good morning now we will start discussion

on a new module tutorial on anaerobic digestion in previous module we have discussed on anaerobic

digestion processes and the production of biogas through this anaerobic digestion methods

in this module we will discuss some numerical problems and we will solve it so the statement

of the first problems is in a high rate biogas plant food waste is anaerobically digested

to produce biogas the slurry contains eight percent of solid food grains the elemental

composition of the food grains on dry basis is carbon fifty eight percent hydrogen eight

percent oxygen twenty six percent nitrogen eight percent on mass basis around eighty

percent of the food grains are converted to biogas and all the converted hydrogen forms

methane if the flow rate of the slurry is four five zero zero liter per day calculate

the rate of biogas that is c o two plus c h four production so this is the problem statement

now we have to solve it see if you see the problem statement we have

given some information and one basis is given here that is four five zero zero liter per

day of slurry handling so we are handling four five zero zero liter per day slurry it

is also given that eight percent solid is present in the slurry so we can calculate

how much solid is present in the slurry then the percentage carbon hydrogen oxygen nitrogen

is also given so we can get amount of elements carbon hydrogen oxygen nitrogen present in

the slurry then it is around eighty percent of the food grains are converted so at eighty

percent of the food grain is converted so all these components will also be converted

eighty percent we can assume then we will see the another condition is given that all

hydrogen forms methane all converted hydrogen forms methane so at first we calculate how much methane

is produced on the basis of hydrogen conversion then from that calculation we get the carbon

used for this methane production and then we will get total carbon and how much carbon

dioxide is remaining and that is that will be used for carbon dioxide production so this

is have the methods which we will use to solve the problem so we have basis four five zero

zero liter per day this is our basis if we assume the density is one kg per liter then

the mass is equal to four five zero zero kg eight percent is slurry so slurry solid is

equal to four five zero zero into zero point zero eight so that is equal to we are getting

three six zero kg so that is equal to three sixty kg then what is our carbon carbon is

fifty eight percent hydrogen is eight percent and oxygen is twenty six percent and then

what is nitrogen that is also eight percent so in the solid how much carbon is present

three sixty into zero point five eight three sixty into zero point five eight so that is

equal to two zero eight point eight kg so what is our hydrogen three sixty into zero

point zero eight three sixty into zero point zero eight so that is equal to twenty eight

point eight kg then what is our oxygen oxygen is twenty six

percent so that will be three sixty into zero point two six [that/eight] kg and that is

equal to we are getting ninety three point six kg and what is our nitrogen three sixty

into zero point zero eight so twenty eight point eight kg so these are the elemental

composition or amount of different components present in the slurry then what we will do

eighty percent is conversion eighty conversion eighty percent conversion so carbon how much

converted two eighty point eight into zero point eight eighty percent so that is equal

to how much that is equal to one sixty seven point zero four then how much hydrogen is

coming twenty eight point eight into zero point eight twenty eight point eight into

zero point eight so that is equal to twenty three eight zero four this equal to two zero

eight is equal to two zero eight point eight so this is this is so now how much oxygen is converted that is

equal to ninety three point six into zero point eight ninety three point six into zero

point eight that is equal to coming seventy four point eight eight kg seventy four point

eight eight kg ok now what we will do we have to find out how much methane is produced how

much hydrogen is converted we have got twenty three point zero four kg now how much methane

will be produced c h four twelve plus four that is equal to sixteen so twelve gram carbon

will give sixteen gram of methane that is why and four gram of hydrogen will give us

sixteen gram of methane so how much methane is produced sixteen divided by four into how

much this conversion is so how much it is twenty three point zero four so into twenty

three point zero four so that is equal to four and that is equal to ninety two point

one six kg ninety two point one six kg so that is equal to ninety two point one six

kg in terms of mole we can get this is divided

by sixteen so that is equivalent to we are getting five point seven six kilo mole five

point seven six kilo mole so this the methane we are getting now how much carbon is used

for the production of this amount of methane we can get carbon used for methane production

that is equal to how much ninety two point one six into twelve by sixteen ninety two

point one six into twelve by sixteen that is equal to sixty nine point one two so how

much carbon we had one sixty seven point zero four conversion and sixty nine point one two

is for methane production the rest carbon will be used for the production of carbon

dioxide so carbon used for carbon dioxide production so carbon used for c o two production

is equal to how much we will be getting one sixty seven point zero four minus sixty nine

point one two kg that is equal to ninety seven point nine two kg that is equal to ninety

seven point nine two kg then how much c o two is produced so c o two

produced is equal to ninety seven point nine two into forty four divided by twelve forty

four divided by twelve and that is equal to three fifty nine point zero four three fifty

nine point zero four kg three fifty nine point zero four kg in how many moles we can get

is equal to divided by forty four so it will be giving us is equal to eight point one six

