FE Civil Water Resources & Env. practice problems

A triangular channel has no bottom width, so $A = zy^2 = 1.5(1.2)^2 = 2.16$ m$^2$ and the wetted perimeter is both sloped sides: $P = 2y\sqrt{1+z^2} = 2(1.2)\sqrt{1+2.25} = 4.327$ m. $R_H = 2.16/4.327 = 0.499$ m. SI units so $K = 1.0$: $Q = (1/0.018)(2.16)(0.499)^{2/3}(0.0016)^{1/2} = 3.02$ m$^3$/s. The 4.47 m³/s distractor uses $P = 2y$ (vertical depth only) instead of the sloped sides, inflating $R_H$. The 0.95 m³/s distractor mistakes the triangular area for $A = zy^2/2 = 1.08$ m$^2$ (treating it as half base times height with the wrong base), which (together with the smaller $R_H$) drops $Q$ to about 0.95 m$^3$/s. The 4.49 m³/s distractor uses $K = 1.486$ (USCS) with SI inputs.

1. Flow area (triangular, $z = 1.5$):
   $A = zy^2 = 1.5(1.2)^2 = 2.16\,\text{m}^2$
2. Wetted perimeter (two sloped sides):
   $P = 2y\sqrt{1+z^2} = 2(1.2)\sqrt{1+2.25} = 4.327\,\text{m}$
3. Hydraulic radius:
   $R_H = \frac{2.16}{4.327} = 0.499\,\text{m}$
4. Manning's equation (SI, $K = 1.0$):
   $Q = \frac{1.0}{0.018}(2.16)(0.499)^{2/3}(0.0016)^{1/2}$
5. Compute:
   $Q = 55.56 \times 2.16 \times 0.629 \times 0.04 = 3.02\,\text{m}^3\text{/s}$

FE Reference Handbook 10.6, p. 303

For a pipe flowing half full: $A = \pi D^2/8 = \pi(4)^2/8 = 6.283$ ft$^2$ and $P = \pi D/2 = 2\pi = 6.283$ ft, so $R_H = A/P = 1.0$ ft. The neat result is that a half-full pipe has the SAME hydraulic radius as a full pipe ($D/4 = 1.0$ ft), but only half the flow area. $Q = (1.486/0.024)(6.283)(1.0)^{2/3}(0.005)^{1/2} = 27.5$ cfs. The 55.0 cfs choice uses the full-pipe area ($\pi D^2/4$), doubling the answer. The 43.7 cfs choice uses $R_H = D/2$ instead of D/4. The 18.5 cfs choice forgets the $K = 1.486$ factor for USCS units.

1. Flow area (half full):
   $A = \frac{\pi D^2}{8} = \frac{\pi(4)^2}{8} = 6.283\,\text{ft}^2$
2. Wetted perimeter (half circumference):
   $P = \frac{\pi D}{2} = 2\pi = 6.283\,\text{ft}$
3. Hydraulic radius:
   $R_H = \frac{A}{P} = \frac{6.283}{6.283} = 1.0\,\text{ft} = \frac{D}{4}$
4. Manning's equation (USCS, $K = 1.486$):
   $Q = \frac{1.486}{0.024}(6.283)(1.0)^{2/3}(0.005)^{1/2}$
5. Compute:
   $Q = 61.92 \times 6.283 \times 1.0 \times 0.0707 = 27.5\,\text{ft}^3\text{/s}$

FE Reference Handbook 10.6, p. 303

Specific energy is depth plus velocity head: $E = y + v^2/(2g)$. First the velocity: $v = Q/A = 20/(4 \times 2.5) = 2.0$ m/s. Then $v^2/(2g) = 4/(2 \times 9.81) = 0.204$ m. $E = 2.5 + 0.204 = 2.70$ m. The 2.50 m choice reports only the depth, forgetting the velocity head entirely. The 2.91 m choice uses $v^2/g$ (omitting the factor of 2 in the velocity head). The 0.20 m choice reports only the velocity head, forgetting to add the depth.

