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Department of Civil Engineering-I.I.T. Delhi

CEL 212: Environmental Engineering Second Semester 2011-2012

Home Work 5 Solution Q1. A settling analysis is run on a type 1 suspension. The column is 1.8 m deep and data are presented

below. What will be the theoretical removal efficiency in a settling basin with a loading rate of 25 m3/d/

(m2)? What parameters need to be changed to increase overall removal?

Time (min) 0 60 80 100 130 200 240 420

Conc. (mg/L) 300 189 180 168 156 111 78 27

Mass fraction

remaining

=189/300

=0.63

=180/300

=0.6

=168/300

=0.56

=156/300

=0.52

=111/300

=0.37

=78/300=

=0.26

=27/300

=0.09

Velocity (m/min) =1.8/60

=0.03

0.0225 0.018 0.0138 0.009 0.0075 0.0043

Solution: As discussed in the class and example given in the text book.

Velocity for which tank need to be designed = 25 m3/d/ (m

2) = 25m/d =0.017 m/min (this velocity lies between 100

min and 130 min of sampling (or between 0.56 and 0.52 mass fraction).

Mass fraction corresponding to 0.017 velocity = (0.56+0.52)/2=0.54 (assume approximate values) (This is X0 value). Now plot velocity versus X (see below):

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.01 0.02 0.03 0.04

vel (m/min)

X (

mass f

racti

on

of

rem

ain

ing

part

icle

s)

Figure 1. X versus velocity (m/min)

Divide 0.54 in five parts, so ∆x=0.54/5=0.108

For every interval determine velocity for mid point (say Vi) and then calculate for remaining portions.

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Q2. Revisit Q1. Consider a scenario where we have four discrete particles in beginning (settling

velocity=0.00017 m/min.). After settling for Z1 depth, two particles come closer and start settling

together (now we have only two discrete particles settling from depth Z1). Further settling of two

discrete particles (approximated for modeling purposes) for Z2 (Z2>Z1) depth from the top of the

column, all particles come closer and started settling together as a single particle only (now we have only

one discrete particle settling starting from depth Z2). Model the settling process of these particles. Is it

possible to model removal of these particles in the settling column and calculate overall removal? What

additional information you would like to have, if any?

Solution: Till Z1 depth from top, all 4 particles settle as type 1 settling and tier overall removal can be determined using

method mentioned in Q1 above. At Z1 depth now we have 2 particles which are formed after aggregation of 4

particles. Now these 2 particles will again settle as discrete particles between Z2 and Z1 depth from top and it

can be modeled as discrete settling. Same is true for depth after Z2 also, where we have one particle settling.

Following information is required:

Diameter of individual discrete particles in beginning; diameter of particles at depth Z1 and Z2 from top;

Water characteristics and their specific gravity information. upon having these information, settling of

particles can be modeled and overall removal of suspension can be determined.

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Q3. Revisit Q1. Determine the theoretical efficiency of the selling basin if vt=0.04x.

Solution:

Instead of using figure, use this relation for calculating removal of particles with velocity < 0.017 m/min.

Other parts remain same as Q1 solution.

Q4. Revisit Q1. Determine the theoretical efficiency of the selling basin if loading rate = 35 m3/d/ (m

2)?

Solution: Instead of using 0.017 m/min design velocity, now use 35 m

3/d/ (m

2) (i.e., 0.024 m/min) and repeat Q1 again. As

we are having high design velocity, X0 will be higher and lies between 0.63 and 0.60. Now use the

methodology used in Q1 for solving this problem.

Q5. What is the basic difference between type-I and type-2 settling?

Solution: As per lecture notes.

Q6. Using the following settling test data, determine the efficiency of a settling tank in removing flocculating

particles if the depth is 8 ft (i.e., 8/3.3 m) and the detention time is 30 minutes? What parameters need to be

changed to increase overall removal?

Solution:

for 30 minute detetion time, say 45% removal line passes from 8ft depth (assumed). this portion will be 100%

removal (so r0=45%). now particles with higher removals will also removed but at ratio of Zi/Z0 along with

verticle line passing through 30 minutes detention time.

∆ r(dfference in

removal)

depth of mid-point for this incremental

removal from top Zi (in ft)

Zi∆ r (in ft)

=50%-45%=5% 7 ft (approximated) =5%*7= 0.35

=60%-50%=10% 4.1 ft =10%*4.1 = 0.41

=70%-60%=10% 3.0 ft =10%*3 = 0.30

=80%-70%=10% 1.5ft =10%*1.5 = 0.15

=100%-80%=20% 0.5 =20%*0.5 = 0.10

total =1.31

removal efficiencty (R) =r0 + remaining portion = 0.45+ (1.31 ft)/(8ft) = 0.45+0.164

=0.614 (i.e., 61.4%) (answer) If I want to increase removal effieicny in this settling, what modifications do I need to do to increase suspended

solids removal?

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Q7. For a flocculants suspension, determine the removal efficiency for a basin 10 ft deep with an overflow

rate V0 equal to 10 ft/h, using the laboratory settling data present in the following table:

Solution:

Use the methodlgy used in above question. Here first one needs to make isoremoval lines similar to the figure

given in Q6 and then repeat steps from Q6 to solve this problem.