So if 1Kg of aluminium takes 396 Kilojoules to melt to liquid: https://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html that means 10Kg will take 3.96 Megajoules of energy. Now if diesel has a specific energy of 44.8 megajoules per kilogram https://neutrium.net/properties/specific-energy-and-energy-density-of-fuels/ and a specific gravity if 0.83 that would be 44.8 x 0.83 = 37.18 megajoules per litre and I burn 15 litres to melt my 10Kg that would be 558 MJ of energy to melt 39.6 MJ worth of aluminium gives you 39.6/558 = 0.071 or 7.1% efficiency. I'd been told of efficiencies of 3% for a particular gas fired furnace so the math may be close but I'm not allowing for the rate of burn either or how fast you can even get heat into the target metal.
Forgive my ignorance of terminology. There are a great number of factors that remove us from the ideal figures given that could, if isolated and optimized, give us better results. Combustion efficiency (burner design and combustion chamber), insulation, mass and initial temperature of sample, initial temp and mass of furnace bore, etc, etc. It would seem to me that the only way to come to a true efficiency percentage is to somehow measure the amount of heat lost from the furnace as exhaust and outside furnace body heat loss. I’d have no idea how to go about that. Pete
The question is did you measure the fuel to both heat up the aluminium and to melt it? Heating things up includes furnace insulation and crucible as well as the aluminium and this takes a lot of fuel. Your calculations looks okay once the metal has reached it's melting point because then you are looking at adding latent heat to the metal plus offsetting the loss of heat through insulation and that would be an indication of furnace efficiency.
It's interesting to contrast the chemical energy of the fuel used to the energy needed to melt the load and assume everything else is a loss of energy, I expected there to be a large difference. That figure I used is the latent heat of fusion, how much energy is used to melt a material: 3.96 MegaJoules for aluminium. The specific heat of aluminium is 0.91 Kilojoules per kilogram per degree C https://www.engineeringtoolbox.com/specific-heat-metals-d_152.html so if we raise 10Kg to 100 degrees over the melting point for pouring that would be 0.91 x 10 x 100 = 9.1 Megajoules extra energy so that would be 39.6 + 9 = 48.6 MJ using 558 MJ of fuel, 48.6 / 558 = 0.0871 or 8.7% efficiency.
Well Jethro Bodeene, all you have to remember is...naught plus naught equals naught. Ooops, you might not be old enough to remember the Clampetts. I figure my efficiency by the number of aluminium melts I can get out of a 20lb propane tank. Around 12 A6's in the small furnace and 10 A10's in the larger one. I'm happy with that.
Bonz, I know all about the clampetts. I just know if it's not hot enough or fast enough, I throw more fuel at it! I tell my guys all the time, fuel is cheap, BURN IT!
I'm also interested in my furnace efficiency. I will have to get some measurements next time I have a casting session. I read that latent heat of fusion might not be the only heat we are interested in. As the link says it is the heat supplied at the melting point to change the state. As Petee points out there must be other variables as well. We first heat the metal and get it to temperatures with what is called Sensible heat. I think it's what Mark calculated on the following post although needs to be measured as the difference between the starting temp and the state change temperature (~660 deg C for aluminum). eg from 25 starting to 660, 635 degrees C. So the total energy needed should be Sensible Heat (energy needed to get to melting temp) + Heat of fusion (energy needed to change state from solid to liquid). Counter that with the energy supplied to the system from the used fuel and bingo. Mark's math looks correct to me as well.
I had to re-read the definition of latent heat of fusion to see that I didn't account for raising the aluminium from room temperature to melting point. According to Wikipedia: https://en.wikipedia.org/wiki/Enthalpy_of_fusion you treat the raising to temperature and the melting of the material as two separate problems, so to correct my calculations it would be: Original calculation with corrected math mistake: so if we raise 10Kg to 100 degrees over the melting point for pouring that would be 0.91 x 10 x 100 = 910 Kilojoules extra energy so that would be 39.6 + 0.91 = 40.51 MJ using 558 MJ of fuel, 40.51 / 558 = 0.0726 or 7.3% efficiency. Corrected calcs to include heating from room temps 25 °C to 760 °C: so if we raise 10Kg by 735 °C to 760 °C for pouring that would be 0.91 x 10 x 735 = 6.688 Megajoules extra energy so that would be 39.6 + 6.688 = 46.288 MJ using 558 MJ of fuel, 46.288 / 558 = 0.0829 or 8.3% efficiency. So correcting for my late night reading and math and as 3Dcasting rightly pointed out the energy to get it hot from room temp to 760 °C plus the energy to melt it to liquid for my furnace gives 8.7% efficiency. A sealed electric furnace is going to be much more efficient to one losing exhaust heat through a hole in the lid.
