# Ask an Expert: Energy output efficiency

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### Energy output efficiency

My son would like to compare various energy forms to see which one is most efficient. We got an energy conversion kit featuring a windmill, hand crank, solar panel, and battery. There are coordinating devices such as an LED, buzzer, and fan. Is simply using a multimeter to measure voltage enough to prove one source is more efficient than the other? Or would we have to apply identical energy (such as shining a 12v flashlight and a 12v fan)? I also had trouble assigning dependent, independent, and constant variables to the project. Any advice or suggestions greatly appreciated.

### Re: Energy output efficiency

Hello jjmt4,

In order to determine the efficiency of an energy form, you divide the energy put into the device by the energy obtained from it and multiply by one hundred in order to obtain a percentage. Therefore, you would divide the energy provided to the devices by the voltage that is displayed on the multimeter before multiplying by one hundred. To see which device is more efficient, you would compare the percentages that you calculated to see which one is closest to one hundred percent.

In this case, the independent variable would be the different devices and the dependent variable would be the amount of volts that the multimeter measures for each device, with the constant variable being the multimeter. Because efficiency is measured by dividing the energy out by the energy in, you do not necessarily have to apply identical energy to the different devices, however if you did it would be another constant variable.

Hope this helps and feel free to reach out with any more questions,

Lili

In order to determine the efficiency of an energy form, you divide the energy put into the device by the energy obtained from it and multiply by one hundred in order to obtain a percentage. Therefore, you would divide the energy provided to the devices by the voltage that is displayed on the multimeter before multiplying by one hundred. To see which device is more efficient, you would compare the percentages that you calculated to see which one is closest to one hundred percent.

In this case, the independent variable would be the different devices and the dependent variable would be the amount of volts that the multimeter measures for each device, with the constant variable being the multimeter. Because efficiency is measured by dividing the energy out by the energy in, you do not necessarily have to apply identical energy to the different devices, however if you did it would be another constant variable.

Hope this helps and feel free to reach out with any more questions,

Lili

### Re: Energy output efficiency

Would it follow that the highest voltage should be considered the most efficient or just the most powerful? I just want to be careful that his hypothesis is valid... thank you!!!!!

### Re: Energy output efficiency

Hi jjmt4,

It would depend on whether the same amount of volts are applied to each device. If so, then the device that produces the most volts would be the most efficient. I wouldn't describe the device as the most powerful as power takes into account time, which is not being measured in this experiment.

Hope this helps,

Lili

It would depend on whether the same amount of volts are applied to each device. If so, then the device that produces the most volts would be the most efficient. I wouldn't describe the device as the most powerful as power takes into account time, which is not being measured in this experiment.

Hope this helps,

Lili

### Re: Energy output efficiency

Hi jjmt4,

lilibaker had the right idea but I'm going to jump in to correct a few things here, as there are very distinct technical meanings to things like power and efficiency in physics, and if you use the wrong units you will get the wrong answer. This is a crash-course in some high-school level physics, but I'll do my best.

Power is energy per unit time, which is measured in joules per second or watts [W]. One joule per second equals 1 watt.

Efficiency is calculated as the output power divided by the input power. It's impossible to get out more than you put in, so this way efficiency is always less than 100% (lilibaker had this backwards).

Next - and this is very important - how you calculate power depends on what type of physical system you are talking about. In an

P = I*V

where power is measured in watts [W], voltage in volts [V], and current in amps [A]. So that's a very important point - power is NOT the same thing as voltage.

In a

P = F*v

where force is measured in newtons [N] and velocity in meters per second [m/s].

For a rotating system, it's power = torque * angular velocity

P = T*w (that should actually be a lowercase Greek letter omega, I just used w instead since this is a plain text editor)

where torque is measured in newton-meters (Nm) and angular velocity is measured in radians per second (rad/s).

So, with that background info, let's get back to your original question - you want to compare the efficiency of a windmill, hand crank, solar panel, and a battery. In each case, the output electrical power is pretty easy to measure - you use a multimeter while it's connected to a load (like your LED, buzzer, or fan) - but as mentioned above, you'd need to measure both the voltage and the current separately and then multiply them to get power, not just measure the voltage.

