Maths - Q&A

Question 1

Water is being pumped from a water source with an elevation of 290ft to an elevation of 365ft. What is the total head if friction or minor head losses are 12ft?

Answer 1

87ft

= (365-290)+12

Question 2

Determine a pump’s total output in million gallons per day if it is pumping 1,550gal/min

Answer 2

2.23MGD

Question 3

Find the total head in feet for a pump with a total static head of 22ft and a head loss of 2ft

Answer 3

24 ft

total (dynamic) head = total static head + loss

Question 4

An aeration tank has a volume of 1.4 million gallons. The MLSS are 2,100 mg/L. What are the mixed liquor suspended solids (MLSS) in pounds under aeration?

a) 1,840
b) 16,490
c) 18,390
d) 24,520

Answer 4

d) 24,520

Question 5

How many gallons of wastewater are contained in 100ft of 10-inch force main

a) 40,800
b) 4,080
c) 408
d) 40.8

Answer 5

c) 408

Question 6

A settling basin has a length of 60ft, width of 20ft, and a depth of 12ft. At a flow rate of 6 mgd. What is the retention time of this basin? (Detention time=(volume, gal)/ (flow, gpd)

a) 15 min
b) 26 min
c) 1.1 hr
d) 3hrs

Answer 6

b) 26 min

… check

Question 7

Using a labor cost of $12.50 man-hour, what is the labor cost of a job that takes two people 10 hours to complete?

Answer 7

$250

$12.50 * 2 persons * 10hrs

Question 8

A sewer has failed and 127 ft (39m) of 8 in. (200mm) pipe must be replaced. How many 5ft (1.5m) sections of pipe will be replaced?

a) 5
b) 16
c) 25
d) 26

Answer 8

d) 26

127 ft / 5 ft = 25.4 ft so 26 ft sections of pipe

Question 9

If a pump discharges 8,750 gal in 2hrs and 45min, how many gallons per minute is the pump discharging?

Answer 9

2hrs 45 min = 2*60min + 45min = 165 min

8,750 gal / 165min =53.03 gal / min = 53 gal/min

Question 10

How long will take it in hours for a pump to discharge 86,400 gal if it is pumping at a rate of 30 gpm?

Answer 10

86,400 gal / 30gpm = 2880 min

2880 min / 60min per hr = 48hrs

Question 11

A pump’s output is averaging 36 gpm. How many gallons will it pump in one day?

Answer 11

1 day = 24 hrs = 24hrs * 60min = 1440min

36gpm * 1440min/day = 51,840 gpd

Question 12

A pump discharges 680 gpm. How many gallons will it discharge in 8hrs?

a) 5,440 gal
b) 130,560 gal
c) 326,400 gal
d) 408,000 gal

Answer 12

c) 326,400 gal

8hrs = 8hrs * 60min = 480min
480min * 680 gpm = 326,400 gal

Question 13

Ho many gallons will a pump discharge if it pumps an average of 65gpm for 1hr and 42 min?

Answer 13

65gpm * (1hr*60min + 42min)

= 65gpm * 102min = 6,630 gal

Question 14

Given the following data, calculate the average in pounds per day for chlorine used.
Monday (34), Tuesday (28), Wed (26), Thursday (29), Friday (31), Saturday (32) and Sunday (35)

a) 26.0 lb/day
b) 30.7 lb/day
c) 107.5 lb/day
d) 214.9 lb/day

Answer 14

(34+28+26+29+31+32+35) / 7 = 30.7 lb/day

Question 15

Determine a pump’s total output in million gallons per day if it is pumping 1,550 gal/min

Answer 15

1 day = 1440min
1,550 gal/min * 1440min = 2,232,000 gpd = 2.23MGD

Note: Check units in question

Question 16

FORMULA
Lbs/day = Conc. (mg/L) x Flow (MGD) x 8.34 Lbs/gallon

How many pounds of suspended solids leave a facility each day if the flow rate is
150,000 gal/day and the concentration of suspended solids is 25 mg/L?

Answer 16

Lbs/day = 25 mg/L x 150,000 gal/day x 8.34 Lbs / gal
————————-
1,000,000 gal/MG
= 25 x 0.15 x 8.34
= 31 Lbs/day

Question 17

Pounds of Volatile Solids in the Aeration Tank

Lbs MLVSS = Volume Aeration Tank, MG X MLVSS, mg/L X 8.34 Lbs/gal

Calculate the pounds of volatile solids in an aeration tank that has a volume of 0.471 MG and the concentration of volatile suspended solids is 1700 mg/L.

Answer 17

Lbs = 0.471 MG X 1700 mg/L X 8.34 lbs/gal

= 6678 lbs MLVSS

Question 18

Food to Microorganism Ratio = Lbs of BOD / Lbs of MLVSS = F / M

The 7-day moving average BOD is 2002 lbs and the mixed liquor volatile suspended solids is 6681 pounds. Calculate the F/M ratio of the process.

Answer 18

F / M = 2002 lbs BOD / 6681 lbs MLVSS = 0.30

Typical Range:
Conventional Activated Sludge - F:M 0.25-0.45
Extended Aeration Activated Sludge - F:M 0.05-0.15

Question 19

Food to Microorganism Ratio Calculations
F/M Ratio is Used to Determine the Lbs of MLVSS Needed at a Particular Loading Rate
F/M = Lbs BOD / Lbs of MLVSS

Suppose F/M of 0.30 is desired
and BOD loading is 1200 lbs/day. What is M = Lbs MLVSS?

Answer 19

F / M = 0.30
F / 0.30 = M
1200 lbs / 0.30 = 4000 lbs MLVSS

Question 20

Food to Microorganism Ratio Calculations
If we Know the Pounds of MLVSS Needed and the Volume of the Aeration Tank We Can Calculate MLVSS, mg/L.

Calculate the MLVSS, mg/L given
an Aeration Tank Volume of 0.20 MG.

Answer 20

4000 lbs = 0.20 MG X 8.34 lbs/gal X ? mg/L

4000 lbs / (0.20 MG X 8.34 lbs/gal) = 2398 mg/L

Question 21

How many pounds of MLVSS should be maintained in an aeration tank with a volume of 0.105 MG receiving primary effluent BOD of 630 lbs/d ? The desired F:M is 0.3.
F/M

Answer 21

F / (F/M) = M
= 630 lbs/d / 0.3
= 2100 lbs MLVSS

Question 22

Cell Residence Time
The Average Length of Time in Days that an Organism Remains in the Secondary Treatment System
Cell Residence Time, CRT

CRT, days = (Total MLVSS, lbs) / (Total MLVSS Wasted, lbs/d)

MLVSS = 6681 lbs
MLVSS Wasted = 835 lbs/d
Calculate the CRT

Answer 22

CRT, days = (Total MLVSS, lbs) / (Total MLVSS Wasted, lbs/d)
CRT = 6681 lbs / 835 lbs/d
CRT = 8.0 days