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Registered Public Utility Engineer - Water Supply & Drainage - independent engineering foundations study materials
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Question 1Water Resources and Environmental engineering models and applications

In the Hardy Cross loop method for water distribution network hydraulic balancing, what is the flow correction formula ($\Delta q$) for a closed loop with pipe head losses $h_i = S_i q_i |q_i|$ and algebraic sum of head losses $\sum h_i$?

  1. $\Delta q = -\frac{\sum h_i}{\sum S_i}$
  2. $\Delta q = -\frac{\sum h_i}{2 \sum \frac{h_i}{q_i}}$ Correct answer.
  3. $\Delta q = +\frac{\sum h_i}{2 \sum h_i q_i}$
  4. $\Delta q = -\frac{\sum q_i}{\sum h_i}$

Explanation

In a closed pipe loop, the sum of head losses must equal zero ($\sum h_i = 0$). By applying a first-order Taylor expansion to $h(q + \Delta q) \approx h(q) + 2 S q \Delta q = h + 2 \frac{h}{q} \Delta q$, setting $\sum h(q + \Delta q) = 0$ yields the Hardy Cross correction: $\Delta q = -\frac{\sum h_i}{2 \sum |h_i / q_i|}$ (for quadratic head loss equations like Darcy-Weisbach or Colebrook-White where exponent $m = 2$).

Question 2Water Resources and Environmental engineering models and applications

In municipal water distribution networks, which condition is the primary contributor to 'water age' (水龄) increase, disinfectant residual decay, and secondary bacterial regrowth?

  1. Operating high flow velocities ($> 2.5\text{ m/s}$) in transmission ring mains
  2. Oversized looped pipes, dead-end branch mains (盲端), and prolonged retention in balancing reservoirs Correct answer.
  3. Installing ductile iron pipes with internal cement mortar lining
  4. Maintaining elevated residual chlorine levels ($> 0.8\text{ mg/L}$) at treatment plant discharge

Explanation

Water age is the cumulative travel time of treated water through pipes and storage tanks. Oversized pipes during early development stages, dead ends, and excessive storage tank capacities cause stagnant or low-velocity flow ($< 0.05\text{ m/s}$), leading to complete disinfectant depletion, biofilm detachment, trihalomethane formation, and taste/odor complaints.

Question 3Water Resources and Environmental engineering models and applications

A high-lift water supply pumping station lifts water from a clear water reservoir (low water level $+2.00\text{ m}$) to an elevated storage tank (high water level $+42.00\text{ m}$). The total friction and minor head losses across suction and discharge piping are $\sum h_f = 6.50\text{ m}$. Including a safety head allowance of $h_{safe} = 1.50\text{ m}$, what is the required design pump total dynamic head ($H$)?

  1. $40.00\text{ m}$
  2. $46.50\text{ m}$
  3. $48.00\text{ m}$ Correct answer.
  4. $52.00\text{ m}$

Explanation

1. Static head ($H_{st}$): $H_{st} = Z_{discharge} - Z_{suction} = 42.00\text{ m} - 2.00\text{ m} = 40.00\text{ m}$.\n2. Total dynamic head ($H$): $H = H_{st} + \sum h_f + h_{safe} = 40.00\text{ m} + 6.50\text{ m} + 1.50\text{ m} = 48.00\text{ m}$.

Question 4Water Resources and Environmental engineering models and applications

When two identical centrifugal water pumps operate in parallel (并联运行) on a steep pipeline characteristic curve ($H = H_{st} + S Q^2$), why is the total combined discharge flow ($Q_{total}$) strictly less than twice the flow of a single pump operating alone ($2 \times Q_1$)?

  1. Parallel pumps interfere electrically with each other's variable frequency drives
  2. As total flow increases, pipe friction loss $\sum h_f = S Q^2$ increases quadratically, shifting the combined operating point to a higher head and lower individual pump discharge Correct answer.
  3. Water density decreases due to turbulence in the shared discharge header
  4. Centrifugal pumps automatically down-throttle when encountering parallel suction headers

Explanation

When two pumps operate in parallel, their combined $Q-H$ curve is obtained by adding flows at each head. However, the system curve is parabolic ($H = H_{st} + S Q^2$). Because pipeline friction increases with the square of total flow, the operating head of the parallel system is higher than that of a single pump operating alone. At this higher head, each pump delivers less than $Q_1$, resulting in $Q_{total} < 2 Q_1$.

Question 5Water Resources and Environmental engineering models and applications

In combined sewer overflow (CSO) pollution control under GB 50014-2021, how is the design intercepted flow rate ($Q_d$, 截流井截流流量) defined using the interception multiple ($n_o$, 截流倍数)?

  1. $Q_d = n_o \times Q_{storm}$
  2. $Q_d = (1 + n_o) \times Q_{dry}$, where $Q_{dry}$ is the dry-weather flow (旱流污水量) Correct answer.
  3. $Q_d = \frac{Q_{dry}}{1 + n_o}$
  4. $Q_d = Q_{storm} - n_o \times Q_{dry}$

Explanation

GB 50014-2021 (Section 4.3.2) specifies that interceptor sewers in combined sewer systems must be designed to carry the intercepted flow: $Q_d = (1 + n_o) Q_{dry}$, where $Q_{dry}$ is the design dry-weather wastewater flow and $n_o$ is the interception multiple (typically $n_o = 1$ to $3$ based on receiving water environmental capacity). Excess wet-weather flow ($Q - Q_d$) overflows the weir into the water body.

Question 6Water Resources and Environmental engineering models and applications

A municipal storm sewer catchment area has a drainage basin of $F = 20.0\text{ ha}$ ($200,000\text{ m}^2$). The comprehensive runoff coefficient is $\psi = 0.65$, and the design rainfall intensity is $q = 180.0\text{ L/(s}\cdot\text{ha})$. Using the rational method ($Q = q \psi F$) per GB 50014-2021, what is the design stormwater flow rate ($Q$)?

  1. $1,170\text{ L/s}$ ($1.17\text{ m}^3/\text{s}$)
  2. $2,340\text{ L/s}$ ($2.34\text{ m}^3/\text{s}$) Correct answer.
  3. $3,600\text{ L/s}$ ($3.60\text{ m}^3/\text{s}$)
  4. $4,680\text{ L/s}$ ($4.68\text{ m}^3/\text{s}$)

Explanation

Applying the rational method formula $Q = q \times \psi \times F$: $Q = 180.0\text{ L/(s}\cdot\text{ha}) \times 0.65 \times 20.0\text{ ha} = 2,340.0\text{ L/s} = 2.34\text{ m}^3/\text{s}$.

