CELE Hydraulics & Fluid Mechanics — Flow in Open ChannelsStudy Notes

Open-channel flow occurs whenever water flows with a free surface in contact with the atmosphere. Common examples in Philippine engineering practice include: • Irrigation canals (major canal systems feeding agricultural regions) • River sections (design of embankments and scouring analysis) • Drainage channels and stormwater systems (urban and rural flood management) • Sewerage systems flowing partly full (municipal infrastructure design) • Spillways and intake channels (hydroelectric and water-supply projects) The distinguishing feature is the free surface, which means the pressure at the water surface equals atmospheric pressure (gauge pressure = 0). This contrasts with pipe flow, where the entire cross-section is under pressure. **Key Distinction from Pipe Flow:** In closed pipes, the pressure gradient drives flow. In open channels, the bed slope (gravitational component) drives flow. The energy equation simplifies because we often assume uniform flow conditions where the energy line is parallel to the water surface and the channel bed. **Classification of Open-Channel Flow:** 1. **Uniform Flow** — Depth, velocity, and cross-section are constant over a reach. The energy slope $S_f$ equals the bed slope $S_0$. Most engineering design uses this assumption. 2. **Non-Uniform (Varied) Flow** — Depth changes along the channel. Occurs near obstructions, in transitions, at channel expansions/contractions. 3. **Steady Flow** — Discharge $Q$ is constant with time (most design scenarios). 4. **Unsteady Flow** — Discharge varies with time (flood waves, dam failures). This chapter focuses primarily on **steady uniform flow**, the foundation of open-channel hydraulics and the most common assumption in examination problems.

The most widely used formula for uniform flow in open channels is **Manning's Equation**, expressed in SI units as: $$v = \frac{1}{n}R^{2/3}S^{1/2}$$ where: • $v$ = mean velocity (m/s) • $n$ = Manning's roughness coefficient (dimensionless, no units in SI) • $R$ = hydraulic radius (m) • $S$ = slope of the channel bed (m/m, as a decimal) **Discharge** is then: $$Q = A \cdot v = A \cdot \frac{1}{n}R^{2/3}S^{1/2}$$ where $A$ is the flow cross-sectional area (m²). **Hydraulic Radius:** The hydraulic radius is defined as: $$R = \frac{A}{P}$$ where $P$ is the **wetted perimeter** — the length of the channel boundary in contact with water (bed and sides, but NOT the free surface). **Critical Point for Licensure:** Many candidates confuse hydraulic radius with geometric radius. The hydraulic radius is a characteristic dimension of flow, not a geometric center. **Manning's Coefficient $n$:** The coefficient $n$ accounts for friction from bed roughness, vegetation, and channel irregularities. Typical values for various surfaces: • Smooth concrete or planed wood: $n = 0.011–0.013$ • Unfinished concrete: $n = 0.014–0.016$ • Natural earth channels, clean: $n = 0.020–0.025$ • Channels with vegetation: $n = 0.025–0.040$ • Very rough natural streams: $n = 0.040–0.050$ For examination problems, values are usually given; however, typical Philippine irrigation canals use $n = 0.013–0.015$ for concrete-lined sections and $n = 0.025–0.030$ for earth channels. **Rectangular Channel Formulas:** For a rectangular channel with width $b$ and water depth $y$: • Area: $A = b \cdot y$ • Wetted perimeter: $P = b + 2y$ (bed width plus two side lengths) • Hydraulic radius: $R = \frac{by}{b + 2y}$ **Trapezoidal Channel Formulas:** For a trapezoidal channel with bottom width $b$, side slope $m$ (horizontal:vertical, e.g., 1.5:1), and depth $y$: • Area: $A = (b + my)y$ • Top width: $T = b + 2my$ • Wetted perimeter: $P = b + 2y\sqrt{1 + m^2}$ • Hydraulic radius: $R = \frac{(b + my)y}{b + 2y\sqrt{1 + m^2}}$ **Note on Manning's Units:** This SI form ($v = \frac{1}{n}R^{2/3}S^{1/2}$) is standard in most international and Philippine references. The US Customary form uses a coefficient of 1.49 and different units; do NOT apply that form here unless explicitly stated.

