CELE Hydraulics & Fluid Mechanics — Flow in Open ChannelsRevision Notes
Final-week revision notes for Flow in Open Channels. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Hydraulics & Fluid Mechanics subtest.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Hydraulics & Fluid Mechanics subtest is marked as "Core" in the official pattern, and Flow in Open Channels appears in position 7th of 10 in the CELE Hydraulics & Fluid Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Flow in Open Channels - Revision Notes
Open-channel flow occurs whenever a liquid flows with a free surface exposed to atmospheric pressure. Unlike pipe flow (which is pressure-driven), open-channel flow is driven by gravity acting on the fluid down a sloping channel bed. Common engineering examples include rivers, irrigation canals, drainage ditches, roadside gutters, and sewers flowing only partially full. For the PRC Civil Engineer Licensure Examination, mastery of Manning's equation, hydraulic radius, specific energy, critical flow, and the hydraulic jump is essential. This chapter consolidates all examinable formulas, concepts, and board-level worked examples in SI units.
Sections
Formulas
Example
Rectangular channel: b = 3 m, y = 1.2 m → A = 3(1.2) = 3.6 m², P = 3 + 2(1.2) = 5.4 m, R = 3.6/5.4 = 0.667 m
Formula
R = A / P
Variables
R = hydraulic radius (m); A = cross-sectional flow area (m²); P = wetted perimeter (m) — the length of the channel boundary in contact with the flowing water (NOT the free surface).
Application
Used in Manning's equation and all hydraulic radius–based discharge calculations.
Example
n = 0.013, R = 0.667 m, S = 0.001: v = (1/0.013)(0.667)^(2/3)(0.001)^(1/2) = 76.92 × 0.763 × 0.03162 = 1.86 m/s
Formula
v = (1/n) · R^(2/3) · S^(1/2)
Variables
v = mean flow velocity (m/s); n = Manning roughness coefficient (dimensionless); R = hydraulic radius (m); S = bed slope (m/m, dimensionless decimal).
Application
Computes mean velocity under uniform flow conditions. This is the SI form — do NOT use the 1.49 multiplier (US customary).
Example
A = 3.6 m², v = 1.86 m/s → Q = 3.6 × 1.86 = 6.68 m³/s
Formula
Q = A · v
Variables
Q = discharge or flow rate (m³/s); A = cross-sectional flow area (m²); v = mean velocity (m/s).
Application
Continuity equation — used to find total discharge once velocity is known.
Exam Tips
- Always write out A and P explicitly before computing R — do not shortcut.
- For a full circular pipe: A = πD²/4, P = πD, so R = D/4. Memorize this result.
- Check unit consistency: R in meters, S dimensionless decimal → v in m/s.
- Common board exam n values: smooth concrete = 0.012, ordinary concrete = 0.013, clean earthen channel = 0.022.
- The product (1/n)·R^(2/3)·S^(1/2) can be computed step by step — avoid combining exponents prematurely on a calculator.
Key Points
- Open-channel flow has a free surface at atmospheric pressure — the key distinction from pipe flow.
- The driving force is the component of gravity along the channel bed slope S (dimensionless, m/m).
- Flow classification by time: Steady (conditions constant with time) vs. Unsteady.
- Flow classification by space: Uniform (depth and velocity constant along the channel) vs. Non-uniform (varied flow).
- Uniform flow requires: constant cross-section, constant slope, and the friction slope equals the bed slope (Sf = S0).
- Non-uniform flow: gradually varied flow (GVF) changes slowly; rapidly varied flow (RVF) changes abruptly (e.g., hydraulic jump).
- The hydraulic radius R = A/P is the single most important geometric parameter in open-channel flow equations.
- Manning's n is a roughness coefficient — lower n means smoother channel and higher velocity.
Definitions
Term
Free Surface
Definition
The interface between the flowing liquid and the atmosphere, where pressure equals atmospheric (gauge pressure = 0). Defines open-channel flow.
Importance
Fundamental distinction from pipe (pressurized) flow. Its presence means pressure is not a variable in most open-channel equations.
Term
Hydraulic Radius (R)
Definition
The ratio of the cross-sectional flow area A to the wetted perimeter P. R = A/P. It quantifies the efficiency of the channel shape in conveying flow.
