Efficiency is the useful output of energy. To calculate efficiency the following formula can be used:

$$Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}$$

### Example 1

An electric motor consumes 100 watts (a joule per second (J/s)) of power to obtain 90 watts of mechanical power. Determine its efficiency.

**Solution:**

Input to the electric motor is in the form of electrical energy and the output is mechanical energy.

Using the efficiency equation:

$$Motor\text{\hspace{0.17em}}Efficiency=\frac{Mechanical\text{\hspace{0.17em}}Power}{Electrical\text{\hspace{0.17em}}Power}=\frac{90\overline{)W}}{100\overline{)W}}=0.9$$

Or efficiency is 90%.

**Caution!**

This is a simple example because both variables are measured in Watts. If the two variables were measured differently, you would need to convert them to equivalent forms before performing the calculation.

#### Practice Problem

Use the following link to generate a random practice problem similar to the Practice 1 example.

The previous example about an electrical motor is very simple because both mechanical and electrical power is given in Watts. Units of both the input and the output have to match; if they do not, you must convert them to similar units.

### Example 2

The United States' power plants consumed 39.5 quadrillion Btus of energy and produced 3.675 trillion kWh of electricity. What is the average efficiency of the power plants in the U.S.?

$$Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}$$

**Solution**:

Total Energy input = 39.5 x 10^15 Btus and the Useful energy output is 3.675 x 10^12 kWh. Recall that both units have to be the same. So we need to convert kWh into Btus. Given that 1 kWh = 3412 Btus:

**Step 1**

$$1\text{}\text{}\text{kWh}=3412\text{}\text{Btus}$$

Therefore:

$$3.675\times {10}^{12}\text{}\text{kWh}=\frac{3.675\times {10}^{12}\text{}\overline{)\text{kWh}}\times 3412\text{}\text{Btus}}{1\text{}\text{}\text{}\overline{)\text{kWh}}}$$

$$=12,539.1\times {10}^{12}\text{Btus}$$

**Step 2**

Use the formula for efficiency.

$$Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}$$

$$=\frac{12,539\times {10}^{12}Btus}{39.5\times {10}^{15}Btus}$$

$$=0.3174$$

$$=31.74\%$$

### Practice 2

**The United States' power plants consumed 39.5 quadrillion Btus of energy and produced 3.675 trillion kWh of electricity. What is the average efficiency of the power plants in the U.S.?**

#### Practice Problem

Use the following link to generate a random practice problem similar to the Practice 2 example.

### Energy Efficiencies

Energy efficiencies are not 100% and sometimes they are pretty low. The table below shows typical efficiencies of some of the devices that are used in day to day life:

Device | Efficiency |
---|---|

Electric Motor | 90 % |

Home Gas Furnace | 95 % |

Home Oil Furnace | 80 % |

Home Coal Stove | 75 % |

Steam Boiler in a Power Plant | 90 % |

Overall Power Plant | 36 % |

Automobile Engine | 25 % |

Electric Bulb: Incandescent | less than 10 % |

Electric Bulb: Fluorescent | 60 % |

Electric Buld: LED | 90 % |

From our discussion on national and global energy usage patterns in Lesson 2, we have seen that:

- about 40% of the US energy is used in power generation;
- about 27% of the US energy is used for transportation.

Yet the energy efficiency of a power plant is about 35%, and the efficiency of automobiles is about 25%. Thus, over 62% of the total primary energy in the U.S. is used in relatively inefficient conversion processes.

Why are power plant and automobile design engineers allowing this? Can they do better?

There are some natural limitations when converting energy from heat to work.