Chapter 13 Biomass Problem Solutions - Stanford...

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 µmoles 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/m2,the uptake is 0.36 µmoles min−1cm−2. 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 µmoles min−1cm−2. 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 CO2

uptake rate, φ, is

φ =r[CO2]aP

1 + rV P

. (1)

From this,

V =rPφ

[CO2]arP − φ.

Under typical conditions, the reaction rate, r, is approximately thesame 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

[CO2] = 330 × 10−6 ×1

24.45= 13.5 × 10−6 kmoles m−3.

At low insolations, unless V is abnormally small,

P <<V

r,

and V has no influence on φ. Thus, to find V from our data, we must useP = 600 W m−2. The corresponding φ (from the problem statement) is0.36 µmoles min−1 cm−2 = 60 × 10−9 kmoles s−1m−2.

V =13.2 × 10−6 × 600 × 60 × 10−9

13.5 × 10−6 × 13.2 × 10−6 × 600 − 60 × 10−9= 0.01 m s−1.

We can now introduce the values of V , r, and [CO]2 into equation 1.

φ =178 × 10−12P

1 + 1.3 × 10−3P. (2)

Solution of Problem 13.1

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

For different power densities, P , we get

P φ φW/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 µmolesper second per cm2. However, measurements show that the actual rate isonly 0.40 µmoles per second per cm2 or 66.7 × 10−9 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 m−2 than at 100 W m−2. The stomatal velocity that fits the 1000 Wm−2 data is

V =13.2 × 10−6 × 1000 × 66.7 × 10−9

13.5 × 10−6 × 13.2 × 10−6 × 1000− 66.7 × 10−9= 0.01 m s−1.

V =1.33 × 10−5 × 1000 × 6.7 × 10−8

1.34 × 10−5 × 1.33 × 10−5 × 1000− 6.7 × 10−8= 0.0079 m s−1.

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 m−2 is79 % of that at 100 W m−2.

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

φ =400

330× 0.0945 = 0.114 µmole min−1cm−2.

A higher carbon dioxide concentration leads to a proportionallyhigher photosynthetic activity. At 100 W m−2, the uptake rateincreases to 0.11 micromoles per minute per square centimeter.

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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

η = 1 − r1−γ ,where r is the compression ratio and γ 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 γ. With hydrogen and carbon monoxide,γ 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.


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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

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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, γ, 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 γ and greater efficiencies.

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

γ γ γ

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 γ of 1.4, the relative kilometrageswould be

Fuel r γ η 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 γ

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

Fuel r γ η 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.


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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 (seedingend), the canal is 0.5 m wide. It expands gradually enough tojust accommodate the growing plants that are slowly swept alongby the current. Assume a current of constant speed such that theplants take 60 days to float from the seeding end to the harvestingend.

a. How wide must the canal be at the harvesting end?.....................................................................................................................

By general dimensional considerations, if the mass of the plant grows5% per day, the surface area should grow approximately 2

35% = 3.33%

day−1. This is because, to a first approximation, surface is proportional tomass2/3. Thus, if i is the day number

mi+1 = mi(1 + α),where m = mass, α = mass growth rate, β = surface growth rate, and

Si+1 = Si(1 + β),

m2/3

i+1 = m2/3

i (1 + α)2/3 → Si+1 = Si(1 + α)2/3,

1 + β = (1 + α)2/3 or (1 + β)3/2 = 1 + α.

But β << 1, thus, by binomial expansion,

(1 + β)3/2 ≈ 1 +3

2β = 1 + α,

β =2

3α.

The surface grows exponentially: S = S0 exp(t/τ).

ln(1 + β) =1

τ= β.

ThusS = S0 exp(βt),

where t is the time in days.After 60 days, S = S0 exp(0.033 × 60) = 7.4S0.Since the length of the daily growth along the direction of the water

flow is constant (the current is of constant speed), the width of the canalat the harvesting end must be 7.4 times the width at the seeding end, i.e.,it must be 0.5 × 7.4 = 3.7 meters.

The width of the canal at the harvesting end must be 3.7 meters.

b. How long must the canal be?

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.....................................................................................................................At any point along the canal, the width is

W = W0 exp(βt).But t = ℓ/V where ℓ is the distance from the seeding end and V is the

water velocity assumed constant.Hence,

W = W0 exp

(

β

Vℓ

)

.

The velocity of the water is V = L/T , where L is the total length ofthe canal and T = 60 days.

The area, A, of the canal surface is

A =

∫ L

0

W exp

(

β

V

)

dℓ

= W0

V

β

[

exp

(

β

VL − 1

)]

=W0L

Tβ

[

exp (βT − 1)]

=0.5L

60 × 0.033

[

exp(0.033× 60) − 1]

= 1.6L = 104 m2

,

L = 6260 m.

The total length of the canal is 6260 meters.

c. How much energy is harvested per day (express this in GJand in barrels of oil)?

.....................................................................................................................Daily harvest is 400 kg of dry matter, because the productivity of water

hyacinth is 400 kg of dry biomass per day per hectare and the total surfaceof the canal is 1 ha. Taking the average energy yield of dry biomass assome 18 MJ kg−1, the energy harvested is 7200 MJ/day. One barrel of oilcorresponds to 6 GJ, hence the energy harvested is 1.2 bbl per day.

