1. Math for APES Reviewing and Practicing the Energy Math Tuesday, March 8 th, 2016

2. Dimensional Analysis Dimensional analysis is a strategy for performing conversions by multiplying by ratios that equal one. Basic Principles of Dimensional Analysis  You can multiply a quantity by 1 without changing its value.  2 equal quantities create a ratio that equals one.

3. The Process of Dimensional Analysis o Set up conversion ratios so that the units being converted cancel out.

4. Calculate Usain Bolt’s speed in miles per hour. He ran a 100 meter dash in 9.58 seconds (10.4 m/s). Conversion factors: 1000 m = 1 km, 1.6 km = 1 mile, 10.4 meters x seconds 1 km x 1000 m 1 mile x 1.6 km 60 seconds x 1 min 60 min =. 1 hr 23.4 miles hours

5. Always include your units and cross off the ones that cancel out so that you can double check your work, making sure you did not miss a step or invert any ratios. For some problems, the same unit may be used for more than one type of measurement. For these cases, you will need to include both the unit and what is being measured. = Tips for Dimensional Analysis

6. Some ratios are normally given in a conventional orientation, such as miles per gallon, so that the information can be more easily interpreted. For example, a higher miles per gallon ratio means better fuel efficiency. However for problem solving purposes these ratios can and may need to be inverted. If a car gets 25 miles per gallon, it also uses 1 gallon per 25 miles.

7. Solve the problem with your table partner : A car has a fuel efficiency of 20 miles per gallon and is used to drive 15,000 miles per year. Determine how many pounds of carbon are released into the atmosphere each year by the car. Each gallon of gasoline burned releases 5 lbs of carbon dioxide. Every lb of carbon dioxide contains 0.27 lbs of carbon 15,000 miles x 1 gallon x 5 lbs CO 2 x 0.27 lbs carbon year 20 miles 1 gallon1 lb CO 2 =1, 012 lb carbon

8.  … is the joule (J). It is a very small unit, so when we are talking about a lot of energy, we use kilojoules (kJ).  Remember your metric conversions…  1000 J = 1 kJ (kilo = 1000)

9.  Power (P) is the rate at which energy is used. ◦ When determining the amount of energy (usually in J or kJ) you must also include a time component.  Power x Time = Energy (P x t = E) ◦ This can be rearranged to determine power as well  P = E/t (Power = Energy /time)

10.  The unit for power is the watt ◦ 1 W = 1 J/sec (1 watt = 1 joule per second)  Therefore a 100 watt light bulb uses 100 J/sec of electrical energy.  If it is 20% efficient (typical for an average light bulb) then the bulb converts 20% of the electrical energy into light, and 80% is lost as waste heat.

11.  Notice that in the previous example, we can see the operation of both the First and Second Laws of Thermodynamics. ◦ The First Law: energy can be converted from one form to another, but none is lost. We have accounted for all the energy, but most of the electrical energy (high quality) was converted to low quality energy (heat). ◦ Therefore we also see the Second Law: in any energy conversion, some energy is converted into lower quality energy (usually heat) and is unable to perform useful work (in this case, light).

12.  Knowing the relationship between energy and power allows us to find the energy used when an appliance of known power (in watts) operates for a known amount of time (in seconds).  Example: How much energy (in kJ) does a 75 watt light bulb use when it is turned on for 25 minutes?  Equation: E = P x t  Conversion: Power (in watts) 1 watt = 1 J/sec

13. How much energy (in kJ) does a 75 watt light bulb use when it is turned on for 25 minutes? E = 75 J60 sec25 min sec min 110,000 J (or 110 kJ)

14.  Some of the information about the current can usually be found.  To find the power (in watts) of any electrical appliance, use the equation P = V x I ◦ P = power, V = voltage, I is current in amps (A).  American household voltage is 100 V (Air conditioners, electric stoves and dryers are 220 V).

15.  The Kilowatt Hour (kwh) is not a unit of power but a unit of energy (because it has a time component). ◦ Notice that a kilowatt is a unit of power and hour is a unit of time. Therefore, E = P x t.  A kilowatt-hour is equal to 1 kw (or 1000 watts) delivered continuously for one hour (3600 sec).  1 kwh = 1000 J/sec x 3600 sec = 3,600,000 J or 3600 kJ  So 1 kwh = 3600 J/sec

16. ◦ 1 calorie (cal) = 4.184 J ◦ 1 BTU = 1.05 kJ ◦ 1 therm = 100,000 BTU

17. Dr. Smith’s power bill shows that his home used 1355 kwh over a 30 day period. ◦ 1 kwh = 3600 kJ a)Find the energy used (in kJ) for the 30-day period. b)Find the energy used in J/day. c)At the rate of $.075/kwh, what is Dr. Smith’s power bill (without tax)?

18. (1355 kwh in 30 days) a) Find the energy used (in kJ) for the 30 day period. 1355 kwh3600 kJ 4.88 x 10 6 kJ 1 kwh Multiplication short-cut: 1355 x 36 (take off the two zeros) = 48780, then put the two zeros back = 4878000. To put into scientific notation, move the decimal six places to the left.

19. b) Find the energy used in J/day. Start with previous calculation 4.88 x 10 6 kJ1000 J 1 month 1.63 x 10 8 J.day Month 1 kJ 30 days 1)4.88 x 10 6 x 1000 is equal to 4.88 x 10 9 2)4.88 x 10 9 3 x 10 1 (this is the same as 30) 3)Calculate in parts: 4.88 / 3 = 1.63, subtract the exponents to get 1.63 x 10 8

20. c) At the rate of $.075/kwh, what is Dr. Smith’s power bill (without tax)? 1355 kwh$.075 $101.63 1 kwh

21. A current through a toaster (110 V) is 8 A. Remember, P = V x I a)What is the power (in watts) of the toaster? b)How much energy (in J) will the toaster use in 5 minutes of operation?

22. A current through a toaster (110 V) is 8 A. a) What is the power (in watts) of the toaster? P = V x I where V = voltage and I = amps P = 110V x 8 A = 880 watts

23. b) How much energy (in J) will the toaster use in 5 minutes of operation? P = E/t so E = P x t (1 watt = 1 joule per second) E = 880 W(1 J/sec)60 sec5 min 264,000 J 1 watt min 2.64 x 10 5 J

24. A 100 watt light bulb is 20% efficient. That means 20% of the energy used is converted to light, while 80% of the energy used is lost as heat. a)How much energy does it use in 12 hours of operation? b)How much energy does the bulb convert into light over the 12-hour period? c)How much energy does the bulb convert into heat over the 12-hour period? d)Convert the total energy use into kwh

25. A 100 watt light bulb is 20% efficient. a) How much energy does it use in 12 hours of operation? E = P x t E = 100 W(1 J/sec)60 sec60 min12 hrs = 4,320,000 J 1 watt min 1 hr 4,320,000 J = 4,320 kJ

26. b) How much energy does the bulb convert into light over the 12-hour period? E = 4.32 x 10 6 J (.20 efficiency) = 864,000 J = 8.64 x 10 5 J Math shortcut: Multiply 4.32 x 0.2 to get 0.864 Move the decimal six places to the right (because 10 6 ) to get 864,000 J (or move the decimal one place to the right and drop the exponent by one to have it in scientific notation)

27. 3c) How much energy does the bulb convert into heat over the 12-hour period? E = 4.32 x 10 6 J (.80 heat) = 3,456,000 J = 3.46 x 10 6 J

28. d) Convert the total energy use into kwh. (the energy calculated in 3a was 4,320 kJ) 1 kwh = 3600 kJ 4320 kJ x 1 kwh = 1.2 kwh 3600 kJ

29. An electric clothes dryer has a power rating of 4000W. Assume that a family does five loads of laundry each week for 4 weeks. Further assume that each load takes one hour. Remember, 1 W = 1 J/sec. a)Find the energy used in both J and kwh b)If the cost of electricity is $.075/kwh, find the cost of operating the dryer for a month (4 weeks).

30. a) Find the energy used in both J and kwh  5 loads x 4 weeks x 1 hr = 20 hrs week load  E = P x t  E = 4000 W x (1 J/sec) x 60 sec x 60 min x 20 hrs = 2.88 x 10 8 J 1 watt min hr  2.88 x 10 5 kJ x 1 kwh = 80 kwh 3600 kJ

31. b) If the cost of electricity is $.075/kwh, find the cost of operating the dryer for a month (4 weeks). 80 kwh x $.075 = $6.00 kwh

32. Dr. Smith’s natural gas bill states that his household used 110 therms of energy over a 30-day period. a)Convert 110 therms to kwh. b)His charge for the energy was $88.78. Find the cost of this natural gas in $/kwh. c)Using the information about electricity costs in the problems above, which form of energy (electricity or natural gas) is more expensive? How many times more expensive is it?

33. 5a) Convert 110 therms to kwh 110 therms x 100,000 BTU x 1.05 kJ x 1 kwh = 3208 kwh 1 therm 1 BTU 3600 kJ

34. 5b) His charge for the energy was $88.78. Find the cost of this natural gas in $/kwh. $88.78 / 3208 kwh = $.028/kwh

35. 5c) Which form of energy (electricity or natural gas) is more expensive? How many times more expensive is it? elec: $.0749 gas: $.028 Electricity is about 2.5 times more expensive