2.83/2.813 SAMPLE Quiz #190 Minutes, Open Book, Open NotesPlease do all of your work in the exam book provided. Cite references if you use numbers supplied by a paper or other source. Make all work clear. State assumptions. Box your answer. Good Luck!1) (15 points) Estimate the reference molar concentration of CO2 in the atmosphere used by Szargut.Solution:From class note we know the change in exergy due to concentration is:refoconcccRTB lnWe are looking for cref !We now from Szargut that for CO2 Bconc = ochb = 19.87 kJ/molc1 = 1 To = 298.15 oKR = 8.314 J/mol oK4)/314.8)(15.298(/870,19103.31KmolJKmolJRTBrefoooconceecc2) (10 points) The governor of California seems proud to drive a GM hummer. If the governor gets 9 miles per gallon, and drives 12,000 miles a year, estimate his ecological footprint in global hectares.Solution:A really similar problem was done in the class slidesgasolineofgalmilesgalmiles 133391000,12  1 gal = 3.785 litters = 3785 cm31Density of gasoline = 0.68 g/cm3Gasoline can be simplified as octane. The reaction for burning octane is:2 C8H18 + 25 O2  16 CO2 + 18 H2OThus:21882188331012.02161141168.0137851333COmolCkgHCmolCOmolgHCmolcmggalcmgasolineofgal = 2,889 kg or 2.89 tons of C haglobalfactorequiCtonshalandtogoesCtons 57.23.1.95.01165.089.2 Eco-footprint = 2.57 global ha3) (20 points) You have a contract to cast aluminum parts. Each finished part weights 0.5 kg. Your client specifies that no more than 80% of the input is recycled aluminum. Since recycled aluminum is cheaper and you want to minimize cost, you use as much as possible. With the current technology, your runner and riser take 0.5 kg of aluminum. With the system you use, you just recuperate 80% of the runner and riser. The rest gets discarded to alandfill.2i. Draw a simplified material/energy flow diagram for casting(with the product, primary material production and secondary mat prod). Solution to 3i:ii. Give the values of and of Solution to 3ii: = 20% x the runner + riser mass = 0.2 x 0.5 = 0.1 kgCasting0.5 kg3Casting0.5 kg0.5 +  = 80% x the runner + riser mass = 0.8 x 0.5 = 0.4 kg = nothing in the problem mentions that our factory sends anything to recycling = 0 kg = the percentage of primary material used = 20% = 0.1 kg = 0.4 kg0 kg = 20%iii. Find the total energy attributed per part. Include both the “manufacturing energy requirement” and the “material energy requirement”. Solution to 3iii:According to class notes (manuf. processes) the energy to cast a metal lies in between 13 and 17 MJ per kg of saleable metal (including fuels for electricity). For this exercise we will assume that this translates to about 9 MJ per kg of metal processed (Note, Em = 9 MJ/kg < Energy per kg saleable product ≈ 15 MJ/kg because you process more metal than you sell due to runner, risers etc). Manufacturing Energy Requirement =MJkgMJEm9)4.01.05.0(/9)5.0( According to class notes (manuf. processes):Primary energy requirement to manufacture Al = Ep = 270 MJ/kg Secondary energy requirement to manufacture Al = Er =16.5 MJ/kg (recycled) Material Energy Requirement =MJEErp32.40)1.05.0)(5.16%80270%20()5.0)()1(( Total = 49.32 MJiv. What is the minimum amount of energy required to do the casting operation? Solution to 3iv:FpHTcEnergyMin  4In the notes it says that for aluminum kgMJHTcFp/95.0. In this case 0.5 kg product x 0.95 MJ/kg = 0.475 MJ 4) (15 points) You have the following reaction in a furnace:2Cu2S + 3O2  2Cu2O + 2SO2How much coal (in kg) do you need to run the reaction assuming 100% efficiency.Solution:Look in Szargut for the exergy values (ochb):Bloss = ( 2 (791.8) + 3 (3.97) ) - ( 2 (124.4) + 2 (313.4) )Bloss = 719.91kJSince the exergy loss is positive the reaction is exothermic and needs no extra energy from the carbon reaction. You can think of the reaction having an extra “heat” term on the right hand side that accounts for the exergy loss. Thus the amount of coal needed is 0 kg5) (25 points) Estimate the energy needed to produce 1 kg of carbon in the ground of a “tropical forest” in a given year. Hint: Think in terms of the carbon cycle. What reaction coverts CO2 into solid carbon and viceversa.5Carbon in groundIf you do not have time to arrive at a numerical answer at least outline the important steps involved and the nature of the calculation using specific equations when appropriate.Solution:Thought Process:Carbon gets to the ground by means of plant sequestration. According to Wackernagel 0.95 tons of C/ha get sequestered by a hectare in a year. This is the same as 0.095 kg/m2.Now we know that the NPP for a tropical forest is 0.83 kg of C/m2 per year. As can be inferred by comparing this number with the sequestration number not all the carbon from NPP gets sequestered. Some of the NPP gets sequestered, some gets used by the plant, and some leaves by means of heterotrophic respiration.How does the plant/tree produce the NPP. It is by means of photosynthesis of the GPP. GPP is approximately twice NPP, or 1.66 kg of C/m2 . The photosynthesis equation together with its exergy loss is:6CO2 + 6H2O  C6H12O6 + 6O2Bloss = ( 6 (19.87) + 6 (0.9) ) - (2928.8 + 6 (3.97) )Bloss = -2828kJThe structure of your calculation should be:So in order to sequester 1 kg of C you need “x” meters squared of forest, which produce acertain NPP, that requires a certain GPP. The carbon of this GPP, is imbedded the in the CO2 term of the photosynthesis reaction. Running this reaction with this number of CO2 moles requires a “y” amount of energy that must be absorbed from the sun. Actual Calculation: 6forestofmCkgmCkgx225.10095.011  GPP = 1.66 kg/m2 of CMJCOmolkJCmolCOmolCkgCmolmCkgmy 6856282811012.01166.15.102222At least 685 MJ of sunlight is needed. (Remember that the absorption of sunlight is not 100% efficient so in reality much moreis