The daily energy harvest is 6.2 GJ.This corresponds to 1.2 barrels of oil.

d. How many kilograms of wet plants must be used daily asseed?

.....................................................................................................................Let M0 be the daily seeding. Then

400 = M0 exp(0.05 × 60)

because the mass grows 5% per day.

M0 =400

e3= 19.9 kg of dry matter..


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This is about 5% of the collected matter. Since 94% of the wet materialis water, the daily re-seeding must consist of

19.9

1 − 0.94= 322 kg of wet plants.

320 kg of wet plants must be used daily for re-seeding.

e. Assume that 50% of the biomass energy is converted intomethane through a digestion process. Estimate the methaneyield in cubic meters per day.

.....................................................................................................................

0.5 × 7200 = 3600 MJ/day.

The heat of combustion of methane is 55.6 MJ/kg. Hence the methaneproduction will be 3600/55.6 = 64.8 kg/day.

Molecular mass of methane is 16 daltons, hence its density (RTP) is16 kg/kmole

24.5 m3/kmole= 0.65.

The daily production is

64.8kg

day×

1

0.65

m3

kg= 99.7 m3 day−1.

Methane production is about 100 m3 per day.

f. If the methane is burned in a gas turbine or in a diesel enginewith 20% efficiency, what is the average electric power thatcan be generated?

.....................................................................................................................The methane power is 3600 MJ/day or 41.7 kW. Withe 20% efficiency, theaverage electric power is

The average electric power is 8.3 kW.

g. Assuming that the average depth of the canal is 1 m, what isthe amount of water that has to flow in daily provided thereis no loss by evaporation and infiltration?

.....................................................................................................................

V = 104 × 1 = 10, 000 m3.

The residence time of the water in the canal is 60 days. Therefore, theflow rate is 10,000/60=166.6 m3/day.

The flow rate of water is 167 m3/day.

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h. Do you have any good ideas of how to insure that the wa-ter velocity is kept reasonably constant notwithstanding theexpanding canal?

.....................................................................................................................The simplest (but not necessarily the most practical) solution is to

make sure the channel cross-section is kept constant regardless of how farit is from the seeding end. In other words,

WD = K

where W is the canal width, d is its depth, and K is a constant to bedetermined.

We saw thatW = 0.5 exp(316 × 10−6)ℓ.

hence

d = 2K exp(−316 × 10−6)ℓ)

The average depth, < d >, is to be 1 meter. It is

< d >=1

L

∫ 6260

0

2K exp(−316 × 10−6)ℓ)dℓ = 1,

−2K

316 × 10−6L

[

exp(−316 × 10−6)ℓ)]6260

0= 1,

2K × 0.862

316 × 10−6 × 6260= 1,

K = 1.15,

d = 2.3 exp(−316 × 10−6)ℓ).

At the seeding end (ℓ = 0, the depth is 2.3 meters and at the harvestingend, it is 0.32 m.

This makes the canal expensively deep at one end and too shallow (theroots of the floating plants will scrape the bottom) at the other end.

Perhaps a better solution is to inject water at different points alongthe canal so that at, the seeding end, the flow is much less than at theharvesting end.


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Prob 13.4 Sugar cane is submitted to an illumination of 500W/m2. Assuming a stomatal velocity of 6 mm/s, what is the pho-tosynthetic efficiency (defined as the ratio of the heat of combus-tion of the dry biomass generated to the incident solar energy)?.....................................................................................................................The carbon dioxide uptake rate is

φ =r[CO2]aP

1 + Pr/V.

From Equation 7 of Chapter 15 of the Textbook,

r[CO2]a =φ

P= 178 × 10−12 kmole/J.

r is 1.33 × 10−5 m3/J.P = 500 W/m2 and V = 6 mm/s or 6× 10−3 m/s, both data from the

problem statement.Then,

φ =178 × 10−12 × 500

1 + 500 × 1.33 × 10−5/6 × 10−3= 42.2 × 10−9 kmole s−1m−2.

The energy fixed for each kilomole of carbon dioxide taken up is about440 MJ (see Textbook).

Thus the rate at which energy is being fixed by the sugar cane is

Pfixed = 42.2 × 10−9 × 440 × 106 = 18.6 W m−2.

The efficiency is

η =18.6

500= 0.037.

Under the conditions of the problem,the efficiency of the sugar cane is 3.7%.

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Prob 13.5 Here is a typical task that an energy consultantmight tackle:

The operator of a large alcohol distillery wants to know ifit makes economic sense to use the leftover bagasse as a fur-ther source of ethanol. In the traditional process, the amount ofbagasse obtained from 1 ton of burned and cropped sugar caneis larger than the amount that has to be burned to drive the dis-tillation process. The excess is either sold or used to generateelectricity for the plant. The question is how much additional al-cohol can be obtained by hydrolyzing all the polysaccharides (cel-lulose and hemicellulose) in the leftover bagasse. We will makethe following simplifying assumptions:

a. The hydrolysis will yield 600 grams of sugars (glucose andpentoses) per kilogram of polysaccharides.

b. The hydrolysis requires no energy (not true!).c. The glucose-to-ethanol and the pentose-to-ethanol yields are

the same as the sucrose-to-ethanol yields of the traditionalprocess.

d.. The data for this problem are those discussed in Section 13.3.2of the Textbook.

