PE Water Resources & Environmental Practice Exam — Free Questions with Worked Solutions

Water Resources & Environmental · problem 1 of 10

A municipal utility estimates evaporation from its water-supply reservoir by multiplying readings from a nearby Class A land pan by a pan coefficient of about 0.7. Which statement best explains why the coefficient is less than 1.0?

A.Rain that falls into the pan is logged as evaporation, so raw pan readings systematically overstate the water lost.
B.The small pan gains heat through its sides and bottom, so its water runs warmer and evaporates faster than the lake.
C.The coefficient converts the pan's depth of loss into an equivalent reservoir depth that allows for its sloping banks.
D.Pan readings also include transpiration from the grass around the pan, which the coefficient removes from the total.

Answer: B — The small pan gains heat through its sides and bottom, so its water runs warmer and evaporates faster than the lake.

Lake evaporation is estimated as $E_L = K_p E_p$ E_L = K_p E_p, with $K_p$ K_p below 1 because a pan evaporates faster than a large water body under the same weather.

E_L = K_pE_p

E_L = K_pE_p · §6.5.8.3 Evaporation (starts p. 404)

A Class A pan is a small, shallow metal tank standing above the ground: sun and warm air heat its sides and bottom, so its few inches of water run warmer than the reservoir surface, and the higher surface temperature raises the vapor pressure that drives evaporation. (Its small area also means the air passing over it is not humidified the way air crossing a large lake is.) The reservoir, deep and large, warms more slowly because it stores heat, so on average its evaporation is only about 70% of the pan's; month by month the ratio varies with that stored heat.

Answer: B

Why the other options are wrong

A.
Rain falling into the pan raises its water level, which would make a raw level reading understate evaporation, not overstate it; pan records are corrected by adding the rain back. The coefficient is not a rainfall correction.
C.
The coefficient converts pan depth to lake depth per unit area; bank slope affects the reservoir's surface area, which is applied separately when a depth of evaporation is turned into a volume.
D.
A pan measures open-water evaporation only; the surrounding grass transpires into the air, not out of the pan. Transpiration belongs to consumptive-use estimates, not to the pan coefficient.

Water Resources & Environmental · problem 2 of 10

During a municipal utility's constant-rate pumping test, drawdown in an observation well in a confined aquifer plots as a Cooper-Jacob straight line against log time. Which condition gives a larger drawdown change per log cycle?

A.A lower aquifer transmissivity, all else unchanged
B.A smaller aquifer storage coefficient, all else unchanged
C.An observation well located farther from the pumped well
D.A longer pumping period before the drawdown is read

Answer: A — A lower aquifer transmissivity, all else unchanged

The slope of the Cooper-Jacob line depends only on Q and T. The straight-line form is

s = \frac{2.303 Q}{4\pi T} \log\left(\frac{2.25 T t}{r^2 S}\right)

s = \frac{2.303 Q}{4\pi T} \log\left(\frac{2.25 T t}{r^2 S}\right)

s = \dfrac{2.303Q}{4\pi T}\log\left(\dfrac{2.25Tt}{r^2S}\right)

s = \dfrac{2.303Q}{4\pi T}\log\left(\dfrac{2.25Tt}{r^2S}\right) · §6.6.3.2 Confined Aquifer (starts p. 421)

so over one log cycle of time ( $\Delta \log t = 1$ \Delta \log t = 1) the drawdown changes by

\Delta s = \frac{2.303 Q}{4\pi T}

\Delta s = \frac{2.303 Q}{4\pi T}

\Delta(\log t) = 1

\Delta(\log t) = 1 (one log cycle) · §6.6.3.2 Confined Aquifer (starts p. 421)

At a constant pumping rate, $\Delta s$ per log cycle is inversely proportional to the transmissivity: a less transmissive aquifer gives a steeper line. The storativity and the distance $r$ appear only inside the logarithm, with $t$ ; they set where the line crosses zero drawdown ( $t_0 = r^2S/2.25T$ t_0 = r^2S/2.25T), shifting the line along the time axis without changing its slope. On the straight-line portion the slope is the same at every time, so reading later does not steepen it.

