I85PREP
⌘ K
Registered Civil Engineer - Water Conservancy & Hydropower (China) - independent engineering foundations study materials
Registered Civil Engineer - Water Conservancy & Hydropower (China) - independent engineering foundations study resources

Registered Civil Engineer - Water Conservancy & Hydropower (China) - independent engineering foundations practice questions

One question at a time. Build your Registered Civil Engineer - Water Conservancy & Hydropower (China) - independent engineering foundations understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Water 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 2Water 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 3Water 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 4Water 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 5Water 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 6Water 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 7Water 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 8Water 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.

Question 9Water Resources and Environmental engineering models and applications

In water resources planning for a large multi-purpose reservoir, what is the standard definition of the Flood Limited Water Level (FLWL, 防洪限制水位)?

  1. The upper limit water level to which the reservoir is allowed to store water during the designated flood season Correct answer.
  2. The maximum water level reached during the passage of the check flood
  3. The lowest water level to which the reservoir may be drawn down during the dry season
  4. The static headwater level corresponding to the minimum firm hydropower capacity

Explanation

Per SL 252 and SL 322, the Flood Limited Water Level (防洪限制水位 / 汛期限制水位) is the maximum permissible water level in the reservoir prior to or during the flood season. Maintaining this level reserves adequate flood control storage to absorb impending design flood inflows.

Question 10Water Resources and Environmental engineering models and applications

A proposed water diversion project requires an annual water delivery of W_demand = 4.50 × 10⁸ m³. If the main canal has a conveyance loss of 15% and the regional field distribution system has an application efficiency of 80%, what total gross annual volume must be diverted at the river intake?

  1. 6.62 × 10⁸ m³ Correct answer.
  2. 5.29 × 10⁸ m³
  3. 5.63 × 10⁸ m³
  4. 7.50 × 10⁸ m³

Explanation

The overall water system efficiency is η_total = η_conveyance × η_distribution = (1 - 0.15) × 0.80 = 0.85 × 0.80 = 0.68 (68%). The required gross intake diversion volume is W_intake = W_demand / η_total = 4.50 × 10⁸ m³ / 0.68 = 6.6176 × 10⁸ m³ ≈ 6.62 × 10⁸ m³.

Question 11Water Resources and Environmental engineering models and applications

In the rational formula for design peak discharge in small ungauged catchments (SL 44-2006), Q_m = 0.278 × α × S × F / τ^n, what physical mechanism does the parameter τ represent?

  1. Catchment basin concentration time (汇流历时), comprising overland flow time and channel flow time Correct answer.
  2. Total return period of the design rainstorm in years
  3. Soil moisture infiltration recovery duration
  4. Reservoir spillway flood routing time lag

Explanation

In the Chinese hydrological rational method for small catchments per SL 44-2006, τ is the total basin catchment concentration time (流域汇流历时, in hours), which represents the sum of the overland slope travel time (坡面汇流历时) and the main stream channel transit time (河网汇流历时).

Question 12Water Resources and Environmental engineering models and applications

A hydrological station analyzes annual extreme 3-day flood volumes (W3d). The mean volume is W̄ = 1.80 × 10⁸ m³, with Cv = 0.50 and Cs = 3.0 Cv = 1.5. For a 100-year design flood (P = 1.0%), the modulus coefficient is K_1% = 2.58. What is the design 3-day flood volume W_3d,1%?

  1. 4.64 × 10⁸ m³ Correct answer.
  2. 3.60 × 10⁸ m³
  3. 5.20 × 10⁸ m³
  4. 2.70 × 10⁸ m³

Explanation

Per SL 44-2006, the design flood volume for a given return period P is obtained from W_P = W̄ × K_P. For P = 1%: W_3d,1% = 1.80 × 10⁸ m³ × 2.58 = 4.644 × 10⁸ m³ ≈ 4.64 × 10⁸ m³.

Question 13Water Resources and Environmental engineering models and applications

When assessing the annual runoff regulation performance of a multi-year regulation reservoir (多年调节水库), which method is traditionally employed to determine the required multi-year active storage capacity (V_long)?

  1. Cumulative mass curve method (Ripley's method / 累积时段均值分析与差积曲线法) Correct answer.
  2. Manning open channel backwater curve calculation
  3. Joukowsky water hammer pressure integration
  4. Bishop slice equilibrium summation

Explanation

In water resources and hydropower planning (SL 322), the cumulative mass curve method (Ripple diagram) or chronological difference integration curve (差积曲线法 / 逐时段水利计算法) is the standard technique used to balance surplus and deficit runoff over multi-year drought-wet hydrological cycles to determine the active multi-year storage V_long.