Question 7Water Resources and Environmental engineering models and applications

In storm sewer design under GB 50014-2021, the design rainfall duration is calculated as $t = t_1 + m t_2$. For a catchment with overland inlet concentration time $t_1 = 10.0\text{ min}$, pipe network travel time $t_2 = 12.0\text{ min}$, and sewer lag retardation factor $m = 1.20$, what is the design rainfall duration ($t$)?

  1. $22.0\text{ min}$
  2. $24.4\text{ min}$ Correct answer.
  3. $26.4\text{ min}$
  4. $32.0\text{ min}$

Explanation

Under GB 50014-2021 (Section 4.2.3), design rainfall duration is $t = t_1 + m t_2$. Here, $t = 10.0\text{ min} + 1.20 \times 12.0\text{ min} = 10.0 + 14.4 = 24.4\text{ min}$. The factor $m = 1.20$ accounts for pipe storage attenuation and non-instantaneous flow routing.

Question 8Water Resources and Environmental engineering models and applications

A concrete storm sewer pipe has a nominal internal diameter of $D = 0.60\text{ m}$. At full-bore flow ($h/D = 1.0$), the hydraulic radius is $R = D/4 = 0.15\text{ m}$. With a Manning roughness coefficient $n = 0.013$ and a pipe slope $S = 0.0040$, what is the flow velocity ($v$) calculated via Manning's equation ($v = \frac{1}{n} R^{2/3} S^{1/2}$)?

  1. $0.68\text{ m/s}$
  2. $1.37\text{ m/s}$ Correct answer.
  3. $2.15\text{ m/s}$
  4. $3.42\text{ m/s}$

Explanation

Applying Manning's formula: $v = \frac{1}{n} R^{2/3} S^{1/2}$.\n1. $R^{2/3} = (0.15)^{2/3} \approx 0.2823\text{ m}^{2/3}$.\n2. $S^{1/2} = (0.0040)^{1/2} \approx 0.06325$.\n3. $v = \frac{1}{0.013} \times 0.2823 \times 0.06325 = 76.923 \times 0.017855 \approx 1.373\text{ m/s} \approx 1.37\text{ m/s}$. This velocity exceeds the self-cleansing minimum of $0.75\text{ m/s}$ and is well below the maximum limit of $5.0\text{ m/s}$.

Question 9Water Resources and Environmental engineering models and applications

What is the primary engineering objective of constructing deep underground stormwater storage and drainage tunnels (深层排水调蓄隧道, e.g., Guangzhou, Shanghai, Wuhan) in mega-cities?

  1. To completely replace all street-level domestic sewer networks
  2. To temporarily capture peak stormwater runoff during extreme convective storm events, prevent urban waterlogging, and capture initial heavy CSO pollution for post-storm pump-out to treatment plants Correct answer.
  3. To provide underground drinking water storage for municipal earthquake emergencies
  4. To generate hydroelectric power during dry-weather flows

Explanation

Deep stormwater storage tunnels (constructed $30\text{–}50\text{ m}$ below ground) provide large volume capacity to: (1) Buffer extreme peak runoff that exceeds shallow drainage pipe capacities during severe cloudbursts, eliminating surface flooding; (2) Intercept heavily polluted initial stormwater and CSO discharges, storing them temporarily and pumping them to wastewater treatment plants at controlled rates during dry weather.

Question 10Water Resources and Environmental engineering models and applications

For public buildings with concentrated simultaneous water usage (such as theaters, stadiums, primary/secondary school recess lavatories) under GB 50015-2019, how is the domestic water design flow rate ($q_g$) calculated?

  1. Using the residential probability formula $q_g = 0.2 \alpha \sqrt{N_g} + k N_g$
  2. Using the simultaneous fixture utilization method $q_g = \sum q_0 n_0 b$, where $q_0$ is rated fixture flow, $n_0$ is fixture count, and $b$ is simultaneous usage percentage Correct answer.
  3. Taking $10\%$ of the municipal fire water flow
  4. Assuming a constant flow rate of $1.0\text{ L/s}$ per story

Explanation

GB 50015-2019 (Section 3.7.8) specifies that for buildings with concentrated peak usage periods (such as schools during class breaks, theaters during intermissions, industrial change rooms, and sports stadiums), the design flow rate is determined by the simultaneous percentage method: $q_g = \sum q_0 n_0 b$, where $b$ is the simultaneous usage percentage (often $100\%$ for urinals and flush valves during recess).

Question 11Water Resources and Environmental engineering models and applications

A residential building houses $m = 200$ residents with a design hot water consumption quota of $q_r = 60.0\text{ L/(person}\cdot\text{d})$ at $t_h = 60^\circ\text{C}$. The design cold water inlet temperature is $t_l = 10^\circ\text{C}$. The water heat capacity is $c = 4.187\text{ kJ/(kg}\cdot^\circ\text{C})$ and density is $\rho = 1.0\text{ kg/L}$. For a continuous 24-hour heating system ($T = 24\text{ h}$), what is the design thermal heating capacity ($Q_h$) under GB 50015-2019?

  1. $12.50\text{ kW}$
  2. $29.08\text{ kW}$ Correct answer.
  3. $58.15\text{ kW}$
  4. $104.68\text{ kW}$

Explanation

1. Total daily hot water mass: $M = m \times q_r \times \rho = 200 \times 60.0\text{ L} \times 1.0\text{ kg/L} = 12,000\text{ kg/d}$.\n2. Temperature differential: $\Delta t = t_h - t_l = 60^\circ\text{C} - 10^\circ\text{C} = 50^\circ\text{C}$.\n3. Total daily heat required: $Q_{total} = M \times c \times \Delta t = 12,000\text{ kg} \times 4.187\text{ kJ/(kg}\cdot^\circ\text{C}) \times 50^\circ\text{C} = 2,512,200\text{ kJ/d}$.\n4. Sizing heating power over $T = 24\text{ h} = 86,400\text{ s}$: $Q_h = \frac{2,512,200\text{ kJ}}{86,400\text{ s}} = 29.076\text{ kW} \approx 29.08\text{ kW}$.

Question 12Water Resources and Environmental engineering models and applications

In a centralized domestic hot water full-circulation system (全循环系统), how is the design circulation flow rate ($q_x$, 循环流量) of the circulation pump calculated under GB 50015-2019?