1. Velocity:
   $v = \frac{Q}{By} = \frac{20}{4 \times 2.5} = 2.0\,\text{m/s}$
2. Velocity head:
   $\frac{v^2}{2g} = \frac{(2.0)^2}{2(9.81)} = 0.204\,\text{m}$
3. Specific energy:
   $E = y + \frac{v^2}{2g} = 2.5 + 0.204 = 2.70\,\text{m}$

FE Reference Handbook 10.6, p. 301

When both conjugate depths are known, the energy loss has a closed form: $\Delta E = (y_2 - y_1)^3/(4 y_1 y_2)$. $\Delta E = (2.5 - 0.5)^3/(4 \times 0.5 \times 2.5) = (2.0)^3/5.0 = 8/5 = 1.60$ m. The 6.40 m option omits the factor of 4 in the denominator: $(2.0)^3/(0.5 \times 2.5) = 6.40$. The 2.00 m option just reports the depth difference $y_2 - y_1 = 2.0$. The 4.00 m option drops $y_2$ from the denominator: $(2.0)^3/(4 \times 0.5) = 4.00$.

1. Energy loss across a hydraulic jump:
   $\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2}$
2. Substitute:
   $\Delta E = \frac{(2.5 - 0.5)^3}{4(0.5)(2.5)}$
3. Compute:
   $\Delta E = \frac{(2.0)^3}{5.0} = \frac{8}{5} = 1.60\,\text{m}$

FE Reference Handbook 10.6, p. 303

Convert $D = 8\text{ in.} = 0.667$ ft, so $A = \pi(0.667)^2/4 = 0.349$ ft$^2$ and $R_H = D/4 = 0.167$ ft. Rearrange Hazen-Williams for the slope: $S_v = [Q/(k_1 C A R_H^{0.63})]^{1/0.54}$. Plugging in: $S_v = [1.8/(1.318 \times 140 \times 0.349 \times 0.167^{0.63})]^{1/0.54} = 0.01073$. Head loss $= S_v L = 0.01073 \times 1{,}200 = 12.9$ ft. The 5.7 ft choice uses $R_H = D/2$, overstating $R_H$ and lowering the required slope. The 24.0 ft choice uses the rougher 20-year cast iron value $C = 100$. The 29.1 ft choice uses the SI coefficient $k_1 = 0.849$ instead of the USCS 1.318.

1. Pipe properties (full, $D = 0.667$ ft):
   $A = \frac{\pi(0.667)^2}{4} = 0.349\,\text{ft}^2, \quad R_H = \frac{D}{4} = 0.167\,\text{ft}$
2. Solve Hazen-Williams for the energy slope:
   $S_v = \left[\frac{Q}{k_1 C A R_H^{0.63}}\right]^{1/0.54}$
3. Substitute:
   $S_v = \left[\frac{1.8}{1.318(140)(0.349)(0.167)^{0.63}}\right]^{1.852} = 0.01073$
4. Head loss:
   $h_f = S_v L = 0.01073 \times 1{,}200 = 12.9\,\text{ft}$

FE Reference Handbook 10.6, p. 303

The 90-degree V-notch formula is $Q = CH^{5/2}$, so $H = (Q/C)^{2/5} = (8.0/2.54)^{0.4} = (3.150)^{0.4} = 1.58$ ft. The 2.01 ft choice uses the SI coefficient $C = 1.40$ instead of the USCS 2.54. The 2.15 ft choice uses the rectangular weir exponent 3/2, solving $H = (Q/C)^{2/3} = 2.15$. The 3.15 ft choice forgets the exponent entirely, computing $H = Q/C = 3.15$.

1. 90-degree V-notch weir (USCS):
   $Q = CH^{5/2}$
2. Solve for the head:
   $H = \left(\frac{Q}{C}\right)^{2/5} = \left(\frac{8.0}{2.54}\right)^{0.4}$
3. Compute:
   $H = (3.150)^{0.4} = 1.58\,\text{ft}$

FE Reference Handbook 10.6, p. 303

First find the composite curve number weighted by area: $CN = 0.70(80) + 0.30(60) = 56 + 18 = 74$. Then $S = 1000/CN - 10 = 1000/74 - 10 = 3.51$ in. Check: $0.2S = 0.70$ in. $< P = 3.5$ in., so runoff occurs. $Q = (P - 0.2S)^2/(P + 0.8S) = (3.5 - 0.70)^2/(3.5 + 2.81) = (2.80)^2/6.31 = 7.84/6.31 = 1.24$ in. The 1.94 in. distractor omits the initial-abstraction subtraction in the numerator, using $P^2/(P + 0.8S) = 1.94$. The 2.80 in. distractor reports the un-squared numerator $(P - 0.2S) = 2.80$. The 3.50 in. distractor assumes all rainfall becomes runoff.