Question 14Water Resources and Environmental engineering models and applications

A hypothetical reservoir starts with50000000m^3. During the stated event it receives32000000m^3 and discharges18000000m^3. Ignore evaporation and seepage. What storage remains at the end? The totals alone do not determine the maximum storage during the event.

  1. 6.40 × 10⁷ m³ Correct answer.
  2. 8.20 × 10⁷ m³
  3. 5.00 × 10⁷ m³
  4. 4.60 × 10⁷ m³

Explanation

The endpoint balance is initial storage plus inflow minus outflow:50000000+32000000-18000000=64000000m^3. Maximum interim storage requires the inflow/outflow timing and the actual routing relationships, which are not supplied.

Question 15Water Resources and Environmental engineering models and applications

A river basin of area F = 1200 km² experiences a design rainstorm with an average total precipitation P = 160 mm. The catchment storage capacity is W_m = 100 mm, and the antecedent soil moisture before the storm is W_0 = 70 mm. Assuming a saturation-excess runoff mechanism (蓄满产流) where all rainfall after soil moisture reaches saturation becomes runoff, what is the total direct runoff depth (R)?

  1. 130 mm Correct answer.
  2. 60 mm
  3. 90 mm
  4. 160 mm

Explanation

In the saturation-excess runoff mechanism: the soil moisture deficit before rainfall is ΔW = W_m - W_0 = 100 mm - 70 mm = 30 mm. Since precipitation P = 160 mm > ΔW, the soil is filled to capacity, absorbing 30 mm, and the remaining precipitation becomes direct runoff: R = P - ΔW = 160 mm - 30 mm = 130 mm.

Question 16Water Resources and Environmental engineering models and applications

An annual maximum flood peak series of length n = 40 years has an arithmetic sample mean x̄ = 1500 m³/s and sample variance S² = 225,000 (m³/s)². What is the sample coefficient of variation (Cv)?

  1. 0.316 Correct answer.
  2. 0.150
  3. 0.474
  4. 0.225

Explanation

The standard deviation is the square root of the variance: σ = sqrt(S²) = sqrt(225,000) = 474.34 m³/s. The coefficient of variation is Cv = σ / x̄ = 474.34 / 1500 = 0.3162 ≈ 0.316.

Question 17Water Resources and Environmental engineering models and applications

A straight rectangular concrete open channel has a bottom width b = 4.0 m, a uniform water depth h = 2.0 m, a longitudinal bed slope S = 0.001 (1‰), and a Manning roughness coefficient n = 0.020. What is the steady uniform discharge (Q) carried by this channel?

  1. 12.65 m³/s Correct answer.
  2. 16.00 m³/s
  3. 9.48 m³/s
  4. 25.30 m³/s

Explanation

Flow area A = b × h = 4.0 × 2.0 = 8.0 m². Wetted perimeter χ = b + 2h = 4.0 + 2(2.0) = 8.0 m. Hydraulic radius R = A / χ = 8.0 / 8.0 = 1.0 m. Using Manning's formula: Q = (1/n) × A × R^(2/3) × S^(1/2) = (1 / 0.020) × 8.0 × (1.0)^(2/3) × sqrt(0.001) = 50 × 8.0 × 1.0 × 0.0316228 = 400 × 0.0316228 = 12.649 m³/s ≈ 12.65 m³/s.

Question 18Water Resources and Environmental engineering models and applications

For rectangular uniform open-channel sections with fixed flow area, slope and Manning roughness, which width-to-depth relation maximizes hydraulic radius and modeled discharge? Do not assume the channel is wide compared with depth.

  1. Channel bottom width b is equal to twice the water depth h (b = 2h), resulting in hydraulic radius R = h / 2 Correct answer.
  2. Channel bottom width b is equal to the water depth h (b = h), resulting in R = h / 3
  3. Channel bottom width b is equal to four times the water depth h (b = 4h)
  4. Channel bottom width b is independent of water depth

Explanation

For a rectangular channel, minimizing the wetted perimeter χ = b + 2h subject to constant area A = b·h yields dχ/dh = -A/h² + 2 = 0, so A = 2h², which means b = 2h. The hydraulic radius is R = A / χ = 2h² / (4h) = h / 2.

Question 19Water Resources and Environmental engineering models and applications

In a rectangular open channel with a unit discharge q = 8.0 m³/(s·m) and gravitational acceleration g = 9.81 m/s², what is the critical water depth (h_c)?

