Serial : 01. PT_CE_A+B_Environmental Engineering_090718 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: E-mail: info@madeeasy.in Ph: 011-451461 CLASS TEST 018-19 CIVIL ENGINEERING Subject : Environmental Engineering Date of test : 09/07/018 Answer Key 1. (d 7. (a 1. (a 19. (a 5. (b. (b 8. (b 14. (d 0. (c 6. (d. (c 9. (c 15. (a 1. (a 7. (c 4. (a 10. (c 16. (b. (d 8. (a 5. (c 11. (d 17. (a. (c 9. (b 6. (b 1. (b 18. (c 4. (d 0. (a
CT-018 CE Environmental Engineering 7 Detailed Explanations 1. (d Trickling filter is an attached growth process i.e. process in which microogranisms responsible for treatment are attached to an inert packing. (e.g. rock, gravel, sand etc. Activated sludge process is a suspended growth process i.e. process in which microorganism metabolize the suspended and soluble organic matter. Oxidation ditch is a particular type of extended aeration process, where aeration tank is constructed in the ditch shape.the activity in the oxidation pond is a complex symbiosis of bacteria and algae, which stabilizes the wastes and reduces pathogens.. (b The dry mass of the solids is m 1000 V.S 1000 kg/m 0.001 m 0.01 0.01 kg Now all of solids are settled and liquid is decanted until the total volume is 500 ml. The new solids M fraction is, S 1000 V 0.01kg 0.0 1000 0.0005 4. (a Suppose specific weight of wood waste is x kg/m 60 x 16.67 kg/m 80 60 40 0 5 75 480 10 + + + 10 10 x 10 5. (c The tolerance limit for BOD 5 in marine environment disposal is 100 mg/l. Minimum efficiency needed, η min 150 100 1 100 100.% 150 7. (a SVI Settled sludge volume 1000 MLSS 50 1000 89.9 m /g 800 l 8. (b Overflow rate m /day/m m/day Percentage particle removal 1000 mm/s 0.55 mm/s 4 60 60 0.1 100 9.% 0.55
8 Civil Engineering 9. (c Let volume of raw sewage V Dilution ratio 00 V Loss of D.O. during 5 days incubation Initial D.O. FInal D.O. 9 5.5.5 mg/l BOD of sewage Loss of D.O. Dilution factor 00 00.5 V V.5 ml 10. (c Head loss per unit length is independent of sewer running full or half full condition. 11. (d (i Lime (CaO required for alkalinity (CaCO Molecular weight of CaO 40 + 16 56 Molecular weight of CaCO 40 + 1 + 16 100 56 CaO required for 150 mg/l alkalinity 150 84 mg/ l 100 6 84 10 kg 84kg l 6 10 l (ii Lime required for MgSO 4 Molecular weight of MgSO 4 4 + + 4 16 10 CaO required for 90 mg/l of MgSO 4 56 4kg 90 4mg/ l 6 10 10 l Total lime required 84 + 4 16 kg/10 6 l Total lime required to treat 10 6 litres of water 16 kg 1. (b Total water filtered in a day 1 (4 60 60 86400 m /day Total surface area of filter required 86400m /day 140m /day/m 617.14 m Area of one filters required 6 8 48 m Total number of filter 617.14 1.86 1 48 As three filters are out of service, number of filters working 1 10 Total surface area of ten filters 48 10 480 m New loading rate 86400 m /day 180 m /day/m 480 m