  1. $q_x = \frac{q_g}{2}$, taking half of the peak design flow
  2. $q_x = \frac{Q_s}{c \rho \Delta t}$, where $Q_s$ is total heat loss of supply/return pipes and $\Delta t = 5^\circ\text{C}$ is allowable temperature drop Correct answer.
  3. $q_x = 10.0\text{ L/s}$ uniform constant
  4. $q_x = \frac{c \rho \Delta t}{Q_s}$

Explanation

GB 50015-2019 (Section 6.5.15) mandates that the circulation flow rate must compensate for the thermal heat dissipation loss ($Q_s$) of the distribution network to ensure water arrives at user taps at $\ge 50^\circ\text{C}$: $q_x = \frac{Q_s}{c \rho \Delta t}$, where $\Delta t$ is the design temperature drop between supply water leaving the heater ($60^\circ\text{C}$) and return water entering the circulation pump ($55^\circ\text{C}$), typically $\Delta t = 5^\circ\text{C}$.

Question 13Water Resources and Environmental engineering models and applications

In a centralized solar water heating system designed for a multi-story hotel per GB 50015-2019, how is the required solar collector aperture area ($A_c$, 太阳能集热器采光面积) calculated?

  1. $A_c = \frac{Q_h \times f}{J_t \times \eta_{cd} \times (1 - \eta_L)}$, where $Q_h$ is daily hot water heat demand, $f$ is solar fraction ($0.4\text{–}0.6$), $J_t$ is solar irradiance, $\eta_{cd}$ is collector efficiency, and $\eta_L$ is piping heat loss factor Correct answer.
  2. $A_c = Q_h \times J_t \times f$
  3. $A_c = \frac{J_t}{Q_h \times \eta_{cd}}$
  4. $A_c = \frac{1}{f \times J_t}$

Explanation

GB 50015-2019 (Section 6.4.4) specifies that solar collector area is sized based on daily solar irradiance and design solar fraction: $A_c = \frac{Q_h \times f}{J_t \times \eta_{cd} \times (1 - \eta_L)}$, where: $Q_h$ is average daily heat demand; $f$ is design solar fraction (太阳能保证率, typically $40\%\text{–}60\%$); $J_t$ is annual average daily solar radiation on tilted surface ($\text{kJ/m}^2\cdot\text{d}$); $\eta_{cd}$ is average daily collector thermal efficiency ($0.45\text{–}0.55$); and $\eta_L$ is pipeline/tank heat loss rate ($0.15\text{–}0.20$).

Question 14Water Resources and Environmental engineering models and applications

For a stipulated nozzle model q=K*sqrt(p), take K=80L/min per sqrt(bar) and p=1bar. What discharge follows inL/min? This exercise does not specify an approved sprinkler operating condition.

  1. $25.3\text{ L/min}$ ($0.42\text{ L/s}$)
  2. $80.0\text{ L/min}$ ($1.33\text{ L/s}$) Correct answer.
  3. $113.1\text{ L/min}$ ($1.89\text{ L/s}$)
  4. $253.0\text{ L/min}$ ($4.22\text{ L/s}$)

Explanation

The standard sprinkler discharge formula is $q = K \sqrt{10 P}$, where $q$ is flow rate in $\text{L/min}$, $P$ is pressure in $\text{MPa}$, and $K$ is the sprinkler discharge coefficient. Substituting $K = 80$ and $P = 0.10\text{ MPa}$: $q = 80 \times \sqrt{10 \times 0.10} = 80 \times \sqrt{1.0} = 80.0\text{ L/min}$. In liters per second: $q = \frac{80.0}{60} \approx 1.333\text{ L/s}$.

Question 15Water supply, wastewater treatment and service boundaries

A rapid mixing tank has an effective volume of $V = 2.5\text{ m}^3$. The mechanical mixer inputs a net power of $P = 1,800\text{ W}$ into the water. At a water temperature of $20^\circ\text{C}$ (dynamic viscosity $\mu = 1.00 \times 10^{-3}\text{ Pa}\cdot\text{s}$), what is the rapid mixing velocity gradient ($G$, 速度梯度)?

  1. $450\text{ s}^{-1}$
  2. $600\text{ s}^{-1}$
  3. $849\text{ s}^{-1}$ Correct answer.
  4. $1,200\text{ s}^{-1}$

Explanation

The Camp-Stein velocity gradient formula is $G = \sqrt{\frac{P}{\mu V}}$. Substituting the given values: $G = \sqrt{\frac{1,800\text{ W}}{(1.00 \times 10^{-3}\text{ Pa}\cdot\text{s}) \times 2.5\text{ m}^3}} = \sqrt{\frac{1,800}{0.0025}} = \sqrt{720,000} \approx 848.53\text{ s}^{-1} \approx 849\text{ s}^{-1}$. This falls squarely within the GB 50013-2018 recommended range for rapid mixing ($G = 700\text{ s}^{-1}$ to $1,000\text{ s}^{-1}$).

Question 16Water supply, wastewater treatment and service boundaries

A drinking water plant treats a maximum daily flow of $Q = 120,000\text{ m}^3/\text{d}$. The plant operates 24 hours continuously with an ordinary rapid sand filter system designed at a normal filtration velocity of $v = 8.0\text{ m/h}$. If the filter station is divided into $N = 8$ equal rectangular filter cells, what is the required net filtration area per cell ($f$)?

  1. $52.1\text{ m}^2$
  2. $62.5\text{ m}^2$
  3. $78.1\text{ m}^2$ Correct answer.
  4. $93.8\text{ m}^2$

Explanation

1. Calculate the hourly design flow rate: $Q_h = \frac{120,000\text{ m}^3/\text{d}}{24\text{ h/d}} = 5,000\text{ m}^3/\text{h}$.\n2. Calculate the total required net filter area: $A = \frac{Q_h}{v} = \frac{5,000\text{ m}^3/\text{h}}{8.0\text{ m/h}} = 625.0\text{ m}^2$.\n3. Sizing into $N = 8$ cells: $f = \frac{A}{N} = \frac{625.0\text{ m}^2}{8} = 78.125\text{ m}^2 \approx 78.1\text{ m}^2$.

Question 17Water supply, wastewater treatment and service boundaries

According to the Chick-Watson law of disinfection ($C^n t = \text{constant}$), a disinfection process achieves 3-log (99.9%) pathogen inactivation with a disinfectant residual of $C_1 = 2.0\text{ mg/L}$ at a contact time of $t_1 = 15\text{ min}$ (assuming $n = 1.0$). If the plant reduces the disinfectant dosage so that residual drops to $C_2 = 0.75\text{ mg/L}$, what contact time ($t_2$) is required to achieve the same 3-log inactivation?

  1. $20\text{ min}$
  2. $30\text{ min}$
  3. $40\text{ min}$ Correct answer.
  4. $60\text{ min}$

Explanation

With $n = 1.0$, the $CT$ value is constant: $C_1 \times t_1 = C_2 \times t_2$. Here, $CT = 2.0\text{ mg/L} \times 15\text{ min} = 30\text{ mg}\cdot\text{min/L}$. Sizing for $C_2 = 0.75\text{ mg/L}$: $t_2 = \frac{CT}{C_2} = \frac{30\text{ mg}\cdot\text{min/L}}{0.75\text{ mg/L}} = 40.0\text{ min}$.

