Chapter 13
Biomass
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
of CO2 . When the light power density is raised to 600 W/m2 ,
the uptake is 0.36 µmoles min−1 cm−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 atmosphere 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−1 cm−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 ?
What is the expected uptake at 100 W/m2 if the CO2 concentration is increased to 400 ppm?
.....................................................................................................................
According to the simple model developed in the Textbook, the CO2
uptake rate, φ, is
r[CO2 ]a P
φ=
.
(1)
1 + Vr P
From this,
rP φ
.
V =
[CO2 ]a rP − φ
Under typical conditions, the reaction rate, r, is approximately the
same for most plants:
r = 13.5 × 10−6 m3 J−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
= 13.5 × 10−6
24.45
kmoles m−3 .
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−1 m−2 .
V =
13.2 × 10−6 × 600 × 60 × 10−9
= 0.01
13.5 × 10−6 × 13.2 × 10−6 × 600 − 60 × 10−9
m s−1 .
We can now introduce the values of V , r, and [CO]2 into equation 1.
178 × 10−12 P
.
(2)
φ=
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
W/m2
φ
µmoles
min−1 cm−2
φ
kmoles
s−1 m−2
50
100
600
1000
50.1 × 10−3
94.5 × 10−3
360 × 10−3
464 × 10−3
8.36 × 10−9
15.8 × 10−9
60.0 × 10−9
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
V =
13.2 × 10−6 × 1000 × 66.7 × 10−9
= 0.01
13.5 × 10−6 × 13.2 × 10−6 × 1000 − 66.7 × 10−9
m s−1 .
V =
1.33 × 10−5 × 1000 × 6.7 × 10−8
= 0.0079
1.34 × 10−5 × 1.33 × 10−5 × 1000 − 6.7 × 10−8
m s−1 .
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,
φ=
400
× 0.0945 = 0.114
330
µmole min−1 cm−2 .
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
090418
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 alcohol. 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 behavior, compare the energy per liter of the fuel with that of gasoline.
The compression ratio is now changed to the maximum compatible 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
CH3 OH → CO + 2H2 ,
(1)
and burns according to
2CH3 OH + 3O2 → 2CO2 + 4H2 O.
(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)
CH3 OH (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/
MASS LITER
kg
Methanol
Ethanol
n-Heptane
iso-Octane
32
46
100
114
0.791
0.789
0.684
0.703
22.7
29.7
48.1
47.8
MJ/
LITER
18.0
23.4
32.9
33.6
MJ/
kg
MJ/
LITER
(rel. to
octane)
(rel. to
octane)
0.475
0.621
1.006
1.000
0.534
0.697
0.979
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.
Methanol
Ethanol
n-Heptane
iso-Octane
MOL.
MASS
32
46
100
114
kg/
LITER
0.791
0.789
0.684
0.703
MJ/
kg
19.9
26.8
45.0
44.7
MJ/
LITER
15.8
21.2
30.8
31.4
Consider gasoline as mostly heptane.
From the table, the volumetric energy densities are 15.8 and 30.8
Solution of Problem
090418
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 insures 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.
γ
γ
γ
r
1.2
1.4
1.7
9
12
16
0.356
0.392
0.426
0.585
0.630
0.670
0.785
0.824
0.856
The kilometrage (kilometers/liter) of a vehicle is proportional to the
volumetric energy density of the fuel multiplied by the engine efficiency.
If all engines operate with an effective γ of 1.4, the relative kilometrages
would be
Fuel
r
γ
η
MJ/l kilometrage
kilometrage
(normalized)
Gasoline
Methanol
Disc. meth.
9
12
16
1.4
1.4
1.4
0.585
0.630
0.670
30.8
15.8
18.9
18.0
10.0
12.7
1
0.552
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 that
the 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
Methanol
Disc. meth.
9
12
16
1.2
1.2
1.7
0.356
0.392
0.856
30.8
15.8
18.9
11.0
6.2
16.2
1
0.518
1.48
Now, the kilometrage of the dissociated methanol engine is far larger
than 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 (Eichhornia crassipes), a floating plant, will grow at such a rate that
their dry biomass increases 5% per day. The total water content
of these plants is high (94%). Nevertheless, 400 kg of dry matter
can be harvested daily from one hectare of plantation.
Consider a plantation with one hectare area consisting of a
long canal (folded upon itself ). At the starting point (seeding
end), the canal is 0.5 m wide. It expands gradually enough to
just accommodate the growing plants that are slowly swept along
by the current. Assume a current of constant speed such that the
plants take 60 days to float from the seeding end to the harvesting
end.
a. How wide must the canal be at the harvesting end?
.....................................................................................................................