The problem is measuring the input power to each device if you want to calculate efficiency.

- For the solar panel, you should be able to look up the amount of power available from the sun per square meter based on where you live, the time of year, and the time of day (this also depends on the angle of the panel relative to the sun). Multiply that by the area of your solar panel in square meters, and that would give you the total solar power input to the panel.

- For the windmill and hand crank, you should be able to measure the angular velocity (e.g. count the rotations per minute and convert to radians per second), but unfortunately there's no easy to way to measure the torque with the equipment you have available, so you can't calculate power.

- I'm not actually sure off the top of my head how you'd measure the efficiency of the battery, since the energy is already stored "inside" the battery and there isn't an input you can measure - some heat gets dissipated inside a battery and it's possible to calculate that, but that would be an even longer post so I'll leave that out for now.

Point being - I'm sorry if this is way more complicated than you'd hoped originally, but you have a bit of an apples and oranges situation in trying to compare the different devices. You had the right idea with applying identical input energy and then measuring output ("such as shining a 12v flashlight and a 12v fan") but then you have another unknown efficiency problem - e.g. a 12V fan does not convert 100% of its input electrical energy into kinetic energy of moving air (same goes for the flashlight and light), so you don't know the actual input to the windmill.

The very important short-term takeaway is that you can't just measure voltage because it's not the same thing as power. Your son could certainly still do a project measuring the power output of the different devices in different conditions (e.g. power output of the hand crank vs. how fast you turn it), but it will be difficult to compare their efficiencies because you can't measure everything you need.

Hope that helps, please write back if you have more questions!

Ben

lilibaker had the right idea but I'm going to jump in to correct a few things here, as there are very distinct technical meanings to things like power and efficiency in physics, and if you use the wrong units you will get the wrong answer. This is a crash-course in some high-school level physics, but I'll do my best.

Power is energy per unit time, which is measured in joules per second or watts [W]. One joule per second equals 1 watt.

Efficiency is calculated as the output power divided by the input power. It's impossible to get out more than you put in, so this way efficiency is always less than 100% (lilibaker had this backwards).

Next - and this is very important - how you calculate power depends on what type of physical system you are talking about. In an

*electrical*system, power (P) is equal to voltage (V) times current (I)P = I*V

where power is measured in watts [W], voltage in volts [V], and current in amps [A]. So that's a very important point - power is NOT the same thing as voltage.

In a

*mechanical*system, how power is calculated depends on whether the system is moving in a straight line or spinning. For something moving in a straight line, it's power = force * velocity:P = F*v

where force is measured in newtons [N] and velocity in meters per second [m/s].

For a rotating system, it's power = torque * angular velocity

P = T*w (that should actually be a lowercase Greek letter omega, I just used w instead since this is a plain text editor)

where torque is measured in newton-meters (Nm) and angular velocity is measured in radians per second (rad/s).

So, with that background info, let's get back to your original question - you want to compare the efficiency of a windmill, hand crank, solar panel, and a battery. In each case, the output electrical power is pretty easy to measure - you use a multimeter while it's connected to a load (like your LED, buzzer, or fan) - but as mentioned above, you'd need to measure both the voltage and the current separately and then multiply them to get power, not just measure the voltage.

The problem is measuring the input power to each device if you want to calculate efficiency.

- For the solar panel, you should be able to look up the amount of power available from the sun per square meter based on where you live, the time of year, and the time of day (this also depends on the angle of the panel relative to the sun). Multiply that by the area of your solar panel in square meters, and that would give you the total solar power input to the panel.

- For the windmill and hand crank, you should be able to measure the angular velocity (e.g. count the rotations per minute and convert to radians per second), but unfortunately there's no easy to way to measure the torque with the equipment you have available, so you can't calculate power.

- I'm not actually sure off the top of my head how you'd measure the efficiency of the battery, since the energy is already stored "inside" the battery and there isn't an input you can measure - some heat gets dissipated inside a battery and it's possible to calculate that, but that would be an even longer post so I'll leave that out for now.

Point being - I'm sorry if this is way more complicated than you'd hoped originally, but you have a bit of an apples and oranges situation in trying to compare the different devices. You had the right idea with applying identical input energy and then measuring output ("such as shining a 12v flashlight and a 12v fan") but then you have another unknown efficiency problem - e.g. a 12V fan does not convert 100% of its input electrical energy into kinetic energy of moving air (same goes for the flashlight and light), so you don't know the actual input to the windmill.

