The links below provide an outline of the material for this lesson. Be sure to carefully read through the entire lesson before returning to Canvas to submit your assignments.
Alright, we are into lesson 3. Lesson 3 basically deals with energy efficiency. Upon completing this lesson you will really get a quantitative feel for what energy efficiency is and also how to calculate energy efficiency with a conversion device. A conversion device can be anything – an automobile or a power plant or any device that converts one form of energy to another form.
And we will also understand the concept of entropy. Entropy is disorder. Because of entropy, our efficiencies are lower than what we normally expect. And we will look at the operating principle of a heat engine. A heat engine is a device that converts heat to work. Particularly, automobiles are all heat engines and they are notoriously inefficient. We will see 2 examples and calculations of why these automobiles are notoriously inefficient.
We can also calculate the efficiency of a whole process from the step efficiencies. For example, if it involves 2 or 3 steps like in a relay race. You know you have 3 or 4 players taking the baton and one lap by each of the athletes. So, what is the overall or team efficiency if we know the efficiency of each of those steps or the efficiency of each of those players? That is a very important concept in this chapter.
While doing this we will also learn about temperature scales; Kelvin scale, Fahrenheit, and also Celsius. So there will be a lot of numerical problems in this lesson. And there are some practice problems that you can do at the end of the chapter and also you can look in the text book for more numerical problems for your practice and you can find the answers there. Some of these problems may also appear on the exam. So watch out for the quantitative examples in this chapter.
After this chapter, there are some questions for review and discussion. You can post some of these questions on the message board – your answers, not just the questions. You can do practice questions as well. They will be scored but will not count for your grade. And after you finish this lesson there is a quiz that will count toward your grade. You can log in only once, and you will have only one opportunity to take that quiz.
Alright! Good luck!
Upon completing this lesson, you should be able to:
Here is your "to do" list for this week.
Step  Activity  Access / Directions 

1  Read  Lesson 3  Energy Efficiency 
2  Watch  Lesson 3  Guided Review (Flash movie) (a printable Review Sheet is also provided) 
3  Read  Lesson 3  Questions for Review and Discussion 
4  Review  Lesson 3  Resources (supplemental materials that are optional...but informative!) 
5  Complete  Lesson 3  StudyMate Activities. (You will obtain feedback for these exercises, but they will not count toward your final course grade.) 
6  Take  Lesson 3  Quiz (graded) The quiz is available in Canvas. 
7  Complete  Home Activity: Residential Energy Consumption [1]

See the calendar in Canvas for due dates/times.
If you have any questions, please post them to the General Course Questions forum in located in the Discussions tab in Canvas. I will check that discussion forum daily to respond. While you are visiting the discussion board, feel free to post your own responses to questions posted by others  this way you might help a classmate!
In the first lesson, we saw that energy can be transformed from one form to another, and during this conversion, all the energy that we put into a device comes out. However, all the energy that we put in may not come out in the desired form.
For example, we put electrical energy into a bulb and the bulb produces light (which is the desired form of output from a bulb), but we also get heat from the bulb (undesired form of energy from an electric bulb).
Therefore, energy flow into and out of any energy conversion device can be summarized in the diagram below:
When all forms of energy coming out of an energy conversion device are added up, it will be equal to the energy that is put into a device. Energy output must be equal to the input. This means that energy can not be destroyed or created. It can only change its form.
In the case of an electric bulb, the electrical energy is converted to light and heat.
The amount of electrical energy put into a bulb = the amount of light energy (desirable form) plus the heat energy that comes out of the bulb (undesirable form).
Say you go to the mall with $100 and you come back with only $10. You need to account for the $90 that was spent. After thinking about it, you come up with the following list:
Gas ($15); Sandwich, fries, and a drink ($8); Lost ($5); New clothes ($62)
So you spent $62 on something useful  the clothes  but you spent additional money for other things that were necessary for your trip to the mall.
Instructions: Identify the useful energy output(s) and undesirable energy output(s) in the energy conversion devices below. Enter your answers in the fields provided, and click the "Check" button to check your work.