  1. 1.87 m Correct answer.
  2. 1.45 m
  3. 2.20 m
  4. 0.98 m

Explanation

For a rectangular channel, the critical depth is calculated by h_c = (q² / g)^(1/3) = (8.0² / 9.81)^(1/3) = (64 / 9.81)^(1/3) = (6.523955)^(1/3) ≈ 1.8686 m ≈ 1.87 m.

Question 20Water Resources and Environmental engineering models and applications

Water flows at a depth h = 2.50 m and mean velocity v = 3.00 m/s in an open channel. Taking the kinetic energy correction factor α = 1.0 and g = 9.81 m/s², what is the specific energy (E, 断面比能) of this cross-section?

  1. 2.96 m Correct answer.
  2. 3.42 m
  3. 2.50 m
  4. 4.10 m

Explanation

The specific energy is defined relative to the channel bottom as E = h + α × v² / (2g). Here, E = 2.50 + 1.0 × (3.00)² / (2 × 9.81) = 2.50 + 9.00 / 19.62 = 2.50 + 0.4587 = 2.9587 m ≈ 2.96 m.

Question 21Water Resources and Environmental engineering models and applications

Water flows in an open channel with a depth h = 1.50 m and flow velocity v = 6.00 m/s. Taking g = 9.81 m/s², what is the Froude number (Fr) and what flow regime does it represent?

  1. Fr = 1.56, representing supercritical flow (急流) Correct answer.
  2. Fr = 0.64, representing subcritical flow (缓流)
  3. Fr = 1.00, representing critical flow (临界流)
  4. Fr = 2.44, representing tranquil flow (层流)

Explanation

The Froude number is Fr = v / sqrt(g·h) = 6.00 / sqrt(9.81 × 1.50) = 6.00 / sqrt(14.715) = 6.00 / 3.836 = 1.564. Since Fr > 1.0, the flow is supercritical (急流).

Question 22Water Resources and Environmental engineering models and applications

In a horizontal rectangular stilling basin, a supercritical jet enters with an initial depth h1 = 0.80 m and initial velocity v1 = 8.00 m/s. Using g = 9.81 m/s², what is the second conjugate water depth (h2, 跃后水深) of the resulting free hydraulic jump?

  1. 2.86 m Correct answer.
  2. 3.52 m
  3. 2.14 m
  4. 4.20 m

Explanation

First, calculate Fr1 = v1 / sqrt(g·h1) = 8.00 / sqrt(9.81 × 0.80) = 8.00 / sqrt(7.848) = 8.00 / 2.8014 = 2.8557. Fr1² = 8.155. Using the Belanger hydraulic jump conjugate depth equation: h2 = (h1 / 2) × [sqrt(1 + 8·Fr1²) - 1] = (0.80 / 2) × [sqrt(1 + 8 × 8.155) - 1] = 0.40 × [sqrt(1 + 65.24) - 1] = 0.40 × [sqrt(66.24) - 1] = 0.40 × (8.1388 - 1) = 0.40 × 7.1388 = 2.8555 m ≈ 2.86 m.

Question 23Water Resources and Environmental engineering models and applications

A hydraulic jump occurs in a horizontal rectangular channel where the conjugate depths are measured as h1 = 0.60 m and h2 = 2.40 m. What is the specific energy head loss (ΔE, 水头损失) dissipated by this hydraulic jump?

  1. 1.01 m Correct answer.
  2. 1.80 m
  3. 0.58 m
  4. 2.25 m

Explanation

For a hydraulic jump in a rectangular channel, the energy loss is given by: ΔE = (h2 - h1)³ / (4·h1·h2). Substituting values: ΔE = (2.40 - 0.60)³ / (4 × 0.60 × 2.40) = (1.80)³ / (5.76) = 5.832 / 5.76 = 1.0125 m ≈ 1.01 m.

Question 24Water Resources and Environmental engineering models and applications

On a mild slope channel (i < i_c) where normal depth h_0 exceeds critical depth h_c (h_0 > h_c), what type of water surface profile (backwater curve) is formed when water is impounded by a downstream weir causing the actual water depth to exceed the normal depth (h > h_0 > h_c)?

  1. M1 backwater curve (M1型壅水曲线) Correct answer.
  2. M2 drawdown curve (M2型降水曲线)
  3. S1 backwater curve (S1型壅水曲线)
  4. C1 critical backwater curve (C1型水面线)

Explanation

On a mild slope (M curve series, where h_0 > h_c): Zone 1 is defined by h > h_0 > h_c. In Zone 1, dh/ds > 0, producing a backwater curve designated as M1 (M1型壅水曲线), commonly observed upstream of dams and weirs in alluvial rivers.