kilo mole so we have now what how much c o two and how

much c h four is produced now so total c h four and c o two is equal to how much

we are getting ah ninety two point one six kg plus three fifty nine point zero four kg

our total is equal to four fifty one point two kg four fifty one point two kg in terms

of mass but in terms of moles five point seven six kilo mole and here we are getting that

is equal to eight point one six kilo mole so these two in terms of kilo mole that is

equal to five point seven six plus eight point one six we are getting total is equal to thirteen

point nine two kilo mole this kilo mole so now the first problem is solved now we

will see the problem number two so in this problem the statement is in the above plant

the effluent from the digester is used for the production of manure through composting

if seventy five percent of organic compounds present in the effluent are converted to manure

what is the rate of manure production that means we have got eighty percent organics

is converted to gas so remaining twenty percent is present in this in the slurry so we had

three sixty kg we had three sixty kg of solids so out of this three sixty kg twenty percent

will be available in the slurry after biogas production so the solids present is equal

to here organic compounds present in the digested effluents is three sixty into zero point two

that is equal to seventy two kg and seventy five percent of it is converted to manure

so seventy five percent means seventy two into point seven five that will be fifty four

kg per day so this will be the manure production here we are assuming that there is no mass

loss or gain due to microbial activities so this is the solution of this problem now we

will solve problem number three the statement is a family having fifteen cows of body weight

two hundred kg each and twenty chicken of body weight one point five kg each wants to

setup a biogas production plant what should be the volume of the anaerobic digestion unit

suitable for this application and mention volume of various zones of the unit assume

discharge per day of each cow is ten kg and total solid that is t s value is sixteen percent

of the discharge similarly for chicken these values are zero point one kg per day and twenty

percent respectively h r t hydraulic retention time is forty days so daily gas production

gas rate is zero point two eight meter cube per kg t s per day or zero point four meter

per digester volume per day if the total volume of the digestion unit that is v is related

with the diameter of the cylindrical portion as d is equal to one point three zero seven

eight v to the power one third and calculate the height of the cylindrical part of the

unit assuming its volume as v three is equal to zero point three one four two d to the

power cube also calculate the volume of sludge zone active digestion zone gas collection

zone and gas storage zone so this is already discussed here various zones of the unit so this is our statement so we have given

the produ the number of cows the number of chickens available in the family and discharge

per cow and per chicken it is given and the k values are also given that which will be

used and it is also given that the t s present in the two discharges v and d relation is

also given see if we have say one digestion plant so it is given this is the sludge zone

sludge zone v sludge this is the gas collection collection this is say storage and this is

active zone v d and we this is your cylindrical portion and this is our h this is our h and

this is equal to v three is equal to this cylindrical portion is equal to v three now v three is zero point one three four two

d to the power to cube and d is equal to one point three zero seven eight is v to the power

half this v is equal to this v is equal to v col collection plus v s t plus gas col gas

storage g s t plus v d plus v sludge and this v three is equal to this portion if this is

our height this is our lower portion see this is our height so this is our v three this

portion is our v three ok these are given here so we have to calculate the value of

v s l v d v s t v col now what is our information given here the total sludge how much slu sludge

we are getting we have fifteen number of cows so fifteen into ten kg per day ten kg per

day plus we have twenty number of we have twenty number of chicken so we have twenty

number of chicken and it has zero point one kg zero point one kg so we are having so this

is your zero point one kg so we are getting one fifty twenty into zero point one kg so

that is equal to one fifty plus two one fifty two kg so this is our discharge per day so the the chicken values are zero point one

kg per day so this is our similar the chicken discharge is zero point one kg so we have

twenty chicken so that is using zero point one kg per day that is why twenty into zero

point one that is why twenty into zero point and then fifteen into twenty ten that is equal

to one fifty plus two so one fifty two kg the total discharge next will be t s in the total discharge what

will be the t s that is equal to one fifty into sixteen percent sixteen percent point

one six that is for cow dung and this is for chicken that is two into zero point two that

is equal to twenty percent that is two kg is coming from the chicken waste so and the

solid content is twenty percent so two into point two that is equal to twenty four point

four so one fifty into zero point one six plus two into point zero point two that is

equal to twenty four point four kg so this is the coming so point one six means sixteen

percent of t s in the cow dung point two means twenty percent of t s in the chicken waste

now what is our condition the slurry will be having eight percent t s so eight percent

t s if the slurry has so eight kg of t s is required to make hundred kg slurry since the

eight percent slurry is required so twenty four [point four kg is produced

daily so what will be the volume of slurry to accommodate this twenty four point four

kg of t s so that is equal to hundred divided by eight into twenty four point four because

to accommodate eight kg we need hundred kg total slurry weight if the density is equal

to one so one hundred liter so one hundred divided by eight into twenty four point four

that is equal to three zero five kg slurry that is three zero five liter when we are

assuming that density is equal one kg per liter now how much material is produced by