Importance
Appears in Manning's, Chezy's, and Darcy-Weisbach equations for open channels. Larger R → less friction per unit area → more efficient flow.
Term
Wetted Perimeter (P)
Definition
The total length of the channel boundary (walls + bottom) that is in contact with the flowing fluid. The free surface is NOT included.
Importance
Critical for computing R. A common board exam error is including the water surface width in P.
Term
Manning's Roughness Coefficient (n)
Definition
An empirical coefficient representing the resistance to flow due to channel surface roughness. Typical values: concrete n = 0.012–0.014; earthen canal n = 0.022–0.025; natural river n = 0.025–0.035.
Importance
Must be memorized or selected correctly from tables. Errors in n directly affect velocity and discharge calculations.
Term
Uniform Flow
Definition
A flow condition where depth, velocity, and cross-section remain constant along the channel length. Energy slope Sf equals the bed slope S0.
Importance
Manning's equation applies only to uniform flow. It is the primary flow condition tested in board exams.
Section Title
1. Fundamentals of Open-Channel Flow
Common Mistakes
- Including the free water surface in the wetted perimeter P — only include wetted solid boundaries.
- Using slope as a percentage (e.g., 0.1%) instead of a decimal (S = 0.001) in Manning's equation.
- Using the US customary form v = (1.49/n)R^(2/3)S^(1/2) instead of the SI form v = (1/n)R^(2/3)S^(1/2).
- Confusing hydraulic radius R = A/P with the pipe radius r — they are completely different quantities.
- Forgetting that Manning's equation is valid only for uniform, steady flow.
Formulas
Example
Design a rectangular channel (n = 0.013, S = 0.001) for Q = 5 m³/s: With b = 2y → A = 2y², P = 4y, R = y/2. Manning: 5 = (2y²)(1/0.013)(y/2)^(2/3)(0.001)^(0.5). Solve numerically for y.
Formula
Rectangular BHS: b = 2y, R = y/2
Variables
b = channel bottom width (m); y = flow depth (m); R = hydraulic radius (m). BHS = Best Hydraulic Section.
Application
Use when designing the most efficient rectangular channel. Substitute b = 2y into area and perimeter formulas, then solve for y using Manning's equation.
Example
A trapezoidal BHS with y = 1.5 m: z = 1/√3 ≈ 0.577, b = 2(1.5)/√3 = 1.732 m, R = 1.5/2 = 0.75 m.
Formula
Trapezoidal BHS: b = (2y/√3), z = 1/√3, R = y/2
Variables
b = bottom width (m); y = depth (m); z = horizontal-to-vertical side slope ratio; R = y/2 for best hydraulic trapezoidal section.
Application
Key result: the hydraulic radius of the best trapezoidal section equals y/2 — same as the rectangular BHS. This simplifies design calculations.
Example
b = 4 m, z = 1.5, y = 1.5 m: A = (4 + 1.5×1.5)(1.5) = (4 + 2.25)(1.5) = 9.375 m²; P = 4 + 2(1.5)√(1 + 2.25) = 4 + 3√3.25 = 4 + 5.408 = 9.408 m; R = 9.375/9.408 = 0.997 m
Formula
A_trap = (b + zy)y; P_trap = b + 2y√(1 + z²)
Variables
A_trap = trapezoidal flow area (m²); P_trap = wetted perimeter (m); b = bottom width (m); y = depth (m); z = side slope (H:V).
Application
General trapezoidal channel geometry — used before applying Manning's equation.
Exam Tips
- Memorize: Rectangular BHS → b = 2y, R = y/2.
- Memorize: Trapezoidal BHS → z = 1/√3 ≈ 0.577, R = y/2.
- Both rectangular and trapezoidal BHS share R = y/2 — this is a powerful simplification for board problems.
- When asked to 'design the most efficient rectangular section' for a given Q, set b = 2y first, then use Manning's to find y.
- For the trapezoidal BHS, the water surface width T = b + 2zy = 2y(1/√3) + 2(1/√3)y = 4y/√3.