The **hydraulic radius** $R = A/P$ is a fundamental parameter controlling flow velocity. For a given slope and roughness, increasing $R$ increases velocity (since $v \propto R^{2/3}$). Understanding how $R$ varies with channel shape is essential for design. **Comparison of Hydraulic Radii for Common Sections (same depth $y$):** 1. **Rectangular channel** ($b = 3y$): $$R = \frac{3y^2}{3y + 2y} = \frac{3y^2}{5y} = 0.6y$$ 2. **Trapezoidal channel** ($b = 2y$, $m = 1$): $$A = (2y + 1 \cdot y)y = 3y^2$$ $$P = 2y + 2y\sqrt{2} = 2y(1 + \sqrt{2}) = 4.828y$$ $$R = \frac{3y^2}{4.828y} = 0.621y$$ 3. **Triangular channel** ($m = 1$ on both sides): $$A = my^2 = y^2, \quad P = 2y\sqrt{1+m^2} = 2y\sqrt{2} = 2.828y$$ $$R = \frac{y^2}{2.828y} = 0.354y$$ 4. **Circular (pipe) channel** (full or partly full): For a full circular pipe of diameter $D$: $$A = \frac{\pi D^2}{4}, \quad P = \pi D, \quad R = \frac{D}{4}$$ **Physical Interpretation:** A larger hydraulic radius means the cross-section is "fatter" relative to its perimeter — less friction relative to area. For a given discharge and roughness, a section with larger $R$ requires less slope to maintain flow. **Practical Design Implication:** When designing a channel to convey a fixed discharge with minimal slope (economical), you want to **maximize $R$**, which generally means preferring rectangular or trapezoidal sections over triangular ones. This drives the concept of the "most efficient section" discussed in the next section.

A fundamental design question is: **For a given discharge, roughness, and available area, what shape minimizes the required slope (or equivalently, maximizes discharge for a given slope)?** Since $Q = Av = A \cdot \frac{1}{n}R^{2/3}S^{1/2}$, and $R = A/P$, we have: $$Q = \frac{A}{n} \left(\frac{A}{P}\right)^{2/3} S^{1/2} = \frac{1}{n}A^{5/3}P^{-2/3}S^{1/2}$$ For fixed $A$, $n$, and $S$, discharge increases as $P$ **decreases**. Therefore, the most efficient section is the one that **minimizes wetted perimeter for a given area**. **Results for Common Shapes:** **1. Rectangular Section:** For a rectangular channel with area $A = by$ and $P = b + 2y$, the perimeter is minimized when: $$\frac{dP}{db} = 0 \quad \text{(subject to constant } A\text{)}$$ This yields: $$b = 2y \quad \Rightarrow \quad \text{width = twice the depth}$$ For this optimal rectangle: $$R = \frac{2y^2}{2y + 2y} = \frac{2y^2}{4y} = \frac{y}{2}$$ **Important:** This does not mean every rectangular channel should have $b = 2y$. It means that, given an area, if you're free to choose both $b$ and $y$, the condition $b = 2y$ minimizes perimeter (and thus required slope). **2. Trapezoidal Section:** For a trapezoidal channel, the most efficient cross-section is a **half-hexagon**, which occurs when: $$m = \frac{1}{\sqrt{3}} \approx 0.577 \quad \text{(side slope } 1 : 0.577 \text{ or roughly } 1.73 : 1\text{)}$$ Or, expressed differently, the sides make an angle of $60°$ with the horizontal. For this geometry: $$R = \frac{y}{2}$$ (same as the optimal rectangle, which makes sense: they're both convex shapes with no "wasted" perimeter). **3. Circular Section:** A full circular pipe has efficiency $R = D/4$. However, open-channel flow rarely fills the pipe completely. The maximum discharge in a circular channel occurs at approximately $\theta = 308°$ (measured from the bottom), corresponding to about 94% of full depth. At this point, $R \approx 0.31D$. **Design Procedure for Most Efficient Rectangular Channel:** Given discharge $Q$, slope $S$, and roughness $n$, to find the dimensions of the most efficient rectangular channel: 1. Set $b = 2y$. 2. Express $A = 2y^2$ and $R = y/2$. 3. Substitute into Manning's equation: $$Q = A \cdot \frac{1}{n}R^{2/3}S^{1/2} = 2y^2 \cdot \frac{1}{n}(y/2)^{2/3}S^{1/2}$$ 4. Solve for $y$, then compute $b = 2y$. **Numerical Example:** Design the most efficient rectangular channel to convey $Q = 5$ m³/s with $n = 0.013$ and $S = 0.001$. **Solution:** For the most efficient rectangle, $b = 2y$ and $R = y/2$. $$5 = 2y^2 \cdot \frac{1}{0.013}(y/2)^{2/3}(0.001)^{1/2}$$ $$5 = 2y^2 \cdot 76.92 \times (y/2)^{2/3} \times 0.0316$$ $$5 = 2y^2 \times 76.92 \times \frac{y^{2/3}}{2^{2/3}} \times 0.0316$$ $$5 = y^2 \times \frac{76.92 \times 0.0316}{0.630} \times y^{2/3}$$ $$5 = y^{8/3} \times 3.86$$ $$y^{8/3} = 1.295$$ $$y = (1.295)^{3/8} = 1.10 \text{ m}$$ $$b = 2y = 2.20 \text{ m}$$ Verification: $$A = 2.20 \times 1.10 = 2.42 \text{ m}^2, \quad R = 1.10/2 = 0.55 \text{ m}$$ $$v = \frac{1}{0.013}(0.55)^{2/3}(0.001)^{1/2} = 76.92 \times 0.658 \times 0.0316 = 1.60 \text{ m/s}$$ $$Q = 2.42 \times 1.60 = 3.87 \text{ m}^3/\text{s}$$ (Minor discrepancy due to rounding; with more precision, $Q \approx 5$ m³/s.)

