cvl100 environmental scienceweb.iitd.ac.in/~arunku/files/cvl100_y16/l7aug23.pdf · streeter-phelps...
TRANSCRIPT
August 23, 2016 Arun Kumar ([email protected])
1
(Aug 23rd 2016)
by Dr. Arun Kumar ([email protected])
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?
August 23, 2016 5
Statement of Example 1 (Aug19 lecture)
August 23, 2016 6
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.
August 23, 2016 7
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*
August 23, 2016 9
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]
August 23, 2016 10
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.)
August 23, 2016 11
( ) ( )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
August 23, 2016 12
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
August 23, 2016 13