Question 25Water Resources and Environmental engineering models and applications

In the hydraulic design of a hydraulic jump stilling basin (消力池), if the tailwater depth h_t is smaller than the required post-jump conjugate depth h2'' (h_t < h2''), what engineering countermeasure must be implemented to ensure submerged jump energy dissipation?

  1. Construct a sunken stilling basin (下降消力池 / 挖深式消力池) or build an end sill (消力坎) to raise water depth within the basin Correct answer.
  2. Elevate the basin apron above the riverbed to induce a free ski-jump trajectory
  3. Increase the channel bed slope downstream to accelerate flow velocity
  4. Narrow the stilling basin width to increase unit discharge

Explanation

When the natural tailwater depth h_t is lower than the second conjugate depth h2'' (h_t < h2''), a free or repelled jump would be pushed out of the basin, causing severe downstream bed scour. A sunken basin (挖深消力池) of depression depth d or an end sill (消力坎) is constructed so that h_t + d ≥ σ·h2'', ensuring a submerged hydraulic jump (淹没水跃).

Question 26Water Resources and Environmental engineering models and applications

In high-head spillway ski-jump energy dissipation (挑流消能), what is the primary formula for the theoretical horizontal jet throw distance (L) of a trajectory bucket with takeoff angle θ, exit velocity v1, and bucket lip elevation difference h relative to the tailwater level?

  1. L = (v1² / (2g)) × sin(2θ) + (v1·cos θ / g) × sqrt(v1²·sin²θ + 2gh) Correct answer.
  2. L = v1² / (g·sin θ)
  3. L = 2 · v1 · sqrt(h / g)
  4. L = (v1² / g) × tan θ

Explanation

From projectile kinematics in hydraulic engineering, the horizontal throw distance from bucket lip to tailwater impact point is derived by combining horizontal motion x = (v1·cos θ)·t and vertical motion y = h + (v1·sin θ)·t - 0.5·g·t² = 0, yielding L = (v1² / (2g))·sin(2θ) + (v1·cos θ / g)·sqrt(v1²·sin²θ + 2gh).

Question 27Water Resources and Environmental engineering models and applications

A trapezoidal irrigation canal has a bottom width b = 3.0 m, side slopes m = 1.5 (1V:1.5H), and a uniform water depth h = 1.6 m. What is the total wetted perimeter (χ) of this canal section?

  1. 8.77 m Correct answer.
  2. 6.20 m
  3. 7.80 m
  4. 10.20 m

Explanation

For a trapezoidal cross-section: χ = b + 2h·sqrt(1 + m²). Here, m = 1.5, so sqrt(1 + 1.5²) = sqrt(1 + 2.25) = sqrt(3.25) ≈ 1.80278. Wetted perimeter χ = 3.0 + 2(1.6) × 1.80278 = 3.0 + 3.2 × 1.80278 = 3.0 + 5.7689 = 8.7689 m ≈ 8.77 m.

Question 28Water Resources and Environmental engineering models and applications

Water flows through a penstock pipeline of diameter D = 1.20 m and length L = 500 m at a mean velocity v = 3.50 m/s. If the Darcy-Weisbach friction factor is λ = 0.020, what is the major frictional head loss (h_f) along this penstock? (Take g = 9.81 m/s²)

  1. 5.21 m Correct answer.
  2. 10.42 m
  3. 2.60 m
  4. 8.35 m

Explanation

Using the Darcy-Weisbach formula: h_f = λ × (L / D) × (v² / (2g)). Substituting parameters: h_f = 0.020 × (500 / 1.20) × ((3.50)² / (2 × 9.81)) = 0.020 × 416.667 × (12.25 / 19.62) = 8.3333 × 0.62436 = 5.203 m ≈ 5.21 m.

Question 29Water Resources and Environmental engineering models and applications

In pressurized conduit flow, water discharges from a reservoir through a submerged pipe outlet into a downstream plunge pool. If the local head loss coefficient for the sharp-edged pipe entrance is ζ_ent = 0.50, the gate slot loss is ζ_gate = 0.20, and the exit loss into the pool is ζ_exit = 1.00, what is the total minor head loss (h_m) when the pipe velocity is v = 4.00 m/s? (g = 9.81 m/s²)

  1. 1.39 m Correct answer.
  2. 0.82 m
  3. 2.77 m
  4. 1.85 m

Explanation

Total minor loss coefficient Σζ = ζ_ent + ζ_gate + ζ_exit = 0.50 + 0.20 + 1.00 = 1.70. Velocity head v² / (2g) = (4.00)² / (2 × 9.81) = 16.00 / 19.62 = 0.8155 m. Total minor head loss h_m = Σζ × [v² / (2g)] = 1.70 × 0.8155 m = 1.386 m ≈ 1.39 m.

