Transcript for:
Environmental and Water Resources Review

all right so let's get started here tonight we're going to be talking about environmental and Water Resources so the overarching a lot of topics here to cover for this section of Fe review so basically we're going to be looking here at three different topics we look at fluid mechanics hydraulics and hydrology and environmental engineering so comprehensively there that makes up about 23% of your FE exam so I just I recently looked up these are kind of the highlights of the points that you can be tested over in each of these different topics and so for fluid mechanics there you can see we could have general fluid properties flow measurements so they're using Bernoulli's equation to measure the velocity within a pipe or the pressure you can have problems on fluid statics energy impulse and momentum equations and we'll go over some examples that cover that hydraulics and hydrology mainly I'm going to cover a couple examples in the hydraulics so here you have basic hydrology basic hydraulics pumping systems municipal water distribution systems reservoirs your a lot of use of Manning's equation and Hazen Williams equation here ok so that's I have an old Fe exam that I've kind of used as a tool here and to pull problems from and mainly there they were testing for the hydraulic section over Hayes and Williams and Manning's equation and then the environmental engineering section here this is primarily going to be water wastewater ok so the old exam when I took five years ago roundabouts it also had air quality of a good amount of air quality and solid and municipal waste management but now the new regulation that looked earlier today at the 2014 exam and those two things were removed so I think they're not as prominent now on the they're not required so here water quality basic test for water and wastewater some tests for a year maybe that'd be about the extent of your air quality that you test it over environmental regulation and then water and wastewater treatment design are the primary things you can be looking at there so let's jump right in I have about eleven examples there in your packet that we're going to cover start first with that fluid mechanics section jump into a problem with the ideal gas law okay so this problem states that you have a volume or you want to calculate what the volume is of one mole of water at 546 degrees Kelvin and one atmosphere so this is likely this is this is probably done with water vapor right because the water itself it's a it's a liquid not a gas so this is probably applying to a water vapor so we want to calculate what that volume is going to be you see there that the equation that we use for the ideal gas all there's three main equations or we can rework modify the equation a little bit generally see that look here at PV is going to be equal to in our T okay so here it's going to be temperature this temperature is an absolute temperature so what that means it's going to be in either ranky or it's going to be in tell me okay so in our problem statement we're giving it in terms of Kelvin we're given that it's equal to 546 Kelvin this r-value here this is going to be our universal gas constant okay so this would be given in your reference material as constant and this has a value of 8.314 joules per mole times Kelvin and your in term here is going to be your number of moles and for our problem statement there we're told that we have one mole of water pressure so this term in this problem is a little tricky so it gives you pressure as one atmosphere okay so one atmosphere is going to be 101.3 kiloPascals right so we have to be careful with our units here because everything else the way the units will cancel we see that we need to convert this over to Pascal's instead of Atilla Pascal's so that's going to be one hundred one thousand three hundred Pascal's okay so we can apply the ideal gas law there to solve for what that volume is going to be so that's going to be volume equals in RT over our pressure end up sitting there that's going to be equal one mole times 8.314 per mole Kelvin times 546 greens Kelvin over 100 1300 so that's to be pascals okay so what we end up getting here is that this volume is equal to 0.04 for 8 meters cubed but we see there that thing with the volume in terms of liters so we know that there's a thousand liters and one cubic meter end up getting a volume there forty four point eight liters so the correct answer there should be D any questions on that problem so you could also be given this in terms let's see here maybe you're given the the density or you're given many baby they want you to solve for the number of moles so basically you can rearrange and rework this equation in the number of different ways and they could ask number of different problems associated with this but you'll probably see at least one ideal gas law problem on that fee alright so moving on their second question here is on hydro static so we're looking at hydro static forces acting on this skate that's going to create a moment a clockwise moment on this skate we want to know what that resultant force are the magnitude of the force required to keep the gate closed okay so let's say we're given that it's a rectangular