PHYS/ENVS 3070 Homework Assignment #6: Problem #1: A car wash needs 1200 gallons of warm water each day heated from 50 degrees Fahrenheit to 110 degrees Fahrenheit. How large a solar collector would be needed to do this job? Give your answer in square feet. Assume the incident solar energy is 1100 BTU per square foot each day, and the collector efficiency is 60%. (2 points) First we can calculate how much energy is needed for the water heating each day. Recall that the definition of 1 BTU is the energy to heat one pound of water by 1 degree Fahrenheit. ()BTUFgallonpoundsgallonsneededE501097.55011013.81200)( ×=−××= If the solar collector is only 60% efficient, it means we need incident solar energy of E(needed) / 0.60 = 1.0 x 106 BTU to do the job. Since incident solar energy is 1100 BTU / ft2, the total size of the collector needs to be: ()squarefeetftBTUBTUftEnergytEnergyInpuArea 900/1100/10)2//(26=== Problem #2: A house is designed with passive solar heating that has energy stored in a concrete floor of 1300 square feet. How thick should the concrete be to store 200,000 BTU with a temperature swing of 25 degrees Fahrenheit? Note you need to use the energy storage capacity of concrete as given in the textbook. (2 points) On page 103 of the E & E textbook, it describes the energy storage properties of different materials. For concrete the value is 22 BTU/ft3/degree Fahrenheit. Thus, if you want to store 200,000 BTU of energy with a temperature swing of 25 degrees Fahrenheit, you can solve for the volume of concrete needed. 300303363)25()()//22(000,200)()()//22(ftVolumeFVolumeFftBTUBTUeSwingTemperaturVolumeFftBTUedEnergyStor=××=××= Now we are told we will have a giant rectangle of concrete with Area = 1300 ft2 and a thickness. Since Area x thickness = Volume, we can solve for the thickness. Thickness = 0.28 feet thick.Problem #3: A reasonable size for overnight heat storage for a solar home is about ____ gallons of water. a. 10 gallons b. 50 gallons c. 1000 gallons d. 5000 gallons (1 point) Water can store heat at 62 BTU/ft3/degree F. The textbook says that "a house in the United States insulated to the usual standards can typically require about 1 million BTU of heat energy on a cold winter way. Thus, as an order of magnitude estimate, say we need to store 0.5 million BTU in the water to be used to keep the house warm overnight. If we assume it is 25 degrees Fahrenheit colder at night, then we can work out the volume of water needed. ()()3036300)25(//62105.0ftVolumeFVolumeFftBTUBTU=××=× Water has a density of 62.4 pounds per cubic foot and there are 8.3 pounds per gallon of water (see the textbook front cover inset). Thus, we would need 300 ft3 x 62.4 pounds/ft3 x (1 gallon / 8.3 pounds) = 2250 gallons of water. This is a rough estimate, but most likely 5000 gallons would be needed. Note that we need more of the heat energy at night than one half since it is colder at night. Problem #4: Only about ____ of the power in the solar spectrum is in the wavelength region to which silicon photovoltaic cells are sensitive. a. 47% b. 57% c. 67% d. 77% e. none of the above (1 point) On page 112 of the E & E textbook it states that about 77% of the solar energy is at wavelengths less than 1.12 microns (which is the sensitivity of the devices). Thus the answer is D. Problem #5: How much electrical energy can be generated by all the water in a lake 2000 meters wide by 8000 meters long by 50 meters deep in all the water falls through a vertical distance of 650 meters? Assume that the generator is 80 percent efficient. Compute your answer in energy units of Joules. (2 points) The gravitational potential energy PE = mgh where m is the mass in kg, g = 9.8 m/s2 the gravitational acceleration at the surface of the earth and h is the height in meters. We are not given the mass of the water, but we can calculatethe volume and then use the density of water 1 gram / cm3 = 0.001 kg / (0.01 m)3= 1000 kg/m3. The volume is 2000 meters x 8000 meters x 50 meters = 8 x 108 m3 And the mass is then m = volume x density = 8 x 1011 kilograms. PE = (8 x 1011 kg) x (9.8 m/s2) x (650 meters) = 5.1 x 1015 Joules. If the efficiency is 80%, then the electrical energy = 4.1 x 1015 Joules. Problem #6: What would the average electric power output be if this lake were drained over a period of one year? Compute your answer in power units of Watts. (2 points) One wants to take the above energy and divide it by the time in seconds in a year. MegaWattsWattssJTimeEnergyPower 140104.1103101.48715=×=××== Problem #7: Decide which of the following statements are true or false and give your reasoning. (5 points - two for each part) a. A realistic turbine in a modern hydroelectric plant will convert gravitational potential energy into electrical energy at an efficiency well below 50%. False. Hydroelectric plants are typically 80-90% efficiency in this regard. See the textbook discussion in Section 5.2. b. If the windspeed doubles, an ideal windmill will produce 16 times the power. False. The Power per area depends on the velocity of the wind to the third power (v3) as detailed in Section 5.3 of the textbook. Thus, if you double the velocity, the power will go up by 23 = 8. c. A realistic turbine in a modern windmill can convert kinetic energy of the air into electrical energy at an efficiency well above 60%. False. Realistic efficiencies are typically of order 30%, though the maximum given by Figure 5.6 of the textbook can be higher (of order 55%). d. With the current corn production in the United States, we could completely replace gasoline with ethanol.Either answer is possible depending on the point of view. If you add up all the crops (of which corn is just one) in the United States, they have a biomass energy of 79 QBTU (see calculation in section 5.5 of the textbook). Transportation represents 27.3% of the U.S. Energy usage (see textbook Table 1.3) and thus about 27 QBTU per year. However, all crops in the U.S. are not corn, but a lot are. Thus on a pure energy basis it might be close (within a factor of 2-4). However, that would mean no corn for food, in particular for feeding livestock. Most realistic estimates do not allow the current corn production to do the job by quite a large margin. e. No countries get the majority of their electricity from water power. False. Many nations like Norway, Nepal,
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