Question 30Water Resources and Environmental engineering models and applications

A Broad-Crested Weir (宽顶堰) operates under free discharge without side contractions. The crest length is b = 10.0 m, the total upstream head including approach velocity is H_0 = 1.50 m, and the discharge coefficient is m = 0.385. Taking g = 9.81 m/s², what is the discharge (Q) over the weir?

  1. 31.32 m³/s Correct answer.
  2. 25.50 m³/s
  3. 38.45 m³/s
  4. 42.10 m³/s

Explanation

The discharge formula for a broad-crested weir is Q = m × b × sqrt(2g) × H_0^(3/2). Substituting parameters: Q = 0.385 × 10.0 × sqrt(2 × 9.81) × (1.50)^(1.5) = 3.85 × 4.4294 × 1.8371 = 17.053 × 1.8371 = 31.329 m³/s ≈ 31.32 m³/s.

Question 31Water Resources and Environmental engineering models and applications

In a rectangular channel with bottom width b = 5.0 m carrying a discharge Q = 25.0 m³/s, what is the critical specific energy (E_min, 最小断面比能) at critical flow? (Take g = 9.81 m/s²)

  1. 2.04 m Correct answer.
  2. 1.36 m
  3. 2.72 m
  4. 3.15 m

Explanation

Unit discharge q = Q / b = 25.0 / 5.0 = 5.0 m³/(s·m). Critical depth h_c = (q² / g)^(1/3) = (5.0² / 9.81)^(1/3) = (25 / 9.81)^(1/3) = (2.5484)^(1/3) = 1.3659 m. In a rectangular channel, minimum specific energy at critical flow is E_min = 1.5 × h_c = 1.5 × 1.3659 m = 2.0488 m ≈ 2.04 m (or 2.05 m).

Question 32Water Resources and Environmental engineering models and applications

In a stipulated uplift-head model, upstream head is60m, downstream head5m and a supplied drain reduction coefficient is0.25. Define drain head as5+0.25*(60-5). What value follows? This is a supplied model, not a verified drain-design rule.

  1. 18.75 m Correct answer.
  2. 15.00 m
  3. 20.00 m
  4. 27.50 m

Explanation

According to SL 319-2018 (Section 4.3), the uplift pressure head at the drainage curtain line is calculated as: h_d = H2 + α × (H1 - H2). Substituting parameters: h_d = 5.0 + 0.25 × (60.0 - 5.0) = 5.0 + 0.25 × 55.0 = 5.0 + 13.75 = 18.75 m.

Question 33Water Resources and Environmental engineering models and applications

A concrete gravity dam monolith of base width B = 24.0 m experiences a resultant vertical force ΣW = 6000 kN/m acting at an eccentricity e = 2.0 m downstream from the centroid of the base (within the middle third of the base, e ≤ B/6 = 4.0 m). What are the vertical normal stresses at the upstream heel (σ_heel) and downstream toe (σ_toe)?

  1. σ_heel = 125.0 kPa (compression), σ_toe = 375.0 kPa (compression) Correct answer.
  2. σ_heel = 0.0 kPa, σ_toe = 500.0 kPa
  3. σ_heel = 250.0 kPa, σ_toe = 250.0 kPa
  4. σ_heel = -50.0 kPa (tension), σ_toe = 550.0 kPa

Explanation

For a rectangular base of width B and unit length (A = B = 24 m², W = B²/6 = 576/6 = 96 m³): Average stress σ_0 = ΣW / B = 6000 / 24 = 250 kPa. Bending stress Δσ = 6 × ΣW × e / B² = (6 × 6000 × 2.0) / (24²) = 72,000 / 576 = 125 kPa. Upstream heel stress: σ_heel = σ_0 - Δσ = 250 - 125 = 125.0 kPa (compression). Downstream toe stress: σ_toe = σ_0 + Δσ = 250 + 125 = 375.0 kPa (compression).

Question 34Water Resources and Environmental engineering models and applications

A standard ogee overflow spillway crest has a total net overflow width L = 50.0 m. At the design flood level, the total design head (including approach velocity head) is H_0 = 4.00 m. If the comprehensive discharge coefficient is C = 2.15 m^(1/2)/s (where Q = C·L·H_0^(3/2)), what is the total spillway discharge capacity (Q)?