Question 18Water supply, wastewater treatment and service boundaries

In water treatment chemistry, what chemical phenomena occur during 'breakpoint chlorination' (折点加氯) when chlorine is added to raw water containing ammonia nitrogen ($\text{NH}_3\text{-N}$)?

  1. Chlorine directly oxidizes turbidity particles into carbon dioxide gas without reacting with ammonia
  2. Chlorine reacts with ammonia to form chloramines up to a peak; further chlorine addition oxidizes chloramines to $\text{N}_2$ gas, reaching a minimum residual (the breakpoint) before free available chlorine appears Correct answer.
  3. Chlorine precipitates ammonia as ammonium chloride crystals which are removed in sedimentation tanks
  4. Ammonia acts as a catalyst that prevents any chlorine consumption until pH drops below 4.0

Explanation

In breakpoint chlorination: (1) Chlorine reacts with ammonia to form monochloramine and dichloramine ($2\text{NH}_3 + 3\text{Cl}_2 \to \text{NH}_2\text{Cl} + \text{NHCl}_2 + 3\text{HCl}$), increasing combined residual chlorine to a peak at a $\text{Cl}_2:\text{NH}_3\text{-N}$ molar ratio of $\approx 1:1$ (mass ratio $5:1$); (2) Additional chlorine oxidizes chloramines to inert nitrogen gas ($\text{NH}_2\text{Cl} + \text{NHCl}_2 + \text{HOCl} \to \text{N}_2\uparrow + 4\text{HCl} + \text{H}_2\text{O}$), driving total residual down to a minimum (the breakpoint, typically at mass ratio $7.6:1$ to $8:1$); (3) Past the breakpoint, any added chlorine exists as free available chlorine ($\text{HOCl}$ and $\text{OCl}^-$).

Question 19Water supply, wastewater treatment and service boundaries

Dissolved Air Flotation (DAF / 气浮池) is designated as the preferred clarification process over gravity sedimentation under GB 50013-2018 when treating which type of raw water?

  1. High-turbidity river water during seasonal monsoon floods ($> 2,000\text{ NTU}$)
  2. Low-temperature, low-turbidity surface water with heavy algal blooms (蓝藻水华) and high organic content Correct answer.
  3. Deep groundwater containing high concentrations of dissolved ferric and manganous ions
  4. Seawater with high total dissolved solids and high salinity

Explanation

GB 50013-2018 (Section 9.4.26) specifies that Dissolved Air Flotation (DAF) is particularly well-suited for low-temperature, low-turbidity raw waters with high organic matter, low floc density, and high algae concentrations ($< 50\text{ NTU}$, density near water). Micro-bubbles ($20\text{–}40\ \mu\text{m}$) attach to the light algal flocs, causing them to float rapidly with surface separation rates of $4\text{–}8\text{ m/h}$.

Question 20Water supply, wastewater treatment and service boundaries

In the standard $A^2/O$ (Anaerobic-Anoxic-Oxic / 厌氧-缺氧-好氧) biological nutrient removal process, what biochemical mechanisms occur in the Anaerobic and Anoxic zones, respectively?

  1. Anaerobic zone: Nitrification of ammonia; Anoxic zone: Aerobic carbon oxidation
  2. Anaerobic zone: PAOs release phosphorus while taking up VFAs; Anoxic zone: Denitrifiers reduce nitrate ($\text{NO}_3^-\text{-N}$) to $\text{N}_2$ gas using influent organic carbon Correct answer.
  3. Anaerobic zone: Total BOD destruction to methane gas; Anoxic zone: Heavy metal precipitation
  4. Anaerobic zone: Sludge aerobic digestion; Anoxic zone: Ozone disinfection

Explanation

In the $A^2/O$ process: (1) In the Anaerobic zone (DO $< 0.2\text{ mg/L}$, $\text{NO}_3^-\text{-N} < 0.5\text{ mg/L}$), Polyphosphate Accumulating Organisms (PAOs) break down intracellular polyphosphate to release orthophosphate into solution while storing volatile fatty acids (VFAs) as polyhydroxyalkanoates (PHAs); (2) In the Anoxic zone (DO $< 0.5\text{ mg/L}$, rich in $\text{NO}_3^-$ from aerobic internal recycle), heterotrophic denitrifiers use raw wastewater COD as electron donors to reduce nitrate to harmless nitrogen gas ($\text{N}_2$).

Question 21Water supply, wastewater treatment and service boundaries

In an explicitly idealized nitrate-recycle exercise, define removal fraction e=(R+r)/(1+R+r), where r=1.0 is supplied return-sludge ratio and R is internal recycle ratio. What R gives e=0.80? This equation is not a complete biological plant-design rule.

  1. $R = 100\%$ ($1.0$)
  2. $R = 200\%$ ($2.0$)
  3. $R = 300\%$ ($3.0$) Correct answer.
  4. $R = 500\%$ ($5.0$)

Explanation

Theoretical total nitrogen removal in pre-denitrification ($A^2/O$) systems is governed by the recycle fraction: $\eta_{TN} = \frac{R + r}{1 + R + r}$. Setting $\eta_{TN} = 0.80$ and $r = 1.0$: $0.80 = \frac{R + 1.0}{1 + R + 1.0} = \frac{R + 1.0}{R + 2.0}$. Solving for $R$: $0.80(R + 2.0) = R + 1.0 \implies 0.80 R + 1.60 = R + 1.0 \implies 0.20 R = 0.60 \implies R = 3.0 = 300\%$.

Question 22Water supply, wastewater treatment and service boundaries

Stoichiometrically, how many grams of alkalinity (as $\text{CaCO}_3$) are consumed per gram of ammonia nitrogen ($\text{NH}_4^+\text{-N}$) fully oxidized to nitrate ($\text{NO}_3^-\text{-N}$) during nitrification, and how much alkalinity is recovered per gram of nitrate nitrogen reduced to $\text{N}_2$ during denitrification?