CT-018 CE Environmental Engineering 9 1. (a Year 190 1940 1950 1960 1970 Total Population 5,000 8,000 4,000 4,000 47,000 Increase in Population 000 6000 8000 5000 000 Average increase in population per decade, x 000 5500 4 Average incremental increase, Incremental increase +000 +000 000 000 y 000 666.67 ( + 55 1 P 00 P1970 + 5x+ y 47000 + 5 5500 + 15 666.67 84500 14. (d Flow rate of river, Q R 600 l/sec Concentration of sodium in river water, C R 17 mg/l Concentration of sodium in waste water channel, C W 50 mg/l Concentration of sodium in the mix water, C mix 50 mg/l Let, flow rate of waste waterq W Concentration of sodium after mixing of waste water with river is given by CQ R R + CWQW C mix QR + QW 17 600 + 50 Qw 50 600 + QW 50 (600 + Q W 1000 + 50 Q W 0000 + 50 Q W 1000 + 50 Q W 19800 00 Q W Q W 99 l/sec Dilution ratio Total discharge after mixing Discharge of waste water QR + QW 600 + 99 7.06 Q 99 W 15. (a Given: BOD 5 L 5 00 mg/l, t 5 days, k 0 0. per day BOD 5 BOD u ( 1 e k 0t k0t L 5 Lu ( 1 e
10 Civil Engineering 0. 5 00 Lu ( 1 e 00 L u 0. 5 1 e k T 0 k T 0 θ 9.67 mg/ l k 15 k 0 ( 1.047 15 0 (θ is generally taken as 1.047 k 15 0. (1.047 5 k 15 0.18 per day k15t L 8 Lu ( 1 e L 8 9.67 (1 e 0.18 8 4.97 mg/l 5 mg/l 16. (b The cloth area required ( 0m /sec 60s/min 100m /min 40m 5.0m/min 5.0m/min The surface area of one bag πdh π 0.4 8 10.05 m Total number of bag 40.87 4bags 10.05 17. (a Efficiency of filter, η 100 1+ 0.0044 Y V F where, Y Total BOD applied to filter unit V Volume in hectare meter 165 m m 0.165 ha.m F Recirculation factor F R 1+ I 0.1R 1+ I 1+ 1.5 F ( 1+ 0.1 1.5 1.89 100 8 Y 1+ 0.0044 1.89 0.165 Y 64.10 kg/day 18. (c Initial oxygen deficit saturation D.O. Initial oxygen content 9. 4. 5 mg/l
CT-018 CE Environmental Engineering 11 f Critical time, t c KR 0. K 0.1 D 1 D0 log 1 ( f 1 f KD ( f 1 L 1 5 log 1 ( 1 0.1( 1 100 t c.7875 days.788 days L f Critical deficit, D C [ 10] 100 10 [ ] KD tc D C 6.1 mg/l.788 0.1 19. (a (1 When 10 ml, 1 ml, and 0.1 ml dilution are considered, the combination of positive tubes is 5-4 -. MPN value for this combination from the table is 80. ( When 1 ml, 0.1 ml and 0.01 ml dilutions are considered, the combination of positive tubes is 4 - -. MPN value for this combination from the table is 9. This value must be multiplied by 10, since surface water volume used are 10 times lesser than the standard values. MPN value is 9 10 90. ( When 0.1 ml, 0.01 ml and 0.001 ml dilution are considered, the combination of positive tubes is - - 1. MPN value for this combination from the table is 17. This value must be multiplied by 100. Therefore, MPN is 17 100 1700. MPN is the maximum of above i.e. 1700. 0. (c Given, d D 0.6 0.6 1 α 1 cos α 0.074 D α 0.6 D d Now, q Q α sinα 60 π / α 60 5/ 0.67 q 0.67 0.09 0.06 m /sec
1 Civil Engineering 1. (a (i Al (SO 4 18HO + Ca(HCO A l(oh + CaSO 4 + 6CO 666 gm ( 16 gm (ii CaCO + HO + CO Ca(HCO 100 gm 16 gm From (i it is clear that 666 gm filter alum can neutralize ( 16 gm of alkalinity as Ca(HCO. Again from (ii it is clear that this ( 16 gm alkalinity as Ca(HCO is equivalent to ( 100 gm of alkalinity as CaCO. Total alkalinity of filter alum as CaCO, for water requiring 5 mg/l of filter alum 100 5 11.6 mg/l 666 Hence, total alkalinity requirement 0 10 6 11.6 mg/day 5. 