CELE Hydraulics & Fluid Mechanics — