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Problem Solutions ChaÔØeÖ ½¿ BiÓÑa×× Problem Solutions Fund. of Renewable Energy Processes Prob. Sol. 13.1 Page 1 of 2 535 Prob 13.1 A plant leaf takes up 0.05 µmoles of CO2 per minute per cm2 of leaf area when exposed to sun light with a power density of 50 W/m2 in an atmosphere containing 330 ppm 2 of CO2. When the light power density is raised to 600 W/m , the uptake is 0.36 µmoles min−1cm−2. Assume that in the above range, the stomata do not change their openings. What is the expected uptake at 100 W/m2 in the same atmo- sphere and the same temperature as above? And at 1000 W/m2? Measurements reveal that the uptake at 1000 W/m2 is only 0.40 µmoles min−1cm−2. This reduction must be the result of a partial closing of the stomata. What is the ratio of the stomatal area at 1000 W/m2 to that at 100 W/m2? 2 What is the expected uptake at 100 W/m if the CO2 con- centration is increased to 400 ppm? ..................................................................................................................... According to the simple model developed in the Textbook, the CO2 uptake rate, φ, is r[CO2]aP φ = r . (1) 1+ V P From this, rP φ V = . [CO2]arP − φ Under typical conditions, the reaction rate, r, is approximately the same for most plants: r = 13.5 × 10−6 m3J−1. The concentration of CO2 in air is some 330 ppm (parts per million). At RTP, the concentration of any gas is 1 kilomole per 24.45 m3. Hence −6 1 −6 −3 [CO2] = 330 × 10 × = 13.5 × 10 kmoles m . 24.45 At low insolations, unless V is abnormally small, V P << , r and V has no influence on φ. Thus, to find V from our data, we must use P = 600 W m−2. The corresponding φ (from the problem statement) is 0.36 µmoles min−1 cm−2 = 60 × 10−9 kmoles s−1m−2. 13.2 × 10−6 × 600 × 60 × 10−9 V = =0.01 ms−1. 13.5 × 10−6 × 13.2 × 10−6 × 600 − 60 × 10−9 We can now introduce the values of V , r, and [CO]2 into equation 1. 178 × 10−12P φ = . (2) 1+1.3 × 10−3P Solution of Problem 13.1 090418 536 Page 2 of 2 Prob. Sol. 13.1 Fund. of Renewable Energy Processes For different power densities, P , we get P φ φ W/m2 µmoles kmoles min−1cm−2 s−1m−2 50 50.1 × 10−3 8.36 × 10−9 100 94.5 × 10−3 15.8 × 10−9 600 360 × 10−3 60.0 × 10−9 1000 464 × 10−3 77.4 × 10−9 The table above predicts a carbon dioxide fixation rate of 0.464 µmoles per second per cm2. However, measurements show that the actual rate is only 0.40 µmoles per second per cm2 or 66.7 × 10−9 kilomoles per second per m2. This discrepancy is probably due to the change in stomatal opening at higher light power densities. Hence, V must be somewhat smaller at 1000 W m−2 than at 100 W m−2. The stomatal velocity that fits the 1000 W m−2 data is 13.2 × 10−6 × 1000 × 66.7 × 10−9 V = =0.01 ms−1. 13.5 × 10−6 × 13.2 × 10−6 × 1000 − 66.7 × 10−9 1.33 × 10−5 × 1000 × 6.7 × 10−8 V = =0.0079 ms−1. 1.34 × 10−5 × 1.33 × 10−5 × 1000 − 6.7 × 10−8 The stomatal velocities are, to first order, proportional to the areas of the stomatal openings, A(1000) V (1000) 0.0079 ≈ = =0.79. A(100) V (100) 0.01 The area of the stomatal opening at 1000 W m−2 is 79 % of that at 100 W m−2. Since φ is proportional to the carbon dioxide concentration, for [CO2]= 400 ppm, 400 φ = × 0.0945 = 0.114 µmole min−1cm−2. 330 A higher carbon dioxide concentration leads to a proportionally higher photosynthetic activity. At 100 W m−2, the uptake rate increases to 0.11 micromoles per minute per square centimeter. Solution of Problem 13.1 090418 Fund. of Renewable Energy Processes Prob. Sol. 13.2 Page 1 of 3 537 Prob 13.2 An automobile can be fueled by dissociated alco- hol. The energy necessary for such dissociation can come from waste exhaust heat. In the presence of catalysts, the process proceeds rapidly at temperatures around 350 C. Consider liquid methanol that is catalytically converted to hydrogen and carbon monoxide. Compare the lower heats of combustion of methanol with those of the products. Do you gain anything from the dissociation? Compare the entropy of gaseous methanol with that of the products. Does this favor the reaction? Assume a gasoline engine with a 9:1 compression ratio fueled by: a. gasoline, b. methanol, and c. dissociated methanol. Assuming