Key Points
- For fixed area A, slope S, and roughness n, discharge Q is maximized when the wetted perimeter P is minimized — this is the most efficient (best hydraulic) section.
- Minimizing P maximizes R = A/P, which maximizes v and Q in Manning's equation.
- The semicircle is the theoretically most efficient cross-section shape for any given area.
- Rectangular best hydraulic section: width = twice the depth (b = 2y), giving R = y/2.
- Trapezoidal best hydraulic section: a half-hexagon where each side equals the bottom width (z = 1/√3 ≈ 0.577, i.e., side slope 1H:√3V or 60° from horizontal), giving R = y/2.
- For trapezoidal: best condition is b = (2y/√3) = 2y·tan(30°), side slope z = 1/√3.
- Circular pipe: maximum discharge occurs at y/D ≈ 0.938 (not full); maximum velocity at y/D ≈ 0.813.
- In design problems, use the best hydraulic section conditions as geometric constraints, then apply Manning's equation to find dimensions.
Definitions
Term
Most Efficient (Best Hydraulic) Section
Definition
The channel cross-section that conveys the maximum discharge for a given flow area, slope, and roughness — achieved by minimizing the wetted perimeter P (and thereby maximizing R).
Importance
Directly tested in board exams as design problems. Results in cost-effective channel construction.
Term
Side Slope (z)
Definition
The ratio of horizontal to vertical distance on the channel side walls, expressed as z:1 (H:V). A slope of 1.5:1 means 1.5 m horizontal per 1 m vertical.
Importance
Used in area and wetted perimeter calculations for trapezoidal channels.
Section Title
2. Most Efficient (Best Hydraulic) Cross-Section
Common Mistakes
- Applying b = 2y to a trapezoidal channel — this condition is only for rectangular channels.
- Forgetting that for both rectangular and trapezoidal BHS, R = y/2 (same result).
- Using √(1 + z²) incorrectly — this is the length factor for sloped sides, not z itself.
- Assuming the circular section carries maximum flow when full — maximum Q occurs at y/D ≈ 0.938, not y/D = 1.0.
Formulas
Example
y = 1.2 m, v = 1.86 m/s: E = 1.2 + (1.86²)/(2×9.81) = 1.2 + 3.460/19.62 = 1.2 + 0.176 = 1.376 m
Formula
E = y + v²/(2g)
Variables
E = specific energy (m); y = flow depth (m); v = mean flow velocity (m/s); g = gravitational acceleration = 9.81 m/s².
Application
Compute the specific energy at any flow section. Can also be written as E = y + Q²/(2gA²).
Example
v = 1.86 m/s, y = 1.2 m: Fr = 1.86/√(9.81×1.2) = 1.86/√11.772 = 1.86/3.431 = 0.542 → subcritical
Formula
Fr = v / √(g·y)
Variables
Fr = Froude number (dimensionless); v = mean velocity (m/s); g = 9.81 m/s²; y = flow depth (m). For non-rectangular channels, replace y with hydraulic depth D_h = A/T.
Application
Classifies flow regime: Fr < 1 subcritical, Fr = 1 critical, Fr > 1 supercritical.
Example
Q = 6 m³/s, b = 3 m: q = 6/3 = 2 m²/s; y_c = (2²/9.81)^(1/3) = (4/9.81)^(1/3) = (0.4077)^(1/3) = 0.742 m
Formula
y_c = (q²/g)^(1/3)
Variables
y_c = critical depth (m); q = unit discharge = Q/b (m²/s or m³/s·m⁻¹ for rectangular channels); g = 9.81 m/s².
Application
Direct formula for critical depth in rectangular channels. Works only for rectangular (prismatic) sections.
Example
y_c = 0.742 m: E_min = (3/2)(0.742) = 1.113 m
Formula
E_min = (3/2)·y_c
Variables
E_min = minimum specific energy (m); y_c = critical depth (m). Valid only for rectangular channels.
Application
Quick calculation of minimum specific energy without computing velocity head separately.
Example
For a trapezoidal channel, substitute A = (b + zy_c)y_c and T = b + 2zy_c, then solve iteratively for y_c.