homogeneous gate with a height of three meters and a width of one meter we're told that it's a frictionless hinge at the bottom okay so we know there's not going to be a moment associated with the friction about the hinge so and we're given the density of the fluid at 1600 kilograms per meter cube and so we know that that hydrostatic force you can look at that pressure distribution acting on that gate so we know that hydro static or that pressure distribution there is going to create a resultant force okay where's that resultant force going to be acting there in reference to the hinge how far from that hinge you guys remember it's going to be a third of that heights right so that moment arm associated with that resultant force is going to be one third of the height okay so let's work through this problem here if we take a sum of the moments about that hinge we want to know when it's what what magnitude of that force we need basically right when it's about to tip okay so we want to know when that's the sum of those moments about that hinge are going to be equal to zero so the force associate or the resultant force associated with that fluid is going to create a clockwise moment and so we know it's going to be at times one third three meters so one third of the height is going to be the moment arm there and the moment associated with that magnitude of the force to keep the gate closed can I have a moment arm there of three meters it's going to be set equal to zero so we can solve rearrange this equation here you say that the force that we're interested in is going to be the resultant force that fluid divided by three okay so that resultant force is going to be equal to the gamma of the fluid remember we're given density so gamma is just going to be Rho times G times H over two so why am i multiplying by H over two there right right so we see there we multiply that by H over 2 times 3 meters times 1 meter so this is a cross-sectional area of that gate we can divide that by 3 so we end up seeing there is that this force is going to equal the density 1600 kilograms per meter cubed is gravity 9.81 m/s^2 in here we have to multiply by gravity to get the gamma of the fluid or the weight of the fluid times 3 meters squared so this is going to be that cross sectional area of the gate times H over 2 which is going to be 1.5 meters again we have to divide all this by 3 so we end up seeing there is that this is going to equal 23,500 Newton's or 23.5 kilonewtons and so 23.5 isn't an answer in your in your problem statement there but it says the the closest one to it is going to be 24 so we round up there and we see that the answer is C see all right so moving on we got a couple more here in the fluid properties or in the fluid mechanics section kind of a conceptual question that they had in the exam says which of the following statements is true of viscosity a it is a ratio of inertial of the viscous forces be it as it always has a large effect on the value of the friction factor see it's a ratio of the shear stress to the rate of shear deformation or D it's usually low and turbulent forces predominate so does anybody remember what Newton's law of viscosity says so we can use Newton's law of viscosity here say that the shear stress is going to equal viscosity times D u dy okay so this is going to be shear stress and this is going to be the rate of shear deformation so we can see there that mu is going to be the ratio of sheer stress to that rate of shear deformation so see that answer is going to be C so a the ratio of inertial to viscous forces anybody know what that is or what that's going to be this is actually Reynolds number okay so VD over the kinematic viscosity your BD there that's going to be your inertial forces so you could easily get a question of which of the following statements is true of Reynolds number and the answer there would be a and this is going to be kinematic viscosity okay any questions on that one all right moving on so obviously there we covered hydrostatics covered a problem there on fluid properties I saw something on the ideal gas law as well some other prime topics that you can see from fluid mechanics would be application of the momentum equation which will do one of those here again Bernoulli's equation and the continuity equation okay so in this example we see that we have a horizontal jet of water with a density of a thousand kilograms per meter cubed just as this would be the density you would normally assume for water if you were working through these problems and it wasn't given in your problem statement it is deflected perpendicularly to the original jet stream by a plate as shown below and once you know the magnitude of the force required to hold the plate in place and so what we see here what we need to do with these types of problems the first thing we want to do we want to draw a control volume for the fluid so we draw a CV for the fluid and in this case you see that our jet is deflected here by this plate so we can choose our control volume here I'll do that in another color to encompass our inputs and outputs all right and so here we're going to have our momentum or our Rho QV coming into the system or into the control volume from the flowing fluid we're going to have this