Lecture 30: Tutorial 6

[Music] [Music] hello everyone now we will have a tutorial session and in this class we will solve some numerical problems based on last four classes the problem one the statement is a drinking water treatment plant uses a circular sedimentation basin to treat 5 mgd of river water mgd stands for million gallons per day and one mgd is equal to zero point zero four three eight meter cube per second the river open carries zero point zero one zero mm silt particles with an average density of two point two gram per centimeter cube and the silt must be removed before the water can be used the plants clarifier is five point five meter deep and twenty five meter in diameter the water is at fifteen degree centigrade the water density and viscosity at fifteen degree centigrade are found to be nine nine nine point one kg per meter cube that is nine ninety nine nine hundred ninety nine point one kg per meter cube and zero point zero zero one one four kg per meter per second respectively so what is the hydraulic detention time of the clarifier and will the clarifier remove all of the silt particles from the river water so these two questions we have to solve now we will see for a part one the hydraulic detention time this theta of a flow through vessel that is equal to v by q b is the volume of the reactor and q is the volumetric flow rate the volume of a circular machine it is given it is a circular machine so volume will be pi r square h or pi d square by four into h so this equal to pi r square h or pi d square h by four so this formula is used here and d v is that diameter of the machine and h is the height and then what will be the theta value theta is equal to this volume divided by q so pi d b square h by four into q now db is given that is equal to 25 meter and h is 5.5 meter and q we have to calculate what is the value of q q is five m g d and one m g d zero point zero four three eight meter cube per second so if i if we want to get the value of flow in meter cube per second so we have to multiply this five mgd with the conversion factor that is zero point zero four three eight so this is the q value we are getting now and by solving this we are getting theta is equal to twelve thousand three hundred twenty one point six three this is in second and if we divide it by three six zero zero that will get the three point four two hour so that is conversion factor and the answer is three point four two hour that is part one is complete and in part two we have to determine we have to verify whether all particles will be settled or not that means we need to calculate the terminal settling velocity of the particles ok and to find the silt particles to be removed we need to calculate that settling velocity and compare it to the critical velocity for the sedimentation machine and we first need the water density rho and the viscosity mu which are provided in this case the water density and biscuit is at fifteen degree centigrade are found to be nine nine nine point one kg per meter cube and zero point zero zero one one four kg per meter per second respectively it is given so will be putting these values in our expression of v t or v s v s is settling velocity or v t is terminal settling velocity both are same in different cases you might have seen different abbreviation so both are same v t and v s so v s is equal to g nine point eight zero seven meter per second square into rho p two thousand two hundred as given here this is two point two gram per c c so that is converted to two thousand two hundred kg per meter cube so minus nine nine nine point one so this is rho p minus rho into dp square so dp is equal to ten to the power minus five meter square so particle diameter yeah the river contains 0.010 mm silt so this is the particle diameter if we convert it into the meter unit so that will be this one ok so eighteen and mu value is given zero point zero zero one one four so by solving this expression we are getting the value of v s as five point seven four into ten to the minus five meter per second and the critical velocity for the sedimentation machine can be found by using this v o so v o can be calculated by using this expression and the values by putting these values we can get this one now we see the cell settling velocity is much less than the machines critical velocity so this is your settling velocity this is our critical velocity therefore it can be concluded that all particles will not settle now problem number two the statement is a sample of ground water has one fifty mg per liter of calcium two plus and two mg per liter of magnesium two plus ion express its hardness in units of mill equivalent per liter and mg per liter adds calcium carbonate molecular rate of calcium 40.1 and magnesium 24.3 equivalent weight of calcium 20 mg per m equivalent and mg for magnesium it is 12.2 mg per ml equivalent it is treated in an adsorption unit and eighty percent hardness is removed calculate the hardness of the treated water so it is given already we have discussed already we have discussed that m equivalent by liter of any substance that is x that is equal to concentration of x mg per liter by equivalent weight of x mg bar mill equivalent the concentration of calcium in this case we have 150 mg