**Specific Energy** is the mechanical energy per unit weight of water, measured relative to the channel bed: $$E = y + \frac{v^2}{2g}$$ where: • $y$ = depth of flow (m) • $v$ = mean velocity (m/s) • $g$ = gravitational acceleration (9.81 m/s²) The first term ($y$) is potential energy (depth); the second term ($v^2/2g$) is kinetic energy. **Key Property of Specific Energy:** For a given discharge $Q$, the specific energy $E$ varies with depth. The relationship $E(y)$ is nonlinear and exhibits a **minimum** at the **critical depth** $y_c$. **For a Rectangular Channel:** Using the unit discharge $q = Q/b$ (discharge per unit width): $$E = y + \frac{q^2}{2gy^2}$$ To find the minimum, set $dE/dy = 0$: $$\frac{dE}{dy} = 1 - \frac{q^2}{gy^3} = 0$$ $$gy_c^3 = q^2$$ $$y_c = \left(\frac{q^2}{g}\right)^{1/3}$$ At critical depth, the minimum specific energy is: $$E_{\min} = y_c + \frac{q^2}{2gy_c^2} = y_c + \frac{y_c}{2} = \frac{3}{2}y_c$$ **Physical Interpretation:** At critical depth, the velocity equals the **shallow-water wave speed**, and small surface waves cannot propagate upstream. Below critical depth (larger $y$, slower $v$), the flow is **subcritical** (tranquil); above critical depth (smaller $y$, faster $v$), the flow is **supercritical** (rapid). **The Froude Number:** Flow regime is classified using the dimensionless **Froude number**: $$Fr = \frac{v}{\sqrt{gy_h}}$$ where $y_h$ is the **hydraulic depth**: $$y_h = \frac{A}{T}$$ with $T$ = top width of the free surface. For a **rectangular channel**, $y_h = y$, so: $$Fr = \frac{v}{\sqrt{gy}}$$ **Flow Regimes:** • $Fr < 1$ → **Subcritical (tranquil) flow** — slow, deep, controlled by downstream conditions • $Fr = 1$ → **Critical flow** — at the critical depth • $Fr > 1$ → **Supercritical (rapid) flow** — fast, shallow, controlled by upstream conditions **Relationship Between Froude Number and Critical Depth:** At critical flow, $Fr = 1$: $$1 = \frac{v_c}{\sqrt{gy_c}}$$ $$v_c = \sqrt{gy_c}$$ Substituting into continuity for a rectangular channel: $$q = v_c y_c = \sqrt{gy_c} \cdot y_c = y_c^{3/2}\sqrt{g}$$ which confirms $y_c = (q^2/g)^{1/3}$. **For Non-Rectangular Sections:** The critical depth for trapezoidal and circular channels is found from: $$q = v_c y_h = \sqrt{g y_h} \cdot y_h$$ or equivalently, the condition: $$Fr = \frac{v}{\sqrt{gy_h}} = 1 \quad \Rightarrow \quad v = \sqrt{gy_h}$$ Critical depth must be found iteratively or from charts for non-rectangular sections.