Self Check:
Energy Conversion Devices
For each of the following examples, determine the types of useful energy and undesired energy for the given energy converter.
Example 1: Lawnmower with a chemical energy input. (Hint: How do you know when your neighbor is mowing the lawn?)
Example 2: Car with a chemical energy input. (Hint: Think about mufflers, tires and generator.)
Example 3: Television with an electrical energy input. (Hint: Have you ever felt the back of your TV after it has been on for a few hours?)
Example 4: Desktop computer with an electrical energy input. (Hint: What’s in your tower and why?)
Answers:
Example 1: The useful energy for a lawnmower is mechanical while the undesired energy is thermal (heat) and radiation (noise).
Example 2: The useful energy for a car is mechanical while the undesired energy is thermal or heat (tail pipe).
Example 3: The useful energy for a TV is radiation (light and sound) and the undesirable energy is heat (from circuits).
Example 4: The useful energy for a computer is radiation (light and sound) and the undesirable energy is heat (circuits – electrons moving through system) and mechanical (fan for cooling).
Efficiency is the useful output of energy. To calculate efficiency the following formula can be used:
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.
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:
$$\begin{array}{l}Efficiency=\frac{Useful\phantom{\rule{.3em}{0ex}}Energy\phantom{\rule{.3em}{0ex}}Output}{Total\phantom{\rule{.3em}{0ex}}Energy\phantom{\rule{.3em}{0ex}}Output}\\ =\frac{90\phantom{\rule{.3em}{0ex}}W}{100\phantom{\rule{.3em}{0ex}}W}\\ =0.9\\ =90\%\end{array}$$
Use the following link to generate a random practice problem [2] 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.
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:
$$Efficiency=\frac{3.675\times {10}^{12}\text{}kWh}{39.5\times {10}^{15}\text{}Btus}\times \frac{3412\text{}Btus}{1\text{}kWh}=0.3174\text{\hspace{0.17em}}or\text{\hspace{0.17em}}31.74\%$$
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.
$$\begin{array}{l}1\text{}\text{}kWh=3412\text{}Btus\\ Therefore\text{\hspace{0.17em}}3.675\times {10}^{12}\text{}kWh=\frac{3.675\times {10}^{12}\text{}kWh\times 3412\text{}Btus}{1\text{}kWh}\\ =12,539.1\times {10}^{12}Btus\end{array}$$
Step 2
Use the formula for efficiency.
$$\begin{array}{l}Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}\\ =\frac{12,539\times {10}^{12}Btus}{39.5\times {10}^{15}Btus}\\ \\ =0.3174\\ \\ =31.74\%\end{array}$$
Use the following link to generate a random practice problem [3] similar to the Practice 2 example.
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  5 % 
Electric Bulb: Fluorescent  20 % 
From our discussion on national and global energy usage patterns in Lesson 2, we have seen that:
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.
Thermal energy is energy associated with random motion of molecules. It is indicated by temperature which is the measure of the relative warmth or coolness of an object.
A temperature scale is determined by choosing two reference temperatures and dividing the temperature difference between these two points into a certain number of degrees.
The two reference temperatures used for most common scales are the melting point of ice and the boiling point of water.
It is important to realize, however, that the temperature of a substance is not a measure of its heat content, but rather, the average kinetic energy of its molecules resulting from their motions.
Below is a 6ounce cup with hot water and a 12ounce cup with hot water at the same temperature.
Instructions: Click the play button to obtain a magnified view of what is happening. Draw your conclusions, enter your answer in the text field provided and then click the link below the video to check your answer.
A six ounce cup and a twelve ounce cup are both filled with 85 degree water.
Conclusion: They do NOT have the same heat content. Since water in the two cups is at the same temperature, the average kinetic energy of the molecules in the cups is the same; however, the 12 ounce cup has twice as many molecules when compared with the 6 ounce cup and thus has the greater total motion or heat energy.
When water molecules freeze at 0°C, the molecules still have some energy compared to ice at 50°C. In both cases, the molecules are not moving, so there is no heat energy.
