August 23, 2016 Arun Kumar (arunku@civil.iitd.ac.in)

1

(Aug 23rd 2016)

by Dr. Arun Kumar (arunku@civil.iitd.ac.in)

CVL100 Environmental Science

Lecture 7: DO sag curve modeling

Objective: To learn about DO sag curvemodeling and its applications

Dissolved Oxygen Sag Curve

Setback distance

Csat=9mg/L

CPCB minimum4mg/L

Streeter-Phelps equation or oxygen sag curve)

• Rate of increase of DO deficit = rate of

deoxygenation – rate of reaeration

• Solution is:

DkLkdt

dD

rtd−=

( ) ( )tk

a

tktk

dr

od

t

rrd eDeekk

LkD

−−−+−

−=

Critical time and Critical DO deficit

−−

−=

ad

dr

a

d

r

dr

c

Lk

kkD

k

k

kkt 1ln

1

( ) crcrcd tk

a

tktk

ar

ad

ceDee

kk

LkD

−−−+−

−=

Critical Point = point where steam conditions are at their worst

D = dissolved oxygen deficit

=Csat-C

Lowest value of DO=C = Csat-Dc

La: ultimate BOD after mixing

Problem #1

• Re-do Example 1(Aug19lecture) and comment

on DO values at t=0.5tc and t=2tc. For policy

making, which DO value should be considered?

After mixing, what is the setback distance (i.e.,

downstream distance along the bank from the

point of mixing) where DO level is at least 70%

of saturated DO level in river water?

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Statement of Example 1 (Aug19 lecture)

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Hint of Problem #1• First calculate value of tc.

• Then calculate DO deficit values at t=0.5tc and t=2tc;

• Calculate DO values= (Csat-DO deficit) and compare

based on their values and also with respect to DO at

critical location.

• For policy making, DO value should be greater than 4

mg/L (as per CPCB) and now see which time gives this

DO value.

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XSetback

Csat=9mg/L

2.7mg/L

Hint of Problem #1 contd..• Say setback distance =X*

• Now time for this distance = X*/(River velocity)

• DO level=0.7 *Csat =0.7*9=6.3mg/L

• So DO deficit at this location = Csat-DO level = 9-

6.3mg/L=2.7mg/L

• Now use DO deficit formula and find out that distance X*

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Problem #2• For a given effluent-stream combination (k1: BOD

reaction rate and stream reaeration rate are 0.26 and

0.42 per day, respectively), initial dissolved oxygen (DO)

deficit is 2 mg/L with ultimate BOD of the mixture equals

to 18 mg/L, discuss the approach for calculating time

(say t*) since mixing of effluent with stream water after

which DO deficit becomes 1% of the initial DO deficit?

[5 points]

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Hint of Problem #2• k1=0.26/day (or kd)

• k2=0.42/day (or kr)

• Da=2mg/L; Ultimate BOD of mixture: L0=18 mg/L

• D(t*)=DO deficit at t* = (1/100)*(2 mg/L)=0.02 mg/L

• Calculate t* from

(Note here that t* will be certainly greater than tc as DO deficit at thistime is 1% of initial DO deficit which can only happen after crossing thecritical point on the DO-sag curve.)

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( ) ( )tk

a

tktk

dr

od

t

rrd eDeekk

LkD

−−−+−

−=

Problem #3

Water body k2 (1/day) (on base e) at 20°C

large streams of low velocity 0.40

sluggish streams 0.10

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If industry of example 1 is discharging its wastewater(without implementing any additional in either of thesestreams with all information being same except k2 valuesas per following table, how does it influence location ofcritical point and magnitude of critical DO deficit?

Hint (Problem #3)

Parameterunit k2=0.3

/dk2=0.4/

dk2=0.1

/d

Time for critical DO Deficit Day

Critical DO deficit mg/L

DO critical mg/L

DO critical (approximated, if DO critical <0)) mg/L

treatment required

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