# AP* Environmental Science Mastering the Math

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5 Part : Determining Energy Efficiency Energy efficiency is simply the ratio of work output to work input. If the problem states that system is less than 0 % efficient, then the energy output must be divided by the stated efficiency to determine how much energy input is needed. Work out Percent Efficiency = 0% Work Our previous sample problem made this easy on us by telling us the efficiency is 0%. Don t always expect that since such machines do not exist! If a machine or process is 75% efficient then you simply multiply anything you calculate as being perfect by 0.75 effectively reducing it to only 75% rather than leaving it at the elusive 0% efficiency. in Sample Problem 3 Answer the questions below regarding the heating of a house in the Midwestern United States. Use the assumptions in the table below to perform the calculations that follow. The house has,000 square feet of living space. 80,000 BTUs of heat per square foot are required to heat the house for the winter. Natural gas is available at a cost of \$5.00 per thousand cubic feet. One cubic foot of natural gas supplies 1,000 BTUs of heat energy. The furnace in the house is 80 percent efficient. (a) Calculate number of cubic feet of natural gas required to heat the house for one winter. Show all the steps of your calculations, including units. 3.0 ft BTU ft ft = ft = ft = 1.6 ft = 1.6 ft BTU 3 3 (8 3) But wait! That s only if the process is 0% efficient. Since the process is only 80% efficient it will require more natural gas to heat the house. So, divide your perfect answer by the efficiency expressed as a decimal value, 0.80 in our case ft 0.80 = ft = ft or 00,000 ft if you prefer (b) Calculate the cost of heating the house for one winter. 5 3 ft \$ ft (5 3) \$ \$ \$1, 000 = = = 5

6 Part 3: Scientific Notation Scientific notation is used to communicate extremely large numbers such as the speed of light or extremely small numbers such as the radius of a blood cell. 1. Converting decimal numbers LESS THAN ONE to Scientific Notation Example: Express the number in scientific notation. Place a decimal after the first nonzero number. We refer to this as the coefficient. Coefficients need to be between 1 and 9. The coefficient is then multiplied by ten raised to an exponent, exponent Determine the exponent on the by counting the number of places you move the decimal point. If you move the decimal to the right, the exponent will be negative. If you move the decimal point to the left, the exponent will be positive. Negative exponents tell us how many times the number has been divided by Positive exponents tell us how many times the number has been multiplied by So, to express the number in scientific notation, we move the decimal 3 places to the right: to give.34-3 What does this really mean mathematically? or 3 or.34 1,000 or.34 3 Essentially, it means the number is small. How small? In the thousandths. Why do this? For years you ve thought was a perfectly good number just the way it is. Why convert it to scientific notation? BECAUSE it allows us to focus on the order of magnitude of measurements or quantities rather than the coefficients. Compare.34 3 to.34 3 It s not that the.34 is the most important part of the number it s that the number is in the thousandths ( 3 ) versus the thousands ( 3 ). So,.34 3 is SIX power of ten s ( ), that s 6, or a million times larger! We say that.34 3 is six orders of magnitude greater than Nuances of Adding and Subtracting Numbers Written in Scientific Notation First, make sure both numbers have the same exponent on the part. It s easiest to convert one number to the greater exponent: becomes or, you may prefer it stacked Either way, ( ) 4 becomes All done! Subtraction works the same way

7 3. Nuances of Multiplying Numbers Written in Scientific Notation: This is much easier! Simply, multiply the coefficients and then add the exponents on the power of. (4 ) ( ) = (4 ) ( + ) becomes 8 ( + ) which simplifies to 8 8 Division works the same way EXCEPT you subtract the exponents on the power of. (4 ) ( ) = (4 ) ( ) becomes ( ) which simplifies to 1 Sample Problem 4 Calculate the potential reduction in petroleum consumption in gallons of gasoline per year that could be achieved in the United States by introducing electric vehicles under the following assumptions: The mileage rate for the average car is 5 miles per gallon of gasoline. The average car is driven,000 miles per year. The United States has 150 million cars. Ten percent of the US cars could be replaced with electric cars. Start with the easy simplification & translate the numbers into scientific notation: % of 150 million cars = 15 million cars = cars now on to the dimensional analysis cars 1 4 mi 1 gal car year 5 mi gal = then simplify to 5 year gal 9 gal = 6 5 year year Part 4: Population Growth Rate Demography is the study of the characteristics of human populations, such as size, growth, density, distribution, and other vital statistics. In demography, population growth is used informally for the more specific term population growth rate (PGR), and is often used to refer specifically to the growth of the human population of the world. The most common way to express population growth is as a percentage, not as a rate. First, we calculate the change in a population using the difference between the crude birth and death rates. What are those? The crude birth rate (CBR) is the total number of births per year per 1,000 people and the crude death rate (CDR) is the total number of deaths per year per 1,000 people. In the formula below we simply divide by ten to report the PGR as a percentage. (We wouldn t need to divide by if the CBR & CDR were expressed per 0 instead of per 1,000.) CBR CDR PGR (as a percentage) = 7

