Lecture 60: Tutorial 12

Hello everyone, now we will have a tutorial session and in this class we will solve some numerical problems based on our discussion made in the last four classes. Problem number one, the statement is the ambient concentration of CO2 in a city is 450 ppm volume by volume on a particular day express its concentration as mg per liter ppm in mass by volume of air assume STP. So this is a very simple question which is associated with the change of unit and the concept of ppm in one case volume by volume and in other case mass by volume so that conversion we have to do. So let us take a basis that is 1 liter air so what will be the carbon dioxide it is 450 ppm volume by volume. So 450 into 10 to the power minus 6 liter of CO2 because the definition of ppm 10 to the minus 6 part so 450 into 10 to the minus 6 liter of CO2 in 1 liter of air by the definition of ppm volume by volume. Now moles of CO2 in 450 into 10 to the minus 6 liter CO2 can be calculated by using ideal gas law PV equal to nRT. So PV equal to nRT so we will be applying this now pressure is 1 atmosphere volume is this much n we have to calculate and R value we know 0.082 liter atmosphere per mole per K into T so 273.15 K so that from this so from this expression by solving the value of n we get n equal to 20.09 into 10 to the minus 6 moles. So now the moles of CO2 we are getting in this 1 liter air that is 450 into 10 to the minus 6 liter of CO2. So mass of CO2 in 1 liter of air that will be this moles into the molecular weight of the CO2 that is 44 that is equal to 20.09 into 10 to the minus 6 into 44 gram that is equal to 883.96 microgram. So that if we take this is gram means this is microgram so this much microgram so if we want to convert it into mg so that will be 0.884 mg. So this is the mass of carbon dioxide present in 1 liter of air. So ppm in mass by volume will be 0.884 mg per liter. So now we are able to solve the problem. Next problem, problem number 2 the statement is carbon monoxide CO affects the oxygen carrying capacity of lungs. Exposure to 50 ppm CO for 90 minute at standard temperature and pressure has been found to impair one's ability to discriminate stopping distance. Thus motorist in heavily polluted areas may be more prone to accidents. Are motorist at a greater risk to accidents if the CO concentration is 65 microgram per meter cube. So this is also a similar type of question where the CO concentration is given in different units. So what is the exposure concentration and at the concentration at which the impact will be visible so those two concentrations are given in different unit. So we have to convert these units into a single one so not in different units so that we have to do. So again we have to take a basis so 1 meter cube air so assume a temperature of 298 K so 25 degree centigrade we are assuming ambient temperature. So pv equal to nRT so this formula will use to calculate the n value or then n by n by v is equal to p by RT n by v equal to p by RT we are getting here. So now which we have 65 microgram per meter cube so 65 microgram per meter cube into 1 gram per 10 to the 6 microgram we are converting it into gram into 1 mole CO by 30 gram CO molecular weight of CO. So we are getting the moles basically n value from this concentration so that is equal to 2.2 into 10 to the minus 6 moles CO per meter cube air. So this we are getting the n value as 2.2 into 10 to the minus 6 and we have one expression n by v equal to p by RT. So then we will be putting these values here n 2.2 into 10 to the minus 6 then we are into RT by p so that is equal to v. So volume is equal to this n into RT by p so that we are using n value we have got from this information then RT by p value will be putting so this is our n value and this is our R this is your 8.025 into 10 to the minus 5 meter cube atmosphere per mole per K so this is and temperature temperature is 298 K and pressure is 1 atmosphere so this value if we put so here what we are getting this is equal to volume so this is equal to volume so that is equal to 5.38 into 10 to the minus 8 meter cube CO per meter cube of air. So now this meter cube if we take that is equal to 10 to the minus 6 into 10 to the minus 2 so 10 to the minus 6 is coming as ppm and due to minus 2 this decimal point will be shifted towards the left so 0.0538 ppm CO ok. So this is we are getting now so what we have 50 ppm CO is the is the limiting concentration at which the impact will be visible but here the concentration is 0.0538 which is less than the 50 ppm of the CO concentration thus motorist are not at a greater risk. So problem number two is solved now we will see problem number three the statement is a refractory combustion unit lined with no water valve and no heat recovery is burning refuse derived fuel consisting of 80% organics 12% water and 8% inorganics or inert at a rate of 19,920 kilogram per day. Determine the temperature of the stack gases assuming the following heating value of the R of the RDF heating value of the RDF is equal to 19,000 kilojoule per kg on dry as free basis stack gas flow rate is equal to 8200 kg per hour 7% of the heat input is lost due to radiation 15% of the RDF remains uncombusted in the ash then as exists the gas exits the combustion chamber at a temperature of 750 degree centigrade specific heat of ash is equal to 0.837 kilojoule per kg