Calculations and reflections on tractors doped with water.
Introduction: why this reflection?
After an unsuccessful experience of passing a tractor on a test bench and given the lack of obvious results, I did a little reflection on the figures given by farmers and published on the Quanthomme site.
Indeed ; The experiment which I carried out on a tractor MF188 of 1978 equipped with a Perkins 4248 engine did not show any difference in efficiency with or without water injection and this for a stable and fixed fixed load. That is to say, with or without water supply, the yield was neither improved nor degraded. This is in itself an astonishing point.
But it should be noted that the conditions were not ideal: old test bench undoubtedly lacking in precision, worn engine (consuming oil: 1 L / 4 h) modifications and measurements carried out hastily, and often under the rain (which is very pleasant!)! Finally, it must be said that the engine had just been modified. I think this may be important given some evidence of improvement over time.
So I decided to look, as a good scientist who became obviously skeptical, on the testimonies of the farmers, and you will see that some figures are astonishing of similarities! It is hard to believe in such coincidences starting from figures announced so different! That is to say that reports would tend to confirm that these testimonies are true. But it is obvious that only a passage on the bench could confirm these figures.
The published figures
This reflection is based on the following montages:
1) 22 assembly, Massey Fergusson tractor from 95 Cv: Click here
2) 23 assembly, Massey Fergusson tractor from 60 Cv:Click here
3) 36 assembly, Deutz D40 tractor, 40 Cv:Click here
4) 42 assembly, Deutz 4006 tractor, 40 Cv:Click here
These are the only montages that give consumption figures (GO and water) before / after modification.
Figures taken before and after modification:
Exploitations and analyzes
1) Estimated average horsepower pulled on the tractor.
Thanks to the original consumption we can calculate the average load drawn on the motor. This is possible by assuming an average mechanical efficiency of 30%, then it suffices to multiply the original consumption by 5 because, at 30% efficiency, 1L of fuel provides energy of 5hp.h. Thus a Diesel engine which consumes 20 L per hour will provide 20 * 5 = 100 hp.h. The average power drawn on this engine is therefore approximately 100 hp.
Average load on these tractors:
We are already seeing overconsumption at the level of the MF of 95 hp but this can be explained by a degraded original output and / or a much more intensive use of the engine (for having visited this farmer and having seen his fields far from be flat, the 2nd hypothesis is plausible)
The other average loads are more consistent: 50% average load.
2) Equivalence, after modification, between water and fuel consumption
Reduction of water consumption and consumption:
We calculate the reduction in consumption in% compared to the original consumption, obviously it is assumed that the working and load conditions are identical. The average reduction in consumption observed is 54%. The average consumption has therefore been divided by 2, it is enormous and only a passage on the bench of one of these tractors would make it possible to really show (or not) a very low specific consumption.
After modification, the ratio of fuel consumption to water consumption varies between 1.43 and 2.5. The average is 1.77. In other words, the water consumption is 1.5 to 2.5 times less than the consumption of diesel.
3) Equivalence between reduced fuel consumption and water consumption
Reduction of water consumption and consumption:
The first column is calculated as follows: (reduction in GO consumption) / (water consumption) = (original GO consumption-GO consumption) / water consumption.
The 2nd column corresponds to the water consumption divided by the original GO consumption. It is a quantity which does not represent anything physical but which
The relative stability of these 2 reports is quite glaring and tends to prove that the numbers advanced by farmers are real. One liter of injected water would therefore lead to a reduction in consumption of 2 L of fuel.
In addition, the stability of water consumption / original consumption can be explained quite easily. The thermal losses of an engine are obviously proportional to the fuel consumption and as it is these losses (30 to 40% in the exhaust) which are used to evaporate the water, it is therefore logical that the quantity of water evaporated is proportional to the original consumption. The stability of this ratio also reflects a constant “heat exchange coefficient” in the various evaporator assemblies.
In the absence of any power test bench, it is impossible to draw an irrefutable conclusion about the figures announced by the farmers. Nevertheless, the stability of certain reports, while the figures announced are very different all the same, tend to prove that the values put forward are real. But it is certain that a larger number of testimonies would make this analysis more reliable.
However, a fact confirming this hypothesis, these are the same values that we observed on our ZxTD assembly: one liter of water consumed leading to a reduction in consumption of 2 L of fuel.
We have chosen not to put the values of the Zx in the comparative tables because, the means of measurement, the load and even the engine technology (indirect injection, turbo engine, etc.) are so different that a comparison could not be made. scientifically acceptable ... but the equivalent reduction in consumption compared to water consumption is, however, the same.
5) Annex: The energy of evaporation of water
The purpose of this annex is to evaluate the energy of evaporation of water and to compare it with the thermal losses at the exhaust to see if the quantities are consistent.
We assume that the water feeding the bubbler arrives at 20 ° C and that it evaporates (under atmospheric pressure) at 100 ° C. This is false since there is a slight depression in the bubbler (0.8 to 0.9 bar), that is to say that in this case, we will obtain an increase in the necessary energy.
Energy required for the evaporation at 100 ° C of X liters of water initially at 20 ° C:
Ev = 4.18 * X * (100-20) + 2250 * X = 334 * X + 2250 * X = 2584 * X.
It is therefore necessary to provide an energy of 2584 kJ per liter of evaporated water.
Exhaust losses represent about 40% of the thermal energy supplied to an engine. (30% being the useful energy and the other 30% in the cooling circuit and in the "accessories": various pumps, etc.)
To obtain the power dissipated at the exhaust, it is therefore simply necessary to apply a correction coefficient to the payload of 4/3: an engine with a load of 10 HP will dissipate 10 * 4/3 HP in thermal form. exhaust is 13.3 hp.
However a Horse = 740 W = 0.74 kW, during one hour this horse (whether thermal or mechanical) will provide an energy of 0.74 kWh.
Gold 1 kWh = 3 600 000 J = 3600 kJ
Above we calculated that it takes an energy of 2584 kJ to evaporate 1 liter of water.
One (1) thermal horse will therefore be able to evaporate 0.74 * 3600/2584 = 1.03 L of water… To simplify the following, we will retain a value of 1.
One (1) mechanical horse will provide 4/3 = 1.33 thermal hp to the exhaust and therefore be able to evaporate 1.33 L of water under the condition of course that 100% of the (thermal) energy of the exhaust gases is recovered.
Conclusion: the water consumption is ridiculously low compared to the thermal losses of tractors with a power of 40, 60 or 95 HP. Under these conditions, it is even astonishing that the water consumption is not higher, but it must be said that the dimensions and shapes of the bubblers do not make them “perfect” gas-liquid exchangers… we are even far from it. Only a small proportion (<5%) of the exhaust heat is therefore recovered for the evaporation of the quantities of water observed ... In addition, this "thermal overpower" at the exhaust probably explains the absence of insulation on most (all?) of the assemblies. For information: 1) a proportion of the energy lost in the exhaust gases is in kinetic form. It is therefore impossible to recover 100% of the losses (thermal + kinetic) at the exhaust. 2) In ideal heating via a boiler, it would take 0.74 kWh or 0.74 / 10 = 0.074 L of GO to evaporate 1 L of water under the same conditions. That is about 80 L for a ton of steam.
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