# Chapter 13 Biomass Problem Solutions - Stanford web. 1.310−3P. (2) Solution of Problem 13.1 090418...

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### Transcript of Chapter 13 Biomass Problem Solutions - Stanford web. 1.310−3P. (2) Solution of Problem 13.1 090418...

Chapter 13BiomassProblem 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 moles of CO2 perminute per cm2 of leaf area when exposed to sun light with apower density of 50 W/m2 in an atmosphere containing 330 ppmof CO2. When the light power density is raised to 600 W/m

2,the uptake is 0.36 moles min1cm2. Assume that in the aboverange, 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 only0.40 moles min1cm2. This reduction must be the result of apartial closing of the stomata. What is the ratio of the stomatalarea at 1000 W/m2 to that at 100 W/m2?

What is the expected uptake at 100 W/m2 if the CO2 con-centration is increased to 400 ppm?.....................................................................................................................

According to the simple model developed in the Textbook, the CO2uptake rate, , is

=r[CO2]aP

1 + rV P. (1)

From this,

V =rP

[CO2]arP .

Under typical conditions, the reaction rate, r, is approximately thesame for most plants:

r = 13.5 106 m3J1.

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

[CO2] = 330 106

1

24.45= 13.5 106 kmoles m3.

At low insolations, unless V is abnormally small,

P

536 Page 2 of 2 Prob. Sol. 13.1 Fund. of Renewable Energy Processes

For different power densities, P , we get

P W/m2 moles kmoles

min1cm2 s1m2

50 50.1 103 8.36 109

100 94.5 103 15.8 109

600 360 103 60.0 109

1000 464 103 77.4 109

The table above predicts a carbon dioxide fixation rate of 0.464 molesper second per cm2. However, measurements show that the actual rate isonly 0.40 moles per second per cm2 or 66.7 109 kilomoles per secondper m2.

This discrepancy is probably due to the change in stomatal opening athigher light power densities. Hence, V must be somewhat smaller at 1000W m2 than at 100 W m2. The stomatal velocity that fits the 1000 Wm2 data is

V =13.2 106 1000 66.7 109

13.5 106 13.2 106 1000 66.7 109= 0.01 m s1.

V =1.33 105 1000 6.7 108

1.34 105 1.33 105 1000 6.7 108= 0.0079 m s1.

The stomatal velocities are, to first order, proportional to the areas ofthe stomatal openings,

A(1000)

A(100)

V (1000)

V (100)=

0.0079

0.01= 0.79.

The area of the stomatal opening at 1000 W m2 is79 % of that at 100 W m2.

Since is proportional to the carbon dioxide concentration, for [CO2]=400 ppm,

=400

330 0.0945 = 0.114 mole min1cm2.

A higher carbon dioxide concentration leads to a proportionallyhigher photosynthetic activity. At 100 W m2, the uptake rateincreases to 0.11 micromoles per minute per square centimeter.

090418

Solution of Problem 13.1

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 fromwaste exhaust heat. In the presence of catalysts, the processproceeds rapidly at temperatures around 350 C.

Consider liquid methanol that is catalytically converted tohydrogen and carbon monoxide. Compare the lower heats ofcombustion of methanol with those of the products. Do you gainanything from the dissociation?

Compare the entropy of gaseous methanol with that of theproducts. Does this favor the reaction?

Assume a gasoline engine with a 9:1 compression ratio fueledby:

a. gasoline,b. methanol, andc. 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

= 1 r1 ,where r is the compression ratio and is the ratio of the specificheats (use 1.4), compare the new energy per liter ratios.

The gasoline and methanol molecules are complex and thisleads to a low value of . With hydrogen and carbon monoxide, is much higher. Change the above calculations using 1.2 forgasoline 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 kmoleof methanol forms 2 kmoles of water. The lower heat of combustion ofmethanol 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 COis 283.0 MJ/kmole. Thus, the dissociation products (2 kilomoles of H2, 1kilomole 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 with766.6/683=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) 237.7CO 197.9H2 130.6

The entropy of the dissociated products is 459.1 kJ/K per kmole ofmethanol. 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 theTextbook from which the table below is transcribed.

Properties of two important alcoholscompared 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. tooctane) octane)

Methanol 32 0.791 22.7 18.0 0.475 0.534Ethanol 46 0.789 29.7 23.4 0.621 0.697n-Heptane 100 0.684 48.1 32.9 1.006 0.979iso-Octane 114 0.703 47.8 33.6 1.000 1.000

The table above lists higher heats of combustion. We can recalculatethe table as shown.

Properties of two important alcoholscompared 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 15.8Ethanol 46 0.789 26.8 21.2n-Heptane 100 0.684 45.0 30.8iso-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

090418

Solution of Problem 13.2

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 volumetricenergy 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 theengine and the ratio of specific heats, , of the fuel air mixture. Here,gasoline is at an disadvantage. The maximum compression ratio a gasolineengine can tolerate is substantially lower than that of a methanol engineand much lower than that of a H2/CO2 engine. In addition, a gasolineengine must operate with richer mixtures (or else it will fire erratically) thanH2/CO2 engines that, owing to the flammability of hydrogen, can operatevery lean. Lean mixtures correspond to large and greater efficiencies.

The table below, computed from, = 1 r1 , shows the theoreticalefficiency of the Otto-cycle operating with different r and .

r 1.2 1.4 1.7

9 0.356 0.585 0.78512 0.392 0.630 0.82416 0.426 0.670 0.856

The kilometrage (kilometers/liter) of a vehicle is proportional to thevolumetric energy density of the fuel multiplied by the engine efficiency.

If all engines operate with an effective of 1.4, the relative kilometrageswould be

Fuel r MJ/l kilometrage kilometrage(normalized)

Gasoline 9 1.4 0.585 30.8 18.0 1Methanol 12 1.4 0.630 15.8 10.0 0.552

Disc. meth. 16 1.4 0.670 18.9 12.7 0.704

Under the above circumstances, gasoline is still the most efficient fuel.If one considers that the H2/CO2 engine can operate with a large

(the problem statement suggests 1.7 which is exaggeratedly high), and thatthe of the gasoline and methanol engines is lower than 1.4 (say, 1.2), then

Fuel r MJ/l kilometrage kilometrage(normalized)

Gasoline 9 1.2 0.356 30.8 11.0 1Methanol 12 1.2 0.392 15.8 6.2 0.518

Disc. meth. 16 1.7 0.856 18.9 16.2 1.48

Now, the kilometrage of the dissociated methanol engine is far largerthan that of the other fuels.

Solution of Problem 13.2

090418

540 Page 1 of 4 Prob. Sol. 13.3 Fund. of Renewable Energy Processes

Prob 13.3 Under proper conditions, water hyacinths (Eich-hornia crassipes), a floating plant, will grow at such a rate thattheir dry biomass increases 5% per day. The total water contentof these plants is high (94%). Nevertheless, 400 kg of dry mattercan be harvested daily from one hectare of plantation.

Consider a plantation with one hectare area consisting of along canal (folded upon itself). At the starting point (se

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