S = \dfrac{2.25Tt_o}{r^2}

S = \dfrac{2.25Tt_o}{r^2} · §6.6.3.2 Confined Aquifer (starts p. 421)

Answer: A

Why the other options are wrong

B.
Treats the storage coefficient (storativity) $S$ as a slope control. A smaller $S$ makes the cone spread faster, so drawdown at a given time is larger and the line shifts to earlier times ( $t_0 = r^2S/2.25T$ t_0 = r^2S/2.25T falls), but $S$ sits inside the logarithm and the change per log cycle, $2.303Q/4\pi T$ 2.303Q/4\pi T, is unchanged.
C.
Treats distance as a slope control. A farther well sees less drawdown at any time and its line shifts to later times ( $t_0$ t_0 grows with $r^2$ r^2), but $r$ sits inside the logarithm with $S$ , so the slope per log cycle is the same at every observation well.
D.
Assumes the line steepens with time. Once the data fall on the Cooper-Jacob straight line, each further log cycle adds the same drawdown, $2.303Q/4\pi T$ 2.303Q/4\pi T; pumping longer extends the line without changing its slope.

Water Resources & Environmental · problem 3 of 10

A developer's subdivision street has a curb-and-gutter section with a 2% cross slope, a 1.5% longitudinal slope, and a Manning's n of 0.016. The gutter flow that produces a spread of 9.0 ft is most nearly:

A.1.83 cfs
B.2.20 cfs
C.2.95 cfs
D.5.86 cfs

Answer: B — 2.20 cfs

Triangular gutter flow (curb and gutter with a uniform cross slope), USCS constant $K_u = 0.56$ K_u = 0.56:

\begin{aligned} Q &= \frac{K_u}{n} S_x^{1.67} S_L^{0.5} T^{2.67} \\ &= \frac{0.56}{0.016}(0.02)^{1.67} (0.015)^{0.5} (9.0)^{2.67} \\ &= 35.0 \times 0.001455 \times 0.1225 \times 353.0 = \mathbf{2.20\ cfs} \end{aligned}

\begin{aligned} Q &= \frac{K_u}{n} S_x^{1.67} S_L^{0.5} T^{2.67} \\ &= \frac{0.56}{0.016}(0.02)^{1.67} (0.015)^{0.5} (9.0)^{2.67} \\ &= 35.0 \times 0.001455 \times 0.1225 \times 353.0 = \mathbf{2.20\ cfs} \end{aligned}

Q = (K_u/n)S_x^{1.67}S_L^{0.5}T^{2.67}

Q = (K_u/n)S_x^{1.67}S_L^{0.5}T^{2.67} · §6.4.10.3 Curb and Gutter as Triangular Channels (starts p. 368)

K_u = 0.56

K_u = 0.56 in USCS units · §6.4.10.3 Curb and Gutter as Triangular Channels (starts p. 368)

Answer: B — 2.20 cfs

Why the other options are wrong

A.
Applies Manning's equation to the whole triangular section with $R = A/P$ R = A/P: $d = 9.0 \times 0.02 = 0.18$ d = 9.0 \times 0.02 = 0.18 ft, $A = 0.81$ A = 0.81 ft², $P = 9.18$ P = 9.18 ft, $Q = (1.49/0.016)(0.81)(0.0882)^{2/3}(0.015)^{0.5} = 1.83$ Q = (1.49/0.016)(0.81)(0.0882)^{2/3}(0.015)^{0.5} = 1.83 cfs, about 1.83 cfs. A single hydraulic radius misrepresents a wide, shallow gutter whose depth varies across the spread; the gutter relation integrates Manning across it.
C.
Applies Manning's equation with the triangular area but the curb depth as the hydraulic radius ( $R \approx d = 0.18$ R \approx d = 0.18 ft, a wide-channel habit): $Q = (1.49/0.016)(0.81)(0.18)^{2/3}(0.015)^{0.5} = 2.95$ Q = (1.49/0.016)(0.81)(0.18)^{2/3}(0.015)^{0.5} = 2.95 cfs, about 2.95 cfs.
D.
Uses Manning's 1.49 in place of the gutter constant $K_u = 0.56$ K_u = 0.56: $Q = (1.49/0.016)(0.02)^{1.67}(0.015)^{0.5}(9.0)^{2.67} = 5.86$ Q = (1.49/0.016)(0.02)^{1.67}(0.015)^{0.5}(9.0)^{2.67} = 5.86 cfs, about 5.86 cfs.