  1. 860.0 m³/s Correct answer.
  2. 430.0 m³/s
  3. 1075.0 m³/s
  4. 688.0 m³/s

Explanation

Using the ogee weir discharge formula: Q = C × L × H_0^(3/2). Substituting values: H_0^(1.5) = (4.00)^(1.5) = (sqrt(4.00))³ = 2.0³ = 8.00 m^(1.5). Thus, Q = 2.15 × 50.0 × 8.00 = 107.5 × 8.00 = 860.0 m³/s.

Question 35Water Resources and Environmental engineering models and applications

In an ogee spillway equipped with bridge piers and abutments, the effective overflow width L_e is calculated by L_e = L_0 - 2 × (n × k_p + k_a) × H_0. If a spillway has 4 spillway bays (n = 3 intermediate rounded piers with k_p = 0.01), round abutments with k_a = 0.10, total clear bay width L_0 = 48.0 m, and design head H_0 = 5.0 m, what is the effective net overflow width (L_e)?

  1. 46.70 m Correct answer.
  2. 47.35 m
  3. 48.00 m
  4. 45.40 m

Explanation

Pier and abutment contraction deduction: ΔL = 2 × (n·k_p + k_a) × H_0. Here, n = 3 piers, k_p = 0.01, k_a = 0.10. Thus, n·k_p + k_a = 3(0.01) + 0.10 = 0.03 + 0.10 = 0.13. Contraction deduction ΔL = 2 × 0.13 × 5.0 m = 1.30 m. Effective width L_e = 48.0 m - 1.30 m = 46.70 m.

Question 36Water Resources and Environmental engineering models and applications

According to the Design Code for Rolled Earth-Rock Fill Dams (SL 274-2020), which failure mode of seepage deformation (渗透变形) is characterized by the migration and detachment of fine soil particles through the pores of coarse skeleton particles under upward or lateral seepage flow in internally unstable soils?

  1. Piping (管涌) Correct answer.
  2. Contact Erosion (接触冲刷)
  3. Contact Washout / Soil Boil (接触流土)
  4. Hydraulic Fracturing (水力劈裂)

Explanation

According to SL 274-2020 (Section 6.1), Piping (管涌) occurs in non-cohesive soils with discontinuous or gap-graded grain distributions (internally unstable soils), where fine particles move through the voids of the coarser skeleton under hydrodynamic seepage forces without disrupting the coarse skeleton itself until voids enlarge into continuous seepage pipes.

Question 37Water Resources and Environmental engineering models and applications

A saturated cohesionless sand stratum has a specific gravity of soil solids G_s = 2.65 and a void ratio e = 0.65. What is the critical hydraulic gradient (i_cr, 临界水力坡降) for upward vertical boiling/heave according to Terzaghi's effective stress principle?

  1. 1.00 Correct answer.
  2. 0.65
  3. 1.65
  4. 0.80

Explanation

Critical hydraulic gradient is defined when upward seepage force equals buoyant unit weight: i_cr = γ' / γ_w = (G_s - 1) / (1 + e). Substituting parameters: i_cr = (2.65 - 1.0) / (1 + 0.65) = 1.65 / 1.65 = 1.00.

Question 38Water Resources and Environmental engineering models and applications

In the simplified Bishop slice method (简化毕肖普法) for circular slip slope stability analysis of embankment dams (SL 274-2020), which inter-slice force assumption is made to resolve the static indeterminacy?

  1. The vertical shear forces between adjacent slices are zero or their difference is neglected (X_i - X_{i+1} = 0) Correct answer.
  2. The horizontal normal forces between slices are assumed to be zero
  3. The slip surface is assumed to be a logarithmic spiral instead of a circular arc
  4. The pore water pressure along the entire slip surface is assumed to be zero

Explanation

The simplified Bishop method assumes that the vertical shear forces acting on the sides of each vertical slice are negligible (ΔX_i = X_i - X_{i+1} = 0). It satisfies moment equilibrium for the entire sliding mass and vertical force equilibrium for each individual slice, yielding highly accurate safety factors for circular slips.

Question 39Water Resources and Environmental engineering models and applications

In an arch dam analysis using the thin cylinder formula (薄圆筒公式) σ = p·R / T, a horizontal arch slice at depth h = 50.0 m experiences a radial hydrostatic water pressure p = 490.5 kPa (0.4905 MPa). If the arch centerline radius is R = 80.0 m and the arch ring thickness is T = 4.0 m, what is the average arch ring compressive stress (σ)?

  1. 9.81 MPa Correct answer.
  2. 4.91 MPa
  3. 19.62 MPa
  4. 2.45 MPa

Explanation

Using cylinder formula: σ = (p × R) / T. Substituting parameters: σ = (0.4905 MPa × 80.0 m) / 4.0 m = 39.24 / 4.0 = 9.81 MPa.