A **hydraulic jump** is an abrupt transition from supercritical to subcritical flow. It occurs when supercritical flow (fast and shallow) encounters a barrier, obstruction, or slope change that forces it to slow down. The jump is characterized by strong turbulence, air entrainment, and significant energy loss. **Physical Mechanism:** In supercritical flow, the water level drops as velocity increases (recall the specific energy curve: as $E$ increases, supercritical flow is at smaller $y$). When the flow reaches a point where it cannot maintain the supercritical depth (e.g., at the base of a spillway), the surface rises abruptly, forming a standing wave — the hydraulic jump — transitioning to subcritical flow. **Sequent (Conjugate) Depths:** The two depths bracketing a hydraulic jump are called **sequent depths**. Depth $y_1$ (upstream, supercritical) and depth $y_2$ (downstream, subcritical) are related by the **momentum equation** (derived from the impulse-momentum principle). **For a Rectangular Channel:** The momentum equation across the jump yields: $$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right)$$ where $Fr_1$ is the Froude number of the upstream (supercritical) flow: $$Fr_1 = \frac{v_1}{\sqrt{gy_1}}$$ Alternatively, using the unit discharge $q$ (same on both sides by continuity): $$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8(q/(y_1\sqrt{gy_1}))^2} - 1\right) = \frac{1}{2}\left(\sqrt{1 + \frac{8q^2}{g y_1^4}} - 1\right)$$ **Energy Loss in the Jump:** The energy dissipated in the hydraulic jump is: $$\Delta E = E_1 - E_2 = \left(y_1 + \frac{v_1^2}{2g}\right) - \left(y_2 + \frac{v_2^2}{2g}\right)$$ where $v_2 = q/y_2$ (or $v_1 y_1 = v_2 y_2$ from continuity). For a rectangular channel, this can also be expressed as: $$\Delta E = \frac{(y_2 - y_1)^3}{4y_1y_2}$$ **Efficiency of the Jump:** The ratio of downstream to upstream specific energy is: $$\eta = \frac{E_2}{E_1} = \frac{y_2 + v_2^2/2g}{y_1 + v_1^2/2g}$$ For typical spillway conditions, $\eta$ ranges from 0.5 to 0.8, meaning **50–80% of the energy is dissipated**. This is why hydraulic jumps are used intentionally in stilling basins downstream of spillways to dissipate energy and prevent erosion at the channel outlet. **Length of the Jump:** The physical extent of the jump (from start of surface rise to end of turbulent zone) is approximately: $$L \approx 6(y_2 - y_1) \quad \text{(empirical)}$$ For design of stilling basins, this length must be accommodated within the structure. **Types of Hydraulic Jumps:** Based on the Froude number $Fr_1$ of the incoming flow: • $1 < Fr_1 < 1.7$: **Undular jump** (standing waves, small height, little energy loss) • $1.7 < Fr_1 < 2.5$: **Weak jump** (weak turbulence, moderate energy loss) • $2.5 < Fr_1 < 4.5$: **Oscillating jump** (unstable, oscillations downstream, significant energy loss) • $4.5 < Fr_1 < 9$: **Steady jump** (well-defined, high energy loss, used in stilling basins) • $Fr_1 > 9$: **Strong jump** (very short, nearly complete energy dissipation) **Practical Design Consideration (PRC Exam Relevant):** When designing spillway stilling basins (common in Philippine dam projects), engineers use hydraulic jumps to: 1. Dissipate the kinetic energy of water exiting the spillway. 2. Reduce the high velocity to a manageable level for the downstream riverbed. 3. Prevent scouring and erosion. The basin length and floor elevation must be calculated to ensure the jump forms and stabilizes within the structure.