Water Resources & Environmental · problem 4 of 10

A rural water district serves 36,000 residents in a county of 49,000 people; the remaining residents rely on private wells. Using the National Board of Fire Underwriters population formula, the needed fire flow for the district's distribution system is most nearly:

A.5,750 gpm
B.6,120 gpm
C.6,640 gpm
D.23,500 gpm

Answer: A — 5,750 gpm

The formula is driven by the population the system serves, in thousands. The National Board of Fire Underwriters relation is

Q = 1{,}020\sqrt{P}\,(1 - 0.01\sqrt{P})

Q = 1{,}020\sqrt{P}\,(1 - 0.01\sqrt{P})

Q = 1{,}020\sqrt{P}\left(1 - 0.01\sqrt{P}\right)

Q = 1{,}020\sqrt{P}\left(1 - 0.01\sqrt{P}\right) · §6.9.1 Drinking Water Distribution Systems (starts p. 463)

$Q$ in gpm, $P$ = population served in thousands.

The district serves 36,000 people, so $P = 36$ P = 36. The county total of 49,000 includes residents on private wells whom the district's mains never supply; it is not used.

\sqrt{P} = 6

\sqrt{P} = 6

Q = 1{,}020 \times 6 \times (1 - 0.06) = 6{,}120 \times 0.94 = 5{,}753\ \text{gpm}

Q = 1{,}020 \times 6 \times (1 - 0.06) = 6{,}120 \times 0.94 = 5{,}753\ \text{gpm}

Answer: A — 5,750 gpm

Why the other options are wrong

B.
Drops the $(1 - 0.01\sqrt{P})$ (1 - 0.01\sqrt{P}) factor and keeps only the leading term: $1{,}020 \times 6 = 6{,}120$ 1{,}020 \times 6 = 6{,}120 gpm. The correction factor is part of the formula and reduces the flow by 6% here.
C.
Uses the county population, 49,000, instead of the population served: $1{,}020 \times 7 \times (1 - 0.07) = 6{,}640$ 1{,}020 \times 7 \times (1 - 0.07) = 6{,}640 gpm. The 13,000 residents on private wells are not served by the district's system.
D.
Enters $P$ without the square roots, as $Q = 1{,}020 P (1 - 0.01 P)$ Q = 1{,}020 P (1 - 0.01 P): $1{,}020 \times 36 \times 0.64 = 23{,}500$ 1{,}020 \times 36 \times 0.64 = 23{,}500 gpm. Both occurrences of $P$ in the formula are under a square root.

Water Resources & Environmental · problem 5 of 10

The schematic shows the steady flows at an industrial plant's lined evaporation pond, which receives cooling-tower blowdown containing 1,500 mg/L of total dissolved solids (TDS) and overflows to a creek. Neglecting rainfall and seepage, and treating TDS as conservative, the TDS concentration of the overflow is most nearly:

A.1,500 mg/L
B.2,080 mg/L
C.2,580 mg/L
D.5,360 mg/L

Answer: D — 5,360 mg/L

Steady-state water and mass balance on an evaporation pond, with TDS treated as conservative.

Draw the control volume around the pond. At steady state, water enters only as blowdown and leaves two ways: as evaporation and as overflow. Evaporation removes water but no dissolved solids, so all of the TDS that enters must leave in the overflow, whose flow is the blowdown minus the evaporation:

Q_{\text{over}} = 0.25 - 0.18 = 0.07\ \text{MGD}

Q_{\text{over}} = 0.25 - 0.18 = 0.07\ \text{MGD}

TDS balance (conservative, steady state, no reaction):

\begin{aligned} C_{\text{in}} Q_{\text{in}} &= C_{\text{over}} Q_{\text{over}} \\ C_{\text{over}} &= \frac{1{,}500 \times 0.25}{0.07} = 5{,}357\ \text{mg/L} \approx 5{,}360\ \text{mg/L} \end{aligned}

\begin{aligned} C_{\text{in}} Q_{\text{in}} &= C_{\text{over}} Q_{\text{over}} \\ C_{\text{over}} &= \frac{1{,}500 \times 0.25}{0.07} = 5{,}357\ \text{mg/L} \approx 5{,}360\ \text{mg/L} \end{aligned}

The section gives the general mass balance, with mass in, mass out and a reaction term, and mass as concentration times volume · §6.7.1 Mass Conservation and Continuity (starts p. 423)

Evaporating 72% of the water concentrates the dissolved solids by a factor of 1/0.28 = 3.6.