Calculate the additional amount of alcohol that can be ob-tained from 1 ton of burned and cropped sugar cane. Comment......................................................................................................................

From the data in the text:1 ton of sugar cane produces 153 kg of sugar which can be transformed

into 80 liters of ethanol. This corresponds to 1.92 kg of sugar per liters ofethanol.

Sugar to ethanol ratio:

Λsug→eth = 0.52 liter/kg. (1)

The same ton of sugar cane also produces 276 kg of wet bagasse or 138kg of dry bagasse. Of these, 52 kg are used to provide the heat to distillthe 80 liters of ethanol produced. 52/80 = 0.65 kg of bagasse are use todistill 1 liter of ethanol:

Λbag→dist = 0.65 kg/liter. (2)

This means that heat needed to distill 1 liter of ethanol can come from1.54 kg of bagasse.

1 kg of polysaccharides yields 0.6 kg of sugar

Λpoly→sug = 0.6. (3)


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Assuming a bagasse composition of 50% cellulose, 30% hemicelluloseand 20% lignin, we have 80% polysaccharides per kg of bagasse:

Λbag→poly = 0.80. (4)

Combining all these ratios,

Λbag→ethhydrolysis= Λbag→poly × Λpoly→sug × Λsug→eth

= 0.80 × 0.60 × 0.52 = 0.25 liter/kg. (5)

This means that 3.70 kg of bagasse can be processed into 1 liter ofethanol, not counting the heat required for distillation. If this energy hasto come from burning bagasse, an additional 1.54 kg of this material is usedup to produce 1 liter of ethanol—a total of 5.24 kg of bagasse per liter ofethanol or 0.19 liters per kg of bagasse.

IIf only the excess bagasse is used (86 kg) then 16.4 liters of additionalethanol is obtained. However if we also use the 140 kg of dry straw (as-suming it behaves like biomass), then the additional ethanol will be 42.9liters.

If only the excess bagasse is hydrolyzed, the extra ethanol amountsto 16.4 liters, but if the straw is included then the additional ethanol

amounts to 42.9 liters.†

† This estimate is wildly optimistic. It totally ignores the substantialamount of energy used up in the hydrolysis of the bagasse into sugar. Addi-tionally, the economics have been totally ignored: if the bagasse is used as rawmaterial for hydrolysis, then, of course, it cannot be used for generating electricityan important source of revenue in a typical Brazilian ethanol plant.

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Prob 13.6 A digester consist of a cylindrical stainless steeltank with a diameter, d, and a height, 2d. The metal is 3 mmthick.

An R-5 (American system) fiberglass blanket completely cov-ers the tank.

The contents of the digester are agitated so that they are,essentially, at a uniform temperature of 37 C. To simplify thisproblem, assume that the influent (the material fed in) is pre-heated to 37 C.

Stainless steel has a thermal conductivity of λ = 60 W m−1

K−1.We desire a net production rate of 1 kW of methane. The

digester must produce, in addition, enough methane to fire aheater that keeps the material in it at a constant temperature (thedigestion process, itself, generates negligible heat). The efficiencyof the heater is 70%.

Loading rate is L = 4 kg of volatile solids per cubic meterof digester per day. Assume that 1 kg of volatile solids produce25 MJ of methane and that 40% of all the volatile solids in theinfluent are digested.

The outside temperature is such that the external walls ofthe digester are at an uniform 20 C.

Estimate the diameter of the digester......................................................................................................................

The digester looses heat at a rate

Pheat loss = A∆T/R

where A is the external surface area of the digester, and R = 0.178×5 = 0.89SI units.

Since the height of the cylinder is 2d, the lateral surface area is 2πd2.Including the top and bottom areas,

A = 2 ×πd2

4+ 2πd2 = 7.85d2.

Intuitively, all the temperature drop must be across the fiberglassblanked, essentially none across the thin highly conductive stainless steelwall. Let us make sure that this is so.

Per unit area the same heat power must flow through the steel as doesthrough the fiberglass:

The heat power density through the steel is

(37 − T )λ

dsteel,


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while that through the fiberglass is

T − 20

R.

In the above, T is the temperature at the steel-to-fiberglass interface,and dsteel is the thickness of the steel wall. Equating these two formulas,

(37 − T )λ

0.003=

T − 20

0.89.

Solving for T , one finds that this temperature is very nearly equal to37 C. Mathematically, it is 36.999 C).

With a ∆T = 37 − 20 = 17,

Pheat loss = 7.85d2 × 17/0.89 = 150d2.

The methane power to make up this loss is

Pheater =150d2

0.7= 214d2 W.

The methane power produced is

Pproduced = 100LV,

where V = πd3/2 is the volume of the digester. Hence

Pproduced = 100 × 4 × π/2 × d3 = 628d3.

The power balance is628d3 − 214d2 = 1000.

This leads to d = 1.17 m.

The diameter of the digester is 1.2 m.

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Prob 13.7 A hypothetical plant has perfectly horizontalleaves. The carbon dioxide uptake rate, φ, depends linearly (inthe usual manner) on the solar light power density, P , providedP ≤ 150 W/m2. Above this value, φ is constant, independently ofP .

