Lecture 10: Tutorial -2

Hello everyone, now we will have a tutorial session and now we will solve some numerical problems based on our discussion during last four classes. Our first problem statement is a confined aquifer 25 meter thick has two monitoring wells spaced 600 meter apart along the direction of groundwater flow. The difference in water level in the wells is 3 meter, the difference in piezometric heat, the hydraulic conductivity is 50 meter per day. Estimate the rate of flow per meter of distance perpendicular to the flow, so we have to estimate the rate of flow per meter of distance perpendicular to the flow, so if datum line 1, 2, okay. So this is our del h and this is our l, this is del h and dl basically l, this is basically l, so that is equal to what is dh, it is given 3 meter, so 3 meter divided by and distance is 600 meter, so we are getting 0.005, so that is the gradient, okay. So now we are getting the gradient, now what we need to calculate the flow rates, that is q, so that is q equal to k into a into dh by l or dl as mentioned here, it is dl, so that we will get it here. So k is given, what is that, 50 meter per day and then dh by dl we have calculated now that is this one. Now we have to get the value of a, how can we get the value of a? It is given that it is 25 meter thick, so if this is the 25 meter thick, this direction say 25 meter thick, so if we imagine a 3-directional plane, then we will see that per meter of distance perpendicular to the flow that will be width and say 25 meter thick will be the height, so 25 into 1 that is equal to cross-sectional area, so that into 0.005 into 50, so that is equal to 6.25 meter cube per day per meter of width. So now the problem is solved, we had to calculate this q value, now we are able to solve it. Now problem number 2, a phosphorous limited lake with surface area equal to 90 into 10 to the power 6 meter square is fed by a 18 meter cube per second stream that has a phosphorous concentration of 0.010 mg per liter. In addition, a plant from a point source adds 0.9 gram per second of phosphorous, the phosphorous settling rate is estimated as 10 meter per year. Estimate the average total phosphorous concentration, question number 1 a, and b part what rate of phosphorous removal at the waste water treatment plant would be required to keep the concentration of phosphorous in the lake at an acceptable level of 0.010 mg per liter. So the problem is like this, so if we have a lake, we have one stream here and it is going out also, so this is our in, in this way it is in and this way it is going out and another point source we have here somewhere that is giving phosphorous. Now if we see at steady state what will be happening, sum of the phosphorous in the input stream minus disappearance due to reactions or settling equal to the sum of phosphorous going out from it. Sum of input we can say here q is the flow rate and concentration in is the concentration of phosphorous here and another source is here we are getting that is we have to calculate, it is already given that 0.9 gram per second, so this is also in mass unit, so we can also get directly put s that is corresponding to the phosphorous, this. And it is given that the phosphorous settling, the phosphorous is settled here and settling rate is 10 meter per year, so that we have to calculate what is the settling rate and what is going out, what is going out that is certainly the, if it is a steady state that will also q, so we are having q, so that is also q, so this phosphorous may come from some solid part, so there is no other volume, so then this q will be there. So what we are getting, so this is our input and our accumulation which is coming at the bottom of it that is cross sectional area A, if we can calculate into velocity of this that is the volumetric flow rate into c that is the c out we can say, c out if we get this minus, so that is equal to q into c out, so this will be the relationship or if we can write it what is the c out that is equal to q c in plus s divided by q plus A v, so this is the relationship we are getting. Now we have to get the value of A, so let us see phosphorous loading from the incoming stream q into c i, what is this q into c i, q is our 18.0 meter cube per second at it is given and the concentration of phosphorous is given here 0.010, so this is our 0.18 gram per second phosphorous is coming from this stream and another phosphorous is coming 0.9 gram per second from the point source. So a total phosphorous input is 0.9 plus 0.18 that is equal to 1.08 gram per second as mentioned here. Now you see another value is given that phosphorous settling rate that is 10 meter per year, so what is the v s that is our v s the v settling this is also v s basically v and v s same, so that v s is equal to 10 meter per year, so 10 divided by 365 into 24 into 3600 meter per second, so that is equal to 3.17 into 10 to the power minus 7 meter per second, so v s we are getting now. So steady state concentration of the total phosphorous would be how much just we have seen that is equal to c q c in plus s by q into A v or A v s, so here by this formula we are getting c is equal to this c means c out, so this equal to c out and c out and you know the concentration of phosphorous in the lake we consider same, so that is c out of c that is equal to q into c in plus s divided by q plus v s into A. So now we have already calculated this is this term is equal to 1.01 1.08, so divided by q is equal to how much we have seen that 18 meter cube per second, so q we have seen that 18 meter cube per second, so this v s into A, so what is our v s, v s we have calculated 3.17 into 10 to the power minus 19. And then what is A, A value is given surface area equal to 19 into 10 to the power 6 meter square it is already given, so we will put this value 19 into 10 to the power 6, so by this by solving this we are getting the c out is equal to 0.023 mg per liter. Now what is the permissible limit that permissible limit is equal to 0.010 or acceptable limit 