Answer: D — 5,360 mg/L

Why the other options are wrong

A.
1,500 mg/L. Ignores evaporation and passes the blowdown concentration straight through. Evaporation leaves the dissolved solids behind, so the same TDS load leaves in a smaller flow and the concentration must rise.
B.
2,080 mg/L. Writes the balance on the wrong outflow, taking the 0.18-MGD evaporation as the stream that carries the TDS away: $1{,}500 \times 0.25/0.18 = 2{,}083$ 1{,}500 \times 0.25/0.18 = 2{,}083 mg/L. The vapor carries no TDS; the overflow is $0.25 - 0.18 = 0.07$ 0.25 - 0.18 = 0.07 MGD.
C.
2,580 mg/L. Raises the concentration by the fraction of water evaporated, $1{,}500 \times (1 + 0.18/0.25) = 2{,}580$ 1{,}500 \times (1 + 0.18/0.25) = 2{,}580 mg/L. The concentration scales with the ratio of inflow to overflow, 0.25/0.07, not with one plus the evaporated fraction.

Water Resources & Environmental · problem 6 of 10

A county road department is staking a simple circular curve with a radius of 900 ft and an intersection angle of 48°00'. The middle ordinate of the curve is most nearly:

A.77.8 ft
B.85.2 ft
C.297.8 ft
D.445.0 ft

Answer: A — 77.8 ft

The middle ordinate runs from the midpoint of the long chord to the midpoint of the curve, along the radius that bisects the central angle, so it is written with the half angle:

M = R\left[1 - \cos(\Delta/2)\right]

M = R\left[1 - \cos(\Delta/2)\right]

M = R\left(1 - \cos\dfrac{\Delta}{2}\right)

M = R\left(1 - \cos\dfrac{\Delta}{2}\right) · §5.2.1 Basic Curve Elements (e.g., Middle Ordinate, Length, Chord, Radius) (starts p. 274)

\begin{aligned} \Delta/2 &= 24^\circ 00', \quad \cos 24^\circ = 0.91355 \\ M &= 900(1 - 0.91355) = 900(0.08645) = 77.8\ \text{ft} \end{aligned}

\begin{aligned} \Delta/2 &= 24^\circ 00', \quad \cos 24^\circ = 0.91355 \\ M &= 900(1 - 0.91355) = 900(0.08645) = 77.8\ \text{ft} \end{aligned}

Check against the external distance. $E = R\left[1/\cos(\Delta/2) - 1\right] = 85.2$ E = R\left[1/\cos(\Delta/2) - 1\right] = 85.2 ft is the PI-to-curve distance, which is always larger than $M$ for the same curve.

Answer: A — 77.8 ft

Why the other options are wrong

B.
85.2 ft: computes the external distance, $E = R\left[1/\cos(\Delta/2) - 1\right] = 900(1/0.91355 - 1) = 85.2$ E = R\left[1/\cos(\Delta/2) - 1\right] = 900(1/0.91355 - 1) = 85.2 ft - the PI-to-curve distance - instead of the chord-to-curve middle ordinate.
C.
297.8 ft: puts the full intersection angle in the cosine, $900(1 - \cos 48^\circ) = 297.8$ 900(1 - \cos 48^\circ) = 297.8 ft. The middle ordinate lies on the radius that bisects the central angle, so it uses $\Delta/2$ .
D.
445.0 ft: both slips together - the external-distance form with the full angle, $900(1/\cos 48^\circ - 1) = 445.0$ 900(1/\cos 48^\circ - 1) = 445.0 ft.

Water Resources & Environmental · problem 7 of 10

A rural water district's gravity main carries water between the two reservoirs shown in the profile. The main flows full, and minor losses and velocity head are negligible. The gauge pressure in the main at the ridge high point is most nearly:

A.8.0 psi
B.12.6 psi
C.17.2 psi
D.35.5 psi

Answer: A — 8.0 psi

Pressure at the high point of a gravity main, from a straight-line hydraulic grade line with the whole available head lost to friction at a uniform rate.

The whole available head is lost to friction, and it is lost at a uniform rate. With minor losses and velocity head negligible, the energy line and the hydraulic grade line coincide, start at the upper water surface and end at the lower one. The main is one diameter and one material and carries one flow, so its friction slope $h_f/L$ h_f/L is the same everywhere: the HGL is a straight line in distance measured along the pipe. The flow rate is never needed.