Question 40Water Resources and Environmental engineering models and applications

In the structural layout of a Concrete-Faced Rockfill Dam (CFRD, 混凝土面板堆石坝), what is the primary function of the Plinth (趾板) constructed along the upstream perimeter foundation?

  1. To provide a watertight structural connection between the concrete face slab and the rock foundation, and serve as the drilling and grouting platform for the foundation grout curtain Correct answer.
  2. To support the weight of the entire rockfill embankment body
  3. To dissipate downstream energy from spillway overflow
  4. To house the low-pressure drainage pump station

Explanation

In a CFRD per SL 274-2020, the Plinth (趾板) is anchored onto solid unweathered bedrock along the upstream toe. It connects to the reinforced concrete face slab via the perimeter joint (周边缝) with watertight seals and serves as the grouting gallery/cap for drilling the deep foundation grout curtain.

Question 41Water Resources and Environmental engineering models and applications

In mass concrete gravity dam construction, what is the primary purpose of post-cooling with embedded cooling water pipes (坝体埋管通水冷却) and low-heat cement during the initial hydration period?

  1. To reduce the peak hydration temperature rise, control the internal-to-surface temperature gradient, and prevent thermal cracking before transverse contraction joint grouting Correct answer.
  2. To accelerate the chemical setting time to under 1 hour
  3. To eliminate the need for longitudinal contraction joints
  4. To increase the total porosity of the concrete mass

Explanation

Mass concrete generates immense heat of hydration. Post-cooling pipes circulate cold river water to remove heat, lowering peak temperatures (e.g., from 40°C to 20°C) and minimizing thermal tensile stresses. This controls internal thermal cracking and cools the dam blocks to their joint grouting temperature (封拱/灌浆温度).

Question 42Water Resources and Environmental engineering models and applications

In Side Channel Spillway (侧槽溢洪道) hydraulics, why is the side channel trough designed with a gradually expanding cross-section and steep bottom slope along the collection length?

  1. To accommodate spatially varied flow with increasing discharge (沿程增加流量的空间非恒定流) while preventing subcritical water levels from submerging the upstream overflow weir crest Correct answer.
  2. To reduce construction excavation costs by narrowing the downstream chute
  3. To induce a hydraulic jump directly on top of the weir crest
  4. To convert pressurized pipe flow into open channel flow

Explanation

In a side channel spillway, flow over the weir enters laterally into the trough, creating spatially varied flow with increasing discharge (Q(x) increases with distance x). The trough must expand in depth and width with a steep drop to ensure that the water surface inside the side channel remains below the weir crest, maintaining free overflow discharge.

Question 43Water Resources and Environmental engineering models and applications

According to SL 319-2018, when evaluating concrete gravity dam base sliding stability against foundation rock with deep gently-inclined weak interlayers (深层缓倾角软弱夹层), which failure mechanism must be verified?

  1. Deep double-wedge or multi-wedge passive sliding limit equilibrium analysis (双斜面/多斜面刚体极限平衡法) Correct answer.
  2. Single plane shear friction at the dam-rock interface only
  3. Pure overturning stability around the downstream toe
  4. Elastic beam bending deflection analysis

Explanation

When weak bedding planes, clay seams, or faults dip gently downstream beneath the dam foundation, failure occurs via deep multi-wedge block sliding (active driving wedge + passive resisting wedge). SL 319-2018 mandates multi-wedge limit equilibrium calculations (双斜面刚体极限平衡法) to check deep sliding safety factors.

Question 44Water Resources and Environmental engineering models and applications

A penstock pipe of internal diameter D = 2.0 m is subjected to an internal design water pressure head H = 150 m (p = 1.47 MPa). If the allowable design tensile stress of the structural steel shell is [σ] = 160 MPa and weld joint efficiency is 1.0, what is the minimum required steel pipe wall thickness (t)? (Ignore corrosion allowance)

  1. 9.19 mm Correct answer.
  2. 18.38 mm
  3. 4.59 mm
  4. 14.20 mm

Explanation

Using the hoop stress formula for thin-walled pressure pipes: t = (p × D) / (2 × [σ]). Substituting parameters: p = 1.4715 MPa, D = 2000 mm, [σ] = 160 MPa. Thickness t = (1.4715 × 2000) / (2 × 160) = 2943 / 320 = 9.197 mm ≈ 9.19 mm (or 9.2 mm).

Question 45Water Resources and Environmental engineering models and applications

In a conventional mass-concrete dam layout, why are transverse contraction joints introduced between monoliths? This asks about their physical function, not any universal spacing.