By general dimensional considerations, if the mass of the plant grows
5% per day, the surface area should grow approximately 32 5% = 3.33%
day−1 . This is because, to a first approximation, surface is proportional to
mass2/3 . 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),
2/3
2/3
mi+1 = mi (1 + α)2/3
1 + β = (1 + α)2/3
→
or
Si+1 = Si (1 + α)2/3 ,
(1 + β)3/2 = 1 + α.
But β << 1, thus, by binomial expansion,
3
(1 + β)3/2 ≈ 1 + β = 1 + α,
2
2
β = α.
3
The surface grows exponentially: S = S0 exp(t/τ ).
1
ln(1 + β) = = β.
τ
Thus
S = 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 canal
at 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?
Solution of Problem
090418
13.3
Fund. of Renewable Energy Processes
Prob. Sol. 13.3
Page 2 of 4
541
.....................................................................................................................
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,
β
W = W0 exp
ℓ .
V
The velocity of the water is V = L/T , where L is the total length of
the canal and T = 60 days.
The area, A, of the canal surface is
L
β
dℓ
A=
W exp
V
0
V
β
W0 L = W0
exp (βT − 1)
L−1
=
exp
β
V
Tβ
0.5L
exp(0.033 × 60) − 1 = 1.6L = 104
=
60 × 0.033
Z
L = 6260
,
m2
m.
The total length of the canal is 6260 meters.
c. How much energy is harvested per day (express this in GJ
and 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 surface
of the canal is 1 ha. Taking the average energy yield of dry biomass as
some 18 MJ kg−1 , the energy harvested is 7200 MJ/day. One barrel of oil
corresponds 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 as
seed?
.....................................................................................................................
Let M0 be the daily seeding. Then
400 = M0 exp(0.05 × 60)
because the mass grows 5% per day.
400
M0 = 3 = 19.9 kg of dry matter..
e
Solution of Problem
13.3
090418
542
Page 3 of 4
Prob. Sol. 13.3
Fund. of Renewable Energy Processes
This is about 5% of the collected matter. Since 94% of the wet material
is water, the daily re-seeding must consist of
19.9
= 322 kg of wet plants.
1 − 0.94
320 kg of wet plants must be used daily for re-seeding.
e. Assume that 50% of the biomass energy is converted into
methane through a digestion process. Estimate the methane
yield in cubic meters per day.
.....................................................................................................................
0.5 × 7200 = 3600
MJ/day.
The heat of combustion of methane is 55.6 MJ/kg. Hence the methane
production will be 3600/55.6 = 64.8 kg/day.
Molecular mass of methane is 16 daltons, hence its density (RTP) is
16 kg/kmole
= 0.65.
24.5 m3 /kmole
The daily production is
1 m3
kg
×
= 99.7 m3 day−1 .
64.8
day 0.65 kg
Methane production is about 100 m3 per day.
f. If the methane is burned in a gas turbine or in a diesel engine
with 20% efficiency, what is the average electric power that
can be generated?
.....................................................................................................................
The methane power is 3600 MJ/day or 41.7 kW. Withe 20% efficiency, the
average electric power is
The average electric power is 8.3 kW.
g. Assuming that the average depth of the canal is 1 m, what is
the amount of water that has to flow in daily provided there
is 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, the
flow rate is 10,000/60=166.6 m3 /day.
The flow rate of water is 167 m3 /day.
Solution of Problem
090418
13.3
Fund. of Renewable Energy Processes
Prob. Sol. 13.3
Page 4 of 4
543
h. Do you have any good ideas of how to insure that the water velocity is kept reasonably constant notwithstanding the
expanding canal?
.....................................................................................................................
The simplest (but not necessarily the most practical) solution is to
make sure the channel cross-section is kept constant regardless of how far
it 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 be
determined.
We saw that
W = 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
Z
1 6260
< d >=
2K exp(−316 × 10−6 )ℓ)dℓ = 1,
L 0
−
6260
2K
exp(−316 × 10−6 )ℓ) 0
= 1,
−6
316 × 10 L
2K × 0.862
= 1,
316 × 10−6 × 6260
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 harvesting
end, it is 0.32 m.
This makes the canal expensively deep at one end and too shallow (the
roots of the floating plants will scrape the bottom) at the other end.
Perhaps a better solution is to inject water at different points along
the canal so that at, the seeding end, the flow is much less than at the
harvesting end.
Solution of Problem
13.3
090418
544
Page 1 of 1
Prob. Sol. 13.4
Fund. of Renewable Energy Processes
Prob 13.4 Sugar cane is submitted to an illumination of 500
W/m2 . Assuming a stomatal velocity of 6 mm/s, what is the photosynthetic efficiency (defined as the ratio of the heat of combustion of the dry biomass generated to the incident solar energy)?
