance around this track on a gallon of diesel. points for speed and distance covered in 24 hours etc.

Reality TV competition #2, the one I am telling you all you ARE playing, like it or not.

Same as above, except your ENTIRE workshop (no gas axe, plasma cutter instead) is either hand tools or electrically powered, and you have 72 hours to build the car from scratch, and everything electrically powered must be driven by energy derived from your precious gallon of diesel, whatever you have left over in the generator after building the car you can tip into the car fuel tank to run the car)

Only after waking up to reality show #2 will you understand efficiency.