How To Calculate Energy In Quadrons

Transcription

1 Problem Examples 1. Suppose the San Antonio population growth rate were 3.5%/year. What would be the doubling time for San Antonio's population? DT = 70/3.5 = 20 years 2. There are 1.2 million people in San Antonio. Suppose the population growth rate for San Antonio is fixed at 3.5%/year. Calculate San Antonio s population in 2041 based on these assumptions. DT = 20 years 40 years / 20 years = 2 doubling times. Population = 2 x 2 x 1.2 million = 4.8 million 3. There are 1.2 million people in San Antonio. Suppose the population growth increase for San Antonio is fixed at 35,000 people/year. Calculate San Antonio s population in 2041 based on these assumptions. Pop = 1.2 million million/year * 40 years = 1.2 million million = 2.6 million This is linear growth and only predicts 2.6 million; whereas, exponential growth predicts 4.8 million. 4. World fossil fuel consumption increased from the equivalent of 3 billion tons of coal in 1950 to 12 billion tons in In other words world fossil fuel consumption increased by a factor of 4 in 36 years. Assuming a constant growth rate, calculate the doubling time and the annual growth rate of fossil fuel consumption. DT = 36/2 = 18 years growth rate = 70 / 18 years = 3.9 % per year 5. If world fossil fuel consumption were to continue to grow at the same constant growth rate from 1986 until 2022, what would be the consumption of world fossil fuel in 2022? DT = 18 years 36 years / 18 years = 2 DT consumption in 2022 = 2 x 2 x 12 billion tons = 48 billion tons

2 Problems 6 through 17 involve the following data: US annual consumption US Resource World annual consumption World Resource 38 Quads/year 400 Quads 131 Quads/year 5,500 Quads 6. Express US annual oil consumption in Gbo units. 38 Quads/year x Gbo/Quad = 6.9 Gbo/year 7. Express US Oil Resources in Gbo units. 400 Quads/year x Gbo/Quad = 72 Gbo 8. Express US annual oil consumption in bbl per capita. 6,900 Mbo/year/285 million people = 24 barrels /person/year 9. Express US Oil Resources in bbl per capita. 72,000 Mbo /285 million people = 254 barrels /person 10. Calculate the lifetime of US oil supply at current rate of consumption. 254 barrels /person / 24 barrels/person/year = 11 years 11. Calculate the lifetime of the US oil supply assuming we import 50% of the oil that we consume. 11 years x 2 = 22 years 12. Express global oil consumption in Gbo units. 131 Quads/year x Gbo/Quad = 24 Gbo/year 13. Express global Oil resources in Gbo units Quads/year x Gbo/Quad = 1000 Gbo 14. Express global annual oil consumption in bbl per capita. 24 Gbo/year/6.2 billion people = 3.8 barrels /person/year 15. Express global Oil Resources in bbl per capita Gbo /6.2 billion people = 160 barrels /person 16. Calculate the lifetime of the world oil supply assuming current rate of consumption worldwide. 160 barrels /person / 3.8 barrels/person/year = 42 years 17. Calculate the lifetime of world oil supply assuming world per capita consumption were to equal US per capita consumption. 160 barrels /person / 23 barrels/person/year = 7 years

3 Problems 18 through 23 involve the following data: (Energy in Quads) Oil Natural Gas Coal fossil Total US annual consumption (C) US Resource (R) ,000 World annual consumption (C) World Resource (R) 5,500 4, ,000 Lifetime = 1/k x ln[(k R/C) + 1] 18. Estimate the lifetime of the US coal supply assuming current rate of coal consumption. Note: "current rate" means with no growth (i.e. k = 0.0). This means that you simply divide the resource (R) by the consumption rate (C): Lifetime = R/C. L = 30,000/22 = 1,360 years 19. The current rate of increase of US coal consumption is about 3.5%. If this growth rate were to remain constant, what would be the US annual coal consumption in 2041? DT = 70/3.5 = 20 years. So, two doublings in 40 years 4 x 22 Q = 88 Quad 20. Estimate the lifetime of the US coal supply using the figures in problem 2. This means that k = 0.035, R = 30,000 Quads, and C = 21 Quads. Lifetime = 1/0.035 x ln[(0.035 x 30,000 / 22) + 1] = 111 years 21. At present consumption rates, US coal should last 1,500 years. However, Dr. Al Bartlett argues that it is probably more reasonable to assume that US coal will last less than 100 more years. Explain the discrepancy between these numbers. Depends on growth of consumption rate. 22. How long will world oil resources last at the present consumption rate (i.e. with k = 0.0)? L = 5,500/131 = 42 years 23.How long would world oil resources last if consumption grew at a fixed rate of 3.5% (i.e. k = 0.035)? Lifetime = 1/0.035 x ln[(0.035 x 5,500 / 131) + 1] = 26 years 24. Global warming might cause the surface of the Earth to warm by 5 C. Express this temperature increase in degrees Fahrenheit. 5ºC x 1.8 F/ºC = 9 F