.....................................................................................................................
The carbon dioxide uptake rate is
φ=
r[CO2 ]a P
.
1 + P r/V
From Equation 7 of Chapter 15 of the Textbook,
r[CO2 ]a =
φ
= 178 × 10−12
P
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
= 42.2 × 10−9
1 + 500 × 1.33 × 10−5 /6 × 10−3
kmole s−1 m−2 .
The energy fixed for each kilomole of carbon dioxide taken up is about
440 MJ (see Textbook).
Thus the rate at which energy is being fixed by the sugar cane is
Pf ixed = 42.2 × 10−9 × 440 × 106 = 18.6
The efficiency is
η=
W m−2 .
18.6
= 0.037.
500
Under the conditions of the problem,
the efficiency of the sugar cane is 3.7%.
Solution of Problem
090418
13.4
Fund. of Renewable Energy Processes
Prob. Sol. 13.5
Page 1 of 2
545
Prob 13.5 Here is a typical task that an energy consultant
might tackle:
The operator of a large alcohol distillery wants to know if
it makes economic sense to use the leftover bagasse as a further source of ethanol. In the traditional process, the amount of
bagasse obtained from 1 ton of burned and cropped sugar cane
is larger than the amount that has to be burned to drive the distillation process. The excess is either sold or used to generate
electricity for the plant. The question is how much additional alcohol can be obtained by hydrolyzing all the polysaccharides (cellulose and hemicellulose) in the leftover bagasse. We will make
the following simplifying assumptions:
a. The hydrolysis will yield 600 grams of sugars (glucose and
pentoses) 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 traditional
process.
d.. The data for this problem are those discussed in Section 13.3.2
of the Textbook.
Calculate the additional amount of alcohol that can be obtained 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 of
ethanol.
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 138
kg of dry bagasse. Of these, 52 kg are used to provide the heat to distill
the 80 liters of ethanol produced. 52/80 = 0.65 kg of bagasse are use to
distill 1 liter of ethanol:
Λbag→dist = 0.65 kg/liter.
(2)
This means that heat needed to distill 1 liter of ethanol can come from
1.54 kg of bagasse.
1 kg of polysaccharides yields 0.6 kg of sugar
Λpoly→sug = 0.6.
Solution of Problem
(3)
13.5
090418
546
Page 2 of 2
Prob. Sol. 13.5
Fund. of Renewable Energy Processes
Assuming a bagasse composition of 50% cellulose, 30% hemicellulose
and 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 of
ethanol, not counting the heat required for distillation. If this energy has
to come from burning bagasse, an additional 1.54 kg of this material is used
up to produce 1 liter of ethanol—a total of 5.24 kg of bagasse per liter of
ethanol or 0.19 liters per kg of bagasse.
IIf only the excess bagasse is used (86 kg) then 16.4 liters of additional
ethanol is obtained. However if we also use the 140 kg of dry straw (assuming it behaves like biomass), then the additional ethanol will be 42.9
liters.
If only the excess bagasse is hydrolyzed, the extra ethanol amounts
to 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 substantial
amount of energy used up in the hydrolysis of the bagasse into sugar. Additionally, the economics have been totally ignored: if the bagasse is used as raw
material for hydrolysis, then, of course, it cannot be used for generating electricity
an important source of revenue in a typical Brazilian ethanol plant.
Solution of Problem
090418
13.5
Fund. of Renewable Energy Processes
Prob. Sol. 13.6
Page 1 of 2
547
Prob 13.6 A digester consist of a cylindrical stainless steel
tank with a diameter, d, and a height, 2d. The metal is 3 mm
thick.
An R-5 (American system) fiberglass blanket completely covers the tank.
The contents of the digester are agitated so that they are,
essentially, at a uniform temperature of 37 C. To simplify this
problem, assume that the influent (the material fed in) is preheated to 37 C.
Stainless steel has a thermal conductivity of λ = 60 W m−1
−1
K .
We desire a net production rate of 1 kW of methane. The
digester must produce, in addition, enough methane to fire a
heater that keeps the material in it at a constant temperature (the
digestion process, itself, generates negligible heat). The efficiency
of the heater is 70%.
Loading rate is L = 4 kg of volatile solids per cubic meter
of digester per day. Assume that 1 kg of volatile solids produce
25 MJ of methane and that 40% of all the volatile solids in the
influent are digested.
The outside temperature is such that the external walls of
the 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 = 0.89
SI units.
Since the height of the cylinder is 2d, the lateral surface area is 2πd2 .
Including the top and bottom areas,
A=2×
πd2
+ 2πd2 = 7.85d2 .