external force that's holding that plate in place so this is going to be the normal force exerted by that plate on the fluid we're going to have this Rho Q V term out of the control volume and this Rho Q V term out of the control volume as well so what the linear we can we can use basically Reynolds transport theorem to derive the linear momentum equation so the linear momentum equation that's telling us is essentially it's applying that conservation of momentum principle to a control volume approach so using that you Larry an approach for conservation of momentum we see that the sum of the forces in the X direction so here we're interested in the X direction because it's an X directional force set is holding that plate in place is going to be equal to our Rho Q V out in the X direction minus Rho Q V in in the X direction okay so this is going to be again our linear momentum equation so let's work this problem looking term by term there so the sir sum of external forces in the X direction are just going to be if we assume a positive x-direction B are positive x-direction we're going to see that we have this negative external force holding that that plate in place or that normal force exerted by that plate on the on the fluid so that's going to be equal to negative F okay does everybody see that our Rho Q V out in the X direction is going to be equal to what so what that that momentum associated with the fluid there from in the X Direction leaving the control volume it's going to be zero right and our Rho Q V in in the X direction is going to be equal or density thousand here we see that Rho Q V coming into the control volume in the X Direction thousand kilograms per meter cube times all right so Q is going to just be velocity times the area right the cross sectional area and we're given in the problem statement there the jet area as 0.01 meters squared so this is going to be 0.01 meters squared then we multiply that by the velocity of 30 meters per second so we ended up seeing here this is going to be equal to 9000 Newton's okay if we go back and plug in our original equation so here we derived each of the individual components for that linear momentum equation we can say here that negative F is going to be equal to zero minus Rho Q V in the X direction okay so basically there we see that we can say that that force is going to be equal those negatives cancel out of course going to be equal to negative ro or positive Rho QV and in the x-direction which equals nine thousand Newtons or nine kilonewtons so there we see that our answer is going to be equal to our going to be B for this question so you've got to be careful there with the units so you see D is ninety kilonewtons so you got to be very careful here when you're working them in check on your units as you go throughout the problem because it'd be very easy to make an order-of-magnitude mistake when you're working through these all right so our last problem here from the fluid mechanics section is going to be an application of Bernoulli's equation so we see there we have a pitot tube it's placed at a point where the velocity is two meters per second so if the velocity in the streams that are in this pipe is two meters per second is it going to change at all as it moves through the pipe so is our velocity here is it going to change obviously here at the the pitot tube we're going to have a stagnation point right but other than that are we going to have any change in our velocity within this pipe why is that so we can assume here that you can neglect frictional losses if you neglect frictional losses so this is just a pure application of Bernoulli's where we assume that friction is negligible okay okay so basically we can use that continuity principle here we see that we don't have a change in area so our Q in is going to be equal to our Q out and so since the area this pipe isn't changing that velocity has to remain constant throughout okay does that make sense okay so we can apply Bernoulli's equation here at two points so we'll call this we have a stream line that runs through the center of this pipe we'll call a stagnation point of this pedo 2.1 and then directly under this other arm of the manometer will be 0.2 so we can apply Bernoulli's equation along that streamline what that says is that P 1 over gamma plus Z 1 what's Z 1 plus V 1 squared over 2g is going to be equal to P 2 over gamma plus Z 2 plus V 2 squared over 2g ok so we can start canceling out terms for the first thing we want to do when we work up a new Lee problem we want to cancel out stuff that is easy and we already know so we we see here this is a horizontal pipe and so along this streamline we're not going to have an elevation difference since we have zero elevation difference we can say Z 1 and Z 2 are equal to 0 or the difference between Z 1 and Z 2 will be 0 ok so point 1 again there was the point in the stream line where we hit the stagnation point so at that point our velocity is going to be equal to 0 it's transferring all that kinetic energy to a pressure head okay and so our velocity and that that point will be 0 we can also say we're so we're given a manometer here and so so we can say p1 is going to be equal we can use our manometry equation here is going to be equal to