Formula
General critical flow condition: A³/T = Q²/g
Variables
A = flow area at critical depth (m²); T = top width of water surface (m); Q = discharge (m³/s); g = 9.81 m/s².
Application
Used for non-rectangular sections (trapezoidal, circular) where the simple y_c formula does not apply. Requires iterative solution.
Exam Tips
- If the problem gives Q and b for a rectangular channel, immediately compute q = Q/b, then y_c = (q²/g)^(1/3).
- Always state the flow regime (subcritical/critical/supercritical) after computing Fr — board exams award marks for this.
- E_min = 1.5·y_c is a 1-line answer for minimum specific energy in rectangular channels.
- At critical flow: v_c = √(g·y_c) for rectangular channels — useful for checking.
- The E-y curve is asymptotic to the line E = y (as y→∞, KE→0) and to the y-axis (as y→0, y→0 but KE→∞).
Key Points
- Specific energy E is the total mechanical energy per unit weight of fluid measured from the channel bottom (datum at channel bed).
- E = y + v²/(2g) = y + Q²/(2gA²) — comprises potential energy (y) and kinetic energy (v²/2g).
- The specific energy diagram (E vs. y for constant Q) shows two possible depths for each E > E_min: subcritical (high y, low v) and supercritical (low y, high v).
- Critical flow occurs at minimum specific energy for a given discharge — only one depth exists (no alternative).
- Froude number Fr classifies flow: Fr < 1 subcritical (tranquil), Fr = 1 critical, Fr > 1 supercritical (rapid/shooting).
- For rectangular channels, unit discharge q = Q/b (m³/s per metre width) is used to find critical depth.
- At critical depth y_c: E_min = (3/2)y_c for rectangular channels.
- Critical depth is independent of channel slope and roughness — it depends only on discharge and channel geometry.
- The hydraulic depth D_h = A/T is used for non-rectangular sections (T = top water surface width).
Definitions
Term
Specific Energy (E)
Definition
Total mechanical energy per unit weight of the flowing fluid, measured with the channel bed as the datum. E = y + v²/(2g). Units: metres (m).
Importance
Central concept for analyzing flow transitions, critical flow, and hydraulic structures. Forms the basis of the E-y diagram.
Term
Critical Depth (y_c)
Definition
The flow depth at which specific energy is minimum for a given discharge. At this depth, Fr = 1 and the flow is critical.
Importance
Reference depth for classifying flow as subcritical (y > y_c) or supercritical (y < y_c). Key in hydraulic jump calculations.
Term
Froude Number (Fr)
Definition
A dimensionless number representing the ratio of inertial forces to gravitational forces: Fr = v/√(gD_h). Controls flow regime classification.
Importance
Board exams frequently ask to 'classify the flow' — always compute Fr and state the regime.
Term
Subcritical Flow (Fr < 1)
Definition
Tranquil or streaming flow where depth is greater than critical depth and velocity is less than critical velocity. Disturbances can propagate both upstream and downstream.
Importance
Most irrigation canals and rivers operate in this regime. Control section is at the downstream end.
Term
Supercritical Flow (Fr > 1)
Definition
Rapid or shooting flow where depth is less than critical depth and velocity exceeds critical velocity. Disturbances can only propagate downstream.
Importance
Occurs on steep slopes and chutes. Control section is at the upstream end.
Term
Unit Discharge (q)
Definition
Discharge per unit width of a rectangular channel: q = Q/b, in units of m²/s (or m³/s per m width).
Importance
Simplifies critical depth calculation to a 1-D formula: y_c = (q²/g)^(1/3).
Section Title
3. Specific Energy and Critical Flow
Common Mistakes
- Using Fr = v/√(gy) for non-rectangular sections — must use hydraulic depth D_h = A/T instead of y.
- Applying y_c = (q²/g)^(1/3) to trapezoidal or circular channels — valid only for rectangular.
- Confusing E_min = (3/2)y_c with E = y_c — the minimum energy is 1.5 times y_c, not equal to it.
- Using flow depth y instead of unit discharge q when computing critical depth.
- Forgetting that critical depth depends only on Q and geometry — NOT on slope or roughness.