  1. Consumed: $3.57\text{ g CaCO}_3 / \text{g N}$; Recovered: $7.14\text{ g CaCO}_3 / \text{g N}$
  2. Consumed: $7.14\text{ g CaCO}_3 / \text{g N}$; Recovered: $3.57\text{ g CaCO}_3 / \text{g N}$ Correct answer.
  3. Consumed: $1.00\text{ g CaCO}_3 / \text{g N}$; Recovered: $1.00\text{ g CaCO}_3 / \text{g N}$
  4. Consumed: $14.28\text{ g CaCO}_3 / \text{g N}$; Recovered: $0.00\text{ g CaCO}_3 / \text{g N}$

Explanation

In biological wastewater treatment: (1) Complete biological nitrification ($\text{NH}_4^+ + 2\text{O}_2 \to \text{NO}_3^- + 2\text{H}^+ + \text{H}_2\text{O}$) consumes $2\text{ moles}$ of $\text{H}^+$ equivalent ($1\text{ mole }\text{CaCO}_3 = 100\text{ g}$ per $14\text{ g N}$), which equals $7.14\text{ g CaCO}_3 / \text{g NH}_4^+\text{-N}$; (2) Denitrification produces hydroxyl ions, recovering exactly $50\%$ of consumed alkalinity ($3.57\text{ g CaCO}_3 / \text{g NO}_3^-\text{-N}$ reduced).

Question 23Water supply, wastewater treatment and service boundaries

In the design of an activated sludge aeration basin under GB 50014-2021, why is the Actual Oxygen Transfer Rate ($AOTR$) under field conditions significantly lower than the Standard Oxygen Transfer Rate ($SOTR$ in clean water at $20^\circ\text{C}$ and $1.0\text{ atm}$)?

  1. Field wastewater contains surfactants/salts lowering the mass transfer coefficient factor $\alpha < 1.0$, lower saturation factor $\beta < 1.0$, and operating DO $C_L > 0\text{ mg/L}$ reduces the driving concentration gradient Correct answer.
  2. Bacteria actively absorb nitrogen gas, blocking oxygen gas diffusion into flocs
  3. Aeration blowers automatically run at half speed when submerged in wastewater
  4. Water pressure increases with depth, preventing bubble formation

Explanation

The conversion formula is $AOTR = SOTR \times \left[ \frac{\beta C^*_{s,T,P} - C_L}{C^*_{s,20,1.0}} \right] \times \alpha \times 1.024^{(T-20)}$. In municipal wastewater: (1) Detergents, dissolved solids, and oils reduce the oxygen mass transfer coefficient ($\alpha \approx 0.80\text{–}0.85$); (2) Dissolved contaminants lower oxygen solubility ($\beta \approx 0.95$); (3) Maintaining operating dissolved oxygen (e.g., $C_L = 2.0\text{ mg/L}$) reduces the driving concentration gradient $(C^*_s - C_L)$ compared to standard zero-DO clean water.

Question 24Water supply, wastewater treatment and service boundaries

An activated sludge aeration basin has an effective volume of $V = 10,000\text{ m}^3$ and operates with a Mixed Liquor Suspended Solids concentration of $X = 3,500\text{ mg/L}$ ($3.5\text{ kg/m}^3$). The daily waste activated sludge mass discharged from the system (including effluent solids loss) is $\Delta X = 2,500\text{ kg/d}$. What is the Solids Retention Time (SRT / 污泥龄 $\theta_c$)?

  1. $4.0\text{ days}$
  2. $10.0\text{ days}$
  3. $14.0\text{ days}$ Correct answer.
  4. $28.0\text{ days}$

Explanation

Solids Retention Time (SRT / $\theta_c$) is the average residence time of biological solids in the reactor: $\theta_c = \frac{V \times X}{\Delta X}$. Total biomass in the reactor is $V \times X = 10,000\text{ m}^3 \times 3.5\text{ kg/m}^3 = 35,000\text{ kg}$. Therefore: $\theta_c = \frac{35,000\text{ kg}}{2,500\text{ kg/d}} = 14.0\text{ days}$. An SRT of $14\text{ days}$ satisfies the nitrification threshold at $12^\circ\text{–}15^\circ\text{C}$.

Question 25Water supply, wastewater treatment and service boundaries

A circular secondary sedimentation clarifier (二沉池) treats an effluent design flow of $Q = 24,000\text{ m}^3/\text{d}$ ($1,000\text{ m}^3/\text{h}$). The return activated sludge (RAS) ratio is $R = 80\%$ ($0.80$), and the aeration basin mixed liquor suspended solids is $MLSS = 3.5\text{ kg/m}^3$. If the total effective surface area of the clarifiers is $A = 1,200\text{ m}^2$, what is the Solids Loading Rate ($SLR$, 固体表面负荷)?

  1. $2.92\text{ kg/(m}^2\cdot\text{h})$
  2. $5.25\text{ kg/(m}^2\cdot\text{h})$ Correct answer.
  3. $7.85\text{ kg/(m}^2\cdot\text{h})$
  4. $10.50\text{ kg/(m}^2\cdot\text{h})$

Explanation

Solids Loading Rate accounts for total solids entering the clarifier from both influent flow $Q$ and return sludge flow $Q_R = R \times Q$: $SLR = \frac{Q (1 + R) \times MLSS}{A}$.\n1. Total flow entering clarifiers: $Q_{total} = 1,000\text{ m}^3/\text{h} \times (1 + 0.80) = 1,800\text{ m}^3/\text{h}$.\n2. Total solids feed rate: $W_{solids} = 1,800\text{ m}^3/\text{h} \times 3.5\text{ kg/m}^3 = 6,300\text{ kg/h}$.\n3. $SLR = \frac{6,300\text{ kg/h}}{1,200\text{ m}^2} = 5.25\text{ kg/(m}^2\cdot\text{h})$. GB 50014-2021 specifies $SLR \le 6.0\text{–}8.0\text{ kg/(m}^2\cdot\text{h})$ for secondary clarifiers, so $5.25\text{ kg/(m}^2\cdot\text{h})$ is fully compliant.

Question 26Water supply, wastewater treatment and service boundaries

In a standard Sequencing Batch Reactor (SBR / CASS / CAST) wastewater treatment facility, what is the chronological operating sequence of phases within each operational cycle?

  1. Settle $\to$ Decant $\to$ Aerate $\to$ Fill $\to$ Sludge Digestion
  2. Fill (进水) $\to$ React/Aerate (反应/曝气) $\to$ Settle (沉淀) $\to$ Decant (滗水) $\to$ Idle/Sludge Wasting (闲置/排泥) Correct answer.
  3. Decant $\to$ Disinfect $\to$ Filter $\to$ Clarify $\to$ Aerate
  4. Anaerobic Fermentation $\to$ Chemical Precipitation $\to$ Air Scour $\to$ Backwash

Explanation

An SBR is a fill-and-draw activated sludge system operating in time sequences rather than spatial tanks. Each complete cycle comprises five distinct sequential steps: (1) Fill (raw wastewater enters the basin); (2) React (aeration and/or mixing for BOD removal, nitrification, and denitrification); (3) Settle (quiescent gravitational solid-liquid separation without hydraulic flow); (4) Decant (clarified supernatant is removed via a motorized surface decanter); (5) Idle/Waste (standby period and scheduled wasting of surplus activated sludge).