Assume the insolation at normal incidence is 1000 W/m2 dur-ing all daylight hours. The latitude is 45N

a. What is the amount of carbon fixed by each square meter ofleaf area during the winter solstice day?

.....................................................................................................................The carbon dioxide uptake rate (in kilomoles per m2 per second) is

given by

φ = 178 × 10−12P for P ≤ 150 W/m2,

φ = 178 × 10−12 × 150 = 26.7 × 10−9 for P ≥ 150 W/m2.

The total uptake over a given period of time is

Φ =

∫ t

0

φdt.

For a horizontal surface, the insolation is

P = PS cosχ = 1000 cosχ.

The zenithal angle is given by

cosχ = sinλ sin δ + cosλ cos δ cosα

= 0.707×−0.380 + 0.707 cosα = −0.281 + 0.649 cosα.

As a consequence,

P = −281 + 649 cosα.

The sunrise and sunset hour angles are

cosαR,S

= − tan δ tanλ = − tan−23.45 tan 45 = 0.434.

αR

= −64.3 (−1.122 rad) tR = 07.71 h, or 27770 s

αR

= +64.3 (+1.122 rad) tS = 16.29 h or 58630 s.

The hour angles when P = 150 W/m2 are

cos150

=150 + 281

649= −0.664.

α150

=

−48.39 (−0.8445 rad) tS = 08.77 h or 31587 s,+48.39 (+0.8445 rad) tS = 15.22 h or 54813 s.


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The situation is symmetrical around noon. It is sufficient to calculatethe uptake from sunrise to noon and then double the results.

Between 07.71 and 08.77 in the morning, the uptake rate depends onthe power density of the sun and is given by

Φ1 = 178 × 10−12

∫ 08.77

07.71

(−281 + 649 cosα)dt

= 178 × 10−12 ×

(

−281

∫ 08.77

07.71

dt + 649

∫ 08.77

07.71

cosαdt

)

≡ Φ11 + Φ12.

Φ11 = −50.0× 10−9

∫ 08.77

07.71

dt = −50.0 × (8.77 − 7.71)× 3600

= −191× 10−6 kmoles/m2.

Note that the time was converted from hours to seconds.

Φ12 = 116 × 10−9

∫ 08.77

07.71

cosαdt.

Let us eliminate α by replacing by

α =2π

86400t − π,

cosα = − cos2π

86400t,

where t is in seconds.

Φ12 = 116 × 10−9

∫ 31587

27770

− cos2π

86400tdt.

Here, again, the time was converted into seconds.

Φ12 =−116× 10−9 86400

2π

[

sin

(

2π

86400× 31587

)

− sin

(

2π

86400× 27770

)]

= −116 × 10−9 × 13751 [sin(2.2966)−sin(2.02160)]

=−116× 10−9 × 13751(0.7480− 0.9001)= 242×10−6 kmoles/m2.

Φ1 = −191 × 10−6 + 242 × 10−6 = 51 × 10−6 kmoles/m2.

Between 08.77 and 12.00, the uptake rate is constant (because it doesnot depend on the solar power density):

Φ2 = 26.7 × 10−9 × (12.00 − 8.77)× 3600 = 310 × 10−6 kmoles/m2.

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The total carbon dioxide uptake during the winter solstice day is

Φ = 2 × (51 + 310)× 10−6 = 722 × 10−6 kmoles/m2.

Each square meter of leaves fixes 722 millimoles of CO2.

b. Estimate the number of kilograms of dry biomass the plantproduces on the winter solstice day per hectare of leaves.

.....................................................................................................................There are several ways to do this estimate and they lead to (roughly)

the same result. Here are two possibilities:

2.a One important end product of photosynthesis is glucose or its poly-mers (starch, cellulose). Lets assume that only glucose, (CH2O)6, isproduced. This has a molecular mass of 6× 12+12× 1+ 6× 16 = 180daltons. Thus, for each 6 × 12 = 72 daltons of carbon fixed, 180 dal-tons of biomass are produced. This is a mass ratio of 180/72 = 2.5 ofbiomass-to-carbon.The CO2 fixation rate is 722 millimoles per square meter per day. Thismeans that 722 millimoles or 722 × 10−6 × 12 = 8.66 × ×10−3 kg ofcarbon a fixed per square meter per day, or 8.66×10−3×2.5×104 = 216kg of biomass per hectare per day.

2.b In the Text it is stated that biomass containing 1 kmole of fixed carbonreleases 440 MJ of energy when burned. This means that 36.7 MJare released per kg of fixed carbon. Taking the heat of combustionof typical biomass as 16 MJ/kg, we get a ratio biomass-to-carbon of36.7/16 = 2.3 which is not too different from the 2.5 obtained earlier.

About 220 kg of dry biomass are produced per hectareon the day of the winter solstice.


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Prob 13.8 A hemispherical, perfectly transparent, containerhas a 5-m radius. The bottom part is covered with moist soil onwhich a bush with horizontal leaves is planted. The leaves arearranged in such a way that they do not shade one another. Thevolume occupied by the plant by the plant is 0.75 m3 and its leafarea is 4 m2.