0.010 mg per liter as per the problem statement part b, but here we are having 0.023, so it is more than that that means the water which is going out from this lake that requires treatment, so this treatment is needed. Now we have to calculate how much of this phosphorous should be removed by this treatment, so we can assume that after this we are getting phosphorous concentration under permissible limit that is 0.010 mg per liter. In that case what should be the value of S, addition of sulphur from the point source can be accommodated without any treatment that we are going to calculate. So if it is S just these expressions will be rearranging, so C equal to C means C out Q C in plus S divided by Q plus V S into A, so now from this expression if we rearrange, so S is equal to coming C into Q plus V S A minus Q into C in, so this is the expression we are getting, so that same expression here. So now we will be putting the C value is equal to this C value what which is going out, so that is equal to 0.010 mg per liter we are assuming. So in that case if we want to get this, so no further treatment is required, then you see the S value will be this into 18 plus this one into this minus 18 into C in that, so that is equal to 0.29 gram per second, but it is presently giving 0.9 gram per second and 0.29 gram per second can be settled here. So we do not need to get any further treatment for this. So what is the requirement for the treatment in the first case? This difference 0.9 minus 0.29, so that has to be removed, so that has to be removed by the treatment. Now what will be the percentage then? If we get the percentage it is converting to 67.8 percent, so almost three times, that is equal to 67.7 percent. So you can get 0.9 minus 0.29 divided by 0.9 into 100, so that we will get this 67.8 percent. So problem 3, CO2 is emitting from a 80 meter high industrial chimney of diameter 0.2 meter to the atmosphere that 1 bar pressure on a clear summer day. Wind speed at 10 meter height from the ground is 4 meter per second. Calculate the effective stack height. Assume B category stability, P value is 0.15. The stack gas velocity is 8 meter per second and the temperature of stack gas and ambient air at 80 meter height are 50 degree centigrade and 25 degree centigrade respectively. So these are the given condition. So we have to calculate the stack height. So how we can do that? So we know that del H is the plume rise that is equal to Vs by U into D 1.5 plus 2.68 into 10 to the minus 2 into P into Ts minus Ta by Ta into D. So that is called Holland's formula we have discussed in our previous classes. So here we have to put the values of this D stack diameter in meter already given. Then U value at 80 meter we have to calculate. It is given as 10 meter is given that is 4 meter per second. We have to calculate this U. Vs is given, stack flue gas velocity it is given and Ts, Ta are given, P is also given. So now we can get the value of del H. How? So del H equal to this expression. So we have to calculate the U value first. So U T is equal to T 1 into Z 2 by Z 1 to the power P. P value is given. So P value is given. We do not need to refer any graph or table. So we will be using this formula directly U T equal to T 1 Z 2 by Z 1 to the power P. So 80 meter and thus 10 meter height velocity is 4. So U T equal to 4 into 80 by 10 to the power 0.15 that is 5.464 meter per second. And then del H value we will put all others. So this is given 5.464 that is equal to U. Vs is also given that is 8 meter per second. And then this is equal to diameter 0.25 duct diameter and 1.5 this is formula 2.68 10 to the power minus 2 into 10 square because this P is atmospheric pressure it is given in bar. So this is bar pressure it is given. And we know that in this empirical relationship our pressure is kilopascal. So we have to put it this atmosphere 2 kilopascal 10 square. So this will be multiplied 323 minus 298 323 because the temperature one is 50 and 25. So 323 minus 298 by this and into diameter stack diameter. So again we are getting 0.568 meter. So effective stack height is equal to how much? So this is our plume rise and then we are having stack height 80 plus this rise in the plume so that is equal to 80.568 meter. Then problem number 4 a 915 megawatt power plant with an efficiency of 40% uses coal as a fuel source. The coal has 1% sulfur content and a calorific value of 30 megajoule per kg. The effective stack height is 520 meter. If neutral conditions prevail find out the maximum ground water concentrations of SO2 at 10 and 100 kilometer from the plant. Assume that u 10 that is wind speed at 10 meter height is equal to 4 meter per second sulfur molecular weight is 32 oxygen 16 exponent for velocity is equal to 0.16 and sigma y and sigma z or S y S z we have also discussed in the previous class. We have discussed in the previous class S y S z and this is our different distance if you consider the S y S z values are given. So now we have to calculate the concentration of the pollutants that is SO2. So we have to use the formula and now we will see how much SO2 is generated what is the rate of release of SO2 from the stack. So 915 megawatt power plant capacity so that is equal to 915 into 10 to the 6 joule per second so this 3294 into 10 to the 3 megajoule per hour. So this conversion joule per second to joule per hour so we have to multiply it 3600 and then coal required equal to how much. So this is our energy required for this but it is told that heating value is 30 megajoule per kg. So this is the kg of coal requirement but 40% usage is there so percent is efficiency is 0.4 so that that has to be divided by 0.4 again so we are getting 274500 that is equal to 274.5 ton per hour. We can calculate the concentration of SO2 by using this formula all terms are defined we have discussed in the previous class. So here we will put the value so C is the concentration of SO2 that is equal to emission rate gram per second so E we have calculated now 274.5 into 0.01 into 64 by 32. So what we have got