Question 27Water supply, wastewater treatment and service boundaries

A wastewater treatment plant produces $W_{dry} = 10,000\text{ kg/d}$ of dry primary and secondary sludge solids. The volatile solids fraction is $VS/TS = 70.0\%$. In a mesophilic anaerobic sludge digester operating at $35^\circ\text{C}$, the volatile solids destruction efficiency is $50.0\%$. If the specific biogas production rate is $0.85\text{ m}^3 / (\text{kg VS destroyed})$, what is the daily biogas production ($V_{gas}$)?

  1. $1,487.5\text{ m}^3/\text{d}$
  2. $2,975.0\text{ m}^3/\text{d}$ Correct answer.
  3. $4,250.0\text{ m}^3/\text{d}$
  4. $5,950.0\text{ m}^3/\text{d}$

Explanation

1. Mass of volatile solids fed to digester: $VS_{feed} = 10,000\text{ kg/d} \times 0.70 = 7,000\text{ kg VS/d}$.\n2. Mass of volatile solids destroyed: $VS_{destroyed} = 7,000\text{ kg VS/d} \times 0.50 = 3,500\text{ kg VS destroyed/d}$.\n3. Daily biogas production: $V_{gas} = 3,500\text{ kg VS destroyed/d} \times 0.85\text{ m}^3/\text{kg} = 2,975.0\text{ m}^3/\text{d}$. The generated biogas typically contains $60\%\text{–}65\%\text{ CH}_4$ and $35\%\text{ CO}_2$.

Question 28Geotechnical engineering models and applications

In dam foundation hydrogeology, what is the standard definition of the Lugeon unit (Lu, 吕荣值) determined from water pressure packer tests (压水试验)?

  1. A water absorption rate of 1.0 Liter per minute per meter of test section (1.0 L/(min·m)) under an effective injection water pressure of 1.0 MPa (10 bar) Correct answer.
  2. A permeability coefficient of exactly 1.0 cm/s under atmospheric pressure
  3. The total groundwater inflow rate into a 100 m deep borehole in m³/hour
  4. The percentage of karst dissolution cavities per cubic meter of rock

Explanation

In water conservancy engineering geology (SL 251-2015), 1 Lugeon (1 Lu) is defined as a water take of 1.0 liter per minute per linear meter of borehole test section under a steady excess water test pressure of 1.0 MPa (10 atmospheres). 1 Lu corresponds approximately to a hydraulic conductivity k ≈ 1.3 × 10⁻⁵ cm/s (10⁻⁷ m/s).

Question 29Geotechnical engineering models and applications

What is the primary engineering function of Consolidation Grouting (固结灌浆) conducted across the shallow surface zone of a dam foundation rock footprint?

  1. To seal shallow blasting-induced fissures, improve rock mass integrity and deformation modulus, reduce differential settlement, and enhance the shear strength of the dam-rock contact zone Correct answer.
  2. To form a deep vertical cutoff barrier penetrating 100 m into impermeable bedrock
  3. To drain high uplift groundwater pressures from beneath the dam
  4. To chemically dissolve karst limestone cavities

Explanation

Consolidation grouting (固结灌浆) is shallow pattern grouting (depths of 3 to 8 m arranged in a 2×2 m to 4×4 m grid) over the entire dam footprint. It fills decompression and excavation blast fissures, stiffens the rock mass, increases deformation modulus, homogenizes stress distribution, and improves contact shear strength.

Question 30Geotechnical engineering models and applications

In dam abutment rock slope engineering, what geological condition is required for a Planar Sliding Failure (平面滑动) to occur kinematically along a single structural discontinuity plane?

  1. The strike of the discontinuity plane is nearly parallel to the slope face (within ±20°), its dip direction is outward toward daylight in the slope, and its dip angle satisfies φ < α < β (where φ is friction angle, α is discontinuity dip, and β is slope inclination) Correct answer.
  2. The discontinuity dips into the hill (opposite to slope face inclination)
  3. The discontinuity dip angle α is strictly less than the internal friction angle φ
  4. The discontinuity is completely perpendicular to the slope face

Explanation

Kinematic planar sliding requires: (1) Discontinuity strike strikes approximately parallel (±20°) to the slope face; (2) Discontinuity dips out of the slope face (daylights: α < β, dip is gentler than slope angle); and (3) Dip angle α exceeds the friction angle φ (α > φ), ensuring gravitational driving shear exceeds frictional resistance.

Question 31Geotechnical engineering models and applications

In karst hydrogeological investigations for reservoir sites (SL 251-2015), which hydrogeological feature indicates the most critical hazard of severe reservoir leakage (水库渗漏)?

  1. The presence of continuous karstified carbonate strata with a regional karst groundwater table located well below the proposed reservoir normal pool level connecting to an adjacent low-lying river valley Correct answer.
  2. The existence of thick impermeable shale formations forming a continuous watershed divide
  3. A karst water table that naturally slopes upwards towards the surrounding mountain ridges above pool level
  4. The complete absence of carbonate soluble rock in the catchment

Explanation

Severe reservoir leakage occurs when soluble carbonate rocks (limestone, dolomite) contain interconnected karst conduits, sinkholes, and underground rivers where the natural karst groundwater divide is lower than the reservoir normal pool level (低地下水位型水库), creating strong hydraulic gradients toward adjacent valleys or downstream river reaches.

Question 32Geotechnical engineering models and applications

What is the primary operational mechanism and geological prerequisite for Reservoir-Induced Seismicity (RIS, 水库诱发地震) occurring after initial reservoir impoundment?

  1. Water impoundment increases pore fluid pressure in pre-existing critically stressed active faults, reducing effective normal stress (σ' = σ - u) and triggering premature shear slip Correct answer.
  2. Water impoundment dissolves all deep granitic tectonic plates instantaneously
  3. Reservoir surface wave action creates tectonic subduction zones
  4. Hydroelectric power transmission creates magnetic fault resonance

Explanation

According to geomechanical principles of fault reactivation (Terzaghi effective stress / Mohr-Coulomb theory): Deep water percolation transmits high hydrostatic pore fluid pressures (u) into fault zones under high tectonic pre-stress. The reduction in effective normal stress (σ' = σ - u) lowers the shear resistance (τ_f = c + σ' tan φ), triggering fault rupture and induced seismic tremors.

Question 33Construction engineering models and applications

In Critical Path Method (CPM) network scheduling, how is Total Float (TF) of an activity calculated from Early Start (ES), Late Start (LS), Early Finish (EF), and Late Finish (LF)?