The “air” inside the container has the following composition:

O2 52 kgN2 208 kgA 2.6 kgH2O 26 kgCO2 0.78 kg

The temperature inside the container is an uniform 298 K.The soil is moist enough to supply all the water needed by

the plant. However no water is ever exchanged directly betweensoil and air.

The plant has the usual value of r (13.2× 10−6 m3/J) and hasa stomatal conductance of 10 mm/s.

Argon has a molecular mass of 40 daltons.The plant has a carbon dioxide uptake rate proportional to

the illumination power density, P , for values of P < 200 W/m2

and independent of P for greater values. The light has the samespectral distribution as the sun.

a. What is the air pressure inside the container?.....................................................................................................................

Converting from kg to kmoles, the air composition is

O2 1.625 kmolesN2 7.429 kmolesA 0.065 kmolesH2O 1.444 kmolesCO2 0.01773 kmoles

Total 10.581 kmoles

The gas pressure is given by the perfect gas law,

p =µRT

V. (1)

The volume of the hemispherical container is

V =2

3πr3 =

2

3π53 = 261.8m3, (2)

Hence, the air pressure is

p =10.58 × 8324× 298

261.8= 100, 2000 Pa. (3)

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The air pressure is almost precisely 1 atmosphere.

b. What is the total carbon dioxide uptake rate (kmoles/s andmg/s) under the above circumstances, when the illuminationpower density is 150 W/m2?

.....................................................................................................................Since the illumination power density is less than 200 W/m2, the carbon

dioxide uptake rate is given by

φ = r|CO2|aP. (4)

We need to calculate the CO2 concentration:

|CO2|a =0.01773 kilomoles

m3= 67.7 × 10−6 kmoles (CO)

2m3. (5)

This is some 5 times the normal concentration in the atmosphere.

φ = 13.2×10−6×67.7×10−6×50 = 134×10−9 kmoles (CO)2

m3s−1. (6)

The total uptake rate (for the given leaf area of 4 m2),

Φ = φA = 134 × 10−9 × 4 = 536.2 × 10−9 kmoles of CO2/s. (7)

Φ = 536.2 × 10−9 × 44 = 23.59 × 10−6 kg of CO2/s. (8)

The plant will remove carbon dioxide from the air at a rateof 536 × 10−9 kilomoles/s or 23.6 mg/s.

c. Make a rough estimate of how long will it take to reduce theCO2 concentration to the level of normal (outside) air.

.....................................................................................................................The normal atmospheric CO2 concentration is 13.5× 10−6 kmoles/m3

which translates to 13.5 × 10−6 × 261.8 × 44 = 0.1555 kg of CO2 in thecontainer. Thus, a total of 0.780 − 0.156 = 0624 kg must be removed tolower the concentration to the “normal” value. Assuming a constant CO2

uptake rate equal to the initial rate of 23.6× 10−6 kg/s, thus will require atime, t, of

t =0.624

23.6 × 10−6= 26, 400 s or 7.33 h. (9)

As a first rough estimate, it will take somewhat over 7 hoursto reduce the carbon dioxide concentration to the normal level.

d. Assume that the leaf area of the plant does not change. Cal-culate more accurately the time required to reduce the carbondioxide concentration to the normal value.


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.....................................................................................................................Let M be the mass of carbon dioxide in the air (expressed in kg), and

Φ be the total carbon dioxide uptake rate (in kg/s). Then, if V is the “air”volume,

M = V [CO2]a (10)

dM = V d[CO2]a = 261.8d[CO2]a = −Φdt, (11)

Φ = rPA[CO2]a = 13.2×10−6×150×4×[CO2]a = 7.92×10−3[CO2]a, (12)

261.8 d[CO2]a = 7.92 × 10−3[CO2]adt, (13)

d[CO2]a[CO2]a

= −30.65× 10−6dt, (14)

ln[CO2]afinal

[CO2]ainitial

= −30.65× 10−6t, (15)

t = −1

30.65× 10−6ln

0.1555

0.78= 53, 300 s or 14.8 h. (16)

Actually, it takes over 14 hours to reduce thecarbon dioxide level to its normal value.

e. What is the composition of the air when the carbon dioxideconcentration reaches its normal level?

.....................................................................................................................Clearly, the amount of nitrogen, argon and water does not change.

The CO2 falls to 0.155 kg. For each kilomole of CO2 used up, 1 kilomole ofoxygen is released. The mass of oxygen is increased by 0.155×32/44 = 0.133kg. So there is a minute increase in the amount of oxygen in the air.

The amount of oxygen in the air is increased to 52.1 kg.

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Prob 13.9 Consult Google for properties of vegetable oils.Tabulate melting point (or cloud points or CFPPs) vs iodine val-ues. Plot melting points versus iodine values, and do a linearregression. What is the degree of correlation between these vari-ables. Does the melting point rise or drop with increasing iodinevalue? Give a reasonable mechanism for such a behavior.

Do this for at least 12 different oils so that you can drawacceptable statistical inferences......................................................................................................................

I found the data below at<http://www.journeytoforever.org/biodiesel yield.html >.