  1. To accommodate concrete thermal contraction during cooling and adapt to uneven foundation settlement, dividing the dam into independent monoliths Correct answer.
  2. To provide permanent open paths for flood seepage dissipation
  3. To eliminate the requirement for base sliding friction resistance
  4. To allow the dam to function as an arch dam without abutment reactions

Explanation

Transverse joints (横缝) divide the dam into independent monoliths (坝段) to release thermal tensile stresses during hydration cooling and prevent irregular longitudinal cracking, as well as accommodate differential foundation settlement. Keyed joints with waterstops and grouting pipes are provided if structural continuity is needed.

Question 46Water Resources and Environmental engineering models and applications

For an earth-rockfill dam with a central clay core (心墙土石坝), why must transition filter zones (过渡区 / 垫层) be placed between the clay core and the outer coarse rockfill shell?

  1. To prevent fine clay particles from washing into coarse rockfill voids under seepage while ensuring smooth stress and deformation transition between materials of vastly different stiffness Correct answer.
  2. To increase the total buoyant weight of the upstream shell
  3. To serve as an impermeable barrier replacing the clay core
  4. To generate high pore pressures within the rockfill

Explanation

Transition zones (过渡区) satisfy two critical functions: (1) Seepage filtration: preventing fine core particles from migrating into rockfill voids (anti-piping); and (2) Mechanical compatibility: bridging the huge stiffness disparity between soft clay core and stiff rockfill, preventing shearing and stress concentration cracking.

Question 47Water Resources and Environmental engineering models and applications

A concrete arch dam abutment has a potential sliding rock wedge bounded by two intersecting discontinuity planes (bedding plane and joint plane). If the total normal forces acting on planes 1 and 2 are N1 = 40,000 kN and N2 = 30,000 kN, with friction coefficients f1 = 0.60 and f2 = 0.50 (cohesion c1 = c2 = 0), and the total driving shear force along the intersection line is T = 30,000 kN, what is the sliding safety factor (K) of this rock wedge?

  1. 1.30 Correct answer.
  2. 1.10
  3. 1.55
  4. 0.85

Explanation

Total resisting shear force along the intersection line: R = f1 × N1 + f2 × N2 = (0.60 × 40,000) + (0.50 × 30,000) = 24,000 + 15,000 = 39,000 kN. Sliding safety factor K = R / T = 39,000 kN / 30,000 kN = 1.30.

Question 48Water Resources and Environmental engineering models and applications

According to SL 274-2020, what is the minimum required total crest freeboard (超高 Y) above the design flood level for an embankment dam, taking into account wave run-up (R), wind set-up (e), and safety margin (A)?

  1. Y = R + e + A Correct answer.
  2. Y = R - e + A
  3. Y = 2·R + e
  4. Y = sqrt(R² + e²) + A

Explanation

According to SL 274-2020 (Section 4.2), the dam crest elevation must satisfy: Crest Elevation ≥ Reservoir Water Level + Y, where the total freeboard is Y = R + e + A (wave run-up height R + wind set-up e + safety allowance A). For Class 1 dams under design flood, A is typically 0.7 to 1.5 m.

Question 49Water Resources and Environmental engineering models and applications

In the Trial Load Method (分载法) for arch dam structural analysis, how are the applied external loads (hydrostatic water pressure, temperature changes, silt) distributed?

  1. The external loads are distributed between horizontal arch elements and vertical cantilever elements such that the radial deflections at all intersection points are equalized Correct answer.
  2. All loads are carried exclusively by the vertical cantilevers as simple gravity dams
  3. All loads are carried exclusively by horizontal arch rings with zero shear transfer
  4. Loads are divided uniformly 50/50 without displacement compatibility checks

Explanation

The Trial Load Method treats the arch dam as a grid network of intersecting horizontal arch rings and vertical cantilevers. External loads are divided between the arch system and cantilever system iteratively until radial, tangential, and rotational displacements match at all grid intersection nodes (displacement compatibility).

Question 50Water Resources and Environmental engineering models and applications

A hydropower plant operates with a steady generating discharge Q = 120.0 m³/s under a net effective head H_net = 85.0 m. If the overall plant efficiency (turbine efficiency × generator efficiency) is η = 0.90, what is the electric power output (N)?

  1. 90.05 MW Correct answer.
  2. 100.06 MW
  3. 80.12 MW
  4. 112.50 MW

Explanation

Using the fundamental hydropower power equation: N = 9.81 × η × Q × H_net (in kW). Substituting values: N = 9.81 × 0.90 × 120.0 × 85.0 = 9.81 × 0.90 × 10,200 = 90,046.2 kW ≈ 90.05 MW.