The energy equation balances the heads at two sections against the head loss between them · §6.2.1.2 Energy Equation (starts p. 308)The energy-line section places the hydraulic grade line one velocity head below the energy line · §6.2.1.5 Energy Line (Bernoulli Equation) (starts p. 309)The Darcy-Weisbach equation gives a pipe's head loss as the friction factor times length over diameter times the velocity head · §6.2.3.1 Head Loss Due to Flow (Darcy-Weisbach Equation) (starts p. 314)

HGL at the high point. The total loss is 1,248.0 − 1,142.0 = 106.0 ft over 7,800 + 5,200 = 13,000 ft of pipe. By the high point, 7,800 ft along, the line has lost

\begin{aligned} h_f &= 106.0 \times (7{,}800/13{,}000) = 63.6\ \text{ft} \\ \text{HGL} &= 1{,}248.0 - 63.6 = 1{,}184.4\ \text{ft} \end{aligned}

\begin{aligned} h_f &= 106.0 \times (7{,}800/13{,}000) = 63.6\ \text{ft} \\ \text{HGL} &= 1{,}248.0 - 63.6 = 1{,}184.4\ \text{ft} \end{aligned}

Pressure. The HGL is the sum of pressure head and elevation head, so the pressure head at the pipe centerline is 1,184.4 − 1,166.0 = 18.4 ft, and

p = \gamma h = \frac{(62.4\ \text{lb/ft}^3)(18.4\ \text{ft})}{144\ \text{in.}^2\text{/ft}^2} = 7.97\ \text{psi} \approx 8.0\ \text{psi (gauge)}

p = \gamma h = \frac{(62.4\ \text{lb/ft}^3)(18.4\ \text{ft})}{144\ \text{in.}^2\text{/ft}^2} = 7.97\ \text{psi} \approx 8.0\ \text{psi (gauge)}

The hydraulic grade line joins the sums of pressure head and elevation head along a conduit · §6.2.1.4 Hydraulic Gradient (Grade Line) (starts p. 309)

\dfrac{P_a}{\gamma} = \text{pressure head}

\dfrac{P_a}{\gamma} = \text{pressure head} · §6.2.1.5 Energy Line (Bernoulli Equation) (starts p. 309)

\gamma_w = 62.4\ \text{lbf/ft}^3

\gamma_w = 62.4\ \text{lbf/ft}^3 (water at standard conditions) · §6.1.2 Density, Specific Volume, Specific Weight, and Specific Gravity (starts p. 293)

Check. The HGL passes above the high point, so the pressure is positive and the main does not drop into a vacuum there. The static head, 82.0 ft, would give 35.5 psi only with the line shut.

Answer: A — 8.0 psi

Why the other options are wrong

B.
12.6 psi. Takes the HGL at the high point as the average of the two water surfaces, $(1{,}248.0 + 1{,}142.0)/2 = 1{,}195.0$ (1{,}248.0 + 1{,}142.0)/2 = 1{,}195.0 ft, as if the high point were halfway along the line: $(1{,}195.0 - 1{,}166.0) \times 62.4/144 = 12.6$ (1{,}195.0 - 1{,}166.0) \times 62.4/144 = 12.6 psi. The HGL falls in proportion to pipe length, and the high point is 60% of the way along, not 50%.
C.
17.2 psi. Applies the downstream leg's share of the length, $5{,}200/13{,}000$ 5{,}200/13{,}000, to the loss: $\text{HGL} = 1{,}248.0 - 106.0 \times 0.40 = 1{,}205.6$ \text{HGL} = 1{,}248.0 - 106.0 \times 0.40 = 1{,}205.6 ft, so $(1{,}205.6 - 1{,}166.0) \times 62.4/144 = 17.2$ (1{,}205.6 - 1{,}166.0) \times 62.4/144 = 17.2 psi. The loss reaching the high point is the loss over the 7,800 ft upstream of it.
D.
35.5 psi. Uses the static head from the upper water surface, $(1{,}248.0 - 1{,}166.0) \times 62.4/144 = 35.5$ (1{,}248.0 - 1{,}166.0) \times 62.4/144 = 35.5 psi, ignoring the friction lost before the high point. That is the pressure with the line full and not flowing.

Water Resources & Environmental · problem 8 of 10

A developer's water reclamation facility supplies irrigation water to a subdivision. To meet the state's reuse limit, it must reduce a herbicide in the filtered effluent from 0.50 mg/L to 0.05 mg/L. Powdered activated carbon (PAC) is added to a completely mixed contact tank, where the carbon reaches equilibrium with the treated water. Jar tests fit a Freundlich isotherm with $K_f$ K_f = 60 $(\text{mg/g})(\text{L/mg})^{1/n}$ (\text{mg/g})(\text{L/mg})^{1/n} and $1/n$ = 0.40. Which statements are correct? Select the two that apply.

