good morning now we will start discussion
on a new module tutorial on fermentation in
the previous modules on fermentation we have
discussed methanol synthesis in different
types of micro organisms and different types
of sugars particularly pentose and hexose
sugars and how they they are converted to
ethanol in this module we will solve some
numerical problems and the statement of the
first problem is during the pre treatment
fifty percent of the mass of a lignocellulosic
waste is converted to monosaccharides that
is c six h twelve o six and c five h ten o
five and disaccharides c twelve h twenty two
o eleven and c ten h eighteen o nine in the
mass ratio of six is to four the monosaccharides
produce five one zero gram ethanol per kg
monosaccharides while the disaccharides produce
five four zero gram ethanol per kg disaccharide
assume same mole ratio of both monosaccharides
as well as disaccharides as one is to one
in the respective saccharide types calculate
the yield of ethanol with respect to total
sugar and lignocellulosic biomass also calculate
the percentage contribution of each saccharide
on the ethanol production so this is the statement
on the basis of which we have to solve it
we have to calculate the yield of ethanol
with respect to sugar and lignocellulosic
biomass as well as we have to find out the
individual contributions of these monosaccharides
and disaccharides towards ethanol production
so to solve these what we will do we will
take one basis first say hundred kg of l c
b is our basis so if lignocellulosic biomass
we have hundred kg so what is the conversion
that is fifty percent of mass of lignocellulosic
waste is converted to monosaccharide and disaccharides
so we have one l c b so we have pre treatment
we are getting sugar we have fermentation
it is giving ethanol so this is the process
now we have hundred kg we will get fifty kg
here fifty percent conversion and then we
will see how much we are getting now the sugar
which we are getting we are getting monosaccharide
as well as disaccharide so how how much it
is six is to four the ratio is six is to four
six monosaccharide and four disaccharide so
fifty kg we are getting so how much monosaccharide
is equal to fifty into zero point six thirty
kg and disaccharide that is equal to fifty
into zero point four that is equal to twenty
kg so thirty and twenty kg monosaccharide
and disaccharides are produced then for the
first part it is very easy because this is
five one zero it is five one zero gram is
produced from one kg of the monosaccharide
so this will give us ethanol is equal to how
much zero point five one kg into thirty so
thirty into zero point five one kg and this
will give us ethanol is equal to twenty into
zero point five four kg so in this case we
are getting is equal to fifteen point three
and ten point eight so fifteen point three
and ten point eight 
so total ethanol which we are getting in this
case so total ethanol we are getting this
plus this so how much it is it is coming equal
to twenty six point one kg so
total ethanol fifteen point three plus ten
point eight so twenty six point one kg so
first part is over now we have to solve the
second part ok then we will go to ethanol
yield so what will be the yield we so we are
getting this twenty six point one kg of ethanol
and what was our substrate if we want to get
the yield with respect to sugar so our fifty
kg is our sugar and one hundred kg is our
l c b so yield is equal to with respect to
sugar that is equal to twenty six point one
divided by fifty twenty six point one divided
by fifty that is zero point five two two kg
ethanol per kg sugar but with respect to your
l c b yield l c b lignocellulosic biomass
that is equal to twenty six point divided
by hundred zero point two six one kg ethanol
by kg l c b so these we are getting so this
is the first part actually this over the second
part we have to see the individual contribution
of the different sugars present in the mixture
on the ethanol production so we have two sugars
monosaccharides and two disaccharides
so two monosaccharides we have c six h twelve
o six and c five h ten o five what is the
condition one is to one molar ratio so we
have to see how much mole average moles are
present in monosaccharide so monosaccharides
how much you have got monosaccharide is equal
to thirty kg so molecular weight of these
two glucose and xylose we will we will see
the glucose molecular weight is equal to c
six h twelve o six whose twelve into six plus
one into twelve plus sixteen into six so one
eighty and similarly for molecular weight
of xylose c five h ten o five that is twelve
into five plus one into ten plus six into
five so one fifty so this is one fifty and
one eighty so as mole ratio is one is to one
present in the solution so average molecular