4 25. In the mid-1950s, the ozone levels in LA reached 0.68 parts per million. Express this concentration in ppb ppm = 680 ppb 26. If the stratosphere starts at 40,000 feet and the surface of the earth is 60 F, what is the temperature of the air at the bottom of the stratosphere? 40,000 ft * -4 F/ 1000 ft = -160 F 60 F 160 F = F 27. Current US emission of CO2 from burning of fossil fuels is 1560 million metric tons of carbon. Global emission of CO2 from burning of fossil fuels is 6210 million metric tons of carbon. Calculate US and global per capita figures and compare them. US: 1580 M tons / 285 million = 5.5 tons per capita world: 6.2 G tons / 6.2 G people = 1.0 tons per capita 28. Reducing US CO2 emissions. US emission of CO2 from burning of fossil fuels was 1355 million metric tons of carbon in 1990 and is projected to be 1792 million metric tons in By what percent would the 2012 value have to be reduced to bring it back to the 1990 level? ( )/1792 = 24% 29. There are 6.2 billion people on Earth. Assuming that the global per capita consumption equaled the US value, 1650 gal per day, calculate the global rate of water use by humans. Compare this value to the available renewable water supply, 3.6x10 15 gal/year gal/day/person x 6.2 billion people x 365 days/year = 3.7 x gal/year. 30. Pumping from the Edwards Aquifer is limited to 450,000 acre feet per year. Approximately 1.8 million people draw water from the Edwards. Express the maximum rate of withdrawal from the in aquifer as a per capita figure in units of gallons/day. Note: 1 acre-foot = 330,000 gallons. 450,000 acre-feet /year x 330,000 gal/acre-foot /1,800,000 people/ 365 days/year = 226 gal / person / day 31. SAWS has1.3 million customers. The average customer consumes 154 gallons of water per day. How many acre-feet of water per year is required to serve these customers? 154 gal/person/day * 1,300,000 people *365 days/year / 330,000 gal/acre-foot = 221,000 AFY

5 32. Suppose an older home in San Antonio is equipped with window unit air-conditioners with EER ratings equal to 5 and that this home consumes 30,000 kwh of electrical power each summer to power these units. Suppose these units are replaced by units with EER rating equal to 12. How much energy is saved during the first year after this air-conditioner upgrade? Assuming electricity costs $0.10/kWh, how money is saved during the first year of operation of this improved cooling plant? Old system: Cost = 30,000 kwh x $0.10 / kwh = $3,000 New system: Energy use = 5/12 x 30,000 kwh = 12,500 kwh Cost = 12,500 kwh x $0.10 / kwh = $1,250 Savings = $3,000 - $1,250 = $1, How much cooling power (Btu per hour) will be delivered by an air-conditioner with an EER rating equal to 12, if the unit is using 2,000 W of power? Delivered Cooling Power = EER x Electrical power = 12 x 2000 W = 24,000 Btu/hour Problems involve the following information: Suppose that 25 incandescent light bulbs rated at 100 Watt are replaced by 25 fluorescent lamps rated at 20 Watts. 34. How much energy (in kwh) is consumed by the 25 incandescent 100 Watt bulbs during 10,000 hours of operation (their lifetime). 25 x 100 W x10,000 h = 25,000 kwh 35. How much energy (in kwh) is consumed by the 25 fluorescent 20 Watt bulbs during 10,000 hours of operation. 25 x 20 W x10,000 h = 5,000 kwh 36. How much energy is saved by replacing the 25 incandescent bulbs by 25 fluorescent bulbs. 25,000 kwh -5,000 kwh = 20,000 kwh 37. How much money is saved, assuming that the cost of electricity equals $0.10/kWh. 20,000 kwh x 0.10 $/kwh = $2, What is the payback period (in terms of hours of lighting) assuming the fluorescent bulbs cost $16 each? $16 x 25 bulbs = $400. Payback = $400 / ($2,000 / 10,000 hours ) = 2,000 hours of operation. So the bulbs pay for themselves during the first 1/5 of their lifetime.