A.The equilibrium loading on the carbon is about 45 mg/g.
B.The required PAC dose is about 10 mg/L.
C.The equilibrium loading on the carbon is about 18 mg/g.
D.The required PAC dose is about 28 mg/L.
E.The required PAC dose is about 25 mg/L.
F.Lowering the target to 0.025 mg/L would roughly double the required PAC dose.

Answers: C — The equilibrium loading on the carbon is about 18 mg/g.; E — The required PAC dose is about 25 mg/L.

In a completely mixed PAC contact tank the carbon leaves in equilibrium with the treated water, so the isotherm is evaluated at $C_e$ C_e.

q_e = K_f C_e^{1/n} = 60 \times (0.05)^{0.40} = 60 \times 0.3017 = 18.1\ \text{mg/g}

q_e = K_f C_e^{1/n} = 60 \times (0.05)^{0.40} = 60 \times 0.3017 = 18.1\ \text{mg/g}

q_e = K_f C_e^{1/n}

q_e = K_f C_e^{1/n} · §6.9.3.1 Freundlich Isotherm (starts p. 467)

A mass balance on the herbicide gives the carbon dose:

\frac{w}{V} = \frac{C_0 - C_e}{q_e} = \frac{0.50 - 0.05}{18.1} = 0.0249\ \text{g/L} = 24.9\ \text{mg/L}

\frac{w}{V} = \frac{C_0 - C_e}{q_e} = \frac{0.50 - 0.05}{18.1} = 0.0249\ \text{g/L} = 24.9\ \text{mg/L}

q_e = \dfrac{(C_0 - C_e)V}{w}

q_e = \dfrac{(C_0 - C_e)V}{w} · §6.9.3.1 Freundlich Isotherm (starts p. 467)

Statement F: at $C_e = 0.025$ C_e = 0.025 mg/L, $q_e = 60 \times (0.025)^{0.40} = 13.7$ q_e = 60 \times (0.025)^{0.40} = 13.7 mg/g and $w/V = 0.475/13.7 = 34.6$ w/V = 0.475/13.7 = 34.6 mg/L, about 1.4 times the dose, not double.

Statements C (≈18 mg/g) and E (≈25 mg/L) are correct. The answers are C and E.

Answer: C and E

Why the other options are wrong

A.
Evaluates the isotherm at the influent concentration: $60 \times 0.50^{0.40} = 45.5 \approx 45$ 60 \times 0.50^{0.40} = 45.5 \approx 45 mg/g. PAC in a completely mixed tank equilibrates with the treated water, $C_e = 0.05$ C_e = 0.05 mg/L.
B.
Uses the loading at the influent concentration (45.5 mg/g) in the mass balance: $0.45/45.5 = 0.0099\ \text{g/L} \approx 10\ \text{mg/L}$ 0.45/45.5 = 0.0099\ \text{g/L} \approx 10\ \text{mg/L}. That loading is not reached when the carbon leaves at $C_e$ C_e.
D.
Divides the whole influent concentration by the loading: $0.50/18.1 = 0.0276\ \text{g/L} \approx 28\ \text{mg/L}$ 0.50/18.1 = 0.0276\ \text{g/L} \approx 28\ \text{mg/L}. The carbon only has to take up the herbicide removed, $C_0 - C_e = 0.45$ C_0 - C_e = 0.45 mg/L.
F.
Treats the dose as inversely proportional to the target. Halving $C_e$ C_e lowers $q_e$ q_e only by $0.5^{0.40}$ 0.5^{0.40} (to 13.7 mg/g) while the removal rises to 0.475 mg/L, so the dose becomes 34.6 mg/L, about 1.4 times, not double.