p2 minus gamma of the fluid times some distance D and so this distance D here essentially defining as distance between that elevation and the streamline so this is point B some arbitrary value we'll see later that this cancels out but I just wanted to give it a definition for now and we subtract gamma of air times H and we can use the fact that the change in pressure with respect to X for that static fluid in the manometer here the pressure at this point will be the same as the pressure at this point so we can move over there in our manometer and then we can say plus gamma of the fluid times h plus gamma of the fluid times d we see here that our gamma fluid times D cancels out and we can say here that gamma air times H the specific weight of air relative to the specific weight of fluid is going to be really small right the density of air is much much less than the density of our fluid so we can say that this term is pretty much zero so what we end up seeing there is that our pressure difference p1 minus p2 is going to be equal to the gamma over fluid times H or H is going to be equal to p1 over gamma minus p2 over gamma okay so if we plug that if we move back into Bernoulli's equation we can see that we can rearrange here with this identity that H is going to equal p1 over gamma minus p2 over gamma and we can say that v2 squared over 2g is going to be equal to H okay so H we're given the velocity as 2 meters per second in that pipe and so H is going to be equal to 4 meters squared per second squared remember that velocity is squared divided by 2 times 9.81 m/s^2 end up seeing here that we have H of 0 point 2 meters okay so there our answer was going to be d all right so any questions on that that pretty much covers fluid mechanics oh and uh you know what I'm not sure that'd be easy enough to calculate so let's see here is anybody have a calculator on them okay so go ahead and multiply that by 9.81 and then being divided by 30 2.2 okay so that's not one of them on there but they will do little little things like that so a lot of the ones you notice you'll note you notice some unit conversions we'll see that later on in the environmental section basically basically if you don't convert your time units to the correct units you can see you would get one of the answers so you will see that pretty often I'm not sure what the tricks are on this one oh do ya well yeah that makes sense so yeah this there's little things here and there that could trip you up along the way so just be very careful when you're working through the problems and check all your units multiple times believe that I read from last year there's there's a hundred and ten questions but if you have six hours to take the exam I'm pretty sure that's still the component is everybody here taking it is it January okay okay all right so let's move in and work a few hydraulics problems so here we're going to be working a couple of Manning's equation problems and Hazen Williams problem so in this problem we're given we want to calculate the velocity using Manning's equation so we're given that we have a 12-inch diameter concrete sanitary sewer with a Manning's n coefficient of zero point zero one three so this is going to be constant with depth we're told that this pipe is flowing half-full and is constructed on a grade of 0.5 percent the flow velocity in this sewer is most nearly going to be what is what it wants to cheat a solve for there so we can use we can go ahead and write down Manning's equation so it'll either give you Manning's as in terms of Q or velocity you know that Q is going to be equal to K over in and this is a constant that K is a constant that accounts for which unit system you're in it's K over N times a times are two the two thirds are there that's hydraulic radius and a is going to be the cross-sectional area of the fluid in that pipe times s to the one-half so this is going to be your bed slope or your friction slope an open channel flow so we can say there the problem statement wants velocity not flow rates so velocity we can just divide by area end up seen it's going to be K over in R to the two-thirds s to the one half okay our that hydraulic radius what's that going to be equal to you remember from hydraulics anybody remember the hydraulic radius is going to be equal to the area divided by the wetted perimeter okay so the wetted perimeter is going to be basically the length that that fluid is in contact with your your sewer wetted perimeter so let's see here for a 1 foot diameter pipe that's flowing half-full have four see that the area is going to be equal to PI R squared times 1/2 again it's flowing half-full so we have to multiply by 1/2 see that this area is going to be equal to 0.3 9 feet squared our wetted perimeter is going to be equal to PI D over 2 so this is going to be the circumference PI D as in your circumference and we have to multiply by 1/2 since its flowing half-full so this is going to end up equaling one point five seven feet okay this is going to be given you can look this up in your reference manual for BG units it's going to be equal to one point four eight six and our slope is given as 0.05 or zero zero five feet per feet so this is an area where you can trip up on this problem because you're given