So what is the temperature at which all the molecules have absolutely zero energy? A temperature scale can be defined theoretically for which zero degree corresponds to zero average kinetic energy. Such a point is called absolute zero, and such a scale is known as an absolute temperature scale. At absolute zero, the molecules do not have any energy.
The Kelvin temperature scale is an absolute scale having degrees the same size as those of the Celsius temperature scale. Therefore, all the temperature measurements related to energy measurements must be made on Kelvin scale.
You can convert a temperature in Celsius (c) to Kelvin (k) with this formula:
$$K=c+273.15$$You can also change a temperature in Kelvin to Celsius:
$$c=k273.15$$To make calculations for this class easier, you may round off and use just 273 in your conversions.
Instructions: Click the "Play" button below and notice what happens to the ice cube. Answer the questions that follow based on your observations.
Energy conversions occurring in an automobile are illustrated below:
Any device that converts thermal energy into mechanical energy—such as an automobile or a power plant—is called a heat engine. In these devices, high temperature heat (thermal energy) produced by burning a fuel is partly converted to mechanical energy to do work and the rest is rejected into the atmosphere, typically as a low temperature exhaust.
A general expression for the efficiency of a heat engine can be written as:
$$\text{Efficiency =}\frac{\text{Work}}{{\text{HeatEnergy}}_{Hot}}$$[5]@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8vrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaake aacaqGfbGaaeOzaiaabAgacaqGPbGaae4yaiaabMgacaqGLbGaaeOB aiaabogacaqG5bGaaeypamaalaaabaGaae4vaiaab+gacaqGYbGaae 4AaaqaaiaabIeacaqGLbGaaeyyaiaabshacaqGGaGaaeyraiaab6ga caqGLbGaaeOCaiaabEgacaqG5bWaaSbaaSqaaiaadIeacaWGVbGaam iDaaqabaaaaaaa@5325@
We know that all the energy that is put into the engine has to come out either as work or waste heat. So work is equal to Heat at High temperature minus Heat rejected at Low temperature. Therefore, this expression becomes:
$$\text{Efficiency=}\frac{{\text{Q}}_{\text{Hot}}{\text{Q}}_{\text{Cold}}}{{\text{Q}}_{\text{Hot}}}$$
Where, Q_{Hot} = Heat input at high temperature and Q_{Cold}= Heat rejected at low temperature. The symbol is often (Greek letter eta) used for efficiency this expression can be rewritten as:
$${\eta}^{\prime}\left(\%\right)=1\frac{{Q}_{Cold}}{{Q}_{Hot}}\times 100$$
The above equation is multiplied by 100 to express the efficiency as percent.
French Engineer Sadi Carnot showed that the ratio of Q_{HighT} to Q_{LowT} must be the same as the ratio of temperatures of high temperature heat and the rejected low temperature heat. So this equation, also called Carnot Efficiency, can be simplified as:
Note: Unlike the earlier equations, the positions of T_{cold} and T_{hot} are reversed.
The Carnot Efficiency is the theoretical maximum efficiency one can get when the heat engine is operating between two temperatures:
In the case of an automobile, the two temperatures are:
Please watch the following 4:40 presentation about how automobile engines work:
Then, why should we operate the automobiles at low efficiencies?
It is not that we cannot achieve high temperatures, but we do not have the engine materials that can withstand the high temperature. As a matter of fact, we do not let the engine gases go the maximum that they can go even now and instead try to keep the engine cool by circulating the coolant.
So we are taking the heat out of the gases (thus lowering the T_{hot}) and making the engine operate at cooler temperatures so that the engine is protected  but lowering the efficiency of an automobile.
It's like Taxes. The more money you earn (heat), the more money is taxed (cold), leaving you with less money to take home (efficiency). However, if you could earn more money (heat) and find a way to have less taxes taken out (better engine material), you would have more money to take home (efficiency).
Below are two temperature scales. The first scale, labeled "HOT," shows the range of temperatures for the combustion of gases in a car engine. The second scale, labeled "COLD," shows the range of temperatures at which gases are exhausted from the car engine.