Water Resources & Environmental · problem 9 of 10

A developer will take fill from a borrow pit laid out as two adjacent 50 ft × 50 ft grid squares. Cut depths are 2.8, 3.4, 3.0, and 2.6 ft at the four outer corners and 7.6 and 6.8 ft at the two corners on the shared edge. The excavated soil swells 20% when loaded. Using the grid method, the in-place volume to be excavated is most nearly:

A.606 yd³
B.728 yd³
C.940 yd³
D.1,128 yd³

Answer: C — 940 yd³

The grid formula is applied square by square. For each grid square the excavated volume is the square's area times the average of its four corner cut depths:

V_{\text{square}} = \frac{1}{4}(a + b + c + d) \times (\text{area of grid square})

V_{\text{square}} = \frac{1}{4}(a + b + c + d) \times (\text{area of grid square})

\text{Volume of material in one grid square} = \dfrac{1}{4}\left(a + b + c + d\right) \times \left(\text{area of grid square}\right)

\text{Volume of material in one grid square} = \dfrac{1}{4}\left(a + b + c + d\right) \times \left(\text{area of grid square}\right)

· §2.1.2.2 Borrow Pit Grid Method (starts p. 48)

The two squares share an edge, so the two shared-edge corners belong to both squares and enter the sum twice; each outer corner belongs to one square and enters once.

Square 1 + square 2. Each square is 50 ft × 50 ft = 2,500 ft²; $\Sigma h$ is the weighted corner cut-depth sum over both squares.

\begin{aligned} \Sigma h &= (2.8 + 3.4 + 3.0 + 2.6) + 2(7.6 + 6.8) = 11.8 + 28.8 = 40.6\ \text{ft} \\ V &= (2{,}500/4)(40.6) = 625 \times 40.6 = 25{,}375\ \text{ft}^3 \\ V &= 25{,}375/27 = 939.8\ \text{yd}^3 \approx 940\ \text{yd}^3 \end{aligned}

\begin{aligned} \Sigma h &= (2.8 + 3.4 + 3.0 + 2.6) + 2(7.6 + 6.8) = 11.8 + 28.8 = 40.6\ \text{ft} \\ V &= (2{,}500/4)(40.6) = 625 \times 40.6 = 25{,}375\ \text{ft}^3 \\ V &= 25{,}375/27 = 939.8\ \text{yd}^3 \approx 940\ \text{yd}^3 \end{aligned}

The 20% swell is not used. Swell describes the loose volume the trucks carry after excavation. The problem asks for the in-place (bank) volume removed from the pit, which is what the grade-rod cut depths measure directly.

Answer: C — 940 yd³

Why the other options are wrong

A.
Counts each shared-edge corner only once, as if the six grid points belonged to one figure: $(2{,}500/4)(11.8 + 14.4) = 625 \times 26.2 = 16{,}375\ \text{ft}^3 = 606\ \text{yd}^3$ (2{,}500/4)(11.8 + 14.4) = 625 \times 26.2 = 16{,}375\ \text{ft}^3 = 606\ \text{yd}^3. The grid formula is applied to each square separately, and the two shared corners are corners of both squares, so they carry weight 2.
B.
Makes both errors of A and D: counts the shared-edge corners once ( $606.5\ \text{yd}^3$ 606.5\ \text{yd}^3) and then applies the 20% swell, $606.5 \times 1.20 = 728\ \text{yd}^3$ 606.5 \times 1.20 = 728\ \text{yd}^3. Neither the corner weighting nor the loose-volume conversion is right for an in-place quantity.
D.
Weights the corners correctly ( $940\ \text{yd}^3$ 940\ \text{yd}^3) but then applies the 20% swell, $939.8 \times 1.20 = 1{,}128\ \text{yd}^3$ 939.8 \times 1.20 = 1{,}128\ \text{yd}^3. That is the loose volume the trucks would haul; the problem asks for the in-place volume removed from the pit, which the cut depths already measure. The swell figure is extraneous here.

Water Resources & Environmental · problem 10 of 10

An industrial plant's treated effluent has a 5-day BOD of 40 mg/L, measured in the laboratory at 20°C, where its base-e BOD rate constant is 0.25 day⁻¹. The effluent is discharged to a river at 26°C. Treating the ultimate BOD as independent of temperature, the BOD the effluent exerts (undiluted basis) during its first 3.0 days in the river is most nearly:

A.25.9 mg/L
B.29.6 mg/L
C.36.3 mg/L
D.44.8 mg/L

Answer: C — 36.3 mg/L

The laboratory result fixes the ultimate BOD; the river temperature fixes only the rate. Three relations run in sequence, and each one must be fed the right rate constant.

