Math - 1

Question 1

The BOD with inhibitor is 7.4 mg/L. The BOD test results without inhibitors are shown below. What is the NBOD?

Sample, ml 30 50 70 100 Blank

Initial DO, mg/L 8.8 8.6 8.9 8.8 8.8

Final DO, mg/L 6.6 7.8 5.9 6.1 8.6

a. 6.9 mg/L
b. 3.4 mg/L
c. 12.4 mg/L
d. 9.9 mg/L
e. 1.2 mg/L

Answer 1

A. 6.9

BOD = DO_I - DO_F (300)

DO_I

DO _uptake = DO_I - DO_F

Conditions:

DO_uptake, blank < 0.2 mg/L

DO _uptake, sample > 2.0 mg/L

DO _Final, sample > 1.0 mg/L

Sample, ml 30 50 70 100 Blank

DO_Initial, mg/L 8.8 8.6 8.9 8.8 8.8

DO_Final, mg/L 6.6 7.8 5.9 6.1 8.6

DO_Uptake,mg/L 2.2 0.8 3.0 2.7 0.2

Condition OK NO OK OK OK

BOD, mg/L 22 —– 12.9 8.1 ——–

(BOD = CBOD + NBOD)

Average = 14.33

BOD is without inhibitor

CBOD is WITH inhibitor

14.32 mg/l = 7.4 mg/l + NBOD

NBOD = 14.32 mg/l - 7.4 mg/l

NBOD = 6.92 mg/

Question 2

35 GPM is pumped for 12 minutes each hour to a 1,000,000 - gallon anaerobic digester. Lab data is as follows:

Primary Slludge TS% = 4.8%

Primary Sludge VS% = 73%

Digester Sludge TS% = 5.1%

Digester Sludge VS% = 53%

What is Organic Loading?

a. 12 lbs/1,000cubid feet
b. 1 lb/cubic feet
c. 44 lbs/cubic feet
d. 22 lbs/1,000 cubic feet
e. 44 lbs/cubic feet

Answer 2

D. 22 PPD/1,000 ft³

lb = (G) (8.34) (SG) (%/100)

PPD _{volatile, feed} = (35 GPM) (12 M/H)(24hr/Day) (8.34lbs/gal) (1) (4.8%/100)(73%/100)

PPD _{volatile, feed} = 2,945.7

V= (1,000,000gpm) (ft³/7.48gal)

V=133,689.8 ft³

OL = 2,945.7 PPD/133,689.8 ft³

OL = 22 PPD_V/1,000ft³

Question 3

1,400 GPM are treated with 0.5 percent polymer at 2 PPM. How much polymer is required

a. 806.9 GPD
b. 102.8 GPD
c. 128.7 GPD
d. 340.6 GPD
e. 450.5 GPD

Answer 3

A. 806.9 GPD

lb=(MG) (8.34) (mg./l)

PPD = (1,400 GPM) (MGD/694 GPM) (8.34) (2 mg/L)

PPD = 33.65 PPD

lb = (G) (8.34) (SG) (5/100)

33.65 PPD = (GPD) (8.34) (1) (0.5%/100)

GPD = 806.9

Question 4

Given the following information about an activated sludge plant, at what concentration should the MLVSS be maintained at?

F/M 0.6 RAS 2.0 MGD

Q 5.0 MGD Prim. Eff. BOD 100 mg/L

Aerator Vol. 1.0 MG Effluent BOD 20mg/L

a. 2,500 mg/L
b. 1,500 mg/L
c. 833 mg/L
d. 2,000 mg/L
e. 500 mg/L

Answer 4

C. 833 mg/L

F/M = 0.6

Lb. = (MG) (8.34) (mg/L)

F = ( 5 MGD )(8.34) (100 mg/L)

F = 4,170 PPD

4,170 PPD / M = 0.6

M = 4,170 PPD/0.6

M=6,950 lbs.

lb=(MG)(8.34)(mg/L)

6,950 lbs.=(1 MG) (8.34)(mg/L)

mg/L = 833.3

Question 5

A 10-acre wastewater pond serves a community of 7,000 people. Wastewater flow is 100 GPD per person. How many inches per day of wastewater are added?

a. 1.12 inches per day
b. 5.67 inches per day
c. 7.20 inches per day
d. 9.67 inches per day
e. 2.58 inches per day

Answer 5

E. 2.58 iches per day

in/Day = (7,000 p) (100 GPD/P)

GPD = 700,000

In/Day = (0.7 MGD) (36.804) / 10 Acres

in/day = 2.576

Question 6

0. 5 MGD is pumped from 1,225 ft with a 2 HP pump. What is the pump efficiency
   a. 49.3%
   b. 76.4%
   c. 54.5%
   d. 87.6%
   e. 97.7%

Answer 6

D. 87.6%

Efficiency = (out/In)(100)

Hp_W = (GPM) (Head,ft) / 3960

GPM _ (500,000 gal/Day)(D/1,440 M)

GPM = 347

HP_W = (347 gpm)(1,225 ft - 1,205 ft) / 3960

HP_W = 1.75

Efficiency = (1.75 Hp/2 HP) (100)

Efficiency = 87.6%

Question 7

Given the following, how many GPD of activated sludge should be wasted to maintain a MCRT of 15 days?

Q 5.0 MGD MLSS 2,200 mg/L

Aeration Vol. 1.5 MG Waste SS 5,000 mg/L

Clarifier Vol. 0.75 MG Effluent SS 15 mg/L

a. 22,000 GPD
b. 18,000 GPD
c. 34,000 GPD
d. 51,000 GPD
e. 12,000 GPD

Answer 7

D. 51,000 GPD

MGD = [lb_ab + lb_cl / MCRT - lb _lost]/ WAS,mg/L)(8.34)

lb. = (MG) (8.34) (mg/L)

LbAB = (1.5 Mg)(8.34)(2,4200 mg/L) = 27,522 lbs.

lbCl=(0.75 Mg)(8.34)(2,200 mg/L) = 13.761 lbs.

lblost = (5.0 MGD)(8.34)(15 mg/L) = 625.5 PPD

MGD = {[27,522 lb+13,761 lb / 15 days] - 625.5 ppd} / (5,000 mg/L) (8.34)

MGD = 0.051

GPD = 51,000