PHYS/ENVS 3070 Homework Assignment #5: Problem #1: An electrically powered heat pump can deliver more energy than it draws from the electrical power line. Is this statement true or false? Explain in detail in relation to conservation of energy. (2 points) True. In the diagram for a heat pump, the energy coming "in" (i.e. arrows pointing inward) are both Qcold and Work. The energy going "out" (i.e. arrows pointing outward) is only Qhot. Thus, conservation of energy dictates that Qcold + Work = Qhot. Thus, it is always the case with a heat pump that Qhot > Work, since once is drawing energy out of the cold air outside. It is important to note that the cold air still has lots of energy in it, but it requires work to move that energy to a higher temperature environment (e.g. inside your house). Problem #2: Compare an ideal heat pump operating between T(hot) = 100 degrees Fahrenheit and T(cold) = 0 degrees Fahrenheit to a second heat pump operating with temperatures of 95 degrees Fahrenheit and 38 degrees Fahrenheit. What is the ratio of the coefficient of performance for the second case to the first case? (2 points) The C.O.P. = Qhot / Work = Qhot / (Qhot - Qcold) = Thot / (Thot - Tcold). Note that the temperatures must be in units of Kelvin. Thus, the following conversions are needed: T1(hot) = 100 degrees Fahrenheit = 311 Kelvin T1(cold) = 0 degrees Fahrenheit = 255 Kelvin T2(hot) = 95 degrees Fahrenheit = 308 Kelvin T2(cold) = 38 degrees Fahrenheit = 276 Kelvin The C.O.P. in case 1 = 5.5 and in case 2 = 9.6, and therefore the ratio is 9.6/5.5 = 1.7 Problem #3: A refrigerator has an energy efficiency ratio EER of 6. For each unit of input work energy (i.e. one BTU of input work) from the electric power company, how many units of heat energy (Qhot) are dumped to the kitchen? Please give your answer in units of BTU. (2 points) The EER = Qcold / Work = 6. If we input (for example) 1 BTU of work, we will move 6 BTU of heat energy out of the refridgerator (Qcold). By conservation of energy Qcold + Work = Qhot, and therefore Qhot = 7 BTU.Problem #4: Compare the following two scenarios for heating a factory. In the first scenario, one has a 40% efficiency natural gas burning power plant to supply electrical to a heat pump having a coefficient of performance of 4. In the second scenario, one just burns the natural gas to generate the heat directly. Calculate quantitatively which of these scenarios is more efficient (i.e. generates more heat energy for the factory per unit of energy input from the natural gas. (2 points) In the first scenario, imagine we have 1 BTU of chemical energy in the natural gas. We convert that to electricity with 40% efficiency, and thus have 0.4 BTU of electrical energy. If we put the electricity to use in a heat pump with C.O.P = Qhot/Work = 4, then we get four times the input electrical work energy as output heat energy. Thus, we get 4 x 0.4 BTU = 1.6 BTU of heat energy. In the second scenario, imagine we have 1 BTU of chemical energy in the natural gas (exactly the same as the above scenario). If we burn it and get all the energy as heat, then we have 1.0 BTU of heat energy. Thus, the heat pump used in scenario one is the better option in terms of heat energy output. This is because the heat pump utilizes the energy in the cold air outside the house (for example). Problem #5: Decide whether each of the following statements are true or false and give the reasoning. (2 points total) A. Good photovoltaic cells typically have an efficiency of 1-5 percent. False. See Table 4.3 of the textbook R & K on page 117. B. The solar spectrum has its maximum intensity at a wavelength corresponding to the visible region. True. See Figure 4.1 of the textbook R & K on page 93. C. The solar constant (2 cal / square cm / minute) is the average solar density at ground level on the earth. False. The textbook states on page 92 that "The power density at the top of the atmosphere on the side of the earth directly facing the sun is 2 cal/min/cm2." Thus it is not at the ground level on the earth. D. Of the solar radiation incident on the earth's upper atmosphere, about 100% on average is available at ground level.False. See above. Problem #6: A hydroelectric power plant is located near a large waterfall. The air temperature is 23 Celsius and the water temperature is slightly lower at 18 Celsius. What is the maximum efficiency for converting the input energy into electrical energy from a thermodynamics points of view? (just the physics limitations, not including engineering issues). (2 points) This is somewhat of a trick question. Hydroelectric power is not from a heat engine. It is just the conversion of gravitational potential energy (of the water at some height) into electrical energy. Thus, there is no limit from the Carnot or thermodynamic efficiency. Problem #7: In the news this week, the Vatican installed for the first time solar panels. The news articles stated that there were 2700 panels installed on the roof of Paul VI auditorium. The installers said "with this plant, if it is working, in about two weeks we avoid 200 tonnes of carbon dioxide, and this is the equivalent to 70 tonnes of oil." The articles did not state the dimensions (size) of each solar panel. Calculate approximately how large these solar panels must be for the statements above to be correct. State your assumptions clearly along with your calculations. Does the statement seem reasonable? (3 points) One tonnes of oil equivalent has a chemical potential stored energy of 42 x 109 Joules (many references including Wikipedia). Thus, 70 tonnes of oil equivalent has 2.9 x 1012 Joules or 800,000 kiloWatt-hours of energy. If one burned this oil in a power plant to generate electricity, if the power plant is 37% efficient, then one gets 1.0 x 1012 Joules or 300,000 kiloWatt-hours of electrical energy. How large a solar panel coverage would be needed to get this amount of electrical energy in two weeks (14 days)? If we use the values in the textbook, we might assume that the solar energy incident (averaged around one year) per day is 5 kiloWatt-hours/ meter2. Thus, in 14 days, that would yield 70 kiloWatt-hours/ meter2. However, the panels have a low efficiency for converting this solar energy into electrical energy. If we assume a 10% efficiency, then we can get 7 kiloWatt-hours/ meter2 of electrical energy. Using this value, one
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