Formulas
Example
y1 = 0.4 m, v1 = 6 m/s: Fr1 = 6/√(9.81×0.4) = 6/1.981 = 3.029; y2 = (0.4/2)(√(1 + 8×3.029²) - 1) = 0.2(√(1 + 73.39) - 1) = 0.2(√74.39 - 1) = 0.2(8.625 - 1) = 0.2(7.625) = 1.525 m
Formula
y2/y1 = (1/2)(√(1 + 8·Fr1²) - 1)
Variables
y1 = upstream supercritical depth (m); y2 = downstream subcritical depth (m); Fr1 = upstream Froude number = v1/√(g·y1).
Application
The primary hydraulic jump formula for rectangular channels. Given y1 and v1 (or Fr1), solve for y2.
Example
y1 = 0.4 m, y2 = 1.525 m: ΔE = (1.525 - 0.4)³/(4 × 0.4 × 1.525) = (1.125)³/(2.44) = 1.4238/2.44 = 0.584 m
Formula
ΔE = E1 - E2 = (y2 - y1)³ / (4·y1·y2)
Variables
ΔE = energy loss in the hydraulic jump (m); y1 = upstream depth (m); y2 = downstream depth (m).
Application
Computes the head loss (energy dissipated) across the hydraulic jump. Used in stilling basin design.
Example
v1 = 6 m/s, y1 = 0.4 m: Fr1 = 6/√(9.81 × 0.4) = 6/1.981 = 3.03 → supercritical (Fr > 1, jump can occur)
Formula
Fr1 = v1 / √(g·y1)
Variables
Fr1 = upstream Froude number; v1 = upstream velocity (m/s); g = 9.81 m/s²; y1 = upstream depth (m).
Application
Always the first step in any hydraulic jump problem — compute Fr1 before applying the sequent depth formula.
Exam Tips
- Step 1: Always compute Fr1 = v1/√(g·y1). Confirm Fr1 > 1 before proceeding.
- Step 2: Substitute Fr1 into y2/y1 = (1/2)(√(1 + 8Fr1²) - 1) to get y2.
- Step 3: If asked for energy loss, use ΔE = (y2 - y1)³/(4y1y2).
- The sequent depth ratio y2/y1 increases with Fr1 — a stronger jump dissipates more energy.
- Board exam shortcut: if given y1 and y2 directly, find Fr1 from y2/y1 = (1/2)(√(1+8Fr1²)−1) → solve for Fr1.
Key Points
- A hydraulic jump is a rapidly varied flow phenomenon where supercritical flow (Fr > 1) abruptly transitions to subcritical flow (Fr < 1) with significant energy loss.
- The jump is driven by momentum balance (not energy) — energy is lost (dissipated as heat and turbulence).
- y1 = upstream (supercritical) depth; y2 = downstream (subcritical) depth — called conjugate or sequent depths.
- The sequent depth ratio y2/y1 depends only on the upstream Froude number Fr1.
- Energy dissipated in the jump: ΔE = E1 - E2 = (y2 - y1)³ / (4·y1·y2).
- Hydraulic jumps are used in engineering: energy dissipators below spillways, stilling basins, and to prevent channel scour.
- Classifications by Fr1: undular jump (1 < Fr1 < 1.7), weak jump (1.7–2.5), oscillating (2.5–4.5), steady (4.5–9.0), strong (Fr1 > 9.0).
- The momentum equation (not Bernoulli) governs conjugate depth relationships.
Definitions
Term
Hydraulic Jump
Definition
An abrupt transition from supercritical to subcritical flow in an open channel, accompanied by significant energy dissipation, surface turbulence, and air entrainment.
Importance
Major energy dissipator in hydraulic structures. Sequent depth formula is a standard board exam problem type.
Term
Conjugate (Sequent) Depths
Definition
The pair of depths y1 (upstream, supercritical) and y2 (downstream, subcritical) that satisfy the momentum equation across a hydraulic jump. They have equal specific momentum (force-momentum function).
Importance
The relationship y2/y1 = f(Fr1) is the core formula — must be memorized exactly.
Term
Energy Loss (ΔE)
Definition
The specific energy dissipated in the hydraulic jump due to turbulence and mixing: ΔE = (y2 - y1)³/(4·y1·y2). Always positive (energy is lost, not gained).
