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January 2010, Vol. 22, No.1
Certification Quiz
Flows and Pumping
Multiple Choice Questions:

Calculate the peakhour flow rate, given the following information.
Population = 50,000 people.
Wastewater generation rate = 85 gal per person/d.
Additional inflow and infiltration = 10 gal per person/d.
Ratio of average daily flow to maximum daily flow = 2.5.
A. 4.75 mgd
B. 11.9 mgd
C. 15 ft3/s
D. 37.5 ft3/s

What is the most accurate way to determine the flow velocity in a pipe?
A. Measure it directly.
B. Calculate it with the Hazen–Williams equation.
C. Calculate it with the Manning equation.
D. Estimate it using d/D ratio tables.

As population increases and the size of the collection area expands, what happens to the peaking factor used to estimate peakhour flows?
A. It decreases.
B. It increases.
C. It remains about the same.
D. It depends on the number of lift stations in the system.

What is the flow velocity in a 15in. sewer line that is flowing half full when the flow rate is 3 ft^{3}/s?
A. 1.8 ft/s.
B. 2.4 ft/s.
C. 4.9 ft/s.
D. 22.5 ft/s.

What is the flow rate, in gallons per minute, of a 16in.diameter force main with a flow velocity of 5 ft/s that is flowing full?
A. 7.05 gal/min.
B. 12.9 gal/min.
C. 3132 gal/min.
D. 3530 gal/min.

What is the flow rate, in gallons per minute, from a pump with a discharge diameter of 6 in. and a velocity of 5 ft/s?
A. 44 gal/min.
B. 198 gal/min.
C. 338 gal/min.
D. 441 gal/min.

Dynamic head — also called headlosses or frictional losses — depends on pipe diameter, pipe material, and what other parameter?
A. Suction head.
B. Velocity of the water through the pipe.
C. Atmospheric pressure.
D. Pump discharge diameter.

If the maximum number of pump on–off cycles per hour desired is three, what capacity, in gal/min, pump should be installed in the following lift station?
Wet well diameter = 8 ft.
Low water level = 2 ft.
High water level = 10 ft.
Rate of water entering the lift station = 10 gal/min.
A. 1012 gal/min.
B. 30 gal/min.
C. 45 gal/min.
D. 160 gal/min.

Wet well sizes must be large enough to minimize what but small enough to do whatelse?
A. Pump cycles per hour — minimize the accumulation of solids in the wet well.
B. Overflow of wastewater — minimize hydrogen sulfide generation.
C. The pump size required — minimize the accumulation of solids in the wet well.
D. Pump cycles per hour — maximize the release of toxic gases.
Questions developed by
Sidney Innerebner
, owner of Indigo Water Group (Littleton, Colo.),and reviewed by the Association of Boards of Certification (Ames, Iowa) Validation and Examination Committee.
Answer Key:
Click here to show or hide the answer key
1. b. Population (wastewater generation rate + inflow and infiltration rate) = average daily flow. Average daily flow × ratio of average to maximum daily flow = maximum daily flow. So, 50,000 (85 + 10) = 4.75 mgd. 4.75 mgd × 2.5 = 11.9 mgd.
2. a.
3. a.
4. c. Flow velocity = flow rate ÷ flow area. To calculate the flow area, find the area of the pipe and divide by 2 (because the pipe is running half full). A = πr^{2} = 3.14 × (7.5 in.)^{2} = 176.6 in.^{2} And, 176.6 in.^{2} ÷ 2 = 88.31 in.^{2} = 0.613 ft^{2}. So, flow velocity = 3 ft^{3}/s ÷ 0.613 ft^{2} = 4.9 ft/s.
5. c. Flow rate = flow velocity × flow area. Area = πr^{2} = 3.14 × (8 in.)^{2} = 200.96 in.^{2} = 1.396 ft^{2}. So, flow rate = 5 ft/s × 1.396 ft^{2} = 6.98 ft^{3}/s. Convert to gal/min: 6.98 ft^{3}/s × (60 s/min) × (7.48 gal/ft^{3}) = 3132 gal/min.
6. d. Flow rate = flow velocity × flow area. Area = πr^{2} = 3.14 × (3 in.)^{2} = 28.27 in.^{2} = 0.196 ft^{2}. So, flow rate = 5 ft/s × 0.196 ft^{2} = 0.982 ft^{3}/s. Convert to gal/min: 0.982 ft^{3}/s × (60 s/min) × (7.48 gal/ft^{3}) = 441 gal/min.
7. b.
8. d. Use the following equation to solve the problem: Cycle time in minutes = storage volume in gal ÷ (pump capacity in gal/min – wet well inflow in gal/min). Storage volume = πr^{2} × depth of the wet well × conversion factor from ft^{3} to gal = (3.14)(4)^{2} × (10 – 2) × 7.48 = 3006 gal. Plug that storage volume into the equation to get the following: 20 min = 3006 ÷ (pump capacity – 10). Solving for pump capacity yields 160 gal/min.
9. a.
References:
California State University–Sacramento (1996). Operation and Maintenance of Wastewater Collection Systems, A Field Study Program, Volumes 1 and 2, Fifth Edition. Sacramento, Calif.: California State University.
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