I need help with the following numbers: 11 (showing calculations) Should you fee
ID: 637897 • Letter: I
Question
I need help with the following numbers: 11 (showing calculations)
Should you feel like answering the other ones then feel free but those four questions identified above are what i need assistance with during this renewable energy topic. Thank you in advance
#3. What are the components of cellulosic biomass? Which one or more kind(s) of these components can convert to sugar efficiently in a biochemical pathway?
#5. What are the differences between the ammonia fiber explosion and steam explosion pretreatments?
#9. What are the major differences between LHW and dilute pretreatments?
#11. Plant Capacity Question:
Hydrolysis Fermentation
Cellulose----------> Glucose ---------->Ethanol
[C6H10O5]n-----> n[C6H12O6] ---->2n[CO2+C2H5OH]
Molecular weight: 162 180 44 46
If a 20 million liter/year ethanol plant is designed, estimate the amount of wheat straw required. Wheat straw contains about 30% (by weight basis) of cellulose. Assume wheat straw has no moisture. The density of ethanol is 0.79 g/L.
Explanation / Answer
3. What are the components of cellulosic biomass? Which one or more kind(s) of these components can convert to sugar efficiently in a biochemical pathway?
cellulosic biomass consist:Cellulose ,Hemicellulose, lignin & ash in some extent .
In biochemical pathway cellulose & hemicellulose can be converted to sugar efficiently .
Q5 What are the differences between the ammonia fiber explosion and steam explosion pretreatments?
steam explosion pretreatments are cost effective ,it show higher efficiency on pretreatments
in ammonia fiber explosion heating is done by high pressure ammonia .99% of ammonia recovered to reuse & remaining serve as downstream nitrogen source for fermentation .
Q9 What are the major differences between LHW and dilute pretreatments?
LWH is physicochemical pretreatment while dilute pretreatments are cjemical pretreatments . now a days dilute pretreatments are getting replaced by LHW.
Q11
plant should be design of capacity of 20 million lier /year
to estimate wheat straw required
assuming wheat straw contain 30% cellulose by wt
density of ethanol=0.79 g/L
ethanol grams in 1 liter of solution = 0.79 g
moles of ethanol in 1 liter = 0.79/46=0.0171739 mol
for production of 2n moles of ethanol there is consumption of 1 mole of cellulose having 'n' units in it
so for
0.0171739 /2n=x/n units
x=8.58695*10-3 units of cellulose
mass of cellulose = 162 *8.58695*10-3 =1.3910859 gram
that means to produce 0.79 gram ethanol we require 1.3910859 gram of cellulose
as wheat straw contain 30 wt% cellulose the wt of straw required= 0.3 W=1.3910859
W=4.636953 gram of straw
therefore
for producing 1 liter of required density ofethanol there is requirement of 4.636953 gram of wheat straw
so for producing 20 million liter in 1 year requirement of straw = 20 *4.636953 million grams of straw per year
so for producing 20 million liter in 1 year requirement of straw =92.73906 million grams of straw per year