Oil Melting point Iodine value(C)

Coconut oil 25 10Palm kernel oil 24 37Mutton tallow 42 40Palm oil 35 54Olive oil -6 81Castor oil -18 85Peanut oil 3 93Rapeseed oil -10 98Cotton seed oil -1 105Sunflower oil -17 125Soybean oil -16 130Tung oil -2.5 168Linseed oil -24 178

This leads to the plot below:

-30 -20 -10 0 10 20 30 40 50 Melting point, Tmelt (C)

Iodi

ne v

alue

, IV

200

150

100

50

0

Tmelt = 97.3 −1.79 IVCorrelation coeff.: 78%

There is a good negative correlation (78%) between the degree of un-saturation (as measured by the iodine value) and the melting point. Highlyunsaturated oils have a low melting point because the molecule has a sharpkink in it making its assembly into a solid more difficult.


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Prob 13.10Draw the structural formula for 2,3-pentanediol and for 2,2,4-

trimethylpentane.Draw also the corresponding condensed structural formulas.

.....................................................................................................................The ending ol in 2,3-pentandiol signifies that it is an alcohol, i.e., that

hydrogens have been replaced by hydroxyls, OH. The preceding syllable, dishows that it is a double alcohol (2 substitutions of H by OH). ane indicatesan alkane and pent indicates that the main chain has 5 carbons. 2, 3 showsthe position of the two substitutions. Thus 2,3-pentanediol is a doublealcohol of pentane in which the OH group occur in positions 2 and 3 (thesecond and the third carbons of the chain).

H H H H H| | | | |

H−−C−−C−−C−−C−−C−−H| | | | |H OH OH H H

The corresponding condensed structural formula is

(CH3)(CHOH)(CHOH)(CH2)(CH3).

2,2,4-trimethylpentane is pentane in which the hydrogens in positions2, 2, and 4 have been replaced by the methyl radical (CH3):

H CH3 H H H| | | | |

H−−C−−C−−C−−C−−C−−H| | | | |H CH3 H CH3 H

The corresponding condensed structural formula is

(CH3)C(CH3)2(CH2)(CHCH3)(CH3)

The empirical formula is C8H18, i.e., it is of the general form CnH2n+2,hence this is octane, however it is not the usual, unbranched n-octane. Itis called isooctane.

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Prob 13.11 What is the main reason to prefer biodiesel tostraight vegetable oils (SVO) as fuel for diesel engines?

What should you do to make to make SVOs more acceptableas diesel fuel (other than making biodiesel out of them)?.....................................................................................................................Straight vegetable oils (SVO) are too viscous and can damage the fuelinjection system of a normal diesel engine. Petrodiesel must have a viscosityof 4 to 5 mm2/s, while most vegetable oils have viscosities between 30 and40 mm2/s. All measures at 40 C.

Since the viscosity falls quickly with rising temperature, one solutionis to preheat the oil to some 65 C.

Another solution is to mix SVO with petrodiesel. The SVO will act asa fuel extender.


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Prob 13.12 During anaerobic digestion, glucose is trans-formed into methane through a series of steps. The overall reac-tion is

(CH2O)6→ 3CH4 + 3CO2.

Calculate the percentage of methane (by both volume andmass) produced......................................................................................................................

The molecular mass of glucose is 180 daltons, that of methane is 16 ,and that of carbon dioxide is 44 daltons. Hence, 1 kmole of glucose produces48 kmoles of methane, a fraction of 48/180 = 0.267 per mass. The biogasproduced consists of 3 kmoles of methane and 3 of carbon dioxide, that is,half methane by volume.

The biogas contains 26,7% methane per mass and 50% by volume.

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Prob 13.13 Biodiesel is produce by transesterification ofvegetable or animal oils. Usually, the glycerine in the oil isreplaced by either methanol or ethanol. Consider the former.Since this amounts to using a fuel (methanol) to produce afuel (biodiesel) one is entitled to question what energy gain isachieved.

Invariably, vegetable oils are a mixture of many differenttriglycerides, but since one of the most common fatty acids foundin vegetable oils is palmitic acid, C16H32O2, we will assume thatour raw material is a pure triglyceride consisting of three palmiticacid groups. Thus, the final product—the biodiesel—will bemethyl palmitate.

Fuel Density Higher heat of comb.

(kg/m2) (MJ/liter)

Diesel 850 40.9Biodiesel 885 36.6Methanol 18.0

a. Stoichiometrically, how many kg of methanol are required foreach kg of vegetable oil?.....................................................................................................................

For each kilomole of the triglyceride (806 kg), 3 kmoles of methanol(3×32 = 96 kg) are needed. The proportions, by mass, are 96/806 = 0.119.Thus 1 kg of oil will require 0.119 kg of methanol.

About 120 g of methanol are needed for each kg of oil.

b. If the methanol, above, were used directly as fuel, how muchenergy would be released?.....................................................................................................................

120 g of methanol release 2.2 MJ of heat when burned.

c. If 1 kg of biodiesel is used as fuel, how much energy is re-leased?

.....................................................................................................................

1 kg of biodiesel is 1/0.885 liters and releases 36.6/0.885 = 41.4 MJ.

1 kg of oil releases 41.4 MJ of heat when burned.

d. What is the methanol-to-biodiesel energy ratio?

The methanol-to-oil energy ratio is 2.2/41.6 = 0.0529.

The energy of the methanol is only 6.52% of the energy of the biodiesel.