Importance
Used to size stilling basins and evaluate the effectiveness of energy dissipators.
Term
Stilling Basin
Definition
An engineered structure designed to contain and stabilize a hydraulic jump downstream of spillways, gates, or chutes, preventing channel scour.
Importance
Practical application context for hydraulic jump calculations in Philippine infrastructure projects.
Section Title
4. The Hydraulic Jump
Common Mistakes
- Applying the sequent depth formula to subcritical incoming flow (Fr1 < 1) — a jump cannot form from subcritical flow.
- Mixing up y1 and y2 — y1 is always the smaller (supercritical) depth upstream of the jump.
- Computing energy loss as E2 - E1 (getting a negative value) instead of E1 - E2 (positive loss).
- Using the energy equation (Bernoulli) across a hydraulic jump — energy is lost, so only the momentum equation applies.
- Forgetting to square Fr1 inside the square root: the formula is √(1 + 8·Fr1²), not √(1 + 8·Fr1).
Formulas
Example
Answer: Q ≈ 17.6 m³/s
Formula
Worked Problem 1 — Trapezoidal Channel Discharge: Q = A·(1/n)·R^(2/3)·S^(1/2)
Variables
b = 4 m, z = 1.5, y = 1.5 m, n = 0.015, S = 0.0008
Application
A = (b + zy)y = (4 + 1.5×1.5)(1.5) = (4 + 2.25)(1.5) = 6.25×1.5 = 9.375 m²; P = b + 2y√(1+z²) = 4 + 2(1.5)√(1+2.25) = 4 + 3√3.25 = 4 + 5.408 = 9.408 m; R = 9.375/9.408 = 0.9965 m; v = (1/0.015)(0.9965)^(2/3)(0.0008)^(0.5) = 66.67 × 0.9977 × 0.02828 = 1.882 m/s; Q = 9.375 × 1.882 = 17.64 m³/s
Example
Answer: y_c = 0.971 m, E_min = 1.457 m
Formula
Worked Problem 2 — Critical depth and E_min: y_c = (q²/g)^(1/3); E_min = (3/2)y_c
Variables
q = 3 m²/s (given unit discharge)
Application
y_c = (3²/9.81)^(1/3) = (9/9.81)^(1/3) = (0.9174)^(1/3) = 0.971 m; E_min = (3/2)(0.971) = 1.457 m
Example
Answer: y2 = 1.525 m, ΔE = 0.583 m
Formula
Worked Problem 3 — Hydraulic jump: y2/y1 = (1/2)(√(1+8Fr1²) − 1)
Variables
y1 = 0.4 m, v1 = 6 m/s
Application
Fr1 = 6/√(9.81×0.4) = 6/1.981 = 3.029; y2 = (0.4/2)(√(1+8×3.029²) − 1) = 0.2(√74.40 − 1) = 0.2(8.626 − 1) = 0.2(7.626) = 1.525 m; ΔE = (1.525−0.4)³/(4×0.4×1.525) = (1.125)³/2.44 = 1.424/2.44 = 0.583 m
Example
Answer: y ≈ 1.20 m, b ≈ 2.40 m
Formula
Worked Problem 4 — Design most efficient rectangular section for Q = 5 m³/s, n = 0.013, S = 0.001
Variables
Set b = 2y (BHS condition), then R = y/2
Application
A = 2y²; R = y/2; Manning: 5 = (2y²)(1/0.013)(y/2)^(2/3)(0.001)^(0.5); 5 = (2y²)(76.92)(0.6300·y^(2/3))(0.03162); 5 = 2y² × 76.92 × 0.6300 × 0.03162 × y^(2/3); 5 = 2y^(8/3) × 1.531; y^(8/3) = 5/(2×1.531) = 5/3.062 = 1.633; y = (1.633)^(3/8) = 1.633^(0.375); ln(1.633) = 0.4900; 0.375×0.4900 = 0.1838; y = e^(0.1838) = 1.202 m; b = 2y = 2.404 m
Exam Tips
- In board exams, show all steps: A, P, R, v, Q — partial credit is awarded for correct setup even if arithmetic is off.