4
Intuitively, all the temperature drop must be across the fiberglass
blanked, essentially none across the thin highly conductive stainless steel
wall. Let us make sure that this is so.
Per unit area the same heat power must flow through the steel as does
through the fiberglass:
The heat power density through the steel is
(37 − T )λ
,
dsteel
Solution of Problem
13.6
090418
548
Page 2 of 2
Prob. Sol. 13.6
Fund. of Renewable Energy Processes
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,
T − 20
(37 − T )λ
=
.
0.003
0.89
Solving for T , one finds that this temperature is very nearly equal to
37 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
= 214d2
0.7
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 is
628d3 − 214d2 = 1000.
This leads to d = 1.17 m.
The diameter of the digester is 1.2 m.
Solution of Problem
090418
13.6
Fund. of Renewable Energy Processes
Prob. Sol. 13.7
Page 1 of 3
549
Prob 13.7 A hypothetical plant has perfectly horizontal
leaves. The carbon dioxide uptake rate, φ, depends linearly (in
the usual manner) on the solar light power density, P , provided
P ≤ 150 W/m2 . Above this value, φ is constant, independently of
P.
Assume the insolation at normal incidence is 1000 W/m2 during all daylight hours. The latitude is 45◦ N
a. What is the amount of carbon fixed by each square meter of
leaf area during the winter solstice day?
.....................................................................................................................
The carbon dioxide uptake rate (in kilomoles per m2 per second) is
given by
φ = 178 × 10−12 P
for P ≤ 150 W/m2 ,
2
φ = 178 × 10−12 × 150 = 26.7 × 10−9
for P ≥ 150 W/m .
The total uptake over a given period of time is
Z t
Φ=
φdt.
0
For a horizontal surface, the insolation is
P = PS cos χ = 1000 cos χ.
The zenithal angle is given by
cos χ = sin λ sin δ + cos λ cos δ cos α
= 0.707 × −0.380 + 0.707 cos α = −0.281 + 0.649 cos α.
As a consequence,
P = −281 + 649 cos α.
The sunrise and sunset hour angles are
cos αR,S = − tan δ tan λ = − 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 =
α150 =
150 + 281
= −0.664.
649
−48.39◦ (−0.8445 rad) tS = 08.77 h or 31587 s,
+48.39◦ (+0.8445 rad) tS = 15.22 h or 54813 s.
Solution of Problem
13.7
090418
550
Page 2 of 3
Prob. Sol. 13.7
Fund. of Renewable Energy Processes
The situation is symmetrical around noon. It is sufficient to calculate
the uptake from sunrise to noon and then double the results.
Between 07.71 and 08.77 in the morning, the uptake rate depends on
the power density of the sun and is given by
Z 08.77
Φ1 = 178 × 10−12
(−281 + 649 cos α)dt
07.71
Z 08.77
Z 08.77
= 178 × 10−12 × −281
dt + 649
cos αdt
07.71
07.71
≡ Φ11 + Φ12 .
Φ11 = −50.0 × 10−9
Z
08.77
dt = −50.0 × (8.77 − 7.71) × 3600
07.71
= −191 × 10−6
2
kmoles/m .
Note that the time was converted from hours to seconds.
Z 08.77
−9
Φ12 = 116 × 10
cos αdt.
07.71
Let us eliminate α by replacing by
α=
2π
t − π,
86400
cos α = − cos
2π
t,
86400
where t is in seconds.
Φ12 = 116 × 10
−9
Z
31587
27770
− cos
2π
tdt.
86400
Here, again, the time was converted into seconds.
86400
2π
2π
Φ12 = −116 × 10−9
sin
× 31587 − sin
× 27770
2π
86400
86400
= −116 × 10−9 × 13751 [sin(2.2966) − sin(2.02160)]
= −116 × 10−9 × 13751(0.7480 − 0.9001) = 242 × 10−6
Φ1 = −191 × 10−6 + 242 × 10−6 = 51 × 10−6
kmoles/m2 .
2
kmoles/m .
Between 08.77 and 12.00, the uptake rate is constant (because it does
not depend on the solar power density):
Φ2 = 26.7 × 10−9 × (12.00 − 8.77) × 3600 = 310 × 10−6
kmoles/m2 .
Solution of Problem
090418
13.7
Fund. of Renewable Energy Processes
Prob. Sol. 13.7
Page 3 of 3
551
The total carbon dioxide uptake during the winter solstice day is
Φ = 2 × (51 + 310) × 10−6 = 722 × 10−6
2
kmoles/m .