  1. TF = LS - ES = LF - EF Correct answer.
  2. TF = ES - LS
  3. TF = EF - ES
  4. TF = LF - LS

Explanation

Total Float is the maximum duration an activity can be delayed from its early start without delaying the project completion date: TF = LS - ES = LF - EF. Activities with zero total float form the Critical Path.

Question 34Construction engineering models and applications

Under an NEC4 Option C Target Cost Contract, the initial Target Price is HK$ 100M. Due to efficiency, the actual final Defined Cost + Fee incurred by the Contractor is HK$ 88M. If the contract pain/gain share arrangement specifies a 50/50 split for cost savings below target, what is the Contractor's Gain Share bonus payout?

  1. HK$ 6M Correct answer.
  2. HK$ 12M
  3. HK$ 44M
  4. HK$ 88M

Explanation

Total cost saving below target = Target Price - Actual Defined Cost = HK$ 100M - HK$ 88M = HK$ 12M. Under the 50/50 gain share rule, the savings are shared equally between Employer and Contractor: Contractor Gain Share = 50% * HK$ 12M = HK$ 6M. Total payout to Contractor = Defined Cost (HK$ 88M) + Gain Share (HK$ 6M) = HK$ 94M.

Question 35Construction engineering models and applications

A Chartered Civil Engineer reviewing site drawings discovers a critical calculation error in a temporary works cofferdam design prepared by another engineering firm that poses an imminent collapse risk. What is the engineer's ethical duty under HKIE Rules of Conduct?

  1. Ignore the error because it was prepared by a competing firm
  2. Immediately notify the site project team, contractor, and designer to rectify the hazard before work proceeds Correct answer.
  3. Wait until the cofferdam fails before reporting
  4. Sell the information to local news media anonymously

Explanation

HKIE Code of Ethics mandates that if a member identifies a structural hazard or error that threatens life safety, they must take immediate action to notify responsible parties (contractor, designer, and relevant authorities) to stop unsafe work and prevent collapse. Public safety overrides commercial courtesies.

Question 36Construction engineering models and applications

An employer explicitly instructs a HKIE member engineer to sign and certify a structural submission containing unverified calculation assumptions that violate the HK Concrete Code. What is the engineer's proper ethical course of action?

  1. Sign the submission immediately to keep their job
  2. Refuse to certify the non-compliant submission, explain the safety non-compliance in writing, and escalate if pressure continues Correct answer.
  3. Sign the submission but add a hidden footnote in fine print
  4. Delegate the signing to an unqualified junior intern

Explanation

Under HKIE Code of Ethics, a member must not sign or seal documents that do not comply with engineering standards or statutory safety codes. If pressured by an employer or client, the engineer must refuse certification, document the technical non-compliance in writing, and if necessary report the matter to HKIE or relevant regulatory authorities.

Question 37Construction engineering models and applications

A construction project requires excavating 5,000 cubic yards of soil in its bank (in-place) state. The soil has a swell factor of 1.4 (1 bank cy expands to 1.4 loose cy when excavated). How many cubic yards of loose soil must be hauled away?

  1. 3,571 cy
  2. 5,000 cy
  3. 6,000 cy
  4. 7,000 cy Correct answer.

Explanation

Loose volume = bank volume × swell factor = 5,000 × 1.4 = 7,000 cubic yards. Soil expands (swells) when excavated from its compacted in-place (bank) state to a loose hauled state; the swell factor (loose/bank) is typically 1.1–1.4 for most soils.

Question 38Construction engineering models and applications

In a critical path method (CPM) schedule, Activity A has a duration of 5 days, Activity B has a duration of 8 days, and Activity C has a duration of 4 days. Activity C can only start after both A and B are complete. What is the early start (ES) for Activity C?

  1. 5 days
  2. 8 days Correct answer.
  3. 9 days
  4. 13 days

Explanation

Activity C requires both A and B to be complete. Activity A finishes at day 5, Activity B finishes at day 8. Activity C can start when the longer predecessor (B) finishes, so ES = 8 days. This is determined by the maximum of all predecessor early finish times.

Question 39Construction engineering models and applications

A concrete mix design requires 350 lbs of cement per cubic yard. For a 100 cubic yard pour with 8% waste factor, how many 94-lb sacks of cement are required?

  1. 371 sacks
  2. 372 sacks
  3. 402 sacks
  4. 403 sacks Correct answer.

Explanation

Total concrete needed = 100 × 1.08 = 108 cy. Total cement = 350 × 108 = 37,800 lbs. Number of sacks = 37,800 ÷ 94 = 402.13 sacks. Round up to 403 sacks (next whole sack). Always round up cement quantities for construction ordering.

Question 40Construction engineering models and applications

What is the primary purpose of a slump test for fresh concrete?

  1. To determine compressive strength
  2. To measure workability and consistency Correct answer.
  3. To test for air content
  4. To verify cement content

Explanation

The slump test (ASTM C143) measures the workability and consistency of fresh concrete. A cone is filled with concrete, lifted, and the settlement (slump) is measured. Typical specifications require 2-4 inches for most structural work. Compressive strength is tested with cylinders cured for 28 days.

Question 41Construction engineering models and applications

A temporary excavation support system uses soldier piles with timber lagging. The lagging spans 8 feet horizontally between soldier piles. If the lateral earth pressure is 600 psf and the lagging is 3-inch thick Douglas Fir-Larch (Fb = 1,500 psi), is the lagging adequate?

  1. Yes, stress is 768 psi
  2. Yes, stress is 1,200 psi
  3. No, stress is 3,200 psi Correct answer.
  4. No, stress is 2,400 psi

Explanation

For a uniformly loaded simple span (1-ft vertical strip of lagging): M = wL²/8 = (600 psf × 1 ft) × (8 ft)² / 8 = 4,800 ft-lb = 57,600 in-lb. For 3" lagging (S = bd²/6 = 12 × 3² / 6 = 18 in³), f = M/S = 57,600/18 = 3,200 psi. This exceeds 1,500 psi allowable, so the lagging is not adequate.

Question 42Construction engineering models and applications

A horizontal curve on a construction site has a radius of 500 feet and a central angle of 30°. What is the length of the curve?

  1. 131 ft
  2. 262 ft Correct answer.
  3. 524 ft
  4. 785 ft

Explanation

Curve length L = R × Δ × (π/180) = 500 × 30 × (π/180) = 500 × 0.5236 = 261.8 ft ≈ 262 ft. This is the arc length along the curve centerline.

Question 43Water Resources and Environmental engineering models and applications

A river basin has an observed annual runoff series with a sample mean runoff x̄ = 850 mm and a coefficient of variation Cv = 0.35. What is the standard deviation (σ) of this runoff series?

