Wastewater Collection Practice Test #3 Page 1 of 26 - Sewer Geek

[Pages:26]Wastewater Collection Practice Test #3

Page 1 of 26

1) A 54 in. storm sewer flowing half full, at a velocity of 1.35 Ft./sec., will discharge how much flow into a creek in MGD?

a) 13.85 MGD

b) 10.73 MGD

c) 1.85 MGD

X

d) 6.92 MGD

Right

FORMULAS NEEDED;

Area of Pipe = D2 x .785

Ft3/sec. = 1.55 x MGD

Calculate the Area of the Pipe;

54 in. pipe = 5 ft.

D2 x .785

5 ft.

***** Half Full ******* 1.4 ft./sec.

5 ft. x 5 ft. x .785 = 15.8963 ft2

Divide ft3/sec. by 1.55 to convert to MGD

x 1.35 f/s 21.46 ft3/sec

Multiply by the Velocity to get flow in ft3/sec

21.46 ft3/sec 1.55

=

13.85 MGD

This is the full pipe flow. This pipe is only half full. Divide the flow in

half.

13.85 MGD

2

=

6.92 M GD = "D"

2) Shoring must protude ________ above the top of the excavation.

A) 3 feet B) 24 inches Right X C) 18 inches D) 1 foot

Wastewater Collection Practice Test #3

Page 2 of 26

3) A degreasing agent is added to a 16.0 ft. diameter wet well that is 18.4 ft. deep. 4.5 lbs. is required for every 1 ft2 of surface area. If the degreaser weighs 8.5 lbs. per gallon and has a concentration of

13.8 mg/l , how many lbs. Of chemical must be added to the well?

a) 16,639.5 lbs. b) 0.78 lbs. c) 6,764.3 lbs.

X d) 904.3 lbs.

Right!.

FORMULAS NEEDED; area of a circle = D2 x .785

1) Calculate the surface area of the well; area of a circle = D2 x .785

= 16.0 ft. x 16.0 ft. x

x .785 = 201.0 ft 2

2) Multiply the required dosage by the surface area;

201.0 ft. 2 x

4.5 lbs./ft. 2 = 904.3 lbs. ="D"

16.0 ft. (201.0 ft. 2)

18.4 ft.

None of the other information is needed

4) In a trench deep enough to require a ladder(s), the worker must not be required to travel more than _______ to get to the ladder

A) Three steps B) 10 feet Right X C) 25 feet D) 15 feet

Wastewater Collection Practice Test #3

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5) What is the detention time in hours in a tank measuring 312 ft. x 97 ft. x 86 ft. , if the tank receives 945,023 GPH?

a) 22.97 Hours

b) 2.75 Hours

X

c) 20.60 Hours

d) 12.36 Hours

Right

FORMULAS NEEDED; 1 ft3 = 7.48 Gallons

Flow =

Volume Time

945,023 Gal./Hr.

Inflow or

97 ft.

312 ft.

Convert from gallons to ft3;

Calculate tank Volume;

945,023 Gal./Hr.

7.48 gal./ft.3

= 126,340 cu.ft./hr.

Use flow formula to calculate hours;

Flow =

Volume Time

Volume = L x W x H

= 312 ft. x 97 ft. x

=

2,602,704 ft. 3

86 ft.

126,340 ft.3/hr. =

2,602,704 ft. Time

Time =

2,602,704 ft. 126,340 ft. 3/hr.

= 20.60 Hrs. "C"

86 ft.

6) Any excavation over _______ must have a ladder for the worker to get in and out of the trench

A) 25 feet long Right X B) 4 feet deep

C) 8 feet deep D) 3 feet wide

Wastewater Collection Practice Test #3

Page 4 of 26

7) A wet well is 9 feet deep by 21 feet in diameter. When the pump is not running, the water rises 33.4 in. in 3 min. 14 sec. If the level falls 4.5 in. in 10.3 min. while the pump is running, what is the pump rate in GPM?

a) 2,135 Gal./Min.

X

b) 2,323 Gal./Min. Right

c) 2,380 Gal./Min.

d) 6,801 Gal./Min.

2229 Off GPM 94.28 On GPM

346.2 7207

3.23 971

FORMULAS NEEDED; Volume of Cylinder = D2 x .785 x Depth

1 ft.3 = 7.48 Gal.

Flow =

Volume Time

Simplify; 33.4 in. 4.5 in.

= 2.8 ft. = 0.4 ft.

3 min, +

14 sec. 60 sec/min

= 3.23 min.

Calculate inflow with the pump off; Volume of Cylinder = D2 x .785 x Depth

= 21 ft. x 21 ft. x .785 = 963.55 ft.3

Convert to gallons; = 963.55 ft.3 x 7. 48 gal/ft3

x 2.8 ft. = 7,207.34 Gal.

Flow =

Volume Time

Flow =

7,207.34 Gal. 3.23 min.

= 2,229 Gal./Min.

Calculate change in volume with the pump on; Volume of Cylinder = D2 x .785 x Depth

= 21 ft. x 21 ft. x .785 x 0.4 ft. = 129.82 ft.3

Convert to gallons; = 129.82 ft.3 x 7. 48 gal/ft3 = 971.05 Gal.