weight will be one eighty plus one fifty divided
by two one is to one mole ratio so that is
equal to one sixty five so the number of moles
of total monosaccharide is equal to how much
we have thirty kg into one thousand this will
be gram divided by one sixty five average
molecular weight that is gram so one sixty
five
so one eighty two one eighty two moles of
monosaccharide are present in the solution
one eighty two moles of monosaccharides are
present in the solution then what will be
the moles of c six h twelve o six and c two
c five h ten o five one is to one so divided
by two one eighty two divided by two so ninety
one so ninety one moles of is here glucose
and ninety one moles of xylose we are getting
then what will be the mass of glucose molecular
weight of glucose into moles that is ninety
one moles into one eighty gram divided by
one thousand so sixteen point three eight
kg so this is the mass of glucose present
in the solution or that is that is produced
after pretreatment after after pretreatment
we are getting this much of sixteen point
three eight kg of glucose and after pretreatment
we are getting xylose as c five h ten o five
that is equal to ninety one into one fifty
divided by one thousand so that is equal to
thirteen point six five kg so these two sugars
mono saccharides we are getting then we will
see the ethanol production so ethanol productions
from these c six h twelve o six will be sixteen
point three eight kg into point five one as
it is given that five one five one zero gram
ethanol is produced per kg of the monosaccharides
similarly first this also xylose also thirty
point six five into zero point five one so
it is becoming six point nine six two kg so
these two different amount of alcohol is produced
by two different monosaccharides similarly
for disaccharide you will at first calculate
the molecular weight so we have c twelve h
twenty two o eleven so twelve into twelve
plus twenty two into one plus sixteen into
eleven so that is it is becoming three forty
two and the molecular weight of not and the
molecular weight of c ten h eighteen o nine
that is equal to twelve into ten plus one
into eighteen plus sixteen into nine so two
eighty two so again here the mole ratio is
one is to one so average molecular weight
in this case will be three forty two plus
two eighty two divided by two that is three
one two so how many moles we have only twenty
kg so twenty kg into one thousand divided
by three one two that is equal to sixty four
moles now
so how many moles of this cellobiose and how
many of these xylobiose so we will have to
divided sixty four by two because one is to
one mole ratio so sixty four by two that is
equal to thirty two both of these two disaccharides
are having thirty two moles each so what will
be the amount of this these one c twelve h
twenty two o eleven cellobiose and xylobiose
so thirty two into three forty two divided
by one thousand so that is equal to ten point
nine four four kg and for the xylobiose this
is equal to thirty two that is moles into
molecular weight of xylobiose two eighty two
divided by one thousand that is gram to kg
conversion so nine point zero two four so
this is the kg now you see we have got four
different saccharides two monosaccharides
two disaccharides and these different saccharides
produced different amount of alcohol so here
this is the this is the cellobiose production
so the ethanol from the cellobiose will be
ten point nine four four into point five four
so that is equal to five point nine one kg
and similarly for xylobiose is equal to nine
point zero two four into zero point five four
so that is equal to four point eight seven
three kg so these are different amount of
ethanol produced by these two disaccharides
so on percentage contribution basis c six
h twelve o six is giving us eight point three
how much it is giving it is giving us ok so
eight point three five four kg and this is
so cellulose giving eight point three five
four kg ethanol and xylose it is giving us
six point nine six two six point nine six
two kg ethanol and then cellobiose 
it is giving us this much of five point nine
one kg five point nine one kg ethanol and
xylosebiose it is giving us four point eight
seven three kg four point eight seven three
kg of ethanol then will sum it of will get
the total of this and then we will get the
percentage of it
so the total is twenty six point one so in
this case the contribution percentage contribution
of glucose that is equal to eight point three
five four divided by twenty six point one
so eight point three five four divided by
twenty six point one into hundred xylose six
point nine six two divided by twenty six point
one onto hundred for these five point nine
one by twenty six point one into hundred and
for xylobiose four point eight seven three