  1. 297.5 mm Correct answer.
  2. 242.8 mm
  3. 350.0 mm
  4. 148.8 mm

Explanation

In hydrological statistics per SL 44-2006, the coefficient of variation is defined as Cv = σ / x̄, where σ is the standard deviation and x̄ is the mean. Therefore, σ = Cv × x̄ = 0.35 × 850 mm = 297.5 mm.

Question 44Water Resources and Environmental engineering models and applications

When plotting empirical hydrological frequencies for a continuous series of n = 49 years of annual peak flood data per SL 44-2006, what is the empirical exceedance probability (P) of the 5th largest historical flood (m = 5) using the standard Weibull plotting position formula P = m / (n + 1)?

  1. 10.0% Correct answer.
  2. 10.2%
  3. 8.0%
  4. 12.5%

Explanation

According to SL 44-2006 (Section 3.2), the empirical frequency of hydrological data arranged in descending order is calculated using the unbiased mathematical expectation formula (Weibull formula): P = [m / (n + 1)] × 100%. Here, P = [5 / (49 + 1)] × 100% = 5 / 50 × 100% = 10.0%.

Question 45Water Resources and Environmental engineering models and applications

A 1-hour Unit Hydrograph (1h UH, 单位线) for a 10 mm effective rainfall excess has runoff ordinates: q1 = 15 m³/s at t = 1h, q2 = 40 m³/s at t = 2h, q3 = 25 m³/s at t = 3h, and q4 = 10 m³/s at t = 4h. If a storm produces two consecutive 1-hour net rainfall pulses of R1 = 20 mm in the first hour and R2 = 30 mm in the second hour, what is the total direct runoff discharge Q2 at t = 2h?

  1. 125 m³/s Correct answer.
  2. 80 m³/s
  3. 145 m³/s
  4. 170 m³/s

Explanation

The unit hydrograph ordinates are scaled by rainfall excess factors (relative to the 10 mm unit): for R1 = 20 mm, multiplier = 2.0; for R2 = 30 mm, multiplier = 3.0. By linear convolution, at t = 2h: Q2 = (R1/10) × q2 + (R2/10) × q1 = 2.0 × 40 m³/s + 3.0 × 15 m³/s = 80 + 45 = 125 m³/s.

Question 46Water Resources and Environmental engineering models and applications

In reservoir water resources regulation per SL 322, which storage capacity is specifically defined as the volume between the Dead Water Level (死水位) and the Normal Pool Level (正常蓄水位)?

  1. Active Storage / Usable Storage (兴利库容 / 有效库容) Correct answer.
  2. Flood Control Storage (防洪库容)
  3. Dead Storage (死库容)
  4. Surcharge / Total Flood Regulation Storage (调洪库容)

Explanation

Active Storage (兴利库容 / 有效库容, V_a) is the reservoir storage volume situated between the Dead Water Level (Z_d) and the Normal Pool Level (Z_n), utilized for seasonal and multi-year regulation of water supply, irrigation, and hydropower generation.

Question 47Water Resources and Environmental engineering models and applications

During a reservoir flood routing calculation over a time step Δt = 1 hour (3600 s), the inflow at the beginning and end of the step are I1 = 500 m³/s and I2 = 900 m³/s, respectively. The outflow at the start is Q1 = 300 m³/s and at the end is Q2 = 420 m³/s. If the initial storage was V1 = 1.200 × 10⁷ m³, what is the reservoir storage V2 at the end of the time step?

  1. 1.3224 × 10⁷ m³ Correct answer.
  2. 1.2612 × 10⁷ m³
  3. 1.4160 × 10⁷ m³
  4. 1.1784 × 10⁷ m³

Explanation

Using the continuity equation for reservoir flood routing: ΔV = V2 - V1 = [(I1 + I2)/2 - (Q1 + Q2)/2] × Δt. Average inflow = (500 + 900)/2 = 700 m³/s; Average outflow = (300 + 420)/2 = 360 m³/s. Net storage change ΔV = (700 - 360) m³/s × 3600 s = 340 × 3600 = 1,224,000 m³ = 1.224 × 10⁶ m³. Thus, V2 = 1.200 × 10⁷ + 0.1224 × 10⁷ = 1.3224 × 10⁷ m³.

Question 48Water Resources and Environmental engineering models and applications

A watershed has a drainage area of F = 500 km². A 24-hour design storm produces an average precipitation depth of P = 120 mm. If the total catchment loss (infiltration, interception, depression storage) is 40 mm, what is the total design runoff volume (W) produced by this storm event?

  1. 4.00 × 10⁷ m³ Correct answer.
  2. 6.00 × 10⁷ m³
  3. 2.00 × 10⁷ m³
  4. 4.00 × 10⁶ m³

Explanation

Net runoff depth R = Precipitation (P) - Losses (L) = 120 mm - 40 mm = 80 mm = 0.080 m. Catchment area F = 500 km² = 500 × 10⁶ m². Total runoff volume W = F × R = 500 × 10⁶ m² × 0.080 m = 40 × 10⁶ m³ = 4.00 × 10⁷ m³.

Question 49Water Resources and Environmental engineering models and applications

In the Horton infiltration model f(t) = fc + (f0 - fc) e^(-k t), the initial infiltration capacity is f0 = 40 mm/h, the ultimate steady infiltration capacity is fc = 10 mm/h, and the decay constant is k = 1.5 h⁻¹. What is the infiltration capacity f(t) at t = 2.0 hours?

  1. 11.49 mm/h Correct answer.
  2. 10.00 mm/h
  3. 14.48 mm/h
  4. 25.00 mm/h

Explanation

Substituting the parameters into Horton's equation: f(2) = 10 + (40 - 10) × e^(-1.5 × 2.0) = 10 + 30 × e^(-3.0). Since e^(-3.0) ≈ 0.049787, f(2) = 10 + 30 × 0.049787 = 10 + 1.4936 = 11.49 mm/h.

Question 50Water Resources and Environmental engineering models and applications

A river catchment has a 30-year systematic runoff record. Statistical fitting yields a mean annual discharge Q̄ = 120 m³/s, Cv = 0.40, and Cs = 2.5 Cv = 1.0. From the standard Pearson Type III modulus table (Kp table), the modulus coefficient for P = 1% design flood is K_1% = 2.25. What is the design annual flood discharge Q_1%?

  1. 270 m³/s Correct answer.
  2. 216 m³/s
  3. 300 m³/s
  4. 168 m³/s

Explanation

In Pearson Type III frequency analysis per SL 44-2006, the design hydrological quantity for exceedance probability P is calculated as Q_P = Q̄ × K_P, where K_P is the statistical modulus coefficient. Thus, Q_1% = 120 m³/s × 2.25 = 270 m³/s.