The very important short-term takeaway is that you can't just measure voltage because it's not the same thing as power. Your son could certainly still do a project measuring the power output of the different devices in different conditions (e.g. power output of the hand crank vs. how fast you turn it), but it will be difficult to compare their efficiencies because you can't measure everything you need.

Hope that helps, please write back if you have more questions!

Ben

### Re: Energy output efficiency

First of all thank you so much for taking the time to explain all of this!! Last night we had a family friend over (he is an engineer)... he actually tried to explain a lot of the same concepts and formulas. But my son is only in 4th grade so we are limited by his grade level. We will have to revise the hypothesis and scrap the word “efficiency”. Instead now we are thinking about measuring and comparing voltage of the solar panel on a sunny day versus cloudy day, the windmill with 2 different fan speeds, and the hand crank with a slower motion vs faster motion. Would each producer have to be connected to a consumer (load) or could we do a direct measure of voltage without the load? And since voltage does not measure power then I’m not even sure what the hypothesis would be. I thought he should do the fruit battery experiment but I think he got dazzled by the cool hand crank and windmill and now we’re stuck trying to find an experiment out of it!

### Re: Energy output efficiency

Hi jjmt4,

So, if your son is only in 4th grade, my guess is that it would be quite sufficient to just examine ONE of the four devices that you have. It's great if he's excited about using all of them, but as we discussed it's sort of hard to compare them directly at this level. So if he's doing a science project, I would recommend picking just one for the project, and basing your hypothesis and conclusions etc. on that single device. Of course, he can still experiment with the other ones for fun!

You're also correct that you could measure the voltage of any of these devices directly, without connecting it to a load. This is called the "open circuit voltage." Technically the voltage will go down "under load" (when you connect it to something like a fan or LED etc) but in 4th grade I don't think you really need to worry about that. I also don't think you really need to worry about the distinction between voltage and power at this level - since again, students don't usually get introduced to the units for those different quantities until middle or high school. The point is to understand the relationship between two different variables, as you suggested - e.g. what do you expect to happen to the voltage of the solar panel on a sunny day vs cloudy day, or to the hand crank when you spin it faster?

You could also add one of the "output" devices if you need to take pictures or he needs to bring a demonstration to the fair, since that can be more interesting than just taking voltage measurements. It also gives you another question to ask/answer - e.g. how fast do you need to spin the crank (or how sunny does it need to be) etc to turn on the LED or make the fan spin etc? Do you have to reach a minimum amount before it will turn on?

Hope that helps, again please don't hesitate to write back with more questions - this is a LOT of information to absorb very quickly!

-Ben

So, if your son is only in 4th grade, my guess is that it would be quite sufficient to just examine ONE of the four devices that you have. It's great if he's excited about using all of them, but as we discussed it's sort of hard to compare them directly at this level. So if he's doing a science project, I would recommend picking just one for the project, and basing your hypothesis and conclusions etc. on that single device. Of course, he can still experiment with the other ones for fun!

You're also correct that you could measure the voltage of any of these devices directly, without connecting it to a load. This is called the "open circuit voltage." Technically the voltage will go down "under load" (when you connect it to something like a fan or LED etc) but in 4th grade I don't think you really need to worry about that. I also don't think you really need to worry about the distinction between voltage and power at this level - since again, students don't usually get introduced to the units for those different quantities until middle or high school. The point is to understand the relationship between two different variables, as you suggested - e.g. what do you expect to happen to the voltage of the solar panel on a sunny day vs cloudy day, or to the hand crank when you spin it faster?

You could also add one of the "output" devices if you need to take pictures or he needs to bring a demonstration to the fair, since that can be more interesting than just taking voltage measurements. It also gives you another question to ask/answer - e.g. how fast do you need to spin the crank (or how sunny does it need to be) etc to turn on the LED or make the fan spin etc? Do you have to reach a minimum amount before it will turn on?

Hope that helps, again please don't hesitate to write back with more questions - this is a LOT of information to absorb very quickly!

-Ben