cedrt_05-07-2015_enviro.pdf

1.1.1.1.1. (c)

2.2.2.2.2. (b)

3.3.3.3.3. (a)

4.4.4.4.4. (b)

5.5.5.5.5. (d)

6.6.6.6.6. (b)

13.13.13.13.13. (d)

14.14.14.14.14. (d)

15.15.15.15.15. (c)

16.16.16.16.16. (c)

17.17.17.17.17. (b)

18.18.18.18.18. (a)

19.19.19.19.19. (d)

20.20.20.20.20. (a)

21.21.21.21.21. (d)

22.22.22.22.22. (b)

23.23.23.23.23. (a)

24.24.24.24.24. (a)

25.25.25.25.25. (a)

26.26.26.26.26. (b)

27.27.27.27.27. (c)

28.28.28.28.28. (d)

29.29.29.29.29. (b)

30.30.30.30.30. (a)

7.7.7.7.7. (c)

8.8.8.8.8. (c)

9.9.9.9.9. (a)

10.10.10.10.10. (c)

11.11.11.11.11. (c)

12.12.12.12.12. (c)

CE Environmental EngineeringDate : 05/07/2015

Civil Engineering

CLASS TEST - 2015

Serial No. : CEDRT_050715_Environmental

ANSWERS

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7 Environmental Engineering www.madeeasy.in | Copyright :

CTCE15 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata

EXPLANATIONS

1.1.1.1.1. (c)(c)(c)(c)(c)Peak demand of water = 1.5 Averagehourly consumption of maximum day

= 1.5

= 1.5

= 2.7 Annual average hourly demand.

2.2.2.2.2. (b)(b)(b)(b)(b)Amount of chlorine used

= 15 kg/dayQuantity of water to be treated

= 25 103 m3/dayResidual chlorine after 10 min contact

= 0.2 mg/L

=

day

= 0.2 25 kg/day= 5 kg/day

Chlorine demand = 15 5 = 10 kg/day4.4.4.4.4. (b)(b)(b)(b)(b)

Jackson turbiditimeter can be used tomeasure turbidities in range between 25-1000 mg/litre. So such turbiditimeter islimited in application to measure theturbidity of raw water of natural source only.For treated supplies which has turbidity lessthan 10 mg/litre, Baylis turbiditimeter isused.

5.5.5.5.5. (d)(d)(d)(d)(d)Average quantity of water required

= 200000 150= 30 106 litres per day= 30 MLD

Now, maximum demand= 1.8 Avg. annual demand

= 1.8 30 MLD = 54 MLD

=

= 0.625 m3/secNow, pumps are working for 12 hours a day. Maximum draft required

=

= 1.25 cumecs

6.6.6.6.6. (b)(b)(b)(b)(b)K0 (intrinsic permeability) = k/g

=

= 3.96 105 cm2

7.7.7.7.7. (c)(c)(c)(c)(c)

Vs =

B.L =

!

= "

= 200 m2

10.10.10.10.10. (c)(c)(c)(c)(c)Population equivalent

=

= 2025

13.13.13.13.13. (d)(d)(d)(d)(d)Total BOD/day = 0.07 50,000

= 3500 kg/dayTotal flow of sewage/day

= 140 50,000= 7 106 litre



BOD of sewage =

"

= 500 mg/l

Flow of sewage =

"

= 0.081 m3/secLet BOD of treated sewage = CS mg/l

5 = # # $ $# $

% %

++

5 = # %

+ +

CS = 8.21 mg/l Percentage treatment required

=

= 98.4%

14.14.14.14.14. (d)(d)(d)(d)(d)K30C = K20C [1.047]

T20

= 0.1 [1.047]10

= 0.158BOD5 = L[1 10

kD t] 110 = L[1 100.158 5]

L = 131.3 mg/l BOD5 = L[1 10kD t]