1. Ultimate BOD from the 20°C test. The 5-day result was measured at 20°C, so it is inverted with the 20°C constant:

L_0 = \frac{\text{BOD}_5}{1 - e^{-k_{20} \cdot 5}} = \frac{40}{1 - e^{-0.25 \times 5}} = \frac{40}{0.7135} = 56.06\ \text{mg/L}

L_0 = \frac{\text{BOD}_5}{1 - e^{-k_{20} \cdot 5}} = \frac{40}{1 - e^{-0.25 \times 5}} = \frac{40}{0.7135} = 56.06\ \text{mg/L}

\text{BOD}_t = L_0\left(1 - e^{-kt}\right)

\text{BOD}_t = L_0\left(1 - e^{-kt}\right) · §6.7.3.2 BOD Exertion (starts p. 426)

2. Rate constant at the river temperature. The kinetic temperature correction $k_T = k_{20} \theta^{(T - 20)}$ k_T = k_{20} \theta^{(T - 20)} uses a BOD coefficient that depends on the temperature range: $\theta = 1.135$ \theta = 1.135 for 4–20°C and $\theta = 1.056$ \theta = 1.056 for 21–30°C. At 26°C the second row governs:

k_{26} = 0.25 \times 1.056^6 = 0.25 \times 1.3869 = 0.3467\ \text{day}^{-1}

k_{26} = 0.25 \times 1.056^6 = 0.25 \times 1.3869 = 0.3467\ \text{day}^{-1}

k_T = k_{20}(\theta)^{T - 20}

k_T = k_{20}(\theta)^{T - 20} · §6.8.3.3 Kinetic Temperature Corrections (starts p. 443)

\theta = 1.056

\theta = 1.056 (BOD

k

,

T

= 21–30 °C) · §6.8.3.3 Kinetic Temperature Corrections (starts p. 443)

3. BOD exerted in 3.0 days at 26°C. The ultimate BOD is unchanged by temperature (as the stem directs), so

\text{BOD}_3 = L_0 (1 - e^{-k_{26} \cdot 3.0}) = 56.06 \times (1 - e^{-1.040}) = 56.06 \times 0.6466 = 36.25\ \text{mg/L}

\text{BOD}_3 = L_0 (1 - e^{-k_{26} \cdot 3.0}) = 56.06 \times (1 - e^{-1.040}) = 56.06 \times 0.6466 = 36.25\ \text{mg/L}

Check. Warmer water speeds exertion, so the 3-day value must exceed the 29.6 mg/L that the uncorrected 20°C constant gives, and it must stay below the 56.06 mg/L ultimate. 36.3 mg/L sits between them.

Answer: C — 36.3 mg/L

Why the other options are wrong

A.
Treats the 5-day BOD of 40 mg/L as the ultimate BOD and skips the back-calculation: $40 \times (1 - e^{-0.3467 \times 3.0}) = 40 \times 0.6466 = 25.9$ 40 \times (1 - e^{-0.3467 \times 3.0}) = 40 \times 0.6466 = 25.9 mg/L. The 5-day value is only the part exerted in 5 days at 20°C; the ultimate demand is 56.06 mg/L.
B.
Omits the temperature correction and uses the 20°C constant in the river: $56.06 \times (1 - e^{-0.25 \times 3.0}) = 56.06 \times 0.5276 = 29.6$ 56.06 \times (1 - e^{-0.25 \times 3.0}) = 56.06 \times 0.5276 = 29.6 mg/L. The river is at 26°C, so the rate constant rises to $0.3467\ \text{day}^{-1}$ 0.3467\ \text{day}^{-1}. (Using the 26°C constant to back-calculate the ultimate BOD from the 20°C test gives $48.58 \times 0.6466 = 31.4$ 48.58 \times 0.6466 = 31.4 mg/L, which also lands nearest this option; both routes misplace the temperature.)
D.
Corrects the constant with the 4–20°C coefficient $\theta = 1.135$ \theta = 1.135 instead of the 21–30°C coefficient: $k = 0.25 \times 1.135^6 = 0.5339\ \text{day}^{-1}$ k = 0.25 \times 1.135^6 = 0.5339\ \text{day}^{-1}, so $56.06 \times (1 - e^{-0.5339 \times 3.0}) = 56.06 \times 0.7988 = 44.8$ 56.06 \times (1 - e^{-0.5339 \times 3.0}) = 56.06 \times 0.7988 = 44.8 mg/L. The right table, the wrong row: 26°C falls in the 21–30°C range.

Free PE Civil Water Resources & Environmental practice exam

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