per degree centigrade specific heat of stack gas is 1.0 kg kilojoule per kg per degree centigrade latent heat of vaporization is 2575 kilojoule per kg. So these information are provided we have to determine the temperature of the stack gases. Now this is a problem related to incineration of the waste materials so here we will see this is a combustion chamber and that furnace but in this furnace the boiler part is not included that means the it is a refractory furnace so water tubes are not installed in the wall of this furnace it is given. So the there is no steam formation basically so heat which is generated that will be going through the flue gas so no steam will be generated in this furnace. So here what is happening the RDF is being combusted so what is the rate of RDF that is 19,920 kg per day. So if we convert it into per hour so that will be 19920 by 24 that is 830 kg per hour so this is the feed rate of RDF into this furnace. Now then we will see due to the combustion of this RDF heat will be released and some amount of heat will be lost and rest of the heat will be available in the so this is the concept so simply energy balance we have to do. So in that way we will see from the RDF during the combustion what amount of energy is released. So here we will get the mass and percentage of organics into heating value of RDF dry ash free basis because it is given that 80% organics 12% water 8% inert. So inert will not give any heat 12% water will not give any heat and 80% organics basically will be giving us heat. So but here the heating value is given with respect to dry and ash free basis so 80% organic which is present that will be the actual material mass or RDF mass which will give us the energy. So mass of the RDF into percentage organics into heating value of RDF dry ash free basis so that will be our heat generated through the combustion. So here 830 kg we are getting into 80% so 0.8 into 1900 this kilo joule per kg that was the heating value of this RDF on dry and ash free basis so that is why this is so by multiplication of these three terms we are getting by solving this 12616000 that is kilo joule per hour. So this is our heat energy released and here some ash is generated so ash will be having some energy and radiation loss will also be there and for the vaporization of water that is 12% that heat will also be taken from this liberated heat. So then these three value we will calculate in case of ash say total ash will be mass of inert so which inert was available plus unburned organic because moisture will goes up so remaining will be unburned organics plus this inert. So 830 is the total RDF rate so 830 into 8% inert 0.08 this is inert and unburned organic that is 830 into 0.8 is the organic content and 15% of it unburned so 0.15 again it is multiplied. So this part is the ash material and this ash is having 750 degree centigrade, centigrade temperature so this into multiplied by CP so MCPdT so that is the heat taken up by this ash so that is CP is equal to 0.837 kilo joule per kg per degree centigrade. So by this by solving this expression we are getting this amount of heat will be associated with the ash and that is lost basically and during the radiation process from the combustion into percentage of radiation so it is given that 7% heat is radiated so total heat which is combust which is generated through the combustion 7% of it is lost so this into 0.07 so that is this one this 883120 kilo joule per hour so this much is the loss due to radiation and now we will see the heat which will be required for the vaporization of this water so that we are getting 830 kg RDF so 12% of it is moisture so multiplied by 0.12 and 2575 that is the latent heat. So by multiplying these three terms we are getting 256470 kilo joule per hour so this one this one and this one these three are our loss and this is our generation heat generation. So which will be available with this stack gas that is equal to this minus this minus this minus this so we did 12616000-104206-883120-256470 and as a result we are getting 11372204 kilo joule per hour so this is the heat which is associated with this stack gas stack gas. Now we have been asked to calculate the temperature of the stack gas so M C p dT we will apply here so temperature of the stack gas is given so we will be putting those formula the heat to stack gas is equal to mass into temperature into specific heat of the stack gas so that is 8200 kg per hour so it is given so this is the stack production and T is also given not given T we have to calculate so that is T we have to calculate and we have specific heat of the stack gas it is given 1 kilo joule per kg per degree centigrade so by equating this term with this the heat available at the stack gas which is calculated here so we will be getting this one is equal to 11372204 kilo joule per hour or by solving this we will be getting the value of T so that is equal to 1386.85 degree centigrade. So now we are able to solve this problem next problem is problem number 4 the statement is calculate the air quality index corresponding to a region with 24 hour PM 2.5 concentration of 41 microgram per meter cube 24 hour PM 10 concentration of 58 microgram per meter cube and 24 hour SO 2 concentration of 9.5 microgram per meter cube and here AQI category pollutants and health breakpoints are given so this is a problem