= 131.3 [1 100.1 5]= 89.8 mg/l

16.16.16.16.16. (c)(c)(c)(c)(c)Pills:Pills:Pills:Pills:Pills: They are nothing but a series of lightfloating wooden or hollow rubber balls.When such a ball is rolled into the sewer, itgets struck up at the place where heavydeposition has taken place. It, thusconstricts the passage thereby raising thevelocity of flow in its neighbouring area andthus scouring out the deposited silt.Drop Manholes:Drop Manholes:Drop Manholes:Drop Manholes:Drop Manholes: The manhole, in which avertical pipe is used is called a dropmanhole whereas, the one using an inclinedpipe is called a ramp.Clean Outs:Clean Outs:Clean Outs:Clean Outs:Clean Outs: A clean out is an inclined pipeextending from the ground and connectedto the under ground sewer. A cleanout isused for cleaning sewer pipes.Inverted Siphon:Inverted Siphon:Inverted Siphon:Inverted Siphon:Inverted Siphon: It is a sewer sectionconstructed lower than the adjacent sewersections, and it runs full under gravity withpressure greater than the atmosphere.

17. (b)17. (b)17. (b)17. (b)17. (b)

Molarity of solution A = 10pH = & '

Concentration of H+ for solution A in 800 ml = " &

=

Similarly, concentration of H+ for solution B in 700 ml = "

" &

=

Concentration of H+ in 1.5 l of mixture = " " & "" & + = Concentration of H+ in 1 l of mixture = " & & =

pH of mixture = log 3.412 107.6

= 7.6 0.533 = 7.07



18.18.18.18.18. (a)(a)(a)(a)(a)Quantity of copperas required per litre water= 8 mg Quantity of copperas required per 10million litre of water

= 8 10 106 mg= 80 kg

Now, CaO + H2O Ca(OH)2FeSO4 7H2O + Ca(OH)2 Fe(OH)2+ CaSO4 + 7H2O 1 mole of copperas requires 1 mole oflime 278 mg copperas will require 56 mg ofquick lime 80 kg copperas needs

=

"

of quick lime= 16.12 kg of quick lime

19.19.19.19.19. (d)(d)(d)(d)(d)

(

=

#

!)

0.22 =

!

V = 9333.33 m3

Hydraulic retention time

=!

*&

=

8 hours

20.20.20.20.20. (a)(a)(a)(a)(a)

Power (kW) =+,

Here total head,H = 18 (W.T. depth) + 20 m (draw down)+ 2 m(Lift above ground) + 5% 50 m(Friction losses)

= 18 + 20 + 2 + 2.5 = 42.5Q = 360 l.p.m

=

-

=

= = 0.65w = 9.81 kN/m3

Power =

= 3.85 kW

21. (d)21. (d)21. (d)21. (d)21. (d)NaCl do not require lime or soda for its removal.Removal of Mg (HCORemoval of Mg (HCORemoval of Mg (HCORemoval of Mg (HCORemoval of Mg (HCO33333)))))22222

.,%/ 0 % ./,0+

%%/ ./,0 , / + +1 mole of .,%/ 0 requires 2 moles of % ./,0

% ./,0 required =

&-+*'&1% ./,0%&-''&&1.,%/ 0

&-+*'&1.,%/ 0

="

= lRemoval of CaSORemoval of CaSORemoval of CaSORemoval of CaSORemoval of CaSO44444

%#/ 2 %/+ %%/ 2 #/ +



2 %/ required =

&-+*'&12 %/%&-''&&1%#/

&-+*'&1%#/

=

= lRemoval of MgClRemoval of MgClRemoval of MgClRemoval of MgClRemoval of MgCl22222

( ) % % /,+ ( )

/, %% +

%% 2 %/+ 2% %%/+

Ca (OH)2 required = ( )

&-+*'&1% /,%&-''&&1%

&-+*'&1%

="

= l

2 %/ required =

&-+*'&12 %/%&-''&&1%

&-+*'&1%

=

""