Each square meter of leaves fixes 722 millimoles of CO2 .
b. Estimate the number of kilograms of dry biomass the plant
produces 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 polymers (starch, cellulose). Lets assume that only glucose, (CH2 O)6 , is
produced. This has a molecular mass of 6 × 12 + 12 × 1 + 6 × 16 = 180
daltons. Thus, for each 6 × 12 = 72 daltons of carbon fixed, 180 daltons of biomass are produced. This is a mass ratio of 180/72 = 2.5 of
biomass-to-carbon.
The CO2 fixation rate is 722 millimoles per square meter per day. This
means that 722 millimoles or 722 × 10−6 × 12 = 8.66 × ×10−3 kg of
carbon a fixed per square meter per day, or 8.66×10−3 ×2.5×104 = 216
kg of biomass per hectare per day.
2.b In the Text it is stated that biomass containing 1 kmole of fixed carbon
releases 440 MJ of energy when burned. This means that 36.7 MJ
are released per kg of fixed carbon. Taking the heat of combustion
of typical biomass as 16 MJ/kg, we get a ratio biomass-to-carbon of
36.7/16 = 2.3 which is not too different from the 2.5 obtained earlier.
About 220 kg of dry biomass are produced per hectare
on the day of the winter solstice.
Solution of Problem
13.7
090418
552
Page 1 of 3
Prob. Sol. 13.8
Fund. of Renewable Energy Processes
Prob 13.8 A hemispherical, perfectly transparent, container
has a 5-m radius. The bottom part is covered with moist soil on
which a bush with horizontal leaves is planted. The leaves are
arranged in such a way that they do not shade one another. The
volume occupied by the plant by the plant is 0.75 m3 and its leaf
area is 4 m2 .
The “air” inside the container has the following composition:
O2
N2
A
H2 O
CO2
52 kg
208 kg
2.6 kg
26 kg
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 between
soil and air.
The plant has the usual value of r (13.2 × 10−6 m3 /J) and has
a 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 same
spectral distribution as the sun.
a. What is the air pressure inside the container?
.....................................................................................................................
Converting from kg to kmoles, the air composition is
O2
N2
A
H2 O
CO2
1.625 kmoles
7.429 kmoles
0.065 kmoles
1.444 kmoles
0.01773 kmoles
Total
10.581 kmoles
The gas pressure is given by the perfect gas law,
µRT
p=
.
V
The volume of the hemispherical container is
2
2
V = πr3 = π53 = 261.8m3 ,
3
3
Hence, the air pressure is
10.58 × 8324 × 298
p=
= 100, 2000 Pa.
261.8
Solution of Problem
090418
(1)
(2)
(3)
13.8
Fund. of Renewable Energy Processes
Prob. Sol. 13.8
Page 2 of 3
553
The air pressure is almost precisely 1 atmosphere.
b. What is the total carbon dioxide uptake rate (kmoles/s and
mg/s) under the above circumstances, when the illumination
power 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 |a P.
(4)
We need to calculate the CO2 concentration:
|CO2 |a =
0.01773 kilomoles
= 67.7 × 10−6
m3
kmoles (CO)2 m3 .
(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 m3 s−1 . (6)
The total uptake rate (for the given leaf area of 4 m2 ),
Φ = φA = 134 × 10−9 × 4 = 536.2 × 10−9
Φ = 536.2 × 10−9 × 44 = 23.59 × 10−6
kmoles of CO2 /s.
(7)
kg of CO2 /s.
(8)
The plant will remove carbon dioxide from the air at a rate
of 536 × 10−9 kilomoles/s or 23.6 mg/s.
c. Make a rough estimate of how long will it take to reduce the
CO2 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 the
container. Thus, a total of 0.780 − 0.156 = 0624 kg must be removed to
lower 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 a
time, t, of
0.624
= 26, 400 s or 7.33 h.
(9)
t=
23.6 × 10−6
As a first rough estimate, it will take somewhat over 7 hours
to reduce the carbon dioxide concentration to the normal level.
d. Assume that the leaf area of the plant does not change. Calculate more accurately the time required to reduce the carbon
dioxide concentration to the normal value.
Solution of Problem
13.8
090418
554
Page 3 of 3
Prob. Sol. 13.8
Fund. of Renewable Energy Processes
.....................................................................................................................
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,
Φ = rP A[CO2 ]a = 13.2×10
×150×4×[CO2 ]a = 7.92×10
(11)
−3
[CO2 ]a , (12)
261.8 d[CO2 ]a = 7.92 × 10−3 [CO2 ]a dt,
(13)
d[CO2 ]a
= −30.65 × 10−6 dt,
[CO2 ]a
(14)
[CO2 ]af inal
= −30.65 × 10−6 t,
[CO2 ]ainitial
(15)
ln
t=−
−6
0.1555
1
ln
= 53, 300 s
−6
30.65 × 10
0.78
or 14.8 h.