6 Problems involve the following information: A flat plate solar collector is used as a solar hot water heater. The collector area equals 20 square meters. The collector is located in a location with annual average daily solar irradiance (insolation) equal to 5.0 kwh/square meter/day. 39. Calculate the amount of solar energy incident on this collect each day. 5.0 kwh/square meter/day x 20 square meter = 100 kwh/day 40. Assuming that the efficiency of the solar collector and the rest of the system equals 50%, calculate the average daily energy produced (as hot water) by this system. Express your answer in kwh/day. 100 kwh/day x 0.50 = 50 kwh/day 41. Calculate the amount of energy produced by this system each year. Express your answer in kwh. 50 kwh/day x 365 day = 18,250 kwh/year 42. Assuming that the solar energy replaces the heating of hot water by electric energy and that electric energy cost 10 /kwh, calculate the yearly savings in electricity costs as a result of using the solar hot water system. Express your answer in $/year. 18,250 kwh/year x $0.10 = $1,825/year 43. Suppose this solar hot water system were to cost $4,000 (installed). Calculate the payback period for this system $4,000/ $1,825/year = about 2 years Problems involve the following information: A photovoltaic power system has a collector area equals 100 square meters. The collector is located in a location with annual average daily solar irradiance (insolation) equal to 5.0 kwh/square meter /day. 44. Calculate the amount of solar energy incident on this collect each day. 5.0 kwh/square meter/day x 100 square meter = 500 kwh/day 45. Assuming that the efficiency of the solar collector and the rest of the system equals 10%, calculate the average daily energy produced (as electricity) by this system. Express your answer in kwh/day. 500 kwh/day x 0.10 = 50 kwh/day 46. Calculate the amount of energy produced by this system each year. Express your answer in kwh. 50 kwh/day x 365 day = 18,250 kwh/year 47. Assuming that electric energy cost 10 /kwh, calculate the yearly savings in electricity costs as a result of using the PV system. Express your answer in $/year. 18,250 kwh/year x $0.10 = $1,825/year

7 48. Suppose this PV system were to cost $10,000 (installed). Calculate the payback period for this system. $10,000/ $1,825/year = about 5 years 49. Wind Power Example Calculation: Site A has an average wind speed of 30 mph and site B has an average wind speed of 15 mph. A wind power installation at site B will produce 5.0 MWh annually. How much would that same installation produce at site A? Windpower is proportional to windspeed cubed. Windspeed at A is 2 times the windspeed at B So, Windpower at A is 2 x 2 x 2 = 8 times the Windpower at B. Power at A = 8 x 5.0 MWh/year = 40 MWh/year. 50. For every kwh of coal-based electrical energy consumed, 1/2 Lb of carbon (in the form of CO2) is released into the atmosphere. A typical toaster uses 40 kwh of electricity per year. How much carbon is released into the atmosphere to power this toaster? 1/2 Lb/kWh x 40 kwh/year = 20 Lb of carbon / year. 51. For every gallon of gasoline consumed, 5 Lb of carbon (in the form of CO2) is released into the atmosphere. A car travels 12,000 miles per year and gets 15 miles per gallon. How much carbon in the form of CO2 does it emit? 12,000 mi x 1 /15 miles/gallon x 5 Lb/gallon = 4,000 Lb or 2 tons. 52. The energy use of the average American is equivalent to the emission of 5 tons of carbon (in the form of CO2) into the atmosphere each year. This equivalent to 20 tons/year of CO2. Americans emit 25% of the world's anthropogenic CO2. The total amount of CO2 emitted into the atmosphere by Americans equals 285 million x 20 tons = 6 billon tons/year. The global anthropogenic emission of CO2 into the atmosphere = 24 billion tons/year. 53. Current global power generation equals 10 Terawatts (10 billion kw). Suppose world power generation increased at a constant 1.4% for the next 100 years. What would be world power generation in 2101? DT = 70/1.4% = 50 years. So, in 100 years, power generation would double twice and equal 40 Terawatts in If we are using fossil fuel to generate most of this energy, then world CO2 emission would be 100 billion tons/year.