Hw2 do sag curve
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This document contains homework questions regarding calculating dissolved oxygen (DO) sag curves for wastewater discharge into surface streams. It provides information on stream and effluent characteristics, formulas for calculating DO concentration over time and distance from the discharge point, and asks the student to use the information and formulas to answer multiple questions. These include calculating DO levels over time after mixing, maximum BOD discharge limits to maintain minimum DO standards, critical DO deficit levels and times/locations, and drawing DO sag curves for different BOD loadings from multiple discharge points.
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Department of Civil Engineering-I.I.T. Delhi
CEL 212: Environmental Engineering
Second Semester 2013-14
Home Work: DO Sag Curve
Q1. Problem # 3-19 (Peavy et al. Text Book). A wastewater treatment plant disposes of its effluent in a surface
stream. Characteristics of the stream and effluent are shown below. [5+5+5 points]
Parameter wastewater stream
flow (m3
/s) 0.2 5
Dissolved oxygen, mg/L 1 8
Temperature, °C 15 20.2
BOD5 at 20°C, mg/L 100 2
Oxygen consumption rate (K1 at 20°C) (1/day) 0.2 -
Oxygen reaeration rate (K2 at 20°C) (1/day) - 0.3
(a) What will be the dissolved oxygen conc. in the stream after 2 days?
(b) What will be the lowest dissolved oxygen concentration as a result of the waste discharge?
(c) Also calculate the maximum BOD5 (20°C) that can be discharged if a minimum of 4.0 mg/L of oxygen
must be maintained in the stream?
Answer:
Parameter wastewater
(given)
stream (given) Wastewater and stream water
mixture
flow (m3
/s) 0.2 5 =Qmixture=5+0.2=5.2 m/s
Dissolved oxygen, mg/L 1 8 DOmixture=(0.2*1+8*5)/(5+0.2)
=7.73 mg/L
Temperature, °C 15 20.2 Tempmixture=(0.2*15+20.2*5)/(5+0.2)
=20 deg C (No temp. correction
required)
BOD5 at 20°C, mg/L 100 2 BODmixture=(0.2*100+2*5)/(5+0.2)
=5.77 mg/L
Oxygen consumption rate (K1 at
°C) (1/day)
0.2 0.23 (No temp. correction required)
(assumed for stream water)
Oxygen reaeration rate (K2 at °C)
(1/day)
- 0.3 0.3 (No temp. correction required)
Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]
=(5.77 mg/L)/ [1-exp(-0.23*5)] =8.44 mg/L
Initial DO deficit (D0)
For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L
D0 = 9.17 mg/L-7.73 mg/L = 1.44 mg/L
To get DO after 2 days in stream water after mixing, we need to calculate DO deficit after 2 days first and then
calculate DO (at 2days). DO deficit at 2 days is given by
D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t)
= [0.2*8.44]*[exp (-0.2*2)-exp (-0.3*2)]/ (0.3-0.2) + 1.44 exp (-0.3*2)
= [1.94]*[0.6703-0.5488]/ (0.07) + 0.7903 =3.07 mg/L
D (t=2 days) = DO saturated-DO (2day) = 3.07 mg/L
DO (2day) = 9.17-3.07=6.10 mg/L (answer for part i)
Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]
= 1/ (0.3-0.23)*ln [(0.3/0.23)*(1-1.44 *(0.3-0.23)/ (0.23*8.44))]
= 14.29*ln [1.3*(1-1.44 *0.036)] = 14.29*ln [1.23] =2.95 days
Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)](https://image.slidesharecdn.com/hw2dosagcurve-200527071646/85/Hw2-do-sag-curve-1-320.jpg)
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= (0.23/0.3)*8.44 exp (-0.23*2.95) = 6.47 *0.507 =3.28 mg/L
Dc = DOsaturated-DOcritical = 3.28 mg/L
DOcritical= 9.17-3.28=5.89 mg/L (answer for part ii)
Required minimum DO = 4.0 mg/L in stream water. As DO at critical location is 5.89 mg/L, greater than the
recommended DO level, no modification in wastewater effluent characteristics is required.
To calculate maximum BOD5 in effluent water, calculate allowable DO deficit (i.e., Dallowable)
= DOsaturated-DOminimum = 9.17-4.0 = 5.17 mg/L [Note that 5.17 mg/L DO deficit is allowable and we are having 5.89
mg/L critical DO deficit.]
Now with calculated allowable DO deficit (this is assumed to be the critical deficit now) and calculated tcritical
(assumed to be similar to previous case, i.e., 2.95 days), calculate ultimate BOD in this case. Then calculate 5-day
BOD of the mixture stream water and then calculate 5-day BOD of the effluent which will be the desired
maximum 5-day BOD value.
Dallowable (t=tcritical) =Dcritical,new
=> 5.17 = (0.23/0.30)*L0 [exp (-0.23*2.95)] = 0.77*0.51L0=0.3927L0
=> Ultimate BOD of the mixture water = L0=5.19 mg/L/ (0.3927) =13.17 mg/L
Now 5-day BOD in mixture water is calculated.