1. Composite (area-weighted) curve number:
   $CN = 0.70(80) + 0.30(60) = 74$
2. Maximum retention:
   $S = \frac{1{,}000}{74} - 10 = 3.51\,\text{in.}$
3. Check initial abstraction:
   $0.2S = 0.70\,\text{in.} < P = 3.5\,\text{in.}\;\checkmark$
4. SCS runoff equation:
   $Q = \frac{(3.5 - 0.70)^2}{3.5 + 2.81} = \frac{(2.80)^2}{6.31} = 1.24\,\text{in.}$

FE Reference Handbook 10.6, p. 296

Rearrange $Q = CIA$ to solve for area: $A = Q/(CI) = 90/(0.55 \times 3.0) = 90/1.65 = 54.5$ acres. The Rational Method units work out directly (1 acre-in./hr $\approx$ 1 cfs), so no conversion is needed. The 16.5-acre choice multiplies by $C$ instead of dividing: $QC/I = 90(0.55)/3.0 = 16.5$. The 30.0-acre choice ignores $C$ entirely: $Q/I = 90/3.0 = 30$. The 163.6-acre choice forgets the intensity: $Q/C = 90/0.55 = 163.6$.

1. Rearrange the Rational Method:
   $A = \frac{Q}{CI}$
2. Substitute:
   $A = \frac{90}{0.55 \times 3.0} = \frac{90}{1.65}$
3. Compute:
   $A = 54.5\,\text{acres}$

FE Reference Handbook 10.6, p. 296

Each rainfall pulse produces its own scaled, lagged unit hydrograph; you add the two. The 3-inch pulse starts one hour later than the 2-inch pulse. Tabulate hour by hour and sum. The maximum occurs at hour 3: from the 2-inch pulse $2 \times 150 = 300$, plus from the lagged 3-inch pulse $3 \times 300 = 900$, totaling $1{,}200$ cfs. The 1,500 cfs choice simply scales the UH peak by the total depth ($5 \times 300 = 1{,}500$), which is wrong because the two pulses do not peak in the same hour. The 900 cfs choice counts only the 3-inch pulse on the UH peak ($3 \times 300$). The 600 cfs choice counts only the 2-inch pulse peak ($2 \times 300$) and ignores the lagged 3-inch pulse.

1. Pulse 1 (2 in.) scaled UH peaks at hour 2:
   $2 \times 300 = 600\,\text{cfs at } t = 2$
2. Pulse 2 (3 in.) is lagged 1 hour; its UH peak lands at hour 3:
   $3 \times 300 = 900\,\text{cfs at } t = 3$
3. At hour 3, add both contributions:
   $\underbrace{2 \times 150}_{\text{pulse 1}} + \underbrace{3 \times 300}_{\text{pulse 2}} = 300 + 900$
4. Peak direct runoff:
   $Q_{peak} = 1{,}200\,\text{cfs}$

FE Reference Handbook 10.6, p. 298

Runoff depth is volume divided by area, then converted to inches. In consistent units, depth $= V/A = 1{,}500/6{,}000 = 0.25$ ft. Convert to inches: $0.25 \times 12 = 3.0$ in. The acre cancels between acre-ft and acres, leaving feet, which you multiply by 12. The 0.25 in. choice stops at 0.25 ft but mislabels it as inches (forgot the factor of 12). The 6.0 in. choice doubles the result by an arithmetic slip. The 36.0 in. choice multiplies by 12 twice (applies 144 instead of 12).

1. Depth in feet (acres cancel):
   $d = \frac{V}{A} = \frac{1{,}500\,\text{acre-ft}}{6{,}000\,\text{acres}} = 0.25\,\text{ft}$
2. Convert to inches:
   $d = 0.25\,\text{ft} \times 12\,\frac{\text{in.}}{\text{ft}} = 3.0\,\text{in.}$

FE Reference Handbook 10.6, p. 298

Darcy's law gives the volumetric flow: $Q = KAi$, where the gradient is $i = dh/L = 6/120 = 0.05$. So $Q = (4 \times 10^{-5})(300)(0.05) = 6.0 \times 10^{-4}$ m$^3$/s. Convert to per day: $6.0 \times 10^{-4} \times 86{,}400 = 51.8$ m$^3$/day. The 0.17 m$^3$/day choice forgets the cross-sectional area, using $Ki$ only (times 86,400). The $6{,}221$ m$^3$/day choice forgets to divide the head drop by the path length, using $i = 6$ instead of 0.05, inflating $Q$ by 120 times. The 2.6 m$^3$/day choice mishandles the area, dividing by it instead of multiplying.