8 Part 5: The Rule of 70 The Rule of 70 is used to calculate the time required for a doubling of a population based upon population growth rate, PGR, expressed as a percentage. Resist the urge to express the growth percentage as a decimal value, leave it as a percentage! The Rule of 70 is mathematically expressed as: time doubling = PGR 70 (expressed as a percentage) How long will it take for a city with an annual population growth rate of 5% to double its population? time doubling = years PGR (expressed as a percentage) = 5% = How long will it take for the same city if the growth rate were 7%? time doubling = years PGR (expressed as a percentage) = 7% = Sample Problem 5 The fictional country Industria is tracking its population data. In 1855, the first year vital statistics were reported for the country, the population was 1.6 million, with a crude birth rate of 43 per 1,000. At that time the population of Industria was growing quite slowly, because of the high death rate of 41 per 1,000. In 1875 the population began to grow very rapidly as the birth rate remained at the 1855 level. While the crude death rate dropped dramatically to 0 per 1,000. Population growth continued to increase in the small country into the late 1800 s, even though birth rates began to decline slowly. In 1895 the crude birth rate had dropped to 37, and the crude death rate to 1 per 1,000. A complete census was also conducted and revealed that the population of Industria had grown to.5 million. By 1950 population growth gradually began to decline as the death rate remained at its 1895 level, while the birth rate continued to decline to per 1,000. In 1977 vital statistics revealed that the death rate was per 1,000, and that population growth had slowed even more to an annual rate of 0.4%. By 1990 Industria had reduced its birth rate to that of its now constant, low death rate, and the population transition was complete. (a) Calculate the annual growth rate of Industria in CBR CDR 1 Annual PGR (as a percentage) = = = 1% (b) Calculate the birth rate in Industria in In 1977 CDR was and the PGR was 0.4%. So, solve for CBR: CBR CDR CBR PGR = 0.4% = 4 = CBR CBR = 14 continued on next page 8

9 (c) Assume the population of Industria continues to maintain the population growth rate recorded for the country in Determine the year in which the expected size of the population will have quadrupled. CBR CDR First, calculate the PGR for 1895: PGR1895 = = = =.5% Next, use the Rule of 70 to determine the number of years it takes the population to double: time doubling = years PGR (expressed as a percentage) =.5% = Finally, realize that a quadrupling is doublings, so 56 years have passed by the time the size of the population has doubled. So, additional years = 1951 (Harry S. Truman was president, I Love Lucy made its TV debut, and the Korean War was taking place.) Part 6: Per Capita Per Capita is a Latin term that translates into "by head," basically meaning "average per person." Per capita can even the place of saying "per person" in any number of statistical observances. In most cases the term is used in relation to economic data or reporting, but can also be used in most any other occurrence of population description. Sample Problem 6 Between 1950 and 000, global meat production increased from 5 billion kilograms to 40 billion kilograms. During this period, the global human population increased from.6 billion to 6.0 billion. (a) Calculate the per capita meat production in 1950 and in kg 5 kg (9 8) 1 For 1950: = = kg = kg = 0 kg per capita For 000: 9 40 kg = 40 kg per capita 9 6 (b) Calculate the change in global per capita meat production during this 50-year period. Express your answer as a percentage. No calculation necessary! Since the consumption per capita doubled, that translates to a 0% increase. 9

10 Sample Problem 7 West Fremont is a community consisting of 3,000 homes. A small coal-burning power plant currently supplies electricity for the town. The capacity of the power plant is 1 megawatts (MW) and the average household consumes 8,000 kilowatt hours (kwh) of electrical energy each year. The price paid to the electric utility by West Fremont residents for the energy is \$0. per kwh. The town leaders are considering a plan, the West Fremont Wind Project (WFWP), to generate their own electricity using wind turbines that would be located on the wooded ridges surrounding the town. Each wind turbine would have a capacity of 1. MW and each would cost the town \$3 million to purchase, finance, and operate for 5 years. (a) Assuming that the existing power plant can operate at full capacity for 8,000 hrs/, how many kwh of electricity can be produced by the plant in a year? 8,000 hr 3 7 kwh 1 kw = 9.6 (b) At the current rate of electricity energy per household, how many kwh of energy does the community consume in one year? 8, 000 kwh kwh 7 kwh 3, 000 homes = 4, 000, 000 or.4 (c) Assuming that the electrical energy needs of the community do not change during the 5-year lifetime of the wind turbines, what would be the cost to the community of the electricity supplied by the WFWP over 5 years? Express your answer in dollars/kwh. TWO DIFFERENT METHODS earn full credit: (surprising since the costs are very different!) 7 kwh.4 5 s 8 = 6.0 kwh and turbines 7 7 \$3 \$3 \$3 \$0.05 Cost is = kwh 7 60 kwh = 60 = kwh 6 \$3 turbines = \$3 7 OR 7 kwh s 9 =.4 kwh 7 7 \$3 \$3 \$3 \$ Cost is.4 9 kwh.4 9 kwh 40 kwh = = =

11 Sample Problem 8 The Cobb family of Fremont is looking at ways to decrease their home water and energy usage. Their current electric hot-water heater raises the water temperature to 140 o F, which requires 0.0 kwh/gallon at a cost of \$0./kWh. Each person in the family of four showers once a day for an average of minutes per shower. The shower has a flow rate of 5.0 gallons per minute. (a) Calculate the total amount of water that the family uses per year for taking showers. 4 people 365 days 1 shower min 5 gal gal = 73,000 person day shower min (b) Calculate the annual cost of the electricity for the family showers, assuming that.5 gallons per minute of the water used is from the hot-water heater. So, they are telling us that ½ of the total water used in each shower is water heated by the hot water heater. Fair enough! Simply start this cost analysis using ½ (73,000) = 36,500 gal. So, 36,500 gal 0.0 kwh gal \$0. kwh \$36,500 (0.0 0.) \$ 730 = = (c) The family is considering replacing their current hot-water heater with a new energy-efficient hot-water heater that costs \$1,000 and uses half the energy that their current hot-water heater uses. How many days would it take for the new hot-water heater to recover the \$1,000 initial cost? \$730 \$ \$1 Cost of OLD: = versus Cost of NEW: Half as much 365 days day day So, the family is saving a \$1 a day, thus it will take 1,000 days to save \$1,000 which recoups their original investment! 11

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