per liter the x value and then equivalent weight of x that is calcium that is equal to twenty point zero so this value is coming seven point five mil equivalent per liter so the calcium concentration in mg per liter as calcium carbonate that we will get multiplying it with the equivalent weight of the caco3 that is nothing but 50 mg calcium carbonate by mill equivalent so this multiplied by this factor so seven point five into fifty will be getting that is equal to this much and three seventy five mg per liter as calcium carbonate similarly for hardness due to magnesium two plus we can calculate the same expression here we have concentration is equal to 20 mg per liter so 20 that will put it here and then using the same expression we will be getting the mil equivalent per liter as one point six three and when we will be converting it into mg per liter as calcium carbonate again we have to multiply it with the factor value of fifty fifteen to one point six three is equal to eighty one point nine six now if we take the total hardness one is due to calcium another is due to magnesium so in terms of mill equivalent per liter for calcium seven point five milli equivalent per liter and for magnesium one point six three milligram per liter so total is nine point one three milli equivalent per liter whereas as calcium carbonate mg per liter will add these two values 375 plus 81.96 so then we are getting 456 456.96 mg per liter as calcium carbonate after the treatment the hardness will be how much so this is our total but 80 percent is removed so 20 percent is remaining so this value into 0.20 so that is 91.39 mg per liter as calcium carbonate now problem number three the statement is a town of fifty thousand sends zero point five meter cube of wastewater per person per day to the wastewater treatment plant a conventional circular primary clarifier would be designed to have an average detention time of 2 hours and an average overflow rate of 45 meter per day what would be the dimensions of the conventional clarifier that we have to calculate so you see here in this case at zero point five meter cube per person per day this is the waste water generation so per day waste water generation for fifty thousand people will be fifty thousand into zero point five so that is equal to twenty five thousand meter cube per day and clarifier surface area that will be volumetric flow rate divided by velocity so in this case two five zero zero zero and this velocity in which case this is our classifier so here water is going and it will be going out so what is the v o in this case that will play that is the vertical direction velocity and hinder settling zone as discussed in the previous class so this v o is equal to forty five meter per day meter per day this is basically meter cube per meter square per day ok so then meter square meter square meter per day so this value is there so then we are getting the 555.55 a so this is the cross sectional area and then velocity that is equal to q ok so v o is overflow rate a b is the clarifier surface area and q daily flow of waste water of total population so now we are able to find out the area as five five five point five five meter square and we have to get the dimension so we have to determine the diameter of it so if it is a circular so we know that pi d square by four that is equal to area so d equal to four into a b by pi so this expression root of that will be the d p d v so that is equal to we will put the value here four into five five five point five five by five root of it so it is coming twenty six point six meter the detention time theta is the ratio of volume to flow rate and the volume is area times the depth so we need to determine the dimension so diameter as well as the height we have to calculate so cross sectional area into height that is the volume that is equal to q into theta detention time so that is the volume total volume and this volume will be equal so detention time into volumetric flow rate that is equal to h into a b so in that case q is equal to two five zero zero zero meter cube per day and theta is equal to two hour and a b value we have got that is equal to five five five point six meter cube per day so multiply it by twenty four so that will be ok so then this will be twenty five thousand into two divided by five five five point six into twenty four hour per day this conversion factor so it is converting to three point seven four meter problem four the statement is and aerated activated sludge tank is being operated under following conditions q is equal to four four zero zero meter cube per day mlss three five zero zero mg per liter y equal to zero point five tank volume equal to 770 meter cube endogenous decay rate constant k is equal to 0.09 per day so estimate weight of solids produced per day for the conditions in which vod is reduced from 350 mg per liter to 130 mg per liter but we second problem part we estimate theta c that is cell residence time and estimate f by m ratio so now the first we have to weight of solids produced per day for the conditions in which vod is reduced from 350 mg per liter to 130 mg per liter so we need to calculate d x d t that is equal to y into d s r by d t minus k d x and in this case we will be putting the values zero