that the three fuels lead to identical engine behav- ior, compare the energy per liter of the fuel with that of gasoline. The compression ratio is now changed to the maximum com- patible with each fuel: gasoline: 9:1, methanol: 12:1, dissociated methanol: 16:1. Remembering that the efficiency of a spark-ignition is η =1 − r1−γ , where r is the compression ratio and γ is the ratio of the specific heats (use 1.4), compare the new energy per liter ratios. The gasoline and methanol molecules are complex and this leads to a low value of γ. With hydrogen and carbon monoxide, γ is much higher. Change the above calculations using 1.2 for gasoline and methanol, and 1.7 for the dissociated methanol. ..................................................................................................................... Methanol dissociates according to CH3OH → CO+2H2, (1) and burns according to 2CH3OH+3O2 → 2CO2 + 4H2O. (2) The higher heat of combustion of methanol is given in the CRC as 173.64 kcal/mol=726 MJ/kmole. The heat of vaporization of water is 44.1 MJ/kmole. Each kmole of methanol forms 2 kmoles of water. The lower heat of combustion of methanol is 726 − 2 × 44.1=638 MJ/kmole. Solution of Problem 13.2 090418 538 Page 2 of 3 Prob. Sol. 13.2 Fund. of Renewable Energy Processes The lower heat of combustion of H2 is 241.8 MJ/kmole and that of CO is 283.0 MJ/kmole. Thus, the dissociation products (2 kilomoles of H2, 1 kilomole of CO) have a combined lower heat of combustion of 2 × 241.8+283.0 = 766.6 MJ/kmole of methanol. This is equivalent to 24.0 MJ/kg or 18.9 MJ/liter of methanol. Dissociating methanol results in a mixture of fuels with 766.6/683=1.20 times more energy. This is possible because the dissociation is endothermic. The entropies of interest are (in kJ/K per kilomole) CH3OH(g) 237.7 CO 197.9 H2 130.6 The entropy of the dissociated products is 459.1 kJ/K per kmole of methanol. The entropy of methanol is 237.7 kJ/K per kmole. The products have much more entropy thus favoring dissociation. Additional data to solve this problem can be found in Chapter 4 of the Textbook from which the table below is transcribed. Properties of two important alcohols compared with heptane and octane. Higher heats of combustion for fuels at 25 C. MOL. kg/ MJ/ MJ/ MJ/ MJ/ MASS LITER kg LITER kg LITER (rel. to (rel. to octane) octane) Methanol 32 0.791 22.7 18.0 0.475 0.534 Ethanol 46 0.789 29.7 23.4 0.621 0.697 n-Heptane 100 0.684 48.1 32.9 1.006 0.979 iso-Octane 114 0.703 47.8 33.6 1.000 1.000 The table above lists higher heats of combustion. We can recalculate the table as shown. Properties of two important alcohols compared with heptane and octane. Lower heats of combustion for fuels at 25 C. MOL. kg/ MJ/ MJ/ MASS LITER kg LITER Methanol 32 0.791 19.9 15.8 Ethanol 46 0.789 26.8 21.2 n-Heptane 100 0.684 45.0 30.8 iso-Octane 114 0.703 44.7 31.4 Consider gasoline as mostly heptane. From the table, the volumetric energy densities are 15.8 and 30.8 Solution of Problem 13.2 090418 Fund. of Renewable Energy Processes Prob. Sol. 13.2 Page 3 of 3 539 MJ/liter for respectively methanol and gasoline. As seen, dissociation in- sures a 20% energy gain over the original methanol. It raises the volumetric energy density of this fuel from its normal 15.8 MJ/liter to 19.0 MJ/liter. Even then, gasoline has, by far, the largest volumetric energy density. Now, we must factor in the performance of the fuels in an Otto engine. This performance is a function of both the compression ratio, r, of the engine and the ratio of specific heats, γ, of the fuel air mixture. Here, gasoline is at an disadvantage. The maximum compression ratio a gasoline engine can tolerate is substantially lower than that of a methanol engine and much lower than that of a H2/CO2 engine. In addition, a gasoline engine must operate with richer mixtures (or else it will fire erratically) than H2/CO2 engines that, owing to the flammability of hydrogen, can operate very lean. Lean mixtures correspond to large γ and greater efficiencies. The table below, computed from, η =1 − r1−γ , shows the theoretical efficiency of the Otto-cycle operating with different r and γ.
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