(16)
Actually, it takes over 14 hours to reduce the
carbon dioxide level to its normal value.
e. What is the composition of the air when the carbon dioxide
concentration 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 of
oxygen is released. The mass of oxygen is increased by 0.155×32/44 = 0.133
kg. 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.
Solution of Problem
090418
13.8
Fund. of Renewable Energy Processes
Prob. Sol. 13.9
Page 1 of 1
555
Prob 13.9 Consult Google for properties of vegetable oils.
Tabulate melting point (or cloud points or CFPPs) vs iodine values. Plot melting points versus iodine values, and do a linear
regression. What is the degree of correlation between these variables. Does the melting point rise or drop with increasing iodine
value? Give a reasonable mechanism for such a behavior.
Do this for at least 12 different oils so that you can draw
acceptable statistical inferences.
.....................................................................................................................
I found the data below at
<http://www.journeytoforever.org/biodiesel yield.html >.
Oil
Melting point
(C)
Iodine value
25
24
42
35
-6
-18
3
-10
-1
-17
-16
-2.5
-24
10
37
40
54
81
85
93
98
105
125
130
168
178
Coconut oil
Palm kernel oil
Mutton tallow
Palm oil
Olive oil
Castor oil
Peanut oil
Rapeseed oil
Cotton seed oil
Sunflower oil
Soybean oil
Tung oil
Linseed oil
This leads to the plot below:
200
Tmelt = 97.3 −1.79 IV
Correlation coeff.: 78%
Iodine value, IV
150
100
50
0
-30 -20 -10 0 10 20 30 40
Melting point, Tmelt (C)
50
There is a good negative correlation (78%) between the degree of unsaturation (as measured by the iodine value) and the melting point. Highly
unsaturated oils have a low melting point because the molecule has a sharp
kink in it making its assembly into a solid more difficult.
Solution of Problem
13.9
090418
556
Page 1 of 1
Prob. Sol. 13.10
Fund. of Renewable Energy Processes
Prob 13.10
Draw the structural formula for 2,3-pentanediol and for 2,2,4trimethylpentane.
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, di
shows that it is a double alcohol (2 substitutions of H by OH). ane indicates
an alkane and pent indicates that the main chain has 5 carbons. 2, 3 shows
the position of the two substitutions. Thus 2,3-pentanediol is a double
alcohol of pentane in which the OH group occur in positions 2 and 3 (the
second 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 positions
2, 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 C8 H18 , i.e., it is of the general form Cn H2n+2 ,
hence this is octane, however it is not the usual, unbranched n-octane. It
is called isooctane.
Solution of Problem
090418
13.10
Fund. of Renewable Energy Processes
Prob. Sol. 13.11
Page 1 of 1
557
Prob 13.11 What is the main reason to prefer biodiesel to
straight vegetable oils (SVO) as fuel for diesel engines?
What should you do to make to make SVOs more acceptable
as diesel fuel (other than making biodiesel out of them)?
.....................................................................................................................
Straight vegetable oils (SVO) are too viscous and can damage the fuel
injection system of a normal diesel engine. Petrodiesel must have a viscosity
of 4 to 5 mm2 /s, while most vegetable oils have viscosities between 30 and
40 mm2 /s. All measures at 40 C.
Since the viscosity falls quickly with rising temperature, one solution
is to preheat the oil to some 65 C.
Another solution is to mix SVO with petrodiesel. The SVO will act as
a fuel extender.
Solution of Problem
13.11
090418
558
Page 1 of 1
Prob. Sol. 13.12
Fund. of Renewable Energy Processes
Prob 13.12 During anaerobic digestion, glucose is transformed into methane through a series of steps. The overall reaction is
(CH2 O)6 → 3CH4 + 3CO2 .
Calculate the percentage of methane (by both volume and
mass) 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 produces
48 kmoles of methane, a fraction of 48/180 = 0.267 per mass. The biogas
produced 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.
Solution of Problem
090418
13.12
Fund. of Renewable Energy Processes
Prob. Sol. 13.13
Page 1 of 1
559
Prob 13.13 Biodiesel is produce by transesterification of
vegetable or animal oils. Usually, the glycerine in the oil is
replaced by either methanol or ethanol. Consider the former.
Since this amounts to using a fuel (methanol) to produce a
fuel (biodiesel) one is entitled to question what energy gain is
achieved.
Invariably, vegetable oils are a mixture of many different
triglycerides, but since one of the most common fatty acids found
in vegetable oils is palmitic acid, C16 H32 O2 , we will assume that
our raw material is a pure triglyceride consisting of three palmitic
acid groups. Thus, the final product—the biodiesel—will be
methyl palmitate.
