Physics 111 - Energy On This World and Elsewhere - Fall 2008 Problem Set #2 with solutions Assigned: 22 October 2008, Due: 23:59pm, 29 October 2008 Please nd below homework assignment #2. In general, you should try to show your work. Otherwise, it will not be possible to assign partial credit. Feel free to discuss the homework with other students. Your answers, however, must be in your own words, and your calculations your own. 1. A particular computer system that is in constant use as a web server consumes an average of 250 Watts of power. a. How much energy does the computer use in two hours? Give your answer in Joules. (250 J/s)(7200 s) = 1.8 106 J b. If you are paying $0.10 per kilowatt-hour for electricity, how much does it cost per day to run the computer? (0.25 kW)(24 h)($0.10/kWh) = (6 kWh)($0.10/kWh) = $0.60 2. A heat engine consumes 5000 Joules of energy from its hot reservoir during each cycle, and performs 3000 Joules of work. a. How much energy is transferred to it s cold reservoir during each cycle? From the conservation of energy, we know that the 5000 J coming out of the hot reservoir cannot be destroyed. If 3000 J of work is performed, the remainder, 2000 J, must be transferred to the cold reservoir. Also, on the sixth slide of lecture #8, we had an equation relating QH , QC and W : W = QH QC which implies that QC = QH W = 5000 J 3000 J = 2000 J b. What is the e ciency of the engine? Again on the sixth slide of lecture #8 we had an equation for the e ciency: E ciency = W/QH = 3000 J/5000 J = 0.60 = 60% c. If each cycle takes one second, what is the output power (associated with work performed) of the engine? Give your answer in both Watts and horsepower. Power = 3000 J Energy = = 3000 W = (3000 W) time 1s 1 hp 746 W = 4.02 hp 1 Physics 111 - Energy On This World and Elsewhere - Fall 2008 Problem Set #2 with solutions continued 3. A new high-temperature nuclear reactor is built that maintains a temperature of 1000 C (or 1273 K) in its core. The temperature of the surrounding air going through the reactor s cooling tower is 20 C (or 293 K). What is the thermodynamic upper limit of e ciency with which this reactor might be used to generate electricity? Carnot e ciency = 1 293 K TC = 0.77 = 77% =1 TH 1273 K 4. Diagram 5 on page 221 of the Annual Energy Review 2007 (showing electricity ow) includes a number representing the total energy consumed to generate electricity in the United States during 2007. For this problem, assume that the population of the United States is 300 million people. a. Using the information from Diagram 5, how much energy is consumed by a city of 100,000 people to produce electricity in a single day? Express your answer in BTU s. 42.08 1015 BTU s 100, 000 300, 000, 000 1 365 = 3.84 1010 BTU s b. Assuming this energy were derived from coal, how many tons of coal does this represent per day? 3.84 1010 BTU s = 1537.5 tons 25 106 BTU s/ton c. In freight trains, coal hoppers are used to haul coal. You probably see them traveling through Charlottesville on a regular basis. If a coal hopper can hold 50 tons of coal, how many coal hoppers does it take to supply an electrical generation power plant for a city of 100,000 for one day? Your answer should be an integral number of coal hoppers. There is no such thing as a fractional coal hopper! You will be awarded full credit if you get the right number to within plus or minus one, so don t worry about round-o errors. 1537.5 tons = 30.75 hoppers 50 tons/hopper Rounding up to the nearest integer, we have 31 hoppers. 2 Physics 111 - Energy On This World and Elsewhere - Fall 2008 Problem Set #2 continued 5. Again, looking at Diagram 5 on page 221 of the Annual Energy Review 2007, a number is given for the gross generation of electricity in the United States in 2007. a. What is this number in Joules? (14.94 1015 BTU s)(1055 J/BTU) = 1.576 1019 J b. What is the average electrical power generated in the United States? Give your answer in both Watts and Gigawatts. Power = 1.576 1019 J Energy = = 4.998 1011 W = 499.8 GW time 365 24 3600 s c. Again taking the population of the United States to be 300 million, what is the per capita electrical power consumption (in Watts) in the United States? 4.998 1011 W = 1, 666 W/person = 1.67 kW per capita 300, 000, 000 people 6. A quantity that is often used to characterize the performance of an electrical power generation facility is something called the heat rate . If one considers the amount of electrical energy produced over some period of time, the heat rate is the ratio of the thermal heat required for production to the electrical energy produced. a. A power-generation system involving a gas turbine and an electrical generator is rated as having a heat rate of 10,430 BTU/kWhr. What is the e ciency (in percent) with which the system converts thermal energy into electricity? The quoted heat rate is telling us that 1 kWh of electricity is being produced for every 10,430 BTU s of heat energy that is used. Here we can use the same formula that we used to problem 2b: E ciency = W/QH = (1 103 J/s)(3600 s) 1 kWh = 10, 430 BTU s (10, 430 BTU s)(1055 J/BTU) 3.6 106 J = = 0.327 = 32.7% 1.10 107 J b. Another system is rated at 6,500 kJ/kWh. What is the e ciency of this other system (in percent) with which it converts thermal energy into electricity? E ciency = 3.6 106 J = 0.554 = 55.4% 6.5 106 J 3 Physics 111 - Energy On This World and Elsewhere - Fall 2008 Problem Set #2 with solutions continued 7. For this exercise, we will use Tables 11.13 and 11.15 from the DOE Annual Energy Review 2007. These tables appear on pages 325 and 329 of the document respectively (page numbers refer to the internal page numbering system of the document). a. Using the total recoverable reserves of coal listed in Table 11.13 and the coal consumption rates listed in Table 11.15, how many years of coal does China have? Assume a constant consumption rate at the level that occurred in 2006. 126, 215 <a href="/keyword/million-tons/" >million tons</a> = 48.96 years 2, 578 <a hre...