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Prob 13.14 You have 1000 kg of a vegetable oil which hap-pens to consist of a single triglyceride (not a mixture of triglyc-erides). The three acids in each molecule are all palmitic acid(C16:0).a. What is the molecular mass of the triglyceride?.....................................................................................................................

The molecular mass of propane (C3H8) is 3 × 12 + 8 = 44 daltons.Glycerine is a triple alcohol, that is, three hydrogens have been replace bythree hydroxyls. This adds three oxygens to the molecule which the masses44 + 3 × 16 = 92 daltons.

Palmitic acid is derived from hexadecane, C16H34 (mass: 16×12+34 =226 daltons. The carboxylic group, COOH (mass: 12 + 2 × 16 + 1 = 45daltons) replaces a methyl radical (CH3,mass: 15 daltons) group at the endof the molecule and winds up with 226 + 45− 15 = 256 daltons.

The triglyceride is formed by combining glycerine with 3 palmitic acidmolecules and a loss of 3 water molecule. Its mass is (92+3×256−3×18 =806 daltons.

The molecular mass of the oil is 806 daltons.

b. What is the proper chemical name of this triglyceride?.....................................................................................................................

This is a triple palmitate of glycerine hence it isglyceryl tri-palmitate.

c. Estimate the number of kilograms of glycerine produced whenthe above vegetable oil is transesterified with ethanol......................................................................................................................

Each kilomole of triglyceride releases 1 kilomole of glycerine upon beingesterified. 1000 kg of oil correspond to 1000/806 = 1.24 kilomoles. Hence1.24 kilomoles of glycerine are produced which mass 1.24 × 92 = 114 kg.

114 kg of glycerine are produced.

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Prob 13.15 A carbon-carbon bond can be single, double, ortriple. Which one is shorter: the single or the double?.....................................................................................................................

In a double bond, there is a larger attraction between the two carbonsthan in a single bond and, as a consequence, the double bond is shorter.

The double bond is shorter than the single one.


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Prob 13.16 What is the basic difference between a σ-bondand a π-bond?.....................................................................................................................

In a σ-bond, the molecular orbital is in line between the atoms;in the π-bond, it is above and below the atoms,but there is nothing in the line between them.

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Prob 13.17 Is amyl oleate a saturated or unsaturated ester?Amyl is the same as pentyl, the radical of the 5-carbon alkane,

pentane......................................................................................................................

Oleic acid is C18:1. It has 1 double carbon-carbon bond and is,therefore, monounsaturated. So is the corresponding ester.


090418


Prob 13.18 The molecule below is being proposed as a bio-fuel (at least, some of its isomers are). Give two names for themolecule.

H CH3 H| | |

H – C – C – C – OH| | |H H H

.....................................................................................................................The empirical formula for the above molecule is C4H9OH. This makes

it a 1-alcohol of butane, C4H10, a hydro carbon of the CnN2n+2 (alkane)series, hence the ending “ane”. It is a butanol, actually, it is isobutanol.

You can look at it as a methylated (CH3) derivative of propanol (a3-carbon 1-alcohol). It would then be 2-methyl-propanol because themethyl group is in the second carbon from the end.

The molecule is isobutanol also known as 2-methyl-propanol.

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Prob 13.19 What is the iodine value of glyceryl trilinoleate?The molecular mass of the oil is 878 daltons and the atomic massof iodine is 127 daltons. Linoleic acid is C18:2......................................................................................................................Linoleic acid has two carbon-carbon double bonds. The trilinoleate willhave a total of 6 carbon-carbon double bonds. 2 iodine atoms will bindto each double bond, so, each kilomole of oil will bind to 12 kilomoles ofiodine. This is a ratio, of 12× 127 = 1524 kg to 878 kg = 1.74. The iodinevalue, IV , is the number of grams of iodine that combine with 100 g of oil.In this problem the IV is 174.

The iodine value of glyceryl trilinoleate is 174.


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Prob 13.20 What is the empirical formula of oleic acid?Don’t just look it up in Google. Derive the formula starting fromthe original hydrocarbon and considering the progressive statesof oxidation......................................................................................................................

Oleic acid is C18:1, i.e., it is monounsaturated, thus it must be derivedform an alkene (second series of hydrocarbons), C18H36.

The first stage of oxidation will lead to a 1-alcohol, the result of re-placing one of the hydrogens by an hydroxil, OH. 1 H is lost and 1 OH isgained—a net gain of 1 oxygen atom: C18H35OH.

The next stage of oxidation leads to an aldehyde: The hydrogen in thehydroxyl is dropped. This leaves a double bonded oxygen at the end of thechain. Since carbon is only tetravalent, one adittional end-of-chain carbonmust be eliminated. Net effect is the loss of 2 hydrogens, C18H34O.

Finally, 1 of the hydrogens attached to the carbon bound to the oxygenis replaced by an hydroxyl and a carboxylic acid is formed. Net change isthe acquisition of an oxygen atom, C18H34O2.