Fuel
Diesel
Biodiesel
Methanol
Density
(kg/m2 )
Higher heat of comb.
(MJ/liter)
850
885
40.9
36.6
18.0
a. Stoichiometrically, how many kg of methanol are required for
each 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 much
energy 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 released?
.....................................................................................................................
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.
Solution of Problem
13.13
090418
560
Page 1 of 1
Prob. Sol. 13.14
Fund. of Renewable Energy Processes
Prob 13.14 You have 1000 kg of a vegetable oil which happens to consist of a single triglyceride (not a mixture of triglycerides). 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 (C3 H8 ) is 3 × 12 + 8 = 44 daltons.
Glycerine is a triple alcohol, that is, three hydrogens have been replace by
three hydroxyls. This adds three oxygens to the molecule which the masses
44 + 3 × 16 = 92 daltons.
Palmitic acid is derived from hexadecane, C16 H34 (mass: 16×12+34 =
226 daltons. The carboxylic group, COOH (mass: 12 + 2 × 16 + 1 = 45
daltons) replaces a methyl radical (CH3 ,mass: 15 daltons) group at the end
of the molecule and winds up with 226 + 45 − 15 = 256 daltons.
The triglyceride is formed by combining glycerine with 3 palmitic acid
molecules 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 is
glyceryl tri-palmitate.
c. Estimate the number of kilograms of glycerine produced when
the above vegetable oil is transesterified with ethanol.
.....................................................................................................................
Each kilomole of triglyceride releases 1 kilomole of glycerine upon being
esterified. 1000 kg of oil correspond to 1000/806 = 1.24 kilomoles. Hence
1.24 kilomoles of glycerine are produced which mass 1.24 × 92 = 114 kg.
114 kg of glycerine are produced.
Solution of Problem
090418
13.14
Fund. of Renewable Energy Processes
Prob. Sol. 13.15
Page 1 of 1
561
Prob 13.15 A carbon-carbon bond can be single, double, or
triple. Which one is shorter: the single or the double?
.....................................................................................................................
In a double bond, there is a larger attraction between the two carbons
than in a single bond and, as a consequence, the double bond is shorter.
The double bond is shorter than the single one.
Solution of Problem
13.15
090418
562
Page 1 of 1
Prob. Sol. 13.16
Fund. of Renewable Energy Processes
Prob 13.16 What is the basic difference between a σ-bond
and 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.
Solution of Problem
090418
13.16
Fund. of Renewable Energy Processes
Prob. Sol. 13.17
Page 1 of 1
563
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.
Solution of Problem
13.17
090418
564
Page 1 of 1
Prob. Sol. 13.18
Fund. of Renewable Energy Processes
Prob 13.18 The molecule below is being proposed as a biofuel (at least, some of its isomers are). Give two names for the
molecule.
H
CH3
H
|
|
|
H – C – C – C – OH
|
|
|
H
H
H
.....................................................................................................................
The empirical formula for the above molecule is C4 H9 OH. This makes
it a 1-alcohol of butane, C4 H10 , a hydro carbon of the Cn N2n+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 (a
3-carbon 1-alcohol). It would then be 2-methyl-propanol because the
methyl group is in the second carbon from the end.
The molecule is isobutanol also known as 2-methyl-propanol.
Solution of Problem
090418
13.18
Fund. of Renewable Energy Processes
Prob. Sol. 13.19
Page 1 of 1
565
Prob 13.19 What is the iodine value of glyceryl trilinoleate?
The molecular mass of the oil is 878 daltons and the atomic mass
of iodine is 127 daltons. Linoleic acid is C18:2.
.....................................................................................................................
Linoleic acid has two carbon-carbon double bonds. The trilinoleate will
have a total of 6 carbon-carbon double bonds. 2 iodine atoms will bind
to each double bond, so, each kilomole of oil will bind to 12 kilomoles of
iodine. This is a ratio, of 12 × 127 = 1524 kg to 878 kg = 1.74. The iodine
value, 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.
Solution of Problem
13.19
090418
566
Page 1 of 1
Prob. Sol. 13.20
Fund. of Renewable Energy Processes
Prob 13.20 What is the empirical formula of oleic acid?
Don’t just look it up in Google. Derive the formula starting from
the original hydrocarbon and considering the progressive states
of oxidation.
.....................................................................................................................
Oleic acid is C18:1, i.e., it is monounsaturated, thus it must be derived
form an alkene (second series of hydrocarbons), C18 H36 .
The first stage of oxidation will lead to a 1-alcohol, the result of replacing one of the hydrogens by an hydroxil, OH. 1 H is lost and 1 OH is
gained—a net gain of 1 oxygen atom: C18 H35 OH.