the slope in terms of percent but you need to convert it there to feet per feet okay so that's how it's derived in the equation it's an empirical formula so it's important that you do that and it's not directly going to impact the answer there but it could be a reason why they have D on here this 32.4 the answer ends up being belief C so they might have that 32.4 there to kind of trip you up in terms of that order of magnitude difference associated with the slope and so we can plug our values in here for into Manning's equation we got one point four eight six over zero point zero one three times zero point three nine to the two-thirds over one point five seven to the two-thirds time 0.005 square root of that end up seeing that it equals 3.2 feet per second okay so our answer there is going to be C you have a question okay all right ready to move on so here's another example using Manning's equation this one has a little trick to it so we see here a sanitary sewer delivers flow from a sump to a lift station as shown in the figure below the sewer length is given as 400 feet and the diameter is 30 inches the sewers made a concrete and has a roughness coefficient of 0.01 3 so we want to know for full pipe flow with water surface elevations in the upstream sewer pump and lift station wet well of 105 and 103 point 5 what's the discharge and so here we're saying that the water surface elevation and that that sump well and wet well this is going to be 103.5 this is going to be 105 feet so what's that slope going to be that we use for Manning's equation it's going to be the friction slope right it's defined it's a friction slope so this is going to be based off of the water surface elevation and these two compartments so our friction slope there is going to be 105 feet minus 103.5 feet divided by 400 so they try and trip you up a little bit here because you're given the basically the bottom of the pipe you're given that elevations for that sump invert and the lift station invert where the pipe is coming through there you're given those as 101 and 100 feet it was it was somewhat surprising but if you plug those numbers in that doesn't give you one of these four answers so they don't crush you if you if you mess that up but the correct slope here is going to be based off that that water surface elevation in those two compartments okay so this is a common theme I've seen this on it occurs again on that next problem that we're going to work with the Hazen Williams equation but very important here you're using friction slope and so if we want to calculate flow rate for Manning's equation see we can apply it again it's gonna be one point four eight six over in times the area times are to the two-thirds slope the friction slope to the one-half so again we calculated friction slope there is going to be equal to zero point zero zero three seven five and this is going to be feet per feet okay so we can plug our values in here to Manning's equation at one point four eight six over zero point zero one three times pi okay so this is one of the areas that will trip you up you're given the diameter in terms of inches you need to convert that over to feet so if you plugged in the numbers in terms of inches I think it does give you one of the wrong answers okay so make sure you convert we need to convert over two feet here so this is going to be pi times R squared so fifteen inches divided by twelve inches per foot square that and then multiply by our hydraulic radius to the two-thirds so it's going to be pi times fifteen over twelve squared divided by PI times D so circumference is going to be PI D again remember the pipe is flowing full here it's going to be 30 over twelve this is going to be raised all this will be raised there to the two-thirds and then we can multiply by our slope zero point zero zero three seven five this is going to be raised to the one half so when we solve here we see that we get a Q equal to 25 point 1 cubic feet per second or fee cube per second okay any questions on that one second throw it down there so the correct answer on that one is going to be be all right so let's look at our last one here in the hydraulic section given a word problem here so they don't give you any depiction here of this problem it says that two tanks are connected by a 500 foot link one inch inner diameter PVC pipe the appropriate value for the Hazen Williams coefficient so this kind of triggers you there to use a Hazen Williams equation is going to be 150 so if you have water at 60 degrees Fahrenheit flowing through the pipe at a velocity of 10 feet per second they were give them the velocity of that fluid it within that pipe and the tanks are open to the atmosphere entrance/exit and minor losses are negligible and the difference in water surface elevation between the two tanks is mostly nearly going to be what so we want to calculate the Delta H between those tanks so if we draw this system we can depict this system here for our two tanks given that this is a one inch inner diameter pipe connecting these two tanks we want to calculate what the difference here and our elevations in those two sort or two tanks are going to be okay so what's going to cause this head loss here conceptually you guys saying friction right so it's