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Posted

You are adding losses in your diesel/spring engine. It takes more power to wind the spring than you can get out of it so you add the 30% efficiency of the diesel to the 70% efficiency of the spring and you get around 21% efficiency from the system. You would do better to use the diesel alone.

 

Paul

 

Paul is it possible that the efficiency of the diesel engine might vary so much (ie from one power usage/form of driving to another) that running a diesel at a constant and maximally efficient rpm and storing (admittedly inefficiently) excess power generated in spring would still be more efficient that running the diesel engine at sub-optimal efficiency for a significant portion of the time

Posted

Not unless you find a way to bend or break the laws of physics.

You can run an engine at a "most efficient for mileage" range of RPM but you cannot use it to efficiently wind a spring and then run a car with it. The engine has to put out more torque than the spring produces, otherwise it wouldn't be able to wind it up. The the car running on the spring would have less power than it took to wind it so more loss.

Short version is that it always take more power to operate a system than it produces. Running alternaters on the wheels of an electric car doesn't work because you ave to have enough power to run the alternators and move the car. alternators are 90% efficient (they produce 90% of the watts that it takes to spin them) an electric motor only produces 90(~) % of the power that is fed into it. If you use a gas engine to run an alternator to run an electric motor you have the losses of the gas engine plus the 19% loss in the electrical system. If you are running a controller for the electric motor then it also consumes power.

 

At the end of the day, month, year, decade or century nothing is free. There are always losses. you have to use more energy to produce motion than the motion can provide in energy.

Posted

Not unless you find a way to bend or break the laws of physics.

You can run an engine at a "most efficient for mileage" range of RPM but you cannot use it to efficiently wind a spring and then run a car with it. The engine has to put out more torque than the spring produces, otherwise it wouldn't be able to wind it up. The the car running on the spring would have less power than it took to wind it so more loss.

Short version is that it always take more power to operate a system than it produces. Running alternaters on the wheels of an electric car doesn't work because you ave to have enough power to run the alternators and move the car. alternators are 90% efficient (they produce 90% of the watts that it takes to spin them) an electric motor only produces 90(~) % of the power that is fed into it. If you use a gas engine to run an alternator to run an electric motor you have the losses of the gas engine plus the 19% loss in the electrical system. If you are running a controller for the electric motor then it also consumes power.

 

At the end of the day, month, year, decade or century nothing is free. There are always losses. you have to use more energy to produce motion than the motion can provide in energy.

 

Nope this isn't a simple "in this house we obey thermodynamics" question - you didnt read what I wrote, merely assumed.

 

Imagine an exaggerated example; you have a diesel that has a sharp peak of efficiency at 2500rpm at which point it is producing 120 bhp at 75pct efficiency - at less or more than this optimal point the efficiency is reduced to 50pct. If whilst you are driving gently downhill on good roads requiring nothing like 120bhp, say 60bhp will maintain your speed then your engine is outside of its optimal band - and going up the steep hill on the other side requires 150 bhp. With that system for the downhill & uphill stretch your engine will run at 50 pct. If you have some form of energy store - a flywheel, an electric battery, a spring (although that might be difficult) for every every minute going downhill running the diesel at 120 bhp you store enough energy to go uphill at maximal powerband for a period; in this exaggerated example even at a storage efficiency of 50 pct you would get a minute up hill for each minute downhill. A few quick sums would then show (60/.5+150/.5>120/.75+120/.75) to you that running the engine at max efficiency and storing excess to use later is a worthwhile proposition - even if that storage is not 100 percent efficient (and of course it couldn't be).

 

this is the principle that hybrids use to an extent - and with an energy storage facility you can also use the vehicles own braking to charge the batteries/spin the flywheel (this harvesting of KE in braking is how the KERS system on F1 cars works).

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