Water Resources and Environmental Engineering practice questions

Answer: B) 60 mg·min/L

• AIncorrect. Multiply, don't divide.
• BCorrect. CT = C × t.
• CIncorrect. Multiply, don't add.
• DIncorrect. Multiply, don't divide.

Answer: D) The pump curve crosses the system curve

• AIncorrect. That's shutoff head.
• BIncorrect. The system sets the operating point.
• CIncorrect. That isn't the method.
• DCorrect. Head supplied equals head required.

Answer: C) Decreases toward a constant value

• AIncorrect. It declines over time.
• BIncorrect. It declines.
• CCorrect. As the soil wets.
• DIncorrect. It approaches a constant.

Answer: B) 0.67 m

• AIncorrect. That's the area.
• BCorrect. R = A/P = 4/(4 + 2).
• CIncorrect. That's the depth.
• DIncorrect. Recheck the wetted perimeter.

Answer: D) Cover hourly demand above the steady supply rate, refilling when demand is low

Equalization storage evens out the hourly gap between a near-constant supply rate and changing demand, mainly by meeting peak-hour demand. Fire and emergency storage are separate components.

• AIncorrect. That is the fire-flow storage component, sized from the required fire flow and its duration. It is added on top of equalization storage.
• BIncorrect. That is emergency (reserve) storage, held for supply interruptions rather than normal daily demand swings.
• CIncorrect. Disinfection contact time is provided at the treatment plant, for example in a clearwell. It is not the purpose of equalization storage in the distribution system.
• DCorrect. Equalization storage lets the treatment plant and pumps run near the average (maximum-day) rate. The tank drains during peak hours and refills overnight when demand drops below the supply rate.

Answer: D) 4 mg/L

• AIncorrect. That isn't the calculation.
• BIncorrect. Weight by flow.
• CIncorrect. Use a mass balance.
• DCorrect. (10 × 2 + 1 × 24)/11.

Answer: D) 5 days at 20°C

• AIncorrect. It's 5 days at 20°C.
• BIncorrect. It's 5 days at 20°C.
• CIncorrect. It's 5 days at 20°C.
• DCorrect. A standard measure of biodegradable organics.

Answer: A) Runoff to travel from the hydraulically most distant point to the outlet

Time of concentration is the time runoff takes to travel from the hydraulically most distant point of a watershed to its outlet. Do not confuse it with lag time, rise time, or recession time.

• ACorrect. Time of concentration is the travel time of runoff from the hydraulically most remote point of the watershed to the outlet or point of interest. In the rational method, the design rainfall duration is set equal to it.
• BIncorrect. This is the recession period, the falling limb of the hydrograph. It is not the time of concentration.
• CIncorrect. This is the rise time of the hydrograph. It depends on both the watershed and the storm, and it is not defined as travel time from the most distant point.
• DIncorrect. This is the watershed lag time. Lag is related to time of concentration (in NRCS methods, lag ≈ 0.6 Tc), but it is a different quantity.

Answer: A) Store runoff temporarily and release it through an outlet so the peak outflow is lower

A detention pond temporarily stores stormwater and releases it at a controlled rate, lowering the peak discharge. It does not mainly reduce runoff volume; that is what retention and infiltration facilities do.

• ACorrect. A detention pond temporarily stores inflow and releases it at a controlled rate through an outlet structure. This attenuates the hydrograph, so the peak outflow is lower than the peak inflow and the downstream system is protected.
• BIncorrect. This describes a retention or infiltration basin. A dry detention pond drains completely through its outlet after each storm.
• CIncorrect. Detention spreads the runoff out over a longer time but releases nearly all of the inflow volume. Its purpose is to reduce the peak rate, not the total volume.
• DIncorrect. Detention delays and attenuates runoff. Speeding runoff toward the outlet would raise downstream peaks, the opposite of the goal.

Answer: A) 2.04 in

With CN = 80, S = 2.5 in and Ia = 0.5 in. So Q = (4.0 − 0.5)²/(4.0 + 0.8 × 2.5) ≈ 2.04 in.

• ACorrect. S = 1000/80 − 10 = 2.5 in and Ia = 0.2S = 0.5 in. Q = (P − Ia)²/(P − Ia + S) = 3.5²/6.0 ≈ 2.04 in.
• BIncorrect. This leaves out the initial abstraction: Q = P²/(P + S) = 16/6.5 ≈ 2.46 in.
• CIncorrect. This subtracts Ia in the numerator but not in the denominator: 3.5²/(4.0 + 2.5) ≈ 1.88 in. The correct denominator is P + 0.8S = 6.0 in.
• DIncorrect. This is P − Ia only. It ignores the further retention that the S term accounts for.