5-day BODmixture = L0 *[1-exp (-K1mixture*5)]
= (13.17 mg/L)*[1-exp (-0.23*5)] =9.0 mg/L
5-day BOD in effluent water is calculated now.
BODmixture= (5-day BODeffQeff+5-day BODstreamQstream)/ (5+0.2)
9.0 mg/L = (5-day BODeff*0.2+2*5)/ (5+0.2)
5-day BODeff *0.2+10= 9.0*5.2 = 46.8
=> 5-day BODeff = (46.8-10)/0.2= 184 mg/L (answer for part iii). This is maximum value of 5-day BOD in
wastewater effluent which can be discharged in the stream water without exceeding the minimum required
DO value of 4 mg/L.
Q2. For a given effluent-stream combination, say values of BOD reaction rate and stream reaeration rate are 0.26
and 0.42 per day, respectively and given that 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]
Q3. The WWTP in the “AA” community discharges 10 million gallons/day of secondary effluent into a stream
“Red Cedar” whose minimum flow rate is 100 m3
/s.
WWTP effluent Stream water
Temperature (°C) 20°C 15°C
BOD5 (mg/L) (at 20 °C) 200 1
Dissolved oxygen (mg/L) 2 80% of saturation level
Oxygen consumption rate (k1) (1/day)
Oxygen re-aeration rate (k2)(1/day)
0.3
0.7
Using above information, calculate the following:
1) Temperature, dissolved oxygen (DO) and BOD of the mixture.
2) Initial dissolved oxygen deficit at the place of mixing.
3) Critical oxygen deficit (Dc), time for maximum dissolved oxygen deficit (tmax) and its location from
discharge point (Xc).
4) The dissolved oxygen level and 20°C BOD5 of a sample taken at the critical point.
(Saturated oxygen concentration in stream before discharge is 10.07 mg/L at 15°C. Saturated DO at 15.7°C is 9.9
mg/L. Use temperature coefficients of 1.135 for k1 and 1.024 for k2 for applying temperature corrections for these
two constants.)](https://image.slidesharecdn.com/hw2dosagcurve-200527071646/85/Hw2-do-sag-curve-2-320.jpg)
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Q4. Look at the BOD patterns of three organic compounds (a, b, and c) (k2=0.42/day; D0 =2 mg/L; ultimate
BODmixture(river and wastewater) = 18 mg/L). Which organic matter degradation pattern would determine critical location
for DO deficit and why? How would you calculate BOD5 at location where DO deficit becomes 1% of the initial
DO deficit (list steps)? [4+6=10 points]
D (t) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + [D0 exp (-K2*t)]
Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]
Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)
Q5. Two industries situated at 1 Km apart are discharging wastewater in a river (BOD effluent =500 mg/L for both
cases). Draw a DO sag curve with distance downstream of discharge location? How does DO sag curve change if
BOD effluent becomes 200 mg/L? Redraw it. [10 points]](https://image.slidesharecdn.com/hw2dosagcurve-200527071646/85/Hw2-do-sag-curve-3-320.jpg)
Hw2 do sag curve
- 1. 1 Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering Second Semester 2013-14 Home Work: DO Sag Curve Q1. Problem # 3-19 (Peavy et al. Text Book). A wastewater treatment plant disposes of its effluent in a surface stream. Characteristics of the stream and effluent are shown below. [5+5+5 points] Parameter wastewater stream flow (m3 /s) 0.2 5 Dissolved oxygen, mg/L 1 8 Temperature, °C 15 20.2 BOD5 at 20°C, mg/L 100 2 Oxygen consumption rate (K1 at 20°C) (1/day) 0.2 - Oxygen reaeration rate (K2 at 20°C) (1/day) - 0.3 (a) What will be the dissolved oxygen conc. in the stream after 2 days? (b) What will be the lowest dissolved oxygen concentration as a result of the waste discharge? (c) Also calculate the maximum BOD5 (20°C) that can be discharged if a minimum of 4.0 mg/L of oxygen must be maintained in the stream? Answer: Parameter wastewater (given) stream (given) Wastewater and stream water mixture flow (m3 /s) 0.2 5 =Qmixture=5+0.2=5.2 m/s Dissolved oxygen, mg/L 1 8 DOmixture=(0.2*1+8*5)/(5+0.2) =7.73 mg/L Temperature, °C 15 20.2 Tempmixture=(0.2*15+20.2*5)/(5+0.2) =20 deg C (No temp. correction required) BOD5 at 20°C, mg/L 100 2 BODmixture=(0.2*100+2*5)/(5+0.2) =5.77 mg/L Oxygen consumption rate (K1 at °C) (1/day) 0.2 0.23 (No temp. correction required) (assumed for stream water) Oxygen reaeration rate (K2 at °C) (1/day) - 0.3 0.3 (No temp. correction required) Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)] =(5.77 mg/L)/ [1-exp(-0.23*5)] =8.44 mg/L Initial DO deficit (D0) For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L D0 = 9.17 mg/L-7.73 mg/L = 1.44 mg/L To get