related to the determination of ambient air quality index so we have already discussed in our previous classes for Indian methods how to calculate this AQI so we have also discussed that at least 3 quality parameter has to be considered here 3 are there and out of that one must be PM 2.5 or PM 10 so both PM 2.5 and PM 10 are available here so we can calculate the AQI and for the calculation of that we need some parameters that is IHI, ILO and then BHI, BLO and CP, CP is given here for different pollutants so that is this is for PM 2.5 and this is for PM 10 is also given and for SO 2 also given. So we will be using this and for this we will be calculating the corresponding BHI and BLO and IHI and ILO from this table and then we will get the individual index and we will find out the maximum value of that individual index that will be our overall value of the ambient air quality index. So now you see we will be using this one so individual indices that is index is equal to IHI minus ILO by BHI minus BLO into CP minus BLO plus ILO. So BHI breakpoint concentration greater or equal to given concentration, BLO breakpoint concentration smaller or equal to given concentration, IHI AQI value corresponding to BHI, ILO is AQI value corresponding to BLO and we know that when we are considering ILO so for that for the calculation of ILO we will subtract 1 from the ILO value if LO is greater than 50. So this is already discussed in our previous class the same process we will apply. So in this case for 24 hour PM 2.5 concentration what is our CP value, CP value is 41 microgram per meter cube. So what will be our IHI and ILO let us see so 41 means in this category it is coming so this is IHI and this is ILO but ILO here it is greater than 50 so it is 51 so we will consider it as 50, BLO is 31 and BHI is 60 for this case and CP is 41 and LIO which we will consider that is equal to 51 minus 1 that is equal to 50. So in this expression now we will put the value so for this IP for the PM 2.5 that is IHI 100 minus ILO 50 although it is 51 so we are considering 50 and then BH and 60 and BLO 31 and this is 41 minus 31 and plus 50 that is equal to ILO 51 so value we have considered as 50 so then we are getting 67.24. Similarly for 24 hour PM 10 concentration of 58 microgram per meter cube so in that case 58 means it will also under this category here so BLO 51 BHI 100 here this I category so again ILO 50 and IHI 100 and 58 hour CP so then this value we will put it here in this expression 100 minus 50 by 100 minus 51 into 58 minus 51. So that plus 50 that is equal to 57.14 similarly for SO 2 for the concentration of 9.5 microgram per meter cube so this will fall under this the break points will be so CP will be here so BLO will be 0 and BHI will be 40 and this is IHI will be 50 and ILO will be 0 so this formula will be using now so 50 minus 0 by 40 minus 0 into 90.5 minus 0 plus 0 so that is equal to 11.87 so these are the three different individual quality index we are getting. So for determining overall ambient air quality index we have to consider the highest value so in that case 67.24 so AQI is maximum of IP in this case 67 around so this is a satisfactory one because this falls under this category 61 so this is a under satisfactory one. Now problem number 5 the statement is calculate the following for the sterilization of biomedical waste needles scalpels blades etc using an autoclave. Consider sterilization hold time of 20 minutes at constant 125 degree centigrade what will be the equivalent exposure time that is FO value what does the result signify. Second part if the equivalent exposure time is 30 minutes at constant 110 degree centigrade how long will it take to complete the sterilization what does the result signify. Use the expression FO equal to del T into sum of 10 to the power T minus 121.1 by Z where del T is the time interval between two temperature readings and this is nothing but the duration basically and T is the temperature at time T of the product under sterilization that is sterilization temperature and Z is the temperature coefficient assumed as 10 degree centigrade. So we have discussed this in the previous class as well so here we have to calculate the FO for two different conditions so we will do that. So for the first case the del T is equal to 20 minute and T is equal to 125 degree centigrade so we will put this value Z value is equal to 10 so FO value is coming 20 into 10 to the power 125 minus 121.1 divided by 10 so that is equal to 49.1 minute. So equivalent exposure time is 49.1 minute so what does it signify in place of using 20 minutes at 125 degree centigrade the sterilization can be done at 121 degree centigrade using 49.1 minute of sterilization. Hence 20 minutes of sterilization at 125 degree centigrade is lethal equivalent to 49.1 minute of sterilization at 121.1 degree centigrade as FO value is the equivalent exposure time at 121.1 degree centigrade to that of the actual exposure time at a variable temperature with a temperature coefficient of destructions equal to 10 degree centigrade. So we have explained this in the previous class. Now for the part 2 part b given FO equal to 30 minute T is equal to 110 degree centigrade now we have to calculate the value of del T. So here 110 so we will be getting for this case FO is equal to 386 minute by using the same formula so time required to complete