# Video: Calculating the Power Wasted by a Machine Based on the Input Power and Machine Efficiency

A machine uses input power to do work. Some of the input energy is wasted heating the machine. Which of the following would waste the most power? [A] Input power: 200 kW, machine efficiency: 30% [B] Input power: 200 kW, machine efficiency: 25% [C] Input power: 150 kW, machine efficiency: 30% [D] Input power: 100 kW, machine efficiency: 30% [E] Input power: 150 kW, machine efficiency: 25%

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### Video Transcript

A machine uses input power to do work. Some of the input energy is wasted heating the machine. Which of the following would waste the most power? a) Input power, 200 kilowatts; machine efficiency, 30 percent. b) Input power, 200 kilowatts; machine efficiency, 25 percent. c) Input power, 150 kilowatts; machine efficiency, 30 percent. d) Input power, 100 kilowatts; machine efficiency, 30 percent. e) Input power, 150 kilowatts; machine efficiency, 25 percent.

The question gives us the input power and machine efficiency of five machine setups and asks us which setup wastes the most power. Power is the rate of transfer of energy. And the base unit of power is the watt, which is equivalent to one joule per second. So, when the question tells us that the machine uses input power, it means that there is energy flowing into the machine at some rate.

And the question then tells us that this energy flow is directed into one of two activities, doing work or heating the machine. Work is the energy expended or recovered as something moves in the presence of a force. For example, a pump pushing water up a pipe is doing work. That is, spending energy to move the water against the force of gravity. Since work is energy, its units are joules.

Let’s draw a picture to summaries the energy flow into and out of the machine. The arrow going into the machine represents the input energy flow. We’ll label it 𝑃 i for input power. To keep the units consistent, we have power entering the machine, so we want power leaving the machine. This means we’ll need to convert the work and the heating into power. Work is a kind of energy transfer with units of joules. So, the rate of work is a rate of energy transfer with units of joules per second, which is power. So, we’ll let the arrow coming out of the right side of the machine be the rate at which the machine directs energy to work. And we’ll label it 𝑃 w for power devoted to work.

As for the heating, we’re told that some of the input energy is wasted heating. So, the rate at which that energy is directed towards heating would be the power directed towards heating. We’ll let the arrow coming out of the top of the machine represent the rate at which energy is wasted heating the machine. We’ll label it 𝑃 H for power spent heating. This power is the waste power that the question is asking about. Remember, though, that what we’re looking for is how this waste power relates to the input power and machine efficiency, which are the quantities given in the answer choices.

Our picture has input power flowing into the machine and waste power flowing out of the machine. So, if we can relate what flows into the machine to what flows out of the machine, we’ll be able to relate input power and waste power. Since power is a rate of energy transfer, it would make sense to try to conserve energy. In this case, conservation of energy would dictate that all the energy flowing into the machine must flow out of the machine. We would have that input power is equal to waste power plus power to work, or the rate in is equal to the total rate out.

Since we want waste power, let’s isolate it by subtracting work power from both sides. 𝑃 w minus 𝑃 w is zero. And we’re left with 𝑃 H is equal to 𝑃 i minus 𝑃 w. This relationship is close to what we need. If we can replace work power with input power and machine efficiency, then we’ll have input power and machine efficiency on one side and waste power on the other. And that’s the relationship we need.

The relationship between power to work, machine efficiency, and input power that we need actually comes from the definition of machine efficiency. The machine efficiency is the fraction of input power directed to work. So, a 100-percent-efficient machine would direct all its input power to work, while a 50-percent-efficient machine would direct half its input to work and waste the other half. Symbolically, using the Greek letter 𝜂 for machine efficiency, we would write 𝑃 w, the power directed toward work, is equal to 𝜂, a fraction, times 𝑃 i, the input power. This provides exactly the substitution that we need in our equation for waste power.

Doing the substitution, we get 𝑃 H is equal to 𝑃 i minus 𝜂 times 𝑃 i. Just for neatness, let’s factor out a factor of 𝑃 i from both terms on the right-hand side. We have waste power is equal to one minus efficiency times input power. We put a thick box on this relation because it’s what we need to solve our problem. It relates waste power on one side to input power and machine efficiency on the other. Using this equation, let’s calculate the power wasted by choice a. The efficiency is 30 percent or 0.30, and the input power is 200 kilowatts. One minus 0.30 is 0.70. 0.70 times 200 kilowatts is 140 kilowatts. So, choice a wastes 140 kilowatts of power.

We now perform the same calculation but with the input power and machine efficiency of the other answer choices. Choice b wastes 150 kilowatts. Choice c wastes 105 kilowatts. Choice d wastes 70 kilowatts. And Choice e wastes 112.5 kilowatts. Just by looking at this list, we see that choice b, wasting 150 kilowatts, is the most wasteful machine setup. This approach involved a lot of calculation to find the waste powers. We could’ve used a different approach that compares the setups themselves and avoids calculation entirely.

We start again with waste power is equal to one minus efficiency times input power. Now, we consider what happens when we keep efficiency or input power constant. For the first case, where efficiency is constant, waste power becomes directly proportional to input power. This is because if efficiency is constant, one minus efficiency is also constant. And the result of being directly proportional means that more input means more waste. In other words, for two machines setups with the same efficiency, the one with more input power is more wasteful.

Now, let’s consider a second case where input power is constant. In this case, waste power is directly proportional to one minus the efficiency. Since one minus 𝜂 gets larger as 𝜂 gets smaller. Then, for two machines with the same input power, the one that is less efficient wastes more power. Let’s apply these two results starting with number one. Choices a, c, and d all have the same efficiency but, of them, choice a has the largest input power. That means that by observation one, choice a is the most wasteful.

So, since we’re looking for the most wasteful, choices c and d are not the right answer because they waste less than a. Similarly, choices b and e have the same efficiency, but e has a smaller input power. So, it’s less wasteful than b and isn’t the right answer. We’re left with choices a and b. a and b have different efficiencies but the same input power. Applying our second observation then, we conclude that a must waste less power than b because it’s more efficient. So, a is not the right answer.

Having eliminated choices a, c, d, and e as being less wasteful than other setups. We’re left with choice b, the correct answer, as the most wasteful machine setup. And we did this by applying observations instead of doing calculations. In fact, many physics questions can be answered by using equations to understand relationships instead of doing calculations.