here ton per hour so we have to convert it into gram per second so here we are getting this into this one so unit conversion so we had here ton per hour we are converting it into gram per second so these are the unit conversions and 1.5 kg per second we are getting. So we will be putting this value here as E and we need to calculate the value of U2 and 10 meter height velocity is given 4 and we have to calculate Z2 that is 520 meter height so this 520 by 10 to the power 0.16 it is given so 7.53 we are getting the U. So in this expression now U value we are getting SYSZ given so YSY all values are given now del H is also given so we will put these values here so see 10 0 at 10 kilometer distance what will the see this one so 1.5 into 10 to the power 3 so this one e to the power minus this in this case this Y is equal to 0 so this term this term will be 0 this will not be there so we will be having this one so only this term it is there the 520 H value say we are getting 140 for this SZ value in case of 10 kilometer and in case of 100 kilometer this value will be 520 by 450 other values are will be like this because at that time time this value SZ and SY value will change you see here these values are changed so we will be getting this expression here and by that we are getting this much or 18.08 microgram per meter cube. Now problem number 5 a coal fired power plant converts one-third of the coals energy into electrical energy the electrical output of this plant is 500 megawatt in this plant only one-third of the coals energy is converted to electrical energy and 90 percent of the rest energy of coal is rejected to river through using cooling water drawn from a nearby river and recycling it back to river the river has a upstream flow of 75 meter cube per second and temperature of 20 degree centigrade if the cooling water temperature increases by 5 degree centigrade what flow of cooling water stream is required what should be the temperature of river water just after receiving the heated cooling water so say river flow is there so here we are getting some water and it is in the power plant and again it is coming back so we have to calculate that what is the impact of the temperature of the river here. So now let us see the input power equal to how much that is the electrical output is 500 megawatt and one-third of the coal energy is converted so total amount of energy that is equal to 3 into 500 that is equal to 1500 megawatt that is available in the coal okay so then energy of loss through water equal to how much this 90 percent of the rest energy of coal is rejected so 90 percent of the coal so this into this of the rest amount means 1500 and this is electrical so rest is 1500 minus 500 so into 0.9 that is lost so that is 900 megawatt so this energy is lost. So flow rate of cooling water stream so what will be the flow rate ms theta that will be giving the equivalent heat of this so 900 into 10 to the power 6 megawatt so watt divided by this 4181 that is equal to joule per kilo per kg so that is this much of kg per second cooling water is required okay so that is equal to 43.01 meter cube per second and then increase in river temperature so again the say whatever heat generated all are coming into this and it is mixed with the total water flow of the river. So that 900 into 10 to the power 6 divided by 75 into 10 to the 3 that is 75 meter cube per second into 10 to the power 3 into 4148 that is equal to 2.87 degree centigrade so this is the temperature rise and initially it has 20 degree centigrade so now overall temperature rise here will be 20 will be 2.87 degree centigrade and final temperature will be 20 plus 2 that is 22.87 degree centigrade okay although the 5 degree centigrade temperature rise here with this stream which is coming here through this power plant okay power plant okay so this is our river stream so this is the numerical problems we wanted to solve. Now problem number 6 determine carboxyhemoglobin concentration as percentage saturation for carbon monoxide concentration of 1 to 0.00 ppm after 0.5 hour of exposure. Further determine the minimum time for 80% saturation so we know some empirical relationship to find out the concentration of carboxyhemoglobin in the blood so that is this percentage equal to beta into 1 minus e to the power minus gamma t into CO where CO is the concentration of the carbon monoxide in the air and beta is your 0.15% per ppm CO that is a constant and gamma is also a constant 0.402 per hour and t is the exposure time in hours. So now if we put this formula here so percentage CO Hb that is equal to beta that is equal to 0.15 into 1 minus e to the power minus gamma 0.402 into t equal to how much we have 0.5 hour of expression exposure 0.5 hour so that way we are getting so this is into we have CO CO is equal to 1 to 0.0 so by this we will get the value of 32.78 percentage. Now this is part 1 now further determine the minimum time for 80% saturation for part B so this 80% that means left hand side is 80% 80 and this equal to 0.15 into 1 minus e to the power minus 0.402 into t into 1 to 0.0 so this t value will come from that expression and by solving we will get t equal to 1.46 hour. Now problem number 7 find out the air to fuel ratio required for complete combustion of a liquid fuel average formula C7H13 in an internal combustion engine. So if we take the stoichiometric expression so C7H13 plus 10.25 O2 plus 78 by 21 into 10.25 N2 that is equal to 7 CO2 plus 6.5 H2O plus 38.07 N2 so this is a balanced equation. So now we have this the mass if we take the molecular weight of this so 97 gram so then here this is 10.25 into oxygen that is 328 gram and then nitrogen you know we will getting this much of 38.07 nitrogen that is 1066 gram. So air to fuel ratio how much we will get we will be getting air that means nitrogen plus oxygen nitrogen plus oxygen so 1066 plus 328 and divided by fuel equal to 97 so this equal to 14.37. So now we are able to solve all the numerical problems in this class. Thank you very much for your patience.