westcliffe01
Veteran Member
I would like to question the math used by flusher. Just because an engine is rated 40hp on the dyno does not mean it is loaded to 40 hp in the application.
I probably have a few rounding errors on my back of the napkin calculation, but bear with me:
40hp ~30kw (its actually a tad bit less)
so over 8 hours one consumes 240kWH of power (8h*30kW)
This 240kWh is equal to 240 000Wh (x1000 to get Watts)
Since a Watt is 1Joule/second lets get the timebase to seconds
So 240 000Wh = 860 million Ws (multiply Wh x 3600 to get hours to seconds)
From the definition that 1W = 1Joule/sec I can get the formula 1J = 1Ws
Thus, to plow the field at full rated 40hp we will need 860 million Joules.
The calorific energy of diesel is 38.6million J/liter or approximately 149 million Joules/gal. Now we need to take the thermal efficiency of the engine into account. The most efficient diesels will have a thermal efficiency of about 30%.
Thus to determine the amount of diesel consumed one divides the work required by the calorific energy and then divide again by the thermal efficiency. Thus 860/149/0.3 = 19.3 gal.
That equates to a fuel useage of 2.42 gal/hr. If you use 1gal/hr it is just because you are running the engine at 16.5hp instead of 40hp. Or you are simply not operating for 8 hours. Taking 3 hour long siestas ?
So back to the math: 19.3 gal x $2.50 = $48.32 So at those costs, it would appear that electricity would reduce the operating cost by 50%. If one actually used that amount of power over an extended time (particularly off peak at night) one would expect the utility to cut you a break and get a major discount, thus reducing the bill further.
The other arguments regarding cost of the battery pack are pretty much on target.
Don't forget, there is more to a tractor that a battery and motor ! You still need the tires, wheels, axles. Some sort of frame to hold the whole thing together (and the massive battery box). A charge controller, a motor controller. Steering system. Electric / hydraulic power pack for the hydraulics / 3 point system.
If you convert tractors, a lot of the mechanicals are a given, but if you don't and you don't manufacture those parts you would be buying them from competitors unless you find independant suppliers who build for the aftermarket.
If the machine works in open fields, one can add a canopy of solar panels. You might get up to 3kW/hr at mid day and it will make a nice shade / rain shield. The batteries will never run flat if it stays out in the field, but to get a full charge would take several days of no operating.
This kind of thing has already been done by individuals. But not for plowing fields 8 hours/day. Tractors with dead engines are cheap and particularly if you use a 3 phase motor the control electronics can be quite affordable. No getting around an expensive battery and if it is going to get cold you need a warm place to keep the whole thing and an insulating blanket around the battery pack to keep the cold out while working.
I probably have a few rounding errors on my back of the napkin calculation, but bear with me:
40hp ~30kw (its actually a tad bit less)
so over 8 hours one consumes 240kWH of power (8h*30kW)
This 240kWh is equal to 240 000Wh (x1000 to get Watts)
Since a Watt is 1Joule/second lets get the timebase to seconds
So 240 000Wh = 860 million Ws (multiply Wh x 3600 to get hours to seconds)
From the definition that 1W = 1Joule/sec I can get the formula 1J = 1Ws
Thus, to plow the field at full rated 40hp we will need 860 million Joules.
The calorific energy of diesel is 38.6million J/liter or approximately 149 million Joules/gal. Now we need to take the thermal efficiency of the engine into account. The most efficient diesels will have a thermal efficiency of about 30%.
Thus to determine the amount of diesel consumed one divides the work required by the calorific energy and then divide again by the thermal efficiency. Thus 860/149/0.3 = 19.3 gal.
That equates to a fuel useage of 2.42 gal/hr. If you use 1gal/hr it is just because you are running the engine at 16.5hp instead of 40hp. Or you are simply not operating for 8 hours. Taking 3 hour long siestas ?
So back to the math: 19.3 gal x $2.50 = $48.32 So at those costs, it would appear that electricity would reduce the operating cost by 50%. If one actually used that amount of power over an extended time (particularly off peak at night) one would expect the utility to cut you a break and get a major discount, thus reducing the bill further.
The other arguments regarding cost of the battery pack are pretty much on target.
Don't forget, there is more to a tractor that a battery and motor ! You still need the tires, wheels, axles. Some sort of frame to hold the whole thing together (and the massive battery box). A charge controller, a motor controller. Steering system. Electric / hydraulic power pack for the hydraulics / 3 point system.
If you convert tractors, a lot of the mechanicals are a given, but if you don't and you don't manufacture those parts you would be buying them from competitors unless you find independant suppliers who build for the aftermarket.
If the machine works in open fields, one can add a canopy of solar panels. You might get up to 3kW/hr at mid day and it will make a nice shade / rain shield. The batteries will never run flat if it stays out in the field, but to get a full charge would take several days of no operating.
This kind of thing has already been done by individuals. But not for plowing fields 8 hours/day. Tractors with dead engines are cheap and particularly if you use a 3 phase motor the control electronics can be quite affordable. No getting around an expensive battery and if it is going to get cold you need a warm place to keep the whole thing and an insulating blanket around the battery pack to keep the cold out while working.