by twenty six point one into hundred so we
are getting the percentage so this is the
percentage contribution of different sugars
for the production of ethanol in this mixture
next we will see the problem number two so
the statement is a food waste is digested
through thermochemical as well as biochemical
routes in the first case the energy requirement
is thirty percent higher than the second case
whereas the sugar production per unit mass
of food waste is twenty percent more in first
case than that of second case now we have
to compare the compare the pretreatment energy
efficiency for the two routes so what is the
case in this in this case what is the situation
we are delivery some pretreatment so we have
some waste organic waste organic food wastage
so this food waste we are pre treating 
so pretreatment will give us sugar that will
give us a sugar so we are following two routes
thermochemicals thermal and another is biological
again we will get sugar and we need some energy
here we will be recovering some energy one
we will be recovering some energy here also
energy two and we will be getting sugar one
will be will be getting sugar two so different
sugar we will get now we have to calculate
the energy efficiency so what is the definition
of energy efficiency that is equal to energy
efficiency is equal to as written here t s
y by t e c t s y by t e c what is t s y and
t e c total sugar yield divided by total energy
consumption during pretreatment so t e c total
energy consumption and total soluble sugar
yield that is t s y so i have given here
so this is the pretreatment energy efficiency
now for the first case say t s y is equal
to y that is sugar yield is equal to say y
we are considering and x m mega joule is t
e c is equal to x mega joule and this is is
equal to say kg is equal to kg now the condition
is that the first route fast route uses thirty
twenty percent the first route uses thirty
percent more energy and also produces twenty
percent more sugar these are the conditions
so that will do the boichemcial routes uses
x mega joule of power and produces y kg of
sugar for hundred kg of waste this is you
are assuming so this is the biochemical routes
we are using y kg we have sugar for producing
and using x mega joule of electricity then
what will be the what will be the efficiency
the energy efficiency is equal to in this
case y by x so y by x so for this case what
will be the t s y in this case y into one
point twenty percent more sugar so one point
two one point two into this y and what will
be the t c that is thirty percent more energy
so one point three into x then what will be
the efficiency in this case it will becoming
is equal to y one point two y divided by one
point three x so that is equal to zero point
nine two y by x and here this is equal to
y by x here equal to point ninety two y by
x
so point nine ninety two percent we can say
so the pretreatment energy efficiency for
thermochemical route is equal to ninety two
percent of that of the biochemical route so
this is the solution of this problem next
we will come to problem number three so the
statement is eight percent slurry of monosaccharides
containing glucose and xylose in the mass
ratio of six is to four converted to ethanol
in fermenter having a working volume of hundred
liter if the maximum conversion of glucose
and xylose are ninety percent and seventy
percent respectively after two hours of operation
determine the ethanol yield and productivity
so we have to determine the ethanol yield
and productivity so let us take one bases
that is its say ah hundred liter of this hundred
liter solution this is our basis so hundred
kg solution were assuming density equal to
one so hundred kg solution so we have eight
percent slurry so that is the solid content
that is that is the mono sugars monosaccharides
percent eight percent monosaccharides is percent
monosaccharides is present is equal to hundred
into zero point zero eight or eight kg monosaccharide
so eight kg monosaccharide is present in it
so how many glucose glucose is equal to six
is to four so eight into zero point six and
xylose eight into zero point four then we
are getting four point eight and three point
two kg so that is four point eight kg and
that is three point two kg so this present
in it but now ninety percent glucose and seventy
percent xylose are converted to ethanol after
two hour of reaction after two hour of reaction
the yield of ethanol is maximum
so with time the yield increases so maximum
will get after two hour so in these case how
much ah ethanol will get you see if if we
want to get the value of yield why is that
yield yield is equal to yield is equal to
v into c divided by m fine b is the volume
in this case one hundred liter of the solution
