CEE 370

Fall 2010

Homework #7

BOD Problems

1. Do Problem 9-6 in the D&M text

If the BOD of a municipal wastewater at the end of 7 days is 60.0 mg/L and the ultimate BOD is 85.0 mg/L, what is the rate constant?

Given: 7 day BOD = 60.0; L = 85.0 mg/L.

Solution:

a. Setup Eqn 8-6

BOD_t = L_o(1 – e^-kt)

60.0 = 85.0 (1 ‑ e^‑k(7))

0.7059 = 1 ‑ e^‑k(7)

‑0.2941 = ‑e^‑k(7)

b. Divide through by ‑1 and take ln of both sides

ln (0.2941) = ln (e^‑k(7))

‑1.224 = ‑k(7)

k = 0.1748 d^‑1

2. Do Problem 9-10 in the D&M text.

Assuming that the data in Problem 9-6 were taken at 25°C, compute the rate constant at 16°C.

Given: k = 0.1748 d^‑1 at 25 ºC from Problem 9-6.

Solution:

a. First convert k at 25 ºC to k at 20 ºC

0.1748 = k₂₀ (1.056)^{25 ‑ 20}

0.1748

k₂₀ = ----------- = 0.1331

1.3132

b. Then convert to 16 ºC

k₁₆ = (0.1331)(1.135)^{16 ‑ 20}

k₁₆ = (0.1331)(0.6026) = 0.0802 d^‑1

3. Supplemental Problem #1.

The BOD₅ of the raw Clarksville wastewater is 225 mg/L. A long term BOD test has revealed that this wastewater also has an ultimate carbonaceous BOD of 325 mg/L. Plant effluent testing has shown that the BOD₅ drops by 92% across the plant. Long term tests on the effluent also show that the BOD decay coefficient is half the original value in the raw wastewater.

a. What is the BOD decay coefficient of the raw wastewater?

b. What is the ultimate carbonaceous BOD of the effluent wastewater?

Answer:

k_b = 0.236 d^-1

Answer:

k_b = 0.5*(0.236) = 0.118 d^-1

y₅ = 0.08*(225) = 18 mg/L

4. Do Problem 9-12 in the D&M text; with an additional part

a. What sample size (in percent) is requried for a BOD₅ of 350.0 mg/L if the oxygen consumed is to be limited to 6.00 mg/L?

b. Assume a standard BOD₅ test is being done with a 300 mL sample bottle. Present your answer as both required volume of sample (in mL and as sample size in percent (as asked for in the text).

Given: BOD₅ = 350.0 mg/L; oxygen consumption = 6.00 mg/L.

Solution:

Stream Modeling problems

5. Do Problem 9-19 in the D&M text.

The Waramurngundi tannery with a wastewater flow of 0.011 m³/s and a BOD₅ of 590 mg/L discharges into Djanggawul Creek. The creek has a 10-year, 7-day low flow of 1.7 m³/s. Upstream of the Waramurungundi tannery, the BOD₅ of the creek is 0.6 mg/L. The BOD rate constants (k) are 0.115 day^-1 for the Waramurungundi tannery and 3.7 day^-1 for the creek. Calculate the initial ultimate BOD after mixing.

Given: Tannery Q_w = 0.011 m³/s, BOD₅ = 590 mg/L, Creek Q_r = 1.7 m³/s, BOD₅ upstream of tannery = 0.6 mg/L, k_tannery = 0.115 d^-1, k_creek = 3.7 d^-1.

Solution:

a. Calculate the ultimate BOD of tannery wastewater using Eqn 9-6

590 mg/L 590

L_o = -------------------- = --------------- = 1,349.2 mg/L

1 - e^(-0.115)(5) 1 - 0.56

Thus, L_w (Eqn 8-18) = 1,349.2 mg/L

b. Calculate the ultimate BOD of Djanggawul Creek

0.6 mg/L 0.6

L_o = ----------------- = -------------------- = 0.6 mg/L

1 - e^(-3.7)(5) 1 - 9.24 x 10^-9

Thus, L_r (Eqn 9-18) = 0.6 mg/L

c. Calculate the initial ultimate BOD using Eqn 9-18

Q_wL_w + Q_rL_r

L_a = -------------------

Q_w + Q_r

(0.011 m³/s)(1,349.2 mg/L) + (1.7 m³/s)(0.6 mg/L) 14.84 + 1.02

L_a = ----------------------------------------------------------------- = --------------------