1. Hydraulic gradient:
   $i = \frac{dh}{L} = \frac{6}{120} = 0.05$
2. Darcy discharge:
   $Q = KAi = (4 \times 10^{-5})(300)(0.05) = 6.0 \times 10^{-4}\,\text{m}^3\text{/s}$
3. Convert to per day:
   $Q = 6.0 \times 10^{-4} \times 86{,}400 = 51.8\,\text{m}^3\text{/day}$

FE Reference Handbook 10.6, p. 298

Rearrange Dupuit, $Q = \pi K(h_2^2 - h_1^2)/\ln(r_2/r_1)$, to solve for the well-face head: $h_1 = \sqrt{h_2^2 - Q\ln(r_2/r_1)/(\pi K)}$. The log term: $\ln(300/1) = 5.70$. So $h_1 = \sqrt{80^2 - 0.8(5.70)/(\pi \times 6 \times 10^{-4})} = \sqrt{6400 - 2419} = \sqrt{3981} = 63.1$ ft. The drawdown is about 17 ft. The 73.1 ft choice uses $\log_{10}$ instead of the natural log. The 72.0 ft choice uses $2\pi$ (the confined Thiem coefficient) instead of $\pi$. The 80.0 ft choice forgets to apply any drawdown and just reports the undisturbed level.

1. Dupuit's formula solved for the well-face head:
   $h_1 = \sqrt{h_2^2 - \frac{Q\ln(r_2/r_1)}{\pi K}}$
2. Evaluate the log term:
   $\ln\!\left(\frac{300}{1}\right) = 5.70$
3. Substitute:
   $h_1 = \sqrt{80^2 - \frac{0.8(5.70)}{\pi(6 \times 10^{-4})}} = \sqrt{6{,}400 - 2{,}419}$
4. Compute:
   $h_1 = \sqrt{3{,}981} = 63.1\,\text{ft}$

FE Reference Handbook 10.6, p. 299

BOD remaining is the organic matter not yet decomposed: $L_t = L_0 e^{-kt} = 320\,e^{-0.18 \times 4} = 320\,e^{-0.72} = 320(0.487) = 155.8$ mg/L. Note the BOD exerted ($L_0(1 - e^{-kt}) = 164.2$) plus the remaining (155.8) sum to the ultimate 320. The 164.2 mg/L choice gives the BOD exerted rather than remaining. The 57.6 mg/L choice applies $kt$ as a percentage ($0.18 \times 4 = 0.72$, then $320 \times 0.18 = 57.6$), a formula confusion. The 320 mg/L choice reports the ultimate BOD, as if no decay had occurred.

1. BOD remaining (decay form):
   $L_t = L_0 e^{-kt} = 320\,e^{-0.18 \times 4}$
2. Evaluate the exponent:
   $e^{-0.72} = 0.487$
3. BOD remaining at 4 days:
   $L_t = 320 \times 0.487 = 155.8\,\text{mg/L}$

FE Reference Handbook 10.6, p. 327

Find the critical time, then the critical deficit, then subtract from saturation. With $D_0 = 0$, the critical time is $t_c = \frac{1}{k_r - k_d}\ln(k_r/k_d) = \frac{1}{0.2}\ln(2) = 3.47$ days. The critical deficit is $D_c = \frac{k_d}{k_r}L_a e^{-k_d t_c} = \frac{0.2}{0.4}(20)e^{-0.2(3.47)} = 10(0.5) = 5.0$ mg/L. Minimum DO $= DO_{sat} - D_c = 9.0 - 5.0 = 4.0$ mg/L. The 5.0 mg/L choice reports the critical deficit itself, forgetting to subtract it from saturation. The 9.0 mg/L choice ignores the deficit entirely. The 0.0 mg/L choice wrongly drops the decay factor $e^{-k_d t_c}$, giving $D_c = 10$ and a negative DO clamped to zero.

1. Critical time (with $D_0 = 0$):
   $t_c = \frac{1}{k_r - k_d}\ln\!\left(\frac{k_r}{k_d}\right) = \frac{1}{0.2}\ln(2) = 3.47\,\text{days}$
2. Critical deficit:
   $D_c = \frac{k_d}{k_r}L_a\, e^{-k_d t_c} = \frac{0.2}{0.4}(20)e^{-0.2(3.47)} = 10(0.5) = 5.0\,\text{mg/L}$
3. Minimum dissolved oxygen:
   $DO_{min} = DO_{sat} - D_c = 9.0 - 5.0 = 4.0\,\text{mg/L}$

FE Reference Handbook 10.6, p. 327

Detention time is volume over flow: $\theta = V/Q$. The tank volume is $V = 80 \times 20 \times 12 = 19{,}200$ ft$^3$. Convert the flow to ft$^3$/day: $Q = 1{,}500{,}000\,\text{gpd}/7.48 = 200{,}535$ ft$^3$/day. So $\theta = 19{,}200/200{,}535 = 0.0958$ days $= 0.0958 \times 24 = 2.3$ hours, a typical primary clarifier value. The 0.31 hours choice forgets to divide by 7.48 (treats gallons as cubic feet). The 17.2 hours choice inverts the gallon conversion. The 0.096 choice reports the answer in days (0.096) rather than converting to hours.