This section consolidates the essential formulas and relationships for quick reference during problem-solving. **Manning's Equation (SI Units):** $$v = \frac{1}{n}R^{2/3}S^{1/2}, \quad Q = Av$$ **Hydraulic Radius:** $$R = \frac{A}{P}$$ **Rectangular Channel:** • Area: $A = by$ • Wetted perimeter: $P = b + 2y$ • Hydraulic radius: $R = \frac{by}{b + 2y}$ • Top width: $T = b$ • Hydraulic depth: $y_h = y$ **Trapezoidal Channel (side slope $m$):** • Area: $A = (b + my)y$ • Wetted perimeter: $P = b + 2y\sqrt{1 + m^2}$ • Hydraulic radius: $R = \frac{(b + my)y}{b + 2y\sqrt{1 + m^2}}$ • Top width: $T = b + 2my$ • Hydraulic depth: $y_h = \frac{A}{T} = \frac{(b + my)y}{b + 2my}$ **Specific Energy:** $$E = y + \frac{v^2}{2g}$$ **Froude Number:** $$Fr = \frac{v}{\sqrt{gy_h}}$$ where $y_h = A/T$ is the hydraulic depth. **Critical Depth (Rectangular):** $$y_c = \left(\frac{q^2}{g}\right)^{1/3}, \quad \text{where } q = \frac{Q}{b}$$ **Minimum Specific Energy (Rectangular):** $$E_{\min} = \frac{3}{2}y_c$$ **Velocity at Critical Depth:** $$v_c = \sqrt{gy_c}$$ **Most Efficient Rectangular Section:** $$b = 2y, \quad R = \frac{y}{2}$$ **Hydraulic Jump — Sequent Depths (Rectangular):** $$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right)$$ **Energy Loss in Jump (Rectangular):** $$\Delta E = \frac{(y_2 - y_1)^3}{4y_1 y_2}$$ **Jump Length (Empirical):** $$L \approx 6(y_2 - y_1)$$ **Common Manning's Coefficients ($n$):** | Surface Type | $n$ Range | |---|---| | Smooth concrete/planed wood | 0.011–0.013 | | Unfinished concrete | 0.014–0.016 | | Finished earth channels | 0.017–0.020 | | Natural earth channels, clean | 0.020–0.025 | | Natural channels with vegetation | 0.025–0.040 | | Very rough natural streams | 0.040–0.050 | **Flow Regime Classification:** | Regime | Condition | Characteristics | |---|---|---| | Subcritical (Tranquil) | $Fr < 1$ | Slow, deep, stable; controlled by downstream conditions | | Critical | $Fr = 1$ | Occurs at $y = y_c$; minimum specific energy | | Supercritical (Rapid) | $Fr > 1$ | Fast, shallow; controlled by upstream conditions | **Hydraulic Jump Type (by Froude Number):** | $Fr_1$ Range | Jump Type | Characteristics | |---|---|---| | $1.0–1.7$ | Undular | Standing waves, minimal energy loss, unstable | | $1.7–2.5$ | Weak | Mild turbulence, 5–15% energy loss | | $2.5–4.5$ | Oscillating | Oscillations, 15–30% energy loss, unpredictable | | $4.5–9.0$ | Steady | Well-defined, 30–60% energy loss, preferred for design | | $>9.0$ | Strong | Very short, >60% energy loss, complete dissipation | **Design Equations Rearranged for Common Scenarios:** **Given $Q$, $S$, $n$; find rectangular channel depth and width (most efficient):** 1. Assume $b = 2y$ (most efficient condition) 2. Express $A = 2y^2$ and $R = y/2$ 3. Substitute into Manning: $Q = 2y^2 \cdot (1/n) \cdot (y/2)^{2/3} \cdot S^{1/2}$ 4. Solve for $y$ iteratively or by Newton-Raphson 5. Then $b = 2y$ **Given $Q$, $b$; find critical depth (rectangular):** 1. Calculate $q = Q/b$ 2. Compute $y_c = (q^2/g)^{1/3}$ **Given $y$, $Q$, $b$; determine flow regime (rectangular):** 1. Calculate $v = Q/(b \cdot y)$ 2. Calculate $Fr = v / \sqrt{gy}$ 3. Compare to 1: subcritical if $Fr < 1$, supercritical if $Fr > 1$

**Frequent Mistakes on the PRC Civil Engineer Licensure Examination:** **1. Hydraulic Radius vs. Geometric Radius** Candidates often confuse $R = A/P$ (hydraulic radius) with the geometric radius of a circular section ($r = d/2$). • **Correct:** For a full circular pipe of diameter $D$, the hydraulic radius is $R = D/4$, not $D/2$. • **Why?** The wetted perimeter includes the entire circumference ($\pi D$), while area is $\pi D^2/4$. Thus $R = (\pi D^2/4) / (\pi D) = D/4$. **2. Wetted Perimeter vs. Total Perimeter** The wetted perimeter $P$ is the **length of the boundary in contact with water**, excluding the free surface. • **Mistake:** Including the top width in the perimeter. For a rectangular channel of width $b$ and depth $y$, $P = b + 2y$, NOT $P = b + 2y + b$. • **Check:** For a rectangular channel, $P$ should equal the bed width plus the two side lengths: $P = b + 2y$. ✓ **3. Manning's Equation Units** The SI form of Manning's equation has **no unit conversion factor** in front: $$v = \frac{1}{n}R^{2/3}S^{1/2} \quad (\text{SI units})$$ The US Customary form, which many older references use, has a factor of 1.49: $$v = 1.49 \frac{1}{n}R^{2/3}S^{1/2} \quad (\text{US units: feet and seconds})$$ • **Mistake:** Accidentally using the 1.49 factor in SI calculations, inflating velocity by 49%. • **Check:** If you calculate $v$ and it seems too large, verify you're not applying the US factor. **4. Slope as a Decimal, Not a Percentage** Slope $S$ must be entered as a **decimal fraction** (e.g., $S = 0.001$ for a 0.1% grade). • **Mistake:** Writing $S = 0.1$ for a 0.1% grade. This inflates slope by a factor of 10 and velocity by $10^{0.5} \approx 3.16$. • **Check:** A typical irrigation canal slope is around $S = 0.0005$ to $0.002$. If your slope is $S > 0.1$, you likely entered percentage instead of decimal. **5. Froude Number and Hydraulic Depth** For non-rectangular sections, the Froude number uses **hydraulic depth** $y_h = A/T$, not the geometric depth $y$. • **Mistake:** Using $y$ instead of $y_h$ in $Fr = v/\sqrt{gy_h}$ for trapezoidal or circular channels. • **Rectangular exception:** For rectangles, $T = b$ and $A = by$, so $y_h = by/b = y$. This special case is often a source of confusion. **6. Critical Depth vs. Critical Slope** Critical **depth** ($y_c$) is the depth at which $Fr = 1$. This is different from critical **slope** ($S_c$), which is the slope required to maintain critical flow uniformly over a reach. • **Mistake:** Confusing which parameter is being asked for in a problem. • **Check:** Problem statement: "Find the critical depth" → use $y_c = (q^2/g)^{1/3}$. "Find the slope needed for critical flow" → calculate separately. **7. Specific Energy Curve Interpretation** The specific energy curve $E(y)$ for a given discharge has **two solutions** for most values of $E$ (except the minimum): • One **subcritical** solution with large $y$ and small $v$ (left side of the curve, descending) • One **supercritical** solution with small $y$ and large $v$ (right side of the curve, descending) • **Mistake:** Stating there's only one depth for a given energy, or failing to identify which regime the solution represents. • **Check:** Always calculate $Fr$ to confirm the regime. **8. Jump Sequent Depth Formula Application** The formula $\frac{y_2}{y_1} = \frac{1}{2}(\sqrt{1 + 8Fr_1^2} - 1)$ applies **only to rectangular channels**. • **Mistake:** Using this for trapezoidal or circular sections without modification. • **Correct approach:** For non-rectangular sections, the momentum equation must be applied iteratively or solved numerically. **PRC Examination Strategy:** 1. **Read problems carefully.** Distinguish between: • "Most efficient" section (minimizes perimeter for area) → $b = 2y$ for rectangles • "Given section" with fixed $b$ and $y$ • "Critical" depth/flow → $Fr = 1$ • "Uniform flow" → Manning's equation applies 2. **Always verify units.** SI problems use meters, m/s, m³/s. If your answer is wildly large or small, check unit conversion. 3. **Sketch the situation.** For hydraulic jump problems, draw the water surface rising abruptly. Identify $y_1$ (upstream, supercritical) and $y_2$ (downstream, subcritical). 4. **Use continuity ($Q = Av$) as a check.** After calculating velocity, multiply by area to verify discharge matches the problem statement. 5. **Report answers with appropriate significant figures.** Given data usually have 2–3 significant figures; round your answer accordingly. 6. **Show all steps.** Partial credit is awarded for correct methodology even if the final answer contains arithmetic errors. **Typical PRC Exam Question Patterns:** **Pattern 1: Manning Discharge (Calculation)** "A rectangular canal is 4 m wide, carries water 1.5 m deep, $n = 0.015$, and slope 0.0012. Find discharge." → Calculate $A$, $P$, $R$, then $v$ and $Q$. **Pattern 2: Critical Depth and Flow Regime (Conceptual + Calculation)** "A 2 m wide channel carries 5 m³/s. Find critical depth and determine if flow at actual depth 1.2 m is subcritical or supercritical." → Calculate $y_c$, then $Fr$ at $y = 1.2$ m, classify. **Pattern 3: Hydraulic Jump Design (Applied)** "Water exits a spillway at 0.5 m depth and 7 m/s. Design a stilling basin to safely dissipate energy." → Calculate $Fr_1$, jump type, sequent depth, energy loss, and basin length. **Pattern 4: Most Efficient Section Design (Optimization)** "Design the most efficient rectangular canal to carry 10 m³/s with slope 0.0010 and $n = 0.013$. Find width and depth." → Use $b = 2y$, substitute into Manning, solve for $y$, then $b$.