0.011 m³/s + 1.7 m³/s 1.711

L_a = 9.269 or 9.3 mg/L

6. Modified Parameter Estimation Problem (similar to 9-23 in the D&M text).

a. Compute the deoxygenation rate constant and reaeration rate constant (base e) for the following wastewater and stream conditions

Source	k (day^-1)	Temp (°C)	H (m)	Veocity (m/s)	ɳ
Wastewater	0.25	20
Mill River		20	2.2	0.7	0.4

b. In addition, calculate the values of k_d and k_r if the temperature is 10ºC.

Solution:

a. at 20C

i. Calculate the deoxygenation rate constant using Eqn 9-30

0.7

k_d = 0.25 + ----------- (0.4) = 0.377 d^-1

2.2

ii. Calculate the reaeration rate constant using Eqn 9-26

b. at 10C

i. Adjust the deoxygenation rate constant to 10C

For CBOD Often we use: q=1.047; D&M cite: 1.056 for 20-30C and 1.135 for 4-20C, so:

	EPA	D&M	Units
Ɵ =	1.047	1.135
kd =	0.2383	0.1063	day-1

ii. Adjust the reaeration constant to 10C

for k_r we use: q=1.024

so kr = 0.789 day^-1

7. Do Problem 9-25 in the D&M text.

The initial ultimate BOD after mixing in the Bergelmir River is 12.0 mg/L. The DO in the Bergelmir River after the wastewater and river have mixed is at saturation. The river temperature is 10°C. At 10°C, the deoxygenation rate constant (k_d) is 0.30 day^-1, and the reaeration rate constant (k_r) is 0.40 day^-1. Determine the critical point (t_c) and the citical DO.

Given: L_a = 12 mg/L, DO = saturation, river temp = 10 °C, k_d = 0.30 d^-1, k_r = 0.40 d^-1

Solution:

a. Since the DO in the river is at saturation after the wastewater and river have mixed, the initial deficit (D_a) is 0.0 mg/L.

b. calculate critical travel time from equation 9-38.

t_c = 2.88 days

c. The critical deficit is found using Eqn 9-36 with t = t_c

D_t = 3.78 mg/L

DO = Cs-Dt = 11.33-3.78 = 7.53 mg/L

8. Supplemental Problem #2.

Consider Monkey creek, a free-flowing stream with a mean water velocity of 0.1 ft/s. At milepoint zero, there is a discharge of 5 cfs of the Clarksville WWTP effluent. The total streamflow above this point is 20 cfs. Ignore any BOD in the upstream water. The Secchi depth for Monkey Creek is 3 ft, however, the average depth is 12 ft.

a. What is the BOD₅ of the river water at a point 5 miles downstream of the Clarksville outfall?

b. If the concentration of benzo[a]pyrene just above the Clarksville outfall is 23 x10^-6 mg/L, and there is no benzo[a]pyrene in the WWTP effluent, what will the concentration be 5 miles downstream? Consider that this compound undergoes photolysis (k_p0 = 1.2 d^-1). .

Answer to a:

First, recall that the Clarksville effluent has an ultimate BOD of 40.4 mg/L:

Then consider the point source (Clarksville WWTP) and upstream flows and mix the two:

Upstream Water
	Qu =	20	Cfs
	Lou =	0	mg/L
Point Source
	Qw =	5	Cfs
	Low =	40.42221	mg/L

The combined flows and concentrations are:

Initial Values (from mass balance)
	Lo =	8.084441	mg/L

And the time of travel for 5 miles is:

t* =

x/U =

3.055556

D

So, now use the BOD plug flow model

and converting this ultimate BOD back into BOD₅, we get:

Answer to b:

There are two major approaches to solving this problem. First you can simply treat the photolysis rate constant as a fully-depth averaged value (i.e., all benzo[a]pyrene will degrade at the same average rate regardless of where it happens to be in the stream). This is the same as saying that photolysis occurs throughout the depth of the water at the same rate as it does at the surface (represented by the k_po). This is an oversimplification, but one that is appropriate for a “first level” approximation for the maximum amount of compound loss one could have. Using this method, all you need is the rate constant, and the secchi depth becomes superfluous.