10 6 mg/day Hence, total alkalinity requirement will be 5. 10 6 mg per day as CaCO.. (d Sewage produced 75000 litres/day 5 day BOD of sewage 185 mg/litres BOD of effluent 5 mg/litres BOD removed by pond (185 5 160 mg/litres Sewage solids removed per day 75000 160 117.6 kg It is given that organic loading 65 kg/ha/day Required area 117.6 65 1.81 ha 1.809 ha. (c Flow rate, Q o 575 m /hour Influent BOD, S o 165 mg/litre Effluent BOD, S 1 mg/litre Hydraulic retention time 6 hours 1 4 day Mean-cell resistance time (θ c 88 hours 1 days Volume, V 6000 m Mixed liquor suspended solids, Aeration tank Secondary clarifier Effluent X 500 mg/l F M QS o o 575 165 4 VX 6000 500 Solids wasted
CT-018 CE Environmental Engineering 1 0.1518 kg biomass per day per kg 0.15 kg biomass per day per kg Mean cell residence time, θ c Q w X u VX Q X w C u VX θ 6000 500 10 1 150 kg/day 4. (d Flow of waste water stream, Q w m /sec Ultimate BOD of waste water stream, Y w 95 mg/l 95 gm/m Flow of river, Q R 16 m /sec Ultimate BOD of river, Y R 4 mg/l 4 gm/m BOD of mixture, Y 0 QY w w + QY R R 95+ 16 4 Qw + QR + 16 18.68 gm/m k D 0.44 k 0.44 0.5 0.1098 /day Area of river 60 m Combined flow of river (downstream + 16 19 m /sec 5. (b Stream velocity Discharge Area 19 0.167 m / sec 60 15 1000 Time taken, t 476 sec 0.548 days 0.167 Y t 0 1 10 kd Y t 18.68 [1 10 0.1098 t ] 18.68 [1 10 0.1098 0.548 ] 18.68 0.195.8 mg/l µ g ppm gm/mole mass 10 m L/mole Gram molecular mass of CO 1 + 16 8 gm/mole At 0 C and 1 atm of pressure (760 mm Hg, the volume of the gas is.4 l/mole So, 15µ g 10 m x 1 10 6 1 ppm 6 8g 10 l 10 g x mole m.4l mole 15.4 10 9 8 10 6. (d Carbonate hardness 10 gm equivalent (If Non-Carbonate Hardness is present, sodium alkalinity i.e. NaHCO will be absent. µ g
14 Civil Engineering g 10 50 as CaCO l 150 mg/l as CaCO Non-Carbonate hardness Total hardness Carbonate hardness Total hardness 6 50 mg/l as CaCO [Total hardness is due to Ca + and Mg + only] 00 mg/l or CaCO NCH 00 150 150 mg/l as CaCO 7. (c 100 1 0.60 t Relative Stability, S ( 7 100 1 0.60 t 7 days 96.06 ( 7 t 7 8. (a Velocity gradient, G P µ V Power input per unit volume, P µg V 1 10 (500 750 W Now, power input per unit volume P V 50 W 750 9. (b Flow rate, Q 0 750 m /hour Influent BOD, S 0 10 mg/l Effluent BOD, S 8 mg/l Hydraulic retention time 6 hours 1 day 4 Mean-cell residence time, (θ c 00 00 hours days 1.5days 4 MLSS (X 500 mg/l Volume 5000 m F M QS 0 0 VX 750 10 4 0.178 day 1 5000 500
CT-018 CE Environmental Engineering 15 0. (a Eckenfelder equation for computing the BOD removed by the filter is given as Y t Y 0 [ ] n kd / Q e L...(i where, Y 0 BOD 5 of the influent entering the filter (in mg/l Y t BOD 5 of the effluent getting out of the filter (in mg/l k rate constant per day D depth of filter (in m n depends on flow characteristics and is an empirical value. Q L Hydraulic loading rate per unit area of filter in m /m /day Q A Given: Y 0 150 mg/l; D 1.5 m; k 1.89 d 1 ; n 0.6 Q L 1900 π 0 4 Substituting these values in equation (i, we get.688m /m /day Y t 150 1.89 1.5.688 [ ] 0.6 e Y t 0.09 150 Y t 150 0.09 1.5 mg/l