Serial : IG1_CE_A_Environmental Engineering_100818 CLASS TEST (GATE 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 : 10/08/018 Answer Key 1. (a 7. (a 1. (a 19. (c 5. (b. (b 8. (b 14. (d 0. (c 6. (d. (c 9. (c 15. (a 1. (a 7. (d 4. (a 10. (c 16. (b. (d 8. (c 5. (c 11. (d 17. (a. (c 9. (b 6. (c 1. (b 18. (c 4. (d 0. (a
CT-018 CE Environmental Engineering 7 Detailed Explanations 1. (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. (b 1 gm of aluminium produces 0.4 gm of Al(OH 16 mg/l of aluminium will produce Al(OH 0.4 16.84 mg/l Dry sludge (suspended solids removed 45 10 5 mg/l Total dry sludge.84 + 5 8.84 mg/l 4. (a 750 Specific weight of MSW Volume of food, V 1 Volume of plastics, V Volume of dirt, V Volume of wood waste, V 4 where x is specific weight of wood waste Total Volume Total weight Volume 5 80 10 m 75 60 10 m 480 40 10 m 0 10 m x 0 10 1.1 10 + x ( 80 + 60 + 40 + 0 10 156.6 0 1.1 10 + 10 x x 16.98 17 kg/m
8 Civil Engineering 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 6. (c 100 1 0.60 t Relative Stability, S ( 7 100 1 0.60 t 7 days 96.06 ( 7 t 7 7. (a SVI Settled sludge volume 1000 50 1000 89.9 m /g MLSS 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 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
CT-018 CE Environmental Engineering 9 (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 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
10 Civil Engineering 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 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
CT-018 CE Environmental Engineering 11 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 f KR 0. K 0.1 D Critical time, t c 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.788 0.1 D C 6.1 mg/l 19. (c Sewage flow, Q 0. m /s 0. 4 60 60 7648 m /d Suspended solid in the influent 50 mg/l Suspended solids in the effluent (1 0.6 50 0.4 50 100 mg/l 0.1 kg/m Total suspended solids per day 0.100 7648 764.8 kg/m
1 Civil Engineering So, sludge production per day 764.8 96.5 (100 96.5 1150 764.8 96.5 6686.5 kg/d 66.9 tonnes/d.5 1150 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. (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
CT-018 CE Environmental Engineering 1 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, X 500 mg/l F M QS o o 575 165 4 VX 6000 500 Aeration tank 0.1518 kg biomass per day per kg 0.15 kg biomass per day per kg Secondary clarifier Effluent Solids wasted 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 Stream velocity Time taken, t Y t Discharge Area 19 0.167 m / sec 60 15 1000 476 sec 0.548 days 0.167 0 1 10 kd Y t
14 Civil Engineering 18.68 [1 10 0.1098 t ] 18.68 [1 10 0.1098 0.548 ] 18.68 0.195.8 mg/l 5. (b Particles with velocity above or equal to SOR will be completely removed and those with settling velocity V below SOR are removed in proportion to, where V is the settling velocity. SOR Thus overall removal 100 + (0.5 100 + 0.1 100 + 0.05 100 1 165 mg/l 6. (d Carbonate hardness 10 gm equivalent (If Non-Carbonate Hardness is present, sodium alkalinity i.e. NaHCO will be absent. 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. (d Dry sludge content produced 0.07 40000 880 kg/day Now 97% moisture content means that kg of dry sludge will produce 100 kg of wet sludge 880 kg of dry sludge will produce 8. (c Volume of wet sludge 100 880 96000kg/day Mass of sludge 96000 94.1m /day Density of sludge 1.0 1000 If capacity of digester is V then % of volume V gets filled with fresh sludge. 100 V 94.1 V 17. m Increase in population per year 64500 4500 100 5 Additional discharge required to reach design capacity 7500 500 00 m /d 500 Present average water consumption per head per day 4500 0.154 m Increase in water consumption each year 0.154 100 184.8 m /day Number of years required to reach design capacity 00 11.905 1 years 184.8
CT-018 CE Environmental Engineering 15 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 5000 500 0.178 day 1 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 1900 Q L.688m /m /day π 0 4 Substituting these values in equation (i, we get Y t 150 1.89 1.5.688 [ ] 0.6 Y t 0.09 150 Y t 150 0.09 1.5 mg/l e