I am currently doing a university project where we are designing a device that a car drives over it will engage a rack and pinion mechanism, that will change the force from linear to rotational. The pinion shaft will connect to a generator and produce electricity.
How do I go about calculating my input to output/ efficiencies and all that kind of stuff. Basically I want to put in the report a reasonably accurate answer like "If the device is pressed 10 times an hour this will turn the generator at an average of 200RPM which means that 10kW will be produced per hour" That is just a hypothetical answer I have no idea if that is a realistic answer or if this is how the answer should be written.
Any feedback much appreciated! Kind Regards, Glen 🙂
UPDATE
The generator I am using is rated at 1kW and works at 50RPM.
If a car has a downwards force acting on the rack of 5000N and the travel of the rack is 0.1m then the work done by one car passing over the device will be 500J. If 12 cars pass over this device in 5 minutes then 6kJ of work will have been done. I am not sure what the next step should be. Lets say we have a gearbox that is 500:1 (for simple numbers sakes). One car going over the device will turn the generator 500 times and that will be 1000 times per minutes (if 2 cars go over it per minute) So does that mean (not including gearbox efficiencies and friction etc.) The generator will produce 2kW of energy per minute?
Best Answer
You can start with "force times distance" to get the work done on the generator; then make some assumptions (measurements) about the losses. For example, if you can make the generator turn without taking any power (no load), then see how long it takes to "spin down", you can use that as an estimate of the rotational losses (this underestimates the eddy current losses you will have with the generator under load, as those will depend on the current flow). Similarly, you can see how much work you have to do to spin the generator up to a particular speed to get an estimate of the kinetic energy stored at a particular RPM.
Example: you apply 100 N over 1 m, once a second for ten seconds, to get the generator up to a speed of 60 rpm (1 rev/sec). The total work done is $100\cdot 1 \cdot 10 = 1000~\rm{J}$. The energy stored in the rotation of the generator is $\frac12 I \omega^2$, and it follows that $I = \frac{2000~\rm{J}}{(2\pi)^2}\approx 50 \rm{~kg~m^2}$
If we assume constant (velocity-independent) friction, then the work done against friction by the generator will be proportional to the number of revolutions ("distance traveled"). We can then write an expression for the angular velocity:
$$\frac{d}{dt}\left(\frac12 I \omega^2\right)=-\alpha\\ I\omega\dot\omega = -\alpha\\ \dot\omega = -\frac{\alpha}{I\omega}\\ \omega=\omega_0 e^{-\alpha t/I}$$
In other words, the generator will lose speed exponentially, and from the time constant of the decay we can deduce the friction factor $\alpha$. If the generator took 1 minute to go down to $1/e$ (37%) of its speed, the characteristic decay time $\tau=\frac{I}{\alpha}$ is 60 seconds, and you can compute the friction coefficient
$$\alpha=\frac{1}{I\tau} = 3.3\cdot 10^{-4}$$
This is just for illustration. See if you can work on your problem with these hints.
Incidentally, to get 10 kW of output from a generator requires a considerable amount of force. An average human working out on a rowing machine may generate about 200 W; a professional cyclist may generate bursts of 1 kW or more for short periods. 10 kW requires lots of people pressing actuators continuously; the total energy expended by 200 students climbing one flight of stairs to go to their lectures once an hour would equate to an average (total) power given by
$$P = \frac{N \cdot m \cdot g \cdot h}{t} = \frac{200\cdot 75 \cdot 10 \cdot 3}{3600} = 125 \rm{~W}$$
UPDATE
You give the example where you have a car exerting 5000 N of force on an actuator that moves 0.1 m, and through a rack-and-pinion this work (500 J) is transferred to a generator that is capable of producing 1 kW.
In fact, your car is doing 500 J of work; if you have one car every 30 seconds, that is about 17 W of work done on average. Assuming your setup is 80% efficient, you would get about 13 W of power out. The fact that the generator could produce 1 kW is irrelevant: that just tells us that the windings are capable of carrying a fair amount of current at a certain voltage, but unless you put the power in, you won't get the power out...