The next stage of oxidation leads to an aldehyde: The hydrogen in the
hydroxyl is dropped. This leaves a double bonded oxygen at the end of the
chain. Since carbon is only tetravalent, one adittional end-of-chain carbon
must be eliminated. Net effect is the loss of 2 hydrogens, C18 H34 O.
Finally, 1 of the hydrogens attached to the carbon bound to the oxygen
is replaced by an hydroxyl and a carboxylic acid is formed. Net change is
the acquisition of an oxygen atom, C18 H34 O2 .
The empirical formula of oleic acid is C18 H34 O2
Solution of Problem
090418
13.20
Fund. of Renewable Energy Processes
Prob. Sol. 13.21
Page 1 of 3
567
Prob 13.21 The molecules involved in this problem contain a large
number of carbon and hydrogen atoms. In solving the problem, I rounded
out the atomic masses to the nearest integer (for example, I used 12 for
carbon instead of the more accurate 12.01). If you use the actual molecular
mass, the results may come out slighted different from those I got. Both
approaches are acceptable).
a - For research purposes, somebody synthesized a vegetable oil by combining 40 kg of palmitic acid (C16:0) and 60
kg of linolenic acid (C18:3) with the stoichiometrically correct
amount of propanotriol. All the water was extracted and discarded. What is the mass of the resulting oil?
.....................................................................................................................
This synthesis does not lead to an unique result. Individual triglyceride
molecules 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 the
epirical formula of each of these two triglycerides:
The alcohol, in either case, is glycerol which is C3 H8 O3 which has an
molecula mass of 92 daltons.
The C16:0 acid is saturated, hence it must have 16 carbons with the
formula CH3 14(CH2 ) COOH = C16 H32 O2 . It has a molecular mass of 256
daltons.
Remember that a carboxylic acid has a methyl group, CH3 , at the Ω
end and a COOH radical attached to the α carbon, at the other end. In
between there are sufficient HCH groups to satisfy the number of carbons
in the molecule.
The triple triglyceride, glyceryl tripalmitate, is C3 H8 O3 +3(C16 H32 O2 )
-3H2 O = C51 H98 O6 , 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 produces
0.156/3=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 = C18 H30 O2 . I has a molecular mass of 278 daltons.
Here, since there are 3 double bonds and each one corresponds to 2
CH groups, we have a total of 6(CH) and only 10(CH2 )
The triple triglyceride, glyceryl trilinolenate, is C3 H8 O3 +3(C18 H30 O2 )
-3H2 O = C57 H92 O6 . and has a molecular mass of 872 daltons.
60 kg of C18:3 correspond to 60/278=0.216 kmoles, and produces
0.216/3=0.0719 kmoles or 62.7 kg of triglyceride.
Solution of Problem
13.21
090418
568
Page 2 of 3
Prob. Sol. 13.21
Fund. of Renewable Energy Processes
A mixture of 40 kg of C16:0 and 60 kg of C18:3
produces 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 to
0.372 kilomoles of acid which yield 0.372/3 = 0.124 kilomoles of triglycerides (having used 0.124 kmoles of propanotriol or 11.41 kg of propanotriol). Adding acids and alcohol, we have 111.41 kg. However each
kmole of triglyceride sheds, upon formation, 3 kmoles of water or a total 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 which
is the amount calculated previously.
b - What is the iodine value of this oil?
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Since the C18:0 is saturated, it will take up no iodine. The C18:3 is
triple unsaturated—it has 3 double bonds, each of which takes up 2 iodine
atoms. 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 kilomoles
of iodine which mass 1.30 × 127 = 165 kg of iodine. But 60 kg of C18:3 will
make 104.5 kg of oil, therefore, 165 kg of iodine are taken up by 104.5 kg
of 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 for
biodiesel? 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 to
oxidize and polymerize inside the engine gumming up the injection pump
and nozzles. This gumming up is accelerated by the high combustion temperature.
2 - The stearic component of the oil has a high melting point and the
oil probably will cloud up easily in the winter.
d - How many kg of ethanol do you need to transform 1
kg 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 radical
for each alcohol). Thus we need 0.0021 + 0.0013 = 0.0034 kmoles of ethanol
Solution of Problem
090418
13.21
Fund. of Renewable Energy Processes
Prob. Sol. 13.21
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569
or 0.0034 × 46 = 0.156 kg of ethanol.
0.156 kg of ethanol are required to transform 1 kg of the synthetic oil
into ethyl esters biodiesel.
e - How many kg of glycerine are left over?
.....................................................................................................................
From the previous answer: 0.105 kg of glycerin are left over.
Solution of Problem
13.21
090418
570
Prob. Sol. Chapter 13
Fund. of Renewable Energy Processes