Instructions: Select numbers from the range on the "HOT" scale and enter them (one at a time) into the text box labeled "Hot" below. At the same time, select numbers from the range on the "COLD" scale and enter them (one at a time) into the text box labeled "COLD" below. Try various combinations of hot and cold numbers and observe the graph to see the temperatures' effect on efficiency.
For a coalfired utility boiler, the temperature of high pressure steam (T_{hot})would be about 540°C and T_{cold}, the cooling tower water temperature, would be about 20°C. Calculate the Carnot efficiency of the power plant:
Solution:
Carnot efficiency depends on high temperature and low temperatures between which the heat engine operates. We are given both temperatures. However, the temperatures need to be converted to Kelvin:
$$\begin{array}{l}{T}_{hot}={540}^{o}C+273=813K\\ {T}_{cold}={20}^{o}C+273=293\text{}K\\ \eta =\left[1\frac{{T}_{cold}}{{T}_{hot}}\right]\times 100\%\\ \eta =\left[1\frac{293\text{}K}{813\text{}K}\right]\times 100\%\\ =64\%\end{array}$$
For a coal fired utility boiler, the temperature of high pressure steam would be about 540 degrees C and T_{cold}, the cooling tower water temperature, would be about 20 degrees C. Calculate the Carnot efficiency of the power plant.
Step 1
Convert the high and low temperatures from Celsius to Kelvin:
$$\begin{array}{l}{T}_{hot}={540}^{o}C+273\\ =813K\end{array}$$
$$\begin{array}{l}{T}_{cold}={20}^{o}C+273\\ =293\text{}K\end{array}$$
Step 2
Determine the efficiency using the Carnot efficiency formula:$$\begin{array}{l}\eta =\left[1\frac{{T}_{cold}}{{T}_{hot}}\right]\times 100\%\\ \\ \eta =\left[1\frac{293\text{}K}{813\text{}K}\right]\times 100\%\\ \\ =64\%\end{array}$$
From the Carnot Efficiency formula, it can be inferred that a maximum of 64% of the fuel energy can go to generation. To make the Carnot efficiency as high as possible, either T_{hot} should be increased or T_{cold} (temperature of heat rejection) should be decreased.
Use the following link to generate a random practice problem [6].
Let’s look at the Energy conversions in a power plant. Please watch the following 6:09 video:
Now view a demonstration of a smallscale power plant in the 7:01 video below:
The basic energy conversions in the three main components in a power plant are shown quantitatively in the 4:26 video below:
Below are two temperature scales. The first scale, labeled "HOT," shows the range of temperatures for the combustion of gases in a power plant. The second scale, labeled "COLD," shows the range of temperatures at which gases are exhausted from the power plant.
Instructions: Select numbers from the range on the "HOT" scale and enter them (one at a time) into the text box labeled "HOT" below. At the same time, select numbers from the range on the "COLD" scale and enter them (one at a time) into the text box labeled "COLD" below. Try various combinations of hot and cold numbers and observe the graph to see the temperatures' effect on efficiency.
Using the energy efficiency concept, we can calculate the component and overall efficiency:
$$Overall\text{}Efficiency=\frac{Electrical\text{}Energy\text{}Output}{Chemical\text{}Energy\text{}Input}$$
Here the electrical energy is given in Wh and Chemical Energy in Btus. So Wh can be converted to Btus knowing that there are 3.412 Wh in a Btu.