going to be friction in the pipe is going to cause our head loss here and if we have a one inch inner diameter pipe then there can be a lot of potential head loss in that system okay it's a very small pipe it's going to be moving very fast and very sensitive there to the roughness of that that pipe so if we assume flowing full obviously it has to be flowing full right we can say there that Hazen Williams equations in the velocity is equal to one point three eight times C times R to the zero point six three so you see here this is very similar to Manning's formula again another empirical equation here formula and we can see here just just in terms of the roughness coefficient so here our C value was 150 right but our Manning's in is normally 0.01 to 0.03 something in that range right so you see here that this is multiplied by this roughness coefficient as opposed to divided by that roughness coefficient inner diameter yeah so I D this is going to be inner diameter good question all right so our our our hydraulic radius is going to be area over the wetted perimeter which is going to be equal to PI d squared over 4 times 1 over PI D so in general for these flowing full conditions we can simplify this as d over 4 ok so we can plug back in here to the Hazen Williams equation we say that 10 feet per second or velocity is going to be equal to one point three eight times at coefficient C 150 times 1 over 12 again here we got to convert from inches to feet divided by 4 so it's that diameter of the pipe it's going to be 1 over 12 feet to the zero point 6 3 s to the zero point 5 4 so we can solve for s to get our friction slope okay to get our friction slope what we see here is that s is going to be equal to 0.36 4 feet per foot so this means that basically for each foot of that linear foot of that piping system that's transporting the fluid from tank 1 to tank 2 that you're going to have a head loss of 0.36 4 feet for each foot of that pipe okay so our Delta H then it's going to be equal to 500 feet times are slow 0.36 4 feet per feet two equals 182 feet so it's a lot of head loss in our system and again that's reasonable there because you have a very thin tube and that velocity is very high ten feet per second that's cranking that's moving pretty fast so it's it's reasonable to think that we'd have that much head loss in our system so the answer for this one was si si okay all right moving on any questions there on the hydraulics if you have any questions on on hydrology or specific questions on hydrology or hydraulics there were none in the sample exam that I had to present for you on hydrology please feel free to come or to email me come to my office ask me about them and I can help you out with them the best of my ability so if you're when you're studying you come across something that you don't remember how to do just let me know and we can work through it so alright so let's move on now to environmental engineering the environmental engineering section of the exam first question that we have here we're looking at basically what the oxygen demand is going to be to biodegrade benzene so this is a carbon containing structure organic matter here to co2 and h2o so the question says you're designing an aerobic system to biodegrade that's benzene and it gives you the chemical reaction there as well as the atomic weights of each of the elements if you have a benzene concentration of 500 milligrams per liter it wants to know the concentration or the amount of oxygen that's going to be consumed completely biodegrade the benzene okay so the first process or first step we need to do here is we need to balance this equation so we see c6h6 plus o2 is going to yield co2 carbon dioxide plus h2o so go back into your chemistry toolbox and we're going to balance this reaction so the easiest way to do this here we see that we have carbon containing this one compound on the left hand side and this one compound on the right hand side so we know that we're going to have to have six moles of carbon dioxide per mole of benzene there in order to get the appropriate amount of carbon as well we see that hydrogen is contained here and contained here those are the only two compounds there that contain hydrogen so that means we have to have three moles of water in order to meet that our balance that hydrogen requirement on the left hand side of the equation so finally we see that we have oxygen occurring multiple places here so we have two atoms oxygen on the left hand side and then we have let's see 12 associated with co2 and three that are going to be associated with water so to balance this reaction we see that we're going to have to have seven and a half moles or 15 over two moles of o2 in order to balance the oxygens that are on the right-hand side of the equation so there but I see that that makes sense okay all right so we can use now that we have a balanced reaction we can use the stoichiometry of this problem to answer how much oxygen is going to be required there to completely degrade or biodegrade that benzene so see that it takes seven and a half moles of oxygen per one mole of benzene okay so if we have 500 milligrams per liter milligrams of c6h6 per liter we can convert that over to a number of moles of benzene so in one