initially c is the ethanol concentration now
will see how much ethanol concentration we
are getting in this case and m is the substrate
or sugar how much sugar we are using so m
we can calculate here and then we have to
calculate the value of c then we have to calculate
the value of c then we can get the value of
yield so now for the calculation of c we have
to see what type of reaction is going on and
what will be the production of ethanol by
these two sugar that is four point eight kg
and three point glucose and three point two
kg of xylose
so this is the reactions so c six h twelve
o six that reacts with micro organisms and
gives two c two h five o h plus two c o two
so if we take the molecular weight of this
glucose it is coming equal to one eighty gram
and where two c h five o h the molecular weight
is equal to forty six onto two so ninety two
gram so one eighty gram glucose is giving
us ninety two gram of c two h five o h and
eighty eight gram of c o two this is the reactions
it is going in the media so now we have how
much four point three two that is four point
three two kg four point three two kg glucose
we are getting so four point three two kg
of glucose will produce so one eighty kg glucose
produces ninety two gram one eighty gram glucose
produces ninety two gram ethanol so four point
three two kg glucose will produce ninety two
divided by one eighty into four point three
two kg of ethanol
so that is equal to two point two zero eight
kg of ethanol so glucose which is present
in media that will produce two point two zero
eight ethanol so this this will produce ethanol
equal to how much it is two point two zero
eight kg two point two zero eight kg now if
we consider the fermentation for the xylose
the reaction is this one three c five h ten
o five that will give us five c two h five
o h plus five c o two so is we take the mass
of these so three into sixty plus ten plus
eighty that is one fifty into three four fifty
gram it is giving us five into forty six that
is two thirty gram so have here three point
two kg the conversion xylose conversion will
be seventy percent so these one will not be
there this will be converted to three point
two will become converted to zero point seven
so it will give us two point two four kg so
two point two four kg so two point two four
kg so this is equal to four point eight into
zero point nine so this is these one so here
we getting two point two four kg so that is
two point two four kg xylose is converted
so this two point two four kg xylose will
be producing how much of ethanol that is equal
to two thirty divided by four fifty into two
point two four kg so that is equal to one
point one four five kg of ethanol so this
will give us ethanol this will give us ethanol
equal to one point one four five kg one point
one four five kg then what will do total ethanol
equal to how much now we have got two point
two zero eight from glucose and one point
one four five from xylose so these two it
is giving us a three point three five three
three point three five three so three point
three five three kg so three point three five
three kg total ethanol is produced by these
two sugar now what is the ethanol concentration
we have hundred liter of solution we are having
three point three five three kg
so three point three five three divided by
hundred that is equal to this is in kg per
liter so multiple into one thousand so that
will be gram per liter so thirty three into
five three gram per liter so this is the concentration
now we have ethanol yield equal to c v m as
we written so c is the concentration of ethanol
thirty three point five three gram per liter
into v v is the one hundred liter solution
volume and m is equal to eight kg because
eight percent slurry is there so sugar is
eight percent so out of one hundred into point
eight so eight kg is there therefore the yield
is coming equal to four one nine gram per
kg so yield is equal to four one nine gram
per kg of sugar kg of sugar this is ethanol
per kg of sugar so this is we are getting
the yield the second question was the productivity
so productivity is what productivity is equal
to c v m by t so four one nine divided by
two gram per kg hour because it is given here
in the statement after two hour we are getting
maximum concentration of the ethanol so we
have to divided by two and this concentration
here for the production of productivity that
will be the maximum concentration so here
we are getting maximum concentration as per
the statement so this is four one nine divided
by two kg two ah gram per kg hour or if we
can multiply it into one thousand into four
one nine divided by two m g per kg sugar per
hour so two zero nine five zero zero m g per
kg per hour so now the problem is solved so
up to these in this module
thank you very much for your patience