= lTotal lime required = 152.06 + 116.84 = 268.9 mg/l

Total soda required = 93.53 + 167.37 = 260.9 mg/lLime required per day = 268.9 100000 mg/day

= 26.89 kg/daySoda required per day = 26.09 kg/day

22.22.22.22.22. (b)(b)(b)(b)(b)

Now, V = $ #

self cleansing velocity = 0.8 m/s

V = $ #

0.8 =

#

S =

say 1 in 220



23. (a)23. (a)23. (a)23. (a)23. (a)(1)(1)(1)(1)(1) When 10 ml, 1 ml and 0.1 ml dilutions

are considered, the combination ofpositive tubes is 5 4 4.MPN value for this combination fromthe table is 350.

(2)(2)(2)(2)(2) When 1 ml, 0.1 ml and 0.01 mldilutions are considered, thecombination of positive tubes is4 4 3.MPN for this combination value fromthe table is 54.

This value must be multiplied by 10,since surface water volume used are10 times lesser then the standardvalues.MPN value is 540.

(3)(3)(3)(3)(3) When 0.1 ml, 0.01 ml and 0.001 mldilutions are considered, thecombination of positive tubes is 4 3 2.MPN value for this combination fromthe table is 39.This value must be multiplied by 100,therefore, MPN value is 3900.

24.24.24.24.24.(a)(a)(a)(a)(a)

Leq =

& .0 '=

=

.0 ' =

.0 .0 .0

+ + = 1.053 107 + 0.842 106 + 0.52632 109

= 106 [10.53 + 0.84 + 526.32] = 537.69 106

Leq = 10. log10 (537.69 106) = 87.3 dB25. (a)25. (a)25. (a)25. (a)25. (a)

When city is expanding at faster or compounding rate, then we predict the population using geometricincrease method.

% increase from 1980 to 1990 =

3

=


""3

=


3

=

Predicted % increase from 2010 to 2020 = ""

3 =

Predicted population in 2020 = ( ) '*&

=

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26.26.26.26.26. (b)(b)(b)(b)(b)

QS =

l

= 92.59 l/sDischarge of river = QR = 200 l/sBOD of sewage,

CS = 250 mg/lBOD of river,

CR = 6 mg/lBOD of diluted mixture,

C = # # $ $# $

% %

++

C =

+ +

= 83.21 mg/l

27.27.27.27.27. (c)(c)(c)(c)(c)For unconfined aquifer, as per Thiemsequation

Q =

*.* * 0

&

h2 = d s2 = 25 0.53= 24.47 m

h1 = d s1 = 25 1.11= 23.89 m

r2 = 90 m, r1 = 30 mQ = 5400 litre/minute

= 5.4 m3/min= 0.09 m3/sec

0.09 =

4."0 .0 5

&

k =

& .0

.."0 .0 0

= 1.121 103 m/secSo transmissibility,

T = k.d= 1.121 103 25= 0.028 m2/sec

T = 1.68 m2/min

28.28.28.28.28. (d)(d)(d)(d)(d)For drawdown,

Q =

+

+

.* * 0

&

0.09 = +

4.0 * 5

&

hw = 12.08 mso, drawdown in well

= 25 12.08= 12.92 m

29.29.29.29.29. (b)(b)(b)(b)(b)Total BOD present in sewage

= 3.79 240= 909.6 kg

Now, filter volume

=

*6

= 0.082 ha-mAssume that 35% of BOD is removed inprimary clarifier.BOD applied to the filter

= 0.65 909.6 kg= 591.24 kg

=

!(

+

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F = $

$

+ +

F =

. 0

+ = =+

=

+

=

""3

=+

30.30.30.30.30. (a)(a)(a)(a)(a)Amount of BOD left in effluent

= 591.24 [1 0.7745] kg= 133.32 kg

BOD concentration in effluent

=7&'/8

#+&

=

"

l

= 35.18 mg/l

cedrt_05-07-2015_enviro.pdf

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