mole we're going to have 12 times 6 we need to get that molecular weight there of benzene plus 6 this is going to be times 10 to the 3rd so this is how many milligrams of benzene we're going to have in one mole so that's going to be equal to see that 500 milligrams per liter times 1 over 78 times 10 to the 3rd milligrams per mole end up seeing here that at 500 milligrams per liter we have 0.0064 moles per liter okay so let's see now that we we know how many moles we have and we know how many moles of oxygen are required for one mole of the benzene then we can say that 0.0064 moles per liter of c6h6 times 7.5 moles of oxygen for one mole of benzene and we go ahead and multiply by that molecular weight of oxygen to get it in terms of milligrams per liter so for oxygen it's going to be 32 right because we have an atomic weight of 16 and we have two oxygens so it's going to be 32 times 10 to the 3rd milligrams of o2 per mole end up getting their a concentration of 15 hundred and thirty-eight milligrams per liter and so if you notice in the problem statement it says the amount of oxygen oxygen that will be consumed to completely biodegrade the benzene is most nearly so where we don't have 1538 as an option but we do see the closest answer there that's going to be 1600 so our answer will be d 1600 okay all right any questions on that so you'll probably have something where you'll have to balance reaction and then use stoichiometry of that balanced reaction to determine concentrations so this would fall under kind of that general water quality problem section of environmental engineering all right now let's look at Bo D demand so typically we're looking here and have in rivers but for this question we're looking at wastewater flows that combine in a sewer so for waste water wastewater people in here probably remember this equation it was on remember this question it was very similar to one they saw in their exam so we have three wastewater flows that are combining in a sewer each having flows and Bo D concentrations as depicted here so if we have infiltration which has zero Bo D so this is going to be percolation from the overlying soil into the pipe or into that sewer and it's 10% of the total flow once to know what the resulting Bo D is going to be so this basically this infiltration is going to be diluting the Bo D that makes sense so the way we go about solving this problem when basically we'll be using a weighted average of the different sources so we can say that our flow we want to calculate the flow rate Q it's going to be 4 times 10 to the 6 plus 0.8 times 10 to the 6 plus 0.2 times 10 to the 6 equal 5 times 10 to the 6 liters per day our infiltration is given as 10 percent so 0.1 of that total flow so times 5 times 10 to the 6 liters per day so we see the amount of infiltration that we're going to have is 0.5 times 10 to the 6 liters per day this means our total Q is equal to five point five times ten to the six liters per day okay so our Bo D of the mixture is going to be equal to the sum of the individual Bo D sources vo D sub I times Q I divided by the sum of Q I so our total flow so here this is going to be equal I'm going to drop the 10 to the 6 right now so I don't have to write it over and over again to be equal to 4 times 200 that's going to the concentrations are the Bo D there believe is given in terms of milligrams per liter so 4 times 200 plus 0.8 times 300 plus 0.2 times 500 and to be complete here to be comprehensive I want to include that 0.5 the infiltration component times 0 it has a bo d of 0 and we can divide all this by the total flow of 5.5 okay so we ended up seeing here that we have a bo D of our mixture that's equal to two hundred and seven milligrams per liter and so our answer there is going to be see what if we are be sorry what if we took instead of dividing by 5.5 what if we divided by five so does anybody got a calculator there that they can calculate that multiply it by five point five and then divide by five to twenty eight so I think that's going to be answer C there so if you don't do not include infiltration you could get the wrong answer there okay God be careful with those kind of details all right any other questions on that all right so the last one we have to talk about here it's it's a two-part problem and the trick with this one is that you're given a lot more information than you need so they try to overwhelm you with information here and you got to cipher through what's important what's pertinent to the problems being asked and what's kind of erroneous additional information that we can we can brush aside right now so we're told four questions 15 and 16 that we have a municipal activated sludge wastewater treatment plant with primary clarification so this is going to be primary sedimentation we'll be talking about this in class here in the next couple weeks with the following characteristic so we know that as a flow it's treating five mill or we have a flow rate here through the system of five million gallons per day it gives us our Bo d5 concentration at 200 milligrams per liter we're given suspended solids and then we're given some additional information about the normal operation of the facility so the first question we want to answer here is if we have