- (y/2)^(2/3) = 0.6300·y^(2/3) is a key simplification for rectangular BHS problems — derive it once, memorize it.
- Always verify your answer: plug y and b back into Manning's equation to confirm Q matches the target.
- For board-level accuracy, carry at least 4 significant figures through intermediate steps.
- Practice the hydraulic jump calculation until the three-step sequence (Fr1 → y2 → ΔE) is automatic.
Key Points
- Trapezoidal channel Manning discharge: set up A and P carefully with z correctly applied.
- Critical depth for non-rectangular sections requires iterative or trial-and-error solution.
- Design of most efficient rectangular section: set b = 2y, express Q in terms of y only, then solve.
- Always state: A, P, R, v, Q in sequence for Manning problems.
- Hydraulic jump problems: Fr1 → y2 → ΔE in a clean three-step sequence.
Section Title
5. Board-Level Solved Problems
Common Mistakes
- For trapezoidal channels, using z directly instead of √(1+z²) for the sloped side length in wetted perimeter.
- In the BHS design problem, forgetting to raise (y/2) to the 2/3 power — it becomes (1/2)^(2/3)·y^(2/3) = 0.6300·y^(2/3).
- Not confirming Fr1 > 1 before applying the hydraulic jump formula.
- Rounding intermediate values too early, leading to accumulated error in the final answer.
Connections
- Manning's equation connects to pipe flow via the Darcy-Weisbach equation — for full circular pipes, R = D/4 bridges the two formulations.
- Specific energy (E = y + v²/2g) is derived directly from Bernoulli's equation applied to a streamline at the channel bed — the connection to fluid mechanics fundamentals.
- The Froude number in open-channel flow is analogous to the Mach number in compressible flow — both represent the ratio of flow speed to wave propagation speed.
- Critical flow (Fr = 1) corresponds to minimum specific energy — connects energy analysis to flow classification.
- The hydraulic jump energy loss formula ΔE = (y2−y1)³/(4y1y2) is derived from the combination of continuity, momentum, and energy equations — integrating three fundamental principles.
- The most efficient section concept connects to optimization in engineering — maximizing discharge minimizes excavation and lining costs for irrigation canals (important in Philippine National Irrigation Administration projects).
- Gradually varied flow (GVF) — not covered here but the next topic — uses the specific energy and critical depth concepts as its foundation.
- Hydraulic grade line (HGL) and energy grade line (EGL) from pipe flow carry over to open channels — HGL is the water surface, EGL is HGL + v²/2g above it.
- RA 544 (Civil Engineering Law of the Philippines) requires licensed civil engineers to sign irrigation and drainage design drawings — this chapter's content is directly applicable to professional practice.
- NSCP 2015 references Manning's n values for drainage design in Section on Storm Drainage — knowledge of this chapter is prerequisite for structural drainage provisions.
Exam Strategy
For PRC CE board exams in Hydraulics, open-channel flow problems appear in every examination. Follow this proven strategy: (1) IDENTIFY the flow type immediately — is it uniform flow (Manning's), critical flow analysis, or a hydraulic jump? (2) For MANNING problems: write A, P, R, v, Q in five explicit steps — partial credit is awarded. (3) For CRITICAL FLOW: compute q = Q/b first, then y_c = (q²/g)^(1/3), then E_min = 1.5y_c — three clean steps. (4) For HYDRAULIC JUMPS: the three-step sequence is Fr1 → y2 → ΔE, always starting with Fr1 computation. (5) MEMORIZE exactly: y_c = (q²/g)^(1/3), E_min = 1.5y_c, y2/y1 = (1/2)(√(1+8Fr1²)−1), R_BHS = y/2, b_rect = 2y, z_trap = 1/√3. (6) WATCH the PITFALLS: SI form (no 1.49), S as decimal (not %), P = wetted sides only (no free surface), Fr = v/√(gD_h) for non-rectangular. (7) TIME MANAGEMENT: Manning problems take 3–4 minutes; critical depth and jump problems take 2–3 minutes each. Allocate accordingly. (8) ALWAYS state the flow regime (subcritical/supercritical) when Fr is computed — this is a free point that candidates often omit.