this the cow and chicken that is equal to one fifty two kg so how much water we need

to add so water addition is equal to water addition is equal three zero five minus one

fifty two minus one five two so that is equal to one fifty three one fifty three kg water

we have to add so one fifty three kg water we have to add now what will be the gas storage volume as

we have discussed in the previous module of anaerobic digestion that v s t plus this v

s t v g s t gas storage plus v d that is equal to eighty percent of total volume this is

our assumption that eighty percent of this and and this is equal to and this is equal

to slurry feeding rate into h r t so this is equal to also v g s t plus v d t v d that

is equal to q into h r t what is the retention time what is the feeding rate what is the

feeding and what is the retention time if we multiply this we will get v g s t plus

v d we have discussed in the anaerobic digestion module so here what is the slurry slurry feeding

rate slurry feeding rate is a word three zero five liter per day into h r t h r t is equal

to how much that is equal to h r ts forty days so we have to multiply it so we multiply

it fourteen to three zero five so it is coming twelve thousand two hundred liter so v g s

t plus v d we are getting v g s t plus v d we are getting how much three zero five into

forty that is equal to one two two zero zero liter or twelve point two meter cube we can

get twelve point two meter cube and another way of discussed again that v g s t plus that

is equal to point eight into v so what will be v g s t plus v d is equal to zero point

eight into v so what is v v is equal to twelve point two

divided by zero point eight so that is equal to fifteen point two five meter cube so that

is equal to fifteen point two five meter cube so now we have got the total volume of these

that is equal to fifteen point two five meter cube now we will have get the different zones

now one information is given that d is equal to one what is d this is the d the diameter

of the cylindrical part is equal to d and that d is equal to one point three zero seven

eight into v to the power one third so what is the d d is equal to you can getting here

one point three zero seven eight into fifteen point two five to the power one third and

that is equal to that is equal to how much we are getting three point two four meter

so three point two four meter so this equal to three point two four meter that is equal

to and then what is v three we will put this d value here so zero point three one four

two into three point two four cube now another important information which we

have got here that what is v three v three we are getting from this equation another

is this is v three that is cylindrical part so pi d square by four into h that is also

the v three so here we will put that is equal to pi d square by four into h that is equal

to zero point three one four two into three point two four per cube ah that is equal to

d we can write it here d also d cube now d square d cube cancel so h is equal to zero

point three one four two d into four divided by pi and so h is equal to one point two nine

meter so by this calculation we are getting h is equal to one point two nine meter now

we have got the value of h now we need to calculate the v collection what is the v collection

that is five percent of v what is sludge collection sludge volume that is equal to fifteen percent

of v fifteen percent of v ok so now we have got the v value so what will

be v v collection is equal to zero point zero five into fifteen point two five and that

is equal to zero point seven six two five zero point seven six two five meter cube then

what is our v sludge that is equal to zero point one five into fifteen point two five

and that is equal to two point two eight eight meter cube so that is equal to two point two

eight eight meter cube so we have got this one we have got this one ok we have to calculate

these and this now in the previous module we had seen that v g s t v g s t come here

so v g s t is equal to zero point five into v g s t plus v d plus v s l into k what is

this k k is nothing but zero point four that is the meter cube per meter gas produced meter

cube of gas produced per cube of volume of reactor per day so that is equal to zero point

five into v g s t plus v d plus v sludge into zero point four now what is v g s t plus v d plus v sludge

plus v c o l that is equal to total volume total volume is equal to v c v s l plus v

d plus v g s t plus v c o l v collection so this three is equal to we can get equal to

v minus v col v collection and this v is equal to fifteen point two five and v collection

is equal to zero point seven six two five so these two it is becoming this one fifteen

point two five into zero point seven six two five so this into zero point five into zero

point four that is giving two point eight nine eight meter cube so that is v g s t we

are getting now now alternatively we also know that v g s

t can be calculated on other thumbs rule that is fifty percent of daily gas yield that is

zero point five into t s or total solid into gas production rate that is meter cube per

kg t s per day so these value is equal to point two eight as given in the statement

so zero point two five t s is equal to twenty four point four as per the statement given

we have calculated and then this is equal to three point four one six meter cube so

v g s t one formula is giving us three point four one six meter cube another formula is

giving us two point eight nine eight meter cube so we have to be conservative so we will

be taking the larger value so three point four one six meter cube so v g s t is equal

to three point four one six meter cube so then what is v d v d is equal to v minus

v g s t minus v s l minus v c o l so everything we have got that is fifteen point two five

we have got this one this one this one and this one so i have seen here so fifteen point

two five minus three point four one six minus zero point seven six three minus two point

two eight eight so we are getting eight point seven three eight meter cube so now we have

got the values of all the zones v s l v d v g s t and v c o l as well as the total volume

so the problem is solved so up to this in this module so thank you very much