EGEE 102
Energy Conservation and Environmental Protection

Efficiency of Energy Conversion Devices

PrintPrint

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

Efficiency =
Useful Energy Output / Total Energy Input

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:

MotorEfficiency= MechanicalPower ElectricalPower = 90 W 100 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 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.

Step 1

Input to the electric motor is in the form of electrical energy and the output is mechanical energy. Using the given formula for efficiency:

Efficiency= UsefulEnergyOutput TotalEnergyOutput = 90W 100W =0.9 =90%

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 =
Useful Energy Output / Total Energy Input

Solution:

Total Energy input = 39.5 x 1015 Btus and the Useful energy output is 3.675 x 1012 kWh. Recall that both units have to be the same. So we need to convert kWh into Btus. Given that 1 kWh = 3412 Btus:

Efficiency= 3.675× 10 12 kWh 39.5× 10 15 Btus × 3412Btus 1kWh =0.3174or31.74%

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.?

Step 1

To find the efficiency, both the units of input energy and the output energy have to be same. So we need to convert kWh into Btus.

1kWh=3412Btus Therefore3.675× 10 12 kWh= 3.675× 10 12 kWh×3412Btus 1kWh =12,539.1× 10 12 Btus
Step 2

Use the formula for efficiency.

Efficiency= UsefulEnergyOutput TotalEnergyOutput = 12,539× 10 12 Btus 39.5× 10 15 Btus =0.3174 =31.74%

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:

Typical Efficiencies of Day to Day Devices
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 5 %
Electric Bulb: Fluorescent 20 %

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.