4) A wastewater treatment operator has been fined because the DO at a point 1 km downstream is below the 4.5 mg/L regulation. Assuming the plant can produce a wastewater effluent with a BODu of 10mg/L, what is the lowest DO in the wastewater? DO in wastewater must remain above 4.5 mg/L year round. QR = 50 m3/s Qw = 15 m3/s Tw Summer 25°C R Summer 18°C kaonaar= 0.15 d-1 (assume KBOD is the same for the river and effluent) H =1 m (assume no change due to wastewater flow) V = 0.04 m/s (assume no change due to wastewater flow) DOR = 6 mg/L BODww=10mg/L BODa = 5 mg/L n = 0.3 DOs @ 20. = 9.2 mg/L Downstream Distance = 1km Solution wastewater (given) Stream Given Mixture of WW and River flow (m3/s) 15 50 Qmixture=50 + 15 = 65 65 Dissolved oxygen, mg/L 9.2 6 DOmixture=(9.2* 15 + 6*50)/(50 + 15) = 6.738 mg/L 6.738 Temperature, °C 25 18 Temperature Mixture = (25 x15 + 18 x 50)/(65 19.615 BOD5 at 20°C, mg/L 10 5 BOD mixture = (10 x 15 + 5 x 50)/65 6.153 Oxygen consumption rate (K1 at °C) (1/day) 0.15 0.15 (0.15 x 15 + 0.15 x 50)/65 0.15 Oxygen reaeration rate (K2 at °C) (1/day) 0.83 0.7 Mixture = (.83 x15 + 0.7 x 50)/65 0.73 L = L0e-k1t So, K1s= K2 = 3.9 x v1/2 x [ (1.025) T-20]1/2/H3/2 K2R = 3.9 x (0.04) 1/2 x [ (1.025) 18-20]1/2/H3/2 = 0.7 K2s = 3.9 x (0.04) 1/2 x [ (1.025) 25 -20]1/2/H3/2= 0.83 Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)] =(6.153 mg/L)/ [1-exp(-0.15*5)] = 11.66 mg/L For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L D0 = 9.17 mg/L-6.738 mg/L = 2.432 mg/L Initial deficit = 2.432 mg/L D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t) = (0.15 x 11.66) *[exp (-0.15*13.649)-exp (-0.73*13.649)]/ (0.73- 0.15) + 2.432 exp (- 0.73*13.649) = 0.388 mg/L DO (2day) = 9.17- 0.388 = 8.782 mg/L Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))] = 1/ (0.73-0.15)*ln [(0.73/0.15)*(1-2.432 *(0.73-0.15)/ (0.73* 11.66 ))] = 2.41 days Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc) = ( 0.15/0.73) x 11.66 exp( - 0.15 x 2.41)= 1.669 mg/L DOcritical= 9.17- 1.669 = 7.501 mg/L wastewater (given) Stream Given Mixture of WW and River flow (m3/s) 15 50 Qmixture=50 + 15 = 65 65 Dissolved oxygen, mg/L 9.2 6 DOmixture=(9.2* 15 + 6*50)/(50 + 15) = 6.738 mg/L 6.738 Temperature, °C 25 18 Temperature Mixture = (25 x15 + 18 x 50)/(65 19.615 BOD5 at 20°C, mg/L 10 5 BOD mixture = (10 x 15 + 5 x 50)/65 6.153 Oxygen consumption rate (K1 at °C) (1/day) 0.15 0.15 (0.15 x 15 + 0.15 x 50)/65 0.15 Oxygen reaeration rate (K2 at °C) (1/day) 0.83 0.7 Mixture = (.83 x15 + 0.7 x 50)/65 0.73.

1. 4) A wastewater treatment operator has been fined because the DO at a point 1 km downstream
is below the 4.5 mg/L regulation. Assuming the plant can produce a wastewater effluent with a
BODu of 10mg/L, what is the lowest DO in the wastewater? DO in wastewater must remain
above 4.5 mg/L year round. QR = 50 m3/s Qw = 15 m3/s Tw Summer 25°C R Summer 18°C
kaonaar= 0.15 d-1 (assume KBOD is the same for the river and effluent) H =1 m (assume no
change due to wastewater flow) V = 0.04 m/s (assume no change due to wastewater flow) DOR
= 6 mg/L BODww=10mg/L BODa = 5 mg/L n = 0.3 DOs @ 20. = 9.2 mg/L Downstream
Distance = 1km
Solution
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L

2. 10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)
(1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73
L = L0e-k1t
So, K1s=
K2 = 3.9 x v1/2 x [ (1.025) T-20]1/2/H3/2
K2R = 3.9 x (0.04) 1/2 x [ (1.025) 18-20]1/2/H3/2 = 0.7
K2s = 3.9 x (0.04) 1/2 x [ (1.025) 25 -20]1/2/H3/2= 0.83
Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]
=(6.153 mg/L)/ [1-exp(-0.15*5)] = 11.66 mg/L
For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L
D0 = 9.17 mg/L-6.738 mg/L = 2.432 mg/L
Initial deficit = 2.432 mg/L
D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t)
= (0.15 x 11.66) *[exp (-0.15*13.649)-exp (-0.73*13.649)]/ (0.73- 0.15) + 2.432 exp (-
0.73*13.649)
= 0.388 mg/L
DO (2day) = 9.17- 0.388 = 8.782 mg/L
Time for critical DO deficit
(tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]
= 1/ (0.73-0.15)*ln [(0.73/0.15)*(1-2.432 *(0.73-0.15)/ (0.73* 11.66 ))]
= 2.41 days
Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)

3. = ( 0.15/0.73) x 11.66 exp( - 0.15 x 2.41)= 1.669 mg/L
DOcritical= 9.17- 1.669 = 7.501 mg/L
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L
10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)

4. (1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73