First, you should perform a mass balance at the point of mixing to determine the initial concentration.

C₀ = 18.4 ng/L

Then, use the plug flow model with first order loss (in this case the loss is due only to photolysis).

In a more detailed analysis (of the type we cover in CEE 577), we would need to consider depth of the water and its cloudiness. First we’d be given the more fundamental rate of 0.31d^-1. Then we’ll want to evaluate the photolysis rate constant using the secchi depth and a characteristic stream depth. There are many ways of selecting a characteristic depth, but probably the easiest is to use an average depth (e.g., 6 ft), since half of the B(a)P will be above this depth and half below:

		Benzo(a)pyrene photolysis
	kpo =	31	/day
secchi depth	zs =	3	ft
	alpha =	0.6	/ft
	kp =	0.847035	/day

Note: another approach that would be more accurate would be to divide the stream up into thin layers (say 0.5 ft each) and calculate a separate k_p for each. Then you could average these up.

Next, perform a mass balance at the point of mixing to determine the initial concentration.

C₀ = 18.4 ng/L

Finally, use the plug flow model with first order loss (in this case the loss is due only to photolysis).

9. Do Problem 9-29 in the D&M text. Please use the description below – the book has several typos.

The town of Edinkira has filed a complaint with the state Department of Natural Resources (DNR) that the City of Quamta is restricting its use of the Umvelinqangi River because of the discharge of raw sewage. The DNR water quality criterion for the Umvelinqangi River is 5.00 mg/L of DO. Edinkira is 15.55 km downstream from Quamta.

a. What is the DO at Edinkira?

b. What is the critical DO and

c. Where (at what distance) downstream does the critical DO occur?

d. Is the assimilative capacity of the river restricted?

The following data pertain to the 7-day, 10-year (7Q10 – note the typo in your book) low flow at Quamta.

Parameter	Wastewater	Umvelinqangi River	Units
Flow	0.1507	1.08	m^3/s
BOD5	128		mg/L
BODu		11.4	mg/L
DO	1.00	7.95	mg/L
k at 20°C	0.4375		day^-1
Velocity		0.390	m/s
Depth		2.8	m
Temperature	16	16	°C
Bed-activity coeff		0.200

Given: Table of data; Edinkira is 15.55 km downstream; standard = 5.00 mg/L

Solution: Note that the problem statement in the book is ambiguous at best. The temperature of the water for this problem should be 16C, not 28C. Also, BOD₅ values are always measured at 20C, not 16C. With this clarification, the following solution should be correct.

Part I. DO at Edinkira

a. Calculate t

(15.55 km)(1,000 m/km)

t = ‑‑‑----------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 0.462 d

(0.390 m/s)(86,400 s/d)

b. Calculate k_d

0.390

k_d = 0.4375 + ‑‑---‑‑‑‑ (0.200)

(2.80)

k_d = 0.4375 + 0.02785 = 0.4654 at 20 ºC

At 16 ºC

k_d = 0.4654 (1.135)^{16 ‑ 20}

k_d = 0.4654 (0.6026) = 0.2805 d^‑1

c. Calculate k_r

3.9(0.390)^0.5

k_r = ‑‑--‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 0.5212 at 20 ºC

(2.80)^1.5

At 16 ºC

k_r = 0.5212 (1.024)^{16 ‑ 20}

k_r = 0.5212 (0.9095) = 0.474 d^‑1

d. Calculate D_a

From Table A‑2 @ 16 ºC DO_s = 9.95 mg/L

(0.1507)(1.00) + (1.08)(7.95)

D_a = 9.95 ‑ {‑‑-------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑}

0.1507 + 1.08

D_a = 9.95 ‑ 7.10 = 2.85 mg/L

e. Calculate L_w

BOD_t 128

L_w = ‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = ‑‑----‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

(1 ‑ e^‑(k)(t)) (1 ‑ e^{‑(0.4375)(5)})