point five into y value is given so this expression we can get this value here s o minus s by t into x so that value we are putting here and we need the value of t so t we have to calculate by this expression so hrt hydraulic retention time that is v by q that is seven seven zero divided by four four zero zero that is equal to zero point one seven five date now we will be putting those values here so zero point five into this expression this expression which we have used that is this one so y equal to zero point five into three five zero minus one three zero divided by t zero point one seven five minus k d which we are getting here that is equal to zero point zero nine into x three five zero zero so that will be there so by solving these expressions we are getting d x d t that is equal to three one three point five seven mg per liter per day so then if we want to convert it into kg per day so we have to multiply it with the volume of the reactor so three one three point five seven into v mg per meter cube per liter per day so that it is converting to 241.4 kg per day and the first part we are getting now now we have to estimate the theta c value so theta c value we know the expression is this one theta c is equal to x by del x by del t net so and we will be getting the value here one by theta c into this one by theta c equal to y into f by m minus k d now f by m we can calculate by this formula h o minus s by t by capital x so that is equal to zero point three five nine kg viewed by kg mlss we will be putting this value three fifty minus one thirty divided by zero point one seven five this is the and then x x is the m l s s three five zero zero so if we put the value will be zero point three five nine kg between kg ml ss so this expression will put it here here we can get one by theta c equal to zero point five into f by m we are getting zero point three five nine minus k d k d equal to zero point that value is given zero point zero nine so by solving these expressions will be getting the theta c will be equal to eleven point one seven days now problem five statement is a circular sedimentation tank fitted with mechanical sludge removal unit is to treat 4 million liters of waters per day the detention time of tank is 5 hours if the depth of the tank is to be restricted to 3 meter calculate the diameter of the tank the capacity of the tank can be calculated as volume equal to d square into 0.11 plus 0.785 into h where h is depth and d is diameter of it so this is given so we have to we have to calculate the diameter of the tank so quantity of raw water to be treated during the detention period that is equal to four into ten to the power six into five by twenty four so four into ten to the power six four million so that is ten to the power six is coming into five by twenty four because this is part day so divided by twenty four into four five hours so that we are getting eight thirty three meter cube hence the capacity of the tank is eight thirty three meter cube and capacity of a circular tank of depth h and d is given by this expression so when you will be putting the h value as three because this cannot be more than three so will be putting the value of three and will equate it with three equate it with eight thirty three meter square eight thirty three meter cube then we will be getting eight thirty three equal to d square into zero point one one plus zero point seven eight five into three or diameter of the tank is eighteen point three eight meter by solving this expression we will get the value of d now problem number six statement is for a continuous flow settling tank three meter deep and sixty meter long what would be the flow velocity of water for effective removal of 0.025 mm particles at 25 degree centigrade the specific gravity of the particle is 2.65 and viscosity of water may be taken at zero point zero one centimeter square per second so again we will be assuming that stocks that stokes law is applicable or laminar flow zone and settling velocity is given for this zone that is g into d p minus g into rho p minus rho into dp square by 18 mu so in this case d is equal to zero point zero zero two five centimeter so this is your millimeter zero point zero to five millimeter so zero point zero zero to five centimeter and this is equal to d p basically and v s is equal to nine eighty one into two point six five minus one into zero point zero zero two five square divided by eighteen into zero point zero one so then v s equal to zero point zero five six two centimeter per second then since flow velocity v is equal to v s into l by h so in that case v is equal to zero point zero five six two into sixty by three because l is equal to sixty and h is equal to three given so we will be getting one point one two five centimeter per second so v means horizontal direction velocity and v h v s settling velocity so that we know so this for this case v is equal to one point one to five meter per second therefore for effective removal of particles up to zero point zero to five millimeter the flow velocity in the settling tank should not be more than one point one two five centimeter per second so we have solved all the six problems and up to this in this class thank you very much for your questions [Music] you