The empirical formula of oleic acid is C18H34O2

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Prob 13.21 The molecules involved in this problem contain a largenumber of carbon and hydrogen atoms. In solving the problem, I roundedout the atomic masses to the nearest integer (for example, I used 12 forcarbon instead of the more accurate 12.01). If you use the actual molecularmass, the results may come out slighted different from those I got. Bothapproaches are acceptable).

a - For research purposes, somebody synthesized a veg-etable oil by combining 40 kg of palmitic acid (C16:0) and 60kg of linolenic acid (C18:3) with the stoichiometrically correctamount of propanotriol. All the water was extracted and dis-carded. What is the mass of the resulting oil?

.....................................................................................................................

This synthesis does not lead to an unique result. Individual triglyceridemolecules may be of mixed acids or they may be of three identical acids.Consider the latter, simpler, case: the oil consists of a mixture of 40%glyceryl tristearate and 60% glyceryl trilinolenate. Let us determine theepirical formula of each of these two triglycerides:

The alcohol, in either case, is glycerol which is C3H8O3 which has anmolecula mass of 92 daltons.

The C16:0 acid is saturated, hence it must have 16 carbons with theformula CH3 14(CH2) COOH = C16H32O2. It has a molecular mass of 256daltons.

Remember that a carboxylic acid has a methyl group, CH3, at the Ωend and a COOH radical attached to the α carbon, at the other end. Inbetween there are sufficient HCH groups to satisfy the number of carbonsin the molecule.

The triple triglyceride, glyceryl tripalmitate, is C3H8O3+3(C16H32O2)-3H2O = C51H98O6, and has a molecular mass of 806 daltons. Again,remember that when an ester bond is formed, a water molecule is rejected.

40 kg of C16:0 correspond to 40/256=0.156 kmoles, and produces0.156/3=0.0521 kmoles or 42.0 kg of triglyceride.

The C18:3 acid is triple-unsaturated with the formula CH3 10(CH2)6(CH) COOH = C18H30O2. I has a molecular mass of 278 daltons.

Here, since there are 3 double bonds and each one corresponds to 2CH groups, we have a total of 6(CH) and only 10(CH2)

The triple triglyceride, glyceryl trilinolenate, is C3H8O3+3(C18H30O2)-3H2O = C57H92O6. and has a molecular mass of 872 daltons.

60 kg of C18:3 correspond to 60/278=0.216 kmoles, and produces0.216/3=0.0719 kmoles or 62.7 kg of triglyceride.


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A mixture of 40 kg of C16:0 and 60 kg of C18:3produces a total of 42.0+62.7=104.7 kg of oil.

Check: 0.156 kmoles of C16:0 plus 0.216 kmoles of C18:3 add up to0.372 kilomoles of acid which yield 0.372/3 = 0.124 kilomoles of triglyc-erides (having used 0.124 kmoles of propanotriol or 11.41 kg of propan-otriol). Adding acids and alcohol, we have 111.41 kg. However eachkmole of triglyceride sheds, upon formation, 3 kmoles of water or a to-tal of 3×0.124 = 0.372 kilomoles of water or 0.372×18 = 6.70 kg of water.Thus the mass of the oil produced is 111.41 − 6.67 = 104.7 kg of oil whichis the amount calculated previously.

b - What is the iodine value of this oil?

.....................................................................................................................

Since the C18:0 is saturated, it will take up no iodine. The C18:3 istriple unsaturated—it has 3 double bonds, each of which takes up 2 iodineatoms. Hence each kilomole of C18:3 takes up 6 kilomoles of iodine.

60 kg of C18:3 (0.216 kmoles) will take up 0.216 × 6 = 1.30 kilomolesof iodine which mass 1.30×127 = 165 kg of iodine. But 60 kg of C18:3 willmake 104.5 kg of oil, therefore, 165 kg of iodine are taken up by 104.5 kgof oil or 158 grams of iodine are taken up by 100 grams of oil.

The iodine value of the oil is 158.

c - Do you think the above oil is a good feedstock forbiodiesel? Justify your answer TERSELY.

.....................................................................................................................

On two counts this is not a good feedstock for biodiesel.

1 - The high iodine value shows that this is a drying oil that tends tooxidize and polymerize inside the engine gumming up the injection pumpand nozzles. This gumming up is accelerated by the high combustion tem-perature.

2 - The stearic component of the oil has a high melting point and theoil probably will cloud up easily in the winter.

d - How many kg of ethanol do you need to transform 1kg of the above oil completely into biodiesel?

.....................................................................................................................

Once the ester bond is cleaved, 1 kg of the oil will yield

60/104.5 = 0.574 kg of linolenic acid (0.574/278 = 0.0021 kilomoles),

40/104.5 = 0.383 kg of stearic acid (0.383/284 = 0.0013 kilomoles),

(0.0021 + 0.0013)/3 = 0.0011 kilomoles of propanetriol (0.105 kg).

We want to produce ethyl esters which are monoesters (one acid radicalfor each alcohol). Thus we need 0.0021+0.0013 = 0.0034 kmoles of ethanol

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or 0.0034× 46 = 0.156 kg of ethanol.

0.156 kg of ethanol are required to transform 1 kg of the synthetic oilinto ethyl esters biodiesel.

e - How many kg of glycerine are left over?.....................................................................................................................

From the previous answer: 0.105 kg of glycerin are left over.


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570 Prob. Sol. Chapter 13 Fund. of Renewable Energy Processes

Chapter 13 Biomass Problem Solutions - Stanford...

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