1. Tank volume:
   $V = 80 \times 20 \times 12 = 19{,}200\,\text{ft}^3$
2. Flow in ft$^3$/day:
   $Q = \frac{1{,}500{,}000\,\text{gpd}}{7.48} = 200{,}535\,\text{ft}^3\text{/day}$
3. Detention time:
   $\theta = \frac{V}{Q} = \frac{19{,}200}{200{,}535} = 0.0958\,\text{day}$
4. Convert to hours:
   $\theta = 0.0958 \times 24 = 2.3\,\text{hours}$

FE Reference Handbook 10.6, p. 345

Overflow rate is $v_o = Q/A$, so the required surface area is $A = Q/v_o = 2{,}500{,}000/900 = 2{,}778$ ft$^2$. For a circular tank $A = \pi D^2/4$, so $D = \sqrt{4A/\pi} = \sqrt{4(2{,}778)/\pi} = \sqrt{3{,}537} = 59.5$ ft. The 29.7 ft choice forgets the factor of 4, computing $D = \sqrt{A/\pi} = 29.7$ ft (this is actually the radius). The 88.4 ft choice uses $A = \pi D^2$ without the 1/4. The 2,778 ft choice reports the surface area itself rather than solving for the diameter.

1. Required surface area:
   $A = \frac{Q}{v_o} = \frac{2{,}500{,}000}{900} = 2{,}778\,\text{ft}^2$
2. Solve the circular area for diameter:
   $D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4(2{,}778)}{\pi}}$
3. Compute:
   $D = \sqrt{3{,}537} = 59.5\,\text{ft}$

FE Reference Handbook 10.6, p. 345

CT = C × t₁₀ = 2.0 × 45 = 90 mg·min/L. The 22.5 mg·min/L choice divides instead of multiplying (45/2). The 47 mg·min/L choice adds the two values (2 + 45). The 0.044 mg·min/L choice divides C by t₁₀ (2/45) rather than multiplying.

1. CT relationship:
   $CT = C \times t_{10}$
2. Compute:
   $CT = 2.0 \times 45 = 90\,\text{mg·min/L}$

FE Reference Handbook 10.6, p. 352

Uniformity coefficient UC = d₆₀/d₁₀ = 0.75/0.50 = 1.5. A lower UC means more uniform grain sizes. The 0.67 option inverts the ratio (d₁₀/d₆₀). The 1.25 option uses a wrong pairing. The 0.25 option subtracts the sizes.

1. Uniformity coefficient:
   $UC = \frac{d_{60}}{d_{10}} = \frac{0.75}{0.50}$
2. Compute:
   $UC = 1.5$

FE Reference Handbook 10.6, p. 347

Efficiency = (in − out)/in = (220 − 30)/220 = 190/220 = 0.864 = 86.4%. The 13.6% choice is the fraction remaining (30/220). The 30% choice reports the effluent number as a percent. The 7.3% choice inverts the ratio.

1. Removal efficiency:
   $E = \frac{220 - 30}{220} \times 100\%$
2. Compute:
   $E = \frac{190}{220} = 86.4\%$

Total dynamic head = static + losses = 18 + 6.5 = 24.5 m. Fluid power = γQH = 9,810 × 0.030 × 24.5 = 7,210 W ≈ 7.2 kW. Brake power = 7.2/0.70 = 10.3 kW. The 7.2 kW option is the fluid power (no efficiency). The 5.3 kW option uses only the static lift (9,810 × 0.030 × 18 ≈ 5,300 W). The 5.0 kW option multiplies by 0.70 instead of dividing (7.2 × 0.70 ≈ 5.0).

1. Total dynamic head:
   $H = 18 + 6.5 = 24.5\,\text{m}$
2. Fluid power:
   $\gamma Q H = 9{,}810 \times 0.030 \times 24.5 = 7{,}210\,\text{W}$
3. Brake power:
   $\frac{7{,}210}{0.70} = 10{,}300\,\text{W} \approx 10.3\,\text{kW}$

FE Reference Handbook 10.6, p. 197