Quick Review Questions
A rectangular channel has width b = 5 m and depth y = 2 m. What is the hydraulic radius R?
A = 5 × 2 = 10 m²; P = 5 + 2(2) = 5 + 4 = 9 m; R = A/P = 10/9 = 1.111 m. Wait — recalculate: R = 10/9 = 1.111 m. Actually: P = b + 2y = 5 + 2(2) = 9 m; R = 10/9 = 1.111 m. Correct answer: R = 1.111 m. Key: P includes only the three wetted sides (bottom + two walls), not the free surface.
For a rectangular channel with b = 4 m and Q = 8 m³/s, what is the critical depth?
q = Q/b = 8/4 = 2 m²/s; y_c = (q²/g)^(1/3) = (4/9.81)^(1/3) = (0.4077)^(1/3). Computing: 0.4077^(1/3) = e^(ln(0.4077)/3) = e^(-0.897/3) = e^(-0.299) = 0.741 m. More precisely: 0.742 m. The answer is y_c ≈ 0.742 m. Formula: y_c = (q²/g)^(1/3) for rectangular channels only.
What are the conditions for the most efficient rectangular cross-section?
The most efficient rectangular section minimizes the wetted perimeter for a given area, achieved when the width equals twice the depth (b = 2y). This gives: A = 2y², P = 4y, R = A/P = 2y²/4y = y/2. This maximizes hydraulic radius and thus discharge for the given area and slope.
Flow in a rectangular channel has v = 3.5 m/s at depth y = 0.8 m. Is the flow subcritical or supercritical?
Fr = v/√(gy) = 3.5/√(9.81 × 0.8) = 3.5/√7.848 = 3.5/2.801 = 1.250. Since Fr > 1, the flow is supercritical (rapid or shooting flow). Since depth y < y_c, the flow is also above-critical depth comparison: y_c = (q²/g)^(1/3) where q = 3.5 × 0.8 = 2.8 m²/s → y_c = (7.84/9.81)^(1/3) = (0.799)^(1/3) = 0.928 m > 0.8 m ✓
A hydraulic jump occurs with upstream depth y1 = 0.5 m and upstream velocity v1 = 5 m/s. Find the sequent depth y2.
Step 1: Fr1 = v1/√(gy1) = 5/√(9.81×0.5) = 5/√4.905 = 5/2.215 = 2.258. Step 2: y2/y1 = (1/2)(√(1+8Fr1²)−1) = (1/2)(√(1+8×2.258²)−1) = (1/2)(√(1+40.77)−1) = (1/2)(√41.77−1) = (1/2)(6.463−1) = (1/2)(5.463) = 2.732. Step 3: y2 = y1 × 2.732 = 0.5 × 2.732 = 1.366 m. Note: small variations in intermediate rounding may give ~1.37 m.
What is the specific energy at y = 1.5 m and v = 2.0 m/s?
E = y + v²/(2g) = 1.5 + (2.0²)/(2×9.81) = 1.5 + 4.0/19.62 = 1.5 + 0.2039 = 1.704 m. The kinetic energy head is v²/(2g) = 0.204 m, and the pressure/elevation head from the channel bottom is y = 1.5 m.
For the best hydraulic trapezoidal section, what is the side slope z and what is the hydraulic radius R in terms of depth y?
The best hydraulic trapezoidal section is a half-hexagon. Each side wall makes a 60° angle from the horizontal, giving a horizontal-to-vertical ratio z = cot(60°) = 1/tan(60°) = 1/√3 ≈ 0.577. The resulting hydraulic radius R = y/2 — identical to the rectangular BHS result. This is the section that maximizes Q for a given area, roughness, and slope among trapezoidal shapes.
What is the Manning's equation in SI units, and why must the coefficient 1.49 NOT be used?
Manning's equation in SI: v = (1/n)R^(2/3)S^(1/2), where v is in m/s and R is in metres. The US customary form is v = (1.49/n)R^(2/3)S^(1/2), where v is in ft/s and R is in feet. Using 1.49 with SI units overpredicts velocity by about 49%, a catastrophic error in design. PRC board exams always use SI — the coefficient is 1.0.
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