128

L_w = ‑‑‑‑--‑‑‑‑ = 144.17 mg/L

0.8878

f. Calculate L_a from Eqn 9-18

(0.1507)(144.17) + (1.08)(11.40)

L_a = ‑‑‑‑----------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 27.65 mg/L

0.1507 + 1.08

g. Calculate Deficit

(0.2805)(27.65)

D = ‑-------‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (e^{‑(0.2805)(0.462)} ‑ e^{‑(0.474)(0.462)}) + 2.85(e^{‑(0.474)(0.462)})

0.474 ‑ 0.2805

7.7558

D = ‑‑----‑‑‑‑‑ (0.8785 ‑ 0.8033) + 2.85 (0.8033)

0.1935

D = 3.014 + 2.289 = 5.303 or 5.30 mg/L

g. Calculate DO with Eqn 9-23 and DO_s = 9.95 mg/L

DO = 9.95 ‑ 5.30 = 4.65 mg/L at Edinkira

Part II. Critical DO

. a.Calculate t_c

1 0.474 0.474 ‑ 0.2805

t_c = ‑‑‑‑‑‑‑‑‑‑‑‑-----‑‑ln [----‑‑‑‑‑‑ (1 ‑ 2.85 (‑--------‑‑‑‑‑‑‑‑‑‑‑‑‑‑ ))]

0.474 ‑ 0.2805 0.2805 (0.2805)(27.65)

t_c = 5.168 ln [1.6898 (0.9288)] = 2.33 d

b. Calculate D_c

(0.2805)(27.65)

D_c = ‑-----‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (e^{‑(0.2805)(2.33)} ‑ e^{‑(0.474)(2.33)}) + 2.85(e^{‑(0.474)(2.33)})

(0.474 ‑ 0.2805)

7.7558

D_c = ‑‑----‑‑‑‑‑ (0.520188 ‑ 0.3314) + 2.85 (0.3314)

0.1935

D_c = 7.5669 + 0.9449 = 8.51 mg/L

c. Calculate DO with Eqn 9-23 and DO_s = 9.95 mg/L

DO = 9.95 ‑ 8.51 = 1.44 mg/L at critical point

d. Distance to Critical Point is the product of the speed of the stream and the travel time

x = (v)(t) = (0.390 m/s)(86,400 s/d)(2.33 d)(1.0 x 10^‑3 km/m)

x = 78.5 km downstream from Quamta

e. The assimilative capacity is restricted.

10. Do Problem 9-30 in the D&M text.

Under the provisions of the Clean Water Act, the U.S. Environmental Protection Agency established a requirement that municipalities had to provide secondary tretment of their waste. This was defined to be treatment that resuted in an effluent BOD₅ that did not exceed 30 mg/L. The discharge from Quamta (Problem 9-29) is clearly in violation of that standard. Given the data in Problem 9-29, rework the problem, assuming that Quamta provides treatment to lower the BOD₅ to 30.00 mg/L.

Given: Data in Problem 9-29 but BOD₅ of Quamta wastewater is reduced to 30.00 mg/L

Solution:

Part I. DO at Edinkira

a. From Problem 9-29 the following remain the same: t = 0.462 d; k_d = 0.2805 d^‑1; k_r = 0.474 d^‑1; D_a = 2.85 mg/L

b. Recalculate L_w

BOD_t 30.00

L_w = ‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = ‑‑--------‑‑‑‑‑‑‑‑‑‑‑‑

(1 ‑ e^‑(k)(t)) (1 ‑ e^{‑(0.4375)(5)})

30.00

L_w = ‑‑‑‑‑--‑‑‑ = 33.79 mg/L

0.8878

f. Calculate L_a from Eqn 8-18

(0.1507)(33.79) + (1.08)(11.40)

L_a = ‑‑‑‑---------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 14.14 mg/L

0.1507 + 1.08

g. Calculate Deficit

(0.2805)(14.14)

D = ‑‑-----‑‑‑‑‑‑‑‑‑‑‑‑‑ (e^{‑(0.2805)(0.462)} ‑ e^{‑(0.474)(0.462)}) + 2.85(e^{‑(0.474)(0.462)})

0.474 ‑ 0.2805

3.966

D = ‑---‑‑‑‑‑‑ (0.8785 ‑ 0.8033) + 2.85 (0.8033)

