# AC Generator Output Calculations

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?

• The gear box can change the rotation speed, but it doesn't generate more power! – Floris Mar 30 '16 at 17:30
• I study Design more than Engineering therefore I am not familiar with a lot of this kind of stuff. I thought I would have to make the generator spin to 500RPM to make it produce power, hence why I thought adding a gearbox would help. Can you help? – Glen McMurchy Mar 30 '16 at 17:34
• Yes you need to make the generator spin fast... but that doesn't mean it will generate more power. If you put your bike in a high gear (to make the wheels spin fast) you have to push harder with your legs. When you climb, and your legs are not strong enough, you put the bike in a low gear; now you can make the wheels spin more slowly, but at least you make them spin. But the power output (work done by you on the bike) is determined mostly by the power input (how hard your legs work), regardless of what gear you are in. Does that help? – Floris Mar 30 '16 at 17:37
• Aw okay I think that helps. So the gearbox would be needed to make it spin fast enough but doesn't actually affect the output. Therefore, if 6kJ of energy are put into the system I then have to calculate the total efficiency of the system to figure out how much output there will be? For example, lets say my whole system was 85% efficient. Input = 6kJ, Output = 5.1kJ. – Glen McMurchy Mar 30 '16 at 17:43
• Also, the introduction of a Gearbox to make the generator spin faster is going to increase the force needed to push down the rack isnt it. So will I have to do 5000N minus the force acting against the rack to find out the total force being put into the system? – Glen McMurchy Mar 30 '16 at 17:45

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...

• Okay, it is starting to make more sense. My project is trying to generate electricity from a car driving over an actuator which is connected to a rack and pinion. This then changes the linear force to rotational. The shaft of the pinion is connected to the generator. For simplicity, we have been told that the car will apply a downwards force of 5000N onto the actuator. – Glen McMurchy Mar 30 '16 at 16:24
• How far does the actuator move? – Floris Mar 30 '16 at 16:25
• Sorry I hadnt finished writing the comment and i clicked enter by mistake. I ran out of characters as well. Ill post the full comment as a new answer so you can see it – Glen McMurchy Mar 30 '16 at 16:31
• Okay, it is starting to make more sense. My project is trying to generate electricity from a car driving over an actuator which is connected to a rack and pinion. This then changes the linear force to rotational. The shaft of the pinion is connected to the generator. For simplicity, we have been told that the car will apply a downwards force of 5000N onto the actuator. The rack connected to the actuator has a travel of 0.1m and 1 car travels over it every 30 seconds and I want to get the RPM up to 500 RPM (based on a 1kW generator I looked up online). [Comment truncated.] – Glen McMurchy Mar 30 '16 at 16:32
• Please note you should not use answers for comments. I get 500 Joules (5000 N times 0.1 m) every 30 seconds; in 5 minutes that is 12 cars, so 6 kJ (about 2000x less than your estimate). Also note that if the car moves down, then it has to "climb out" of the hole, and so it will be doing work. In essence, unless you are taking energy from a car that needed to decelerate anyway (perhaps they have to come to a halt at the entry to the car park), you are stealing... – Floris Mar 30 '16 at 17:12