the sterilization is 386 minute. So hence when the temperature of the system under sterilization is maintained at a constant 110 degree centigrade the time of 386 minute would be required to achieve the lethal effect of 121.1 degree centigrade at 30 minutes. So the sterilization has to be made for 386 minute to get the same impact. So it is very clear from this data that if the sterilization temperature is lower than 121 degree centigrade so very prolonged heating is required so that is not feasible so normally the temperature is maintained at 121 degree centigrade and more than that. Now problem number 6 the statement is typical biomedical waste has a moisture content of around 30% and roughly 8% of the dry mass of BMW is hydrogen using the high heating value of BMW 13902 kilo joule per kg estimate the latent heat of vaporized water and the lower heating value. So let in 1 kg of waste there will be 0.3 kg of moisture 30% already given leaving a dry weight of 0.7 kg in that dry mass there will be hydrogen of how much 8% of the dry mass. So 0.7 into 0.08 so 0.056 kg so if we apply the formula for the latent heat of vaporization so QL is equal to 2440 into W plus 9H this formula is actually used. So in that case 2440 into W equal to how much that is 0.3 kg moisture and hydrogen we have to calculate so 9 into hydrogen we have calculated 0.056 kg so we will be putting this value here and by multiplying we are getting 1961.76 kilo joule. So this is the first part and then we have to calculate the lower heating value so lower heating value is equal to HHV minus QL so HHV is given that is 13902 so that minus this QL so that is equal to 11940.3 kilo joule per kg. So applying the conversion factor so this 2.326 kilo joule per kg equal to 1 Btu per pound so the LHV equal to 5133.4 Btu per lb so this is the solution of this problem. Now we are moving to problem number 7 the statement is the composition of flue gas dry basis from an incinerator of a biomedical management facility is as follows 12.3% CO2 5.1% O2 and the rest is nitrogen from this data calculate the weight ratio of hydrogen in the biomedical waste the percent carbon and hydrogen in the dry waste the percent excess air used the moles of exhaust gas discharged from the unit per kilogram dry waste burned. So this is a problem of incineration so we will be solving it so here we take say basis that is 100 mole of dry flue gas so carbon dioxide will be 12.3% and oxygen will be 5.1% so inert will be how much 100-12.3-5.1 so that will be 82.6% in the exit gas. So oxygen available with supplied air will be how much so this much 82.6% if we take 100 moles of dry flue gas so 82.6 moles of nitrogen is present in the flue gas so what was the oxygen available with the supplied air so considering 21% oxygen and 79% nitrogen so 21 by 79 into 82.6 moles so that is equal to 22 moles then oxygen consumed during combustion is equal to how much so this is our additional oxygen which is available as excess so 22-5.1 so that is 16.9 moles that was converted during the combustion and from the flue gas composition it is evident that the waste does not contain sulphur chloride and nitrogen thus 12.3 mole O2 is consumed by carbon during combustion because the flue gas contains 12.3% of that is 12.3 mole of that CO2 and for the production of 1 mole of CO2 1 mole of oxygen is required so that 12.3 mole of oxygen is consumed by same moles of carbon during the combustion. So oxygen consumed by hydrogen is equal to how much then so total this one used in the process so this minus this one will be there so 4.6 mole or we can get is 22-12.3-5.1 that is 4.6 mole therefore hydrogen present in waste is equal to how much 4.6 into 2 that is 9.2 moles then weight of hydrogen in waste is equal to 9.2 into 2 that is 18.4 gram weight of carbon in waste that is 12 into 12.3 that is equal to 146. gram so what will be the molecular weight of the waste it is containing carbon and hydrogen so 18.4 plus 147.6 that is equal to 166 gram so H 2 C ratio in the waste will be 18.4 by 147.6 is equal to 0.125 and percentage CO in dry waste that is equal to this 147.6 divided by 147.6 plus 18.4 into 100 so this equal to 88.92 percent and percentage hydrogen in dry waste that is equal to 100 into 18.4 divided by 147.6 plus 18.4 that is equal to 11.08 percent excess air applied equal to how much we have 5.1 moles now and total moles which was required for the stoichiometric combustion that is 22-5.1 so that excess air is 30.18 percent and H 2 O produced during combustion equal to 2 into 4.6 H 2 O this is the reaction 2 moles of hydrogen giving 2 moles of H 2 O so that will be 2 into 4.6 equal to 9.2 moles and moles of exhaust gas per kg of waste is equal to so this is hydrogen produced so we are taken on dry basis so moisture we have ignored so that will also be available in the flue gas so exhaust gas so that is 100 plus 9.2 divided by 166 so that is equal to 0.658 moles per gram that is equal to 8 that is equal to 658 moles per kg dry biomedical wastes now we have solved all the numerical problems we decided to discuss in this class and with this we are at the end of the course and I hope you will enjoy the course you have enjoyed the course and thank you for taking this course and best wishes for those who will opt for the examination and certificates thank you once again for taking this course thank you very