0.1935

D = 1.541 + 2.289 = 3.83 mg/L

g. Calculate DO with Eqn 9-23 and DO_s = 9.95 mg/L

DO = 9.95 ‑ 3.83 = 6.12 mg/L at Edinkira

Part II. Critical DO

a. Calculate t_c

1 0.474 0.474 ‑ 0.2805

t_c = ‑‑‑‑‑‑-------‑‑‑‑‑‑‑‑ln [ ‑‑---‑‑‑‑ (1 ‑ 2.85 (‑‑‑‑‑------‑‑‑‑‑‑‑‑‑‑ ))]

0.474 ‑ 0.2805 0.2805 (0.2805)(14.14)

t_c = 5.168 ln [1.6898 (0.8610)] = 1.937 d

b. Calculate D_c

(0.2805)(14.14)

D_c = ‑‑‑‑----‑‑‑‑‑‑‑‑‑‑‑ (e^{‑(0.2805)(1.937)} ‑ e^{‑(0.474)(1.937)}) + 2.85(10^{‑(0.474)(1.937)})

(0.474 ‑ 0.2805)

3.966

D_c = ‑‑‑---‑‑‑‑ (0.580732 ‑ 0.399169) + 2.85 (0.399169)

0.1935

D_c = 3.72133 + 1.1376 = 4.8589 or 4.86 mg/L

c. Calculate DO with Eqn 9-23 and DO_s = 9.95 mg/L

DO = 9.95 ‑ 4.86 = 5.09 mg/L at critical point

d. Distance to Critical Point

x = (v)(t) = (0.390 m/s)(86 400 s/d)(1.937 d)(1.0 x 10^‑3 km/m)

x = 65.27 km downstream from Quamta

e. The assimilative capacity is restricted but much less, so at Edinkira (6.12 ‑ 5.00 = 1.12 mg/L above standard).

11. Do Problem 9-34 in the D&M text.

What amount of ultimate BOD (in kg/d) may Quamta (problem 9.29) discharge and still allow Edinkira 1.50 mg/L of DO above the DNR water quality criteria for assimilation of its waste?

Given: Data from Problem 9-29; allow 1.50 mg/L above DNR regulations at Edinkira

Solution:

a. Allowable deficit

D = 9.95 ‑ (1.50 + 5.00) = 3.45 mg/L

b. Allowable initial ultimate BOD

(D ‑ D_ae^(‑kr)(t))(k_r ‑ k_d)

L_a = ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

k_d(e^‑(kd)(t) ‑ e^‑(kr)(t))

(3.45 ‑ 2.85(e^{(‑0.474)(0.462)})(0.474 ‑ 0.2805)

L_a = ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

(0.2805)(e^{(‑0.2805)(0.462)} ‑ e^{(‑0.474)(0.462)})

(3.45 ‑ 2.289)(0.1935)

L_a = ‑‑-------‑‑-‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

(0.2805)(0.07512)

L_a = 10.66 mg/L

c. Allowable ultimate BOD in discharge (solve Eqn 9-18 for L_w)

La(Q_w + Q_r) ‑ Q_rL_r

L_w = ‑‑‑‑----‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

Q_w

10.66(0.1507 + 1.08) ‑ 1.08(11.40)

L_w = ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑-----------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

0.1507

L_w = 5.35 mg/L

d. Allowable Mass Discharge in kg/d (Note: mg/L = g/m³)

Q_wL_w = (0.1507 m³/s)(5.35 mg/L)(86,400 s/d)(1 x 10^‑3 kg/g)

Q_wL_w = 69.67 or 70 kg/d

12. As an add-on to Problems 9-29, 9-30 and 9-34, produce a graph of dissolved oxygen concentration versus distance downstream from the wastewater discharge point, and show 3 plots on the graph, one for each problem.

An efficient method to solve these problems is to create a generic Excel spreadsheet with cells for input of parameters in the problem and then cells for calculating the deficit and DO level as a function of travel time and travel distance. You will also find it convenient to set up your spreadsheet to calculate initial conditions for ultimate BOD and DO in the wastewater/river mixture, allowing for ease in changing loading conditions (as required in Problems 9-30 and 9-34).