a plant B od5 removal efficiency so it can remove 95% of that B OD concentration and the primary clarifier so the primary sedimentation removes 35% of fbod in the amount of Bo d5 removed in the biological reaction so after we go through the primary sedimentation process we're going to send it to activated sludge and those microbes are basically going to break down that Ord Enoch matter and remove that that VOD ok so it wants to know what the amount of Bo D removed during that biological reaction is going to be so the information we need here for this question we're going to need to know the flow rate the concentration of that Bo D 5 that 200 milligrams per liter and we're and I believe that's all we're going to need for that problem from this initial table so we're given that we have a plant efficiency so in total the plant efficiency RP e F F is going to be equal to 0.95 okay the primary clarifier we can call it p dot c its efficiency is going to be equal to 0.35 okay so w will use that term right now it's basically telling us a weight of that Bo d is going to be equal to Q times the concentration at the OD five times the conversion factor so here we have to use this conversion factor that we have eight point three four pounds per million gallon remember here mg D is going to be million gallons per day divided by milligrams per liter okay so we ended up seeing here that this is going to be equal to five million gallons per day times two hundred milligrams per liter go ahead and put that in there that it's units and gallons per day times eight point three four or conversion factor there we ended up seeing we have eighty three hundred and forty pounds pounds per day and so this is a matter the the weight of Bo D coming through that plant in a given day so it's going to be eighty three hundred and forty pounds in a given day so a removal efficiency in that initial step that primary clarifier is going to be thirty five percent alright so for efficiency there is thirty five percent then we can see it's going to be eighty three hundred and forty pounds per day that's what's coming through there times zero point three five actually let's do this let's look at how much is going to be if we have 35% removed that means 65% is going to be moving on to the activated sludge process so we can say 1 minus 0.35 we'll call this go into activated sludge okay so we end up seeing there that this value of what's removed there in the primary clarifier is going to give us 50 421 pounds per day that are going to the activated sludge process okay and we also know that the amount that's leaving the plant if it has a 95 percent removal efficiency that the amount that's going to be leaving the plan is 80 340 times 0.05 so that mean five percent of the Bo D loading is leaving our plan it's going to be equal to four hundred and seventeen pounds per day okay so if we owe that if we do a simple balance there and we say that 5,400 is coming in to the activated sludge and 417 leaving the activated sludge then we can say our removal it's just a simple mass balance approach here and our activated sludge is going to be equal to 50 for 21 pounds per day - 417 pounds per day end up seeing that we have five thousand and four pounds per day removed in that activated sludge process okay any questions on that one so I know you guys haven't seen this yet we'll be talking about that more a little later in the semester all right moving on last one here so we want it also wants this to calculate what the aeration Basin volume is going to be in gallons okay so here if we go back to our original problem statement we say that under this normal operation of the facility that the aeration Basin has a high residence time of 10 hours ok we know what the flow is coming through that Basin so if we know what the flow coming through that Basin is and what the residence time is we can determine what the volume of that Basin is going to be okay so this is pretty pretty much just using the flow rate identity in this problem okay so we see here that flow rate is going to be equal to volume over the residence time so we can rearrange that see that our volume it's going to be equal to Q times T so we know that the flow is 5 million gallons per day and since we're working in gallons let's go ahead and write out we have five million gallons there and we know that the residence time is 10 hours okay so one thing to notice here and if you don't do this she'll end up getting you answered e is that you're given your flow rate in gallons per day residence time in hours you've got to make that conversion okay so this is the easy place to get tripped up on the conversion when you're trying to work fast so we're going to have basically there so one day 24 hours end up getting their a volume of 2.08 3 times 10 to the 6 gallons okay so 2 million 80,000 gallons all right so our answer there is going to be D or C sorry yeah all right any questions on that I think that's all I got for you today so again remember if you have any questions or any specific problems that you're you're struggling with or need some help on feel free or you want an explanation on it feel free to come to my office or shoot me an email and we'll work through them so all right thank you guys good luck on it