(Inflow)

Flow =

Volume Time

Flow =

971.05 Gal. 10.30 min.

= 94 Gal./Min.

21 feet

Pump off

Pump on falls

9 feet

2.8 ft. in 3.23 min. rises

0.4 ft. in 10.3 min. level falls

Add or subtract the change in volume to the inflow The level falls when the pump is on.

This means the pump is keeping up add the to the 2,229 GPM Inflow.

2,229 GPM + 94 GPM

2,323 GPM = "B"

Wastewater Collection Practice Test #3

Page 5 of 26

8) Given the data below, what is the most likely cause of the lift station problem?

DATA:

Wet well inlet is normal Well drops normally when pump #1 is running Well level rises slowly when pump #2 or pump #3 is running Run amperage is the same for all three pumps One of the pump motors turn backwards when off. Level system is reading correctly. Electrical controls are all in automatic.

A) Pump #1 & #2 are air-bound Right X B) Pump #1 check valve stuck open.

C) Either pump #1 or #2 is wired backwards D) Check valve on pump #3 is clogged.

Wastewater Collection Practice Test #3

Page 6 of 26

9) Sewer "A" has 106,000 people at 95 GPCD. Sewer "B" has 94,875 people at 100 GPCD. Sewer "C" has 88,756 people at 90 GPCD. What percent of the flow is due to I&I if the total plant flow is 43.00 MGD?

a) 43.1%

b) 64.1%

c) 51.2%

X

d) 35.9%

Right

FORMULAS NEEDED;

GPCD = Gallons Per Capita Per Day

Add up known flows;

Sewer "A" Sewer "B" Sewer "C"

10,070,000 Gal./Day 9,487,500 Gal./Day

+ 7,988,040 Gal./Day 27,545,540 Gal./Day

Subtract know flows from the plant flow to get to get I & I; 43,000,000 Gal./Day - 27,545,540 Gal./Day 15,454,460 Gal./Day

(I & I)

Divide I & I flow by the plant flow & multiply by 100;

Plant Flow = 43,000,000 Gal./Day (43.00 MGD)

106,000 People x 95 GPCD. Sewer "A"

= 10,070,000 Gal./Day

94,875 People x 100 GPCD. Sewer "B"

= 9,487,500 Gal./Day

88,756 People x 90 GPCD. Sewer "C"

= 7,988,040 Gal./Day

I & I = ?

15,454,460 Gal./Day 43,000,000 Gal./Day

x 100 =

35.9%

="D"

** Before picking your answer, look at your I & I flows, does 15,454,460 Gal./Day I & I look like it might be 35.9% of 43,000,000 Gal./Day plsnt flow? If not, you probably divided by the wrong number.

10) An engineer must approve any trench shoring design above

A) 4 feet deep B) A water line C) 50 feet in length Right X D) 20 feet deep

Wastewater Collection Practice Test #3

Page 7 of 26

11) All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of

A) 1:1 B) 4 feet C) 20 feet Right X D) 1 1/2:1

12) A certain town's household flow rate is measured at 90 GPCD. If the plant receives 34.25 MGD, but 12% of that is inflow & infiltration, then what is the population of the town?

X

a) 334,889 People

Right

b) 45,667 People

c) 3,699,000 People

d) 256,875 People

FORMULAS NEEDED; GPCD = Gallons per capita per day

If you have 12% I&I, then 88% of the flow is from people (Assuming no industry) 34,250,000 Gal./Day x 88% = 30,140,000 Gal./Day (from People)

If each person uses 90 Gal./Day, then

30,140,000 90 GPCD.

= 334,889 People

= 'A'

13) According to "Ten State Standards" When a sewer is installed parallel to a water line, it must be a minimum of _________ away (measured from the outside diameters)

A) 6 feet B) 48 inches C) 36 inches Right x D) 10 feet

14) What is the minimum distance from the edge of the spoils to the edge of the trench

A) 10 feet B) 18 inches Right X C) 2 feet D) 6 feet

Wastewater Collection Practice Test #3

Page 8 of 26

15) What capacity blower is required to ventilate a manhole 54 in. in diameter and 49 feet deep, if 8 air change(s) are required every 60 minutes?

a) 13 Ft3/Min.

X

b) 104 Ft3/Min. Right

c) 6231 Ft3/Min.

d) 249 Ft3/Min.

FORMULAS NEEDED;

Volume Time

= Flow

4.5 ft.

Convert inches to feet;

54 in. 12 in./ft.

= 4.5 ft.

Volume of a Cylinder = D2 x .785 x Depth

4.5 ft. x 4.5 ft. x .785 x 49 ft.

= 778.9 Ft.3

Formula;

Flow =

Volume Time

=

778.9 Ft.3

60 min.

=

13.0 Ft.3

Multiply ft3/min x Number of air changes required;

13.0 Ft.3/Min x 8 Air Changes Req'd = 104

="B"

Ft.3/Min

49 feet

16) A(n) _________ is required for any CSO outfall pipe.

A) Netting facility Right X B) NPDES Permit

X C) Outfall flow meter D) Monthly inspection

17) Shoring must protude ________ above the top of the excavation.

A) 3 feet B) 24 inches wrong X C) 1 foot Right X D) 18 inches

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