This overall efficiency can also be expressed in steps as follows:
$$Overall\text{}Efficiency=\underset{Efficiency\text{}of\text{}the\text{}Boiler}{\underbrace{\left[\frac{Thermal\text{}Energy}{Chemical\text{}Energy}\right]}}\times \underset{Efficiency\text{}of\text{}the\text{}Turbine}{\underbrace{\left[\frac{Mechanical\text{}Energy}{Thermal\text{}Energy}\right]}}\times \underset{Efficiency\text{}of\text{}the\text{}Generator}{\underbrace{\left[\frac{Electrical\text{}Energy}{Mechanical\text{}Energy}\right]}}$$
$$Overall\text{}Efficiency=Boiler\text{}\eta \times Turbine\text{}\eta \times Generator\text{}\eta $$
Applying this method to the above power plant example:
$$\begin{array}{l}Overall\text{}Efficiency=\left[\frac{88\text{}Btus}{100\text{}Btus}\right]\times \left[\frac{36\text{}Btus}{88\text{}Btus}\right]\times \left[\frac{35\text{}Btus}{36\text{}Btus}\right]\\ \\ =0.88\times 0.41\times 0.97\\ \\ =0.35\text{}or\text{}35\%\end{array}$$
It can be seen that the overall efficiency of a system is equal to the product of efficiencies of the individual subsystems or processes. What is the implication of this?
We have been looking at the efficiencies of an automobile or a power plant individually. But when the entire chain of energy transformations is considered—from the moment the coal is brought out to the surface to the moment the electricity turns into its final form—true overall efficiency of the energy utilization will be revealed. The final form at home could be light from a bulb or sound from a stereo. The series of steps as shown in the Figure below are: 1) Production of coal (Mining), 2) Transportation to power plant, 3) Electricity generation, 4) Transmission of electricity, and 5) Conversion of electricity into light (Use). Please watch the following 4:01 video about overall power plant efficiency:
If efficiency of each step is known, we can calculate the overall efficiency of production of light from coal in the ground. The table below illustrates the calculation of overall efficiency of a light bulb.
Step  Step Efficiency  Cumulative Efficiency or Overall Efficiency 

Extraction of Coal  96%  96% 
Transportation  98%  94% = (0.96 x 0.98) * 100 
Electricity Generation  35%  33% = (0.94 x 0.35) * 100 
Transmission of Electricity  95%  31% = (0.33 x 0.95) * 100 
Lighting: Incandescent Bulb 
5%  1.56% = (0.31 x 0.05) * 100 
Lighting: Fluorescent Bulb 
20%  6.2% = (0.31 x 0.20) * 100 
A similar analysis on automobile efficiency is shown in the Figure below.
The table below shows that only about 10% of the energy in the crude oil in the ground is in fact turned into mechanical energy moving people.
Step  Step Efficiency  Cumulative Efficiency or Overall Efficiency 

Extraction of Crude  96%  96% 
Refining  87%  84% 
Transportation  97%  81% 
Engine  25%  20% 
Transmission  50%  10% 
Please watch the 5:30 Lesson 3 Review below:
The questions below are your chance to test and practice your understanding of the content covered in this lesson. In other words, you should be able to answer the following questions if you know the material that was just covered! If you have problems with any of the items, feel free to post your question on the unit message board so your classmates, and/or your instructor, can help you out!
For more information on topics discussed in Lesson 3, see these selected References:
The following practice exercises are designed to help you assess your readiness for the Lesson 3 quiz. You can attempt these activities as many times as you like and it will not count for your course grade.
Try them out and see how well you do!
You must complete a short quiz that covers the reading material in lesson 3. The Lesson 3 Quiz, can be found in the Lesson 3: Energy Efficiency module in Canvas. Please refer to the Calendar in Canvas for specific time frames and due dates.
You must complete the Home Activity: Residential Energy Consumption. The activity can be accessed through the main Home Activities [1] page. You have a limited window of time to access and complete the Home Activity, please refer to the Calendar in Canvas for specific time frames and due dates.
Note: Your grade for the Home Activity will be posted one week after the due date.
Links
[1] https://activity.eeducation.psu.edu/
[2] https://courseware.eeducation.psu.edu/courses/egee102/L03Instruction/activity/check030601.html
[3] https://courseware.eeducation.psu.edu/courses/egee102/L03Instruction/activity/check030801.html
[4] https://courseware.eeducation.psu.edu/courses/egee102/L03Instruction/3_14.html
[5] mailto:MathType@MTEF
[6] https://courseware.eeducation.psu.edu/courses/egee102/L03Instruction/activity/check032001.html