DO after 2 days in stream water after mixing, we need to calculate DO deficit after 2 days first and then calculate DO (at 2days). DO deficit at 2 days is given by D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t) = [0.2*8.44]*[exp (-0.2*2)-exp (-0.3*2)]/ (0.3-0.2) + 1.44 exp (-0.3*2) = [1.94]*[0.6703-0.5488]/ (0.07) + 0.7903 =3.07 mg/L D (t=2 days) = DO saturated-DO (2day) = 3.07 mg/L DO (2day) = 9.17-3.07=6.10 mg/L (answer for part i) Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))] = 1/ (0.3-0.23)*ln [(0.3/0.23)*(1-1.44 *(0.3-0.23)/ (0.23*8.44))] = 14.29*ln [1.3*(1-1.44 *0.036)] = 14.29*ln [1.23] =2.95 days Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)
- 2. 2 = (0.23/0.3)*8.44 exp (-0.23*2.95) = 6.47 *0.507 =3.28 mg/L Dc = DOsaturated-DOcritical = 3.28 mg/L DOcritical= 9.17-3.28=5.89 mg/L (answer for part ii) Required minimum DO = 4.0 mg/L in stream water. As DO at critical location is 5.89 mg/L, greater than the recommended DO level, no modification in wastewater effluent characteristics is required. To calculate maximum BOD5 in effluent water, calculate allowable DO deficit (i.e., Dallowable) = DOsaturated-DOminimum = 9.17-4.0 = 5.17 mg/L [Note that 5.17 mg/L DO deficit is allowable and we are having 5.89 mg/L critical DO deficit.] Now with calculated allowable DO deficit (this is assumed to be the critical deficit now) and calculated tcritical (assumed to be similar to previous case, i.e., 2.95 days), calculate ultimate BOD in this case. Then calculate 5-day BOD of the mixture stream water and then calculate 5-day BOD of the effluent which will be the desired maximum 5-day BOD value. Dallowable (t=tcritical) =Dcritical,new => 5.17 = (0.23/0.30)*L0 [exp (-0.23*2.95)] = 0.77*0.51L0=0.3927L0 => Ultimate BOD of the mixture water = L0=5.19 mg/L/ (0.3927) =13.17 mg/L Now 5-day BOD in mixture water is calculated. 5-day BODmixture = L0 *[1-exp (-K1mixture*5)] = (13.17 mg/L)*[1-exp (-0.23*5)] =9.0 mg/L 5-day BOD in effluent water is calculated now. BODmixture= (5-day BODeffQeff+5-day BODstreamQstream)/ (5+0.2) 9.0 mg/L = (5-day BODeff*0.2+2*5)/ (5+0.2) 5-day BODeff *0.2+10= 9.0*5.2 = 46.8 => 5-day BODeff = (46.8-10)/0.2= 184 mg/L (answer for part iii). This is maximum value of 5-day BOD in wastewater effluent which can be discharged in the stream water without exceeding the minimum required DO value of 4 mg/L. Q2. For a given effluent-stream combination, say values of BOD reaction rate and stream reaeration rate are 0.26 and 0.42 per day, respectively and given that 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] Q3. The WWTP in the “AA” community discharges 10 million gallons/day of secondary effluent into a stream “Red Cedar” whose minimum flow rate is 100 m3 /s. WWTP effluent Stream water Temperature (°C) 20°C 15°C BOD5 (mg/L) (at 20 °C) 200 1 Dissolved oxygen (mg/L) 2 80% of saturation level Oxygen consumption rate (k1) (1/day) Oxygen re-aeration rate (k2)(1/day) 0.3 0.7 Using above information, calculate the following: 1) Temperature, dissolved oxygen (DO) and BOD of the mixture. 2) Initial dissolved oxygen deficit at the place of mixing. 3) Critical oxygen deficit (Dc), time for maximum dissolved oxygen deficit (tmax) and its location from discharge point (Xc). 4) The dissolved oxygen level and 20°C BOD5 of a sample taken at the critical point. (Saturated oxygen concentration in stream before discharge is 10.07 mg/L at 15°C. Saturated DO at 15.7°C is 9.9 mg/L. Use temperature coefficients of 1.135 for k1 and 1.024 for k2 for applying temperature corrections for these two constants.)
- 3. 3 Q4. Look at the BOD patterns of three organic compounds (a, b, and c) (k2=0.42/day; D0 =2 mg/L; ultimate BODmixture(river and wastewater) = 18 mg/L). Which organic matter degradation pattern would determine critical location for DO deficit and why? How would you calculate BOD5 at location where DO deficit becomes 1% of the initial DO deficit (list steps)? [4+6=10 points] D (t) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + [D0 exp (-K2*t)] Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))] Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc) Q5. Two industries situated at 1 Km apart are discharging wastewater in a river (BOD effluent =500 mg/L for both cases). Draw a DO sag curve with distance downstream of discharge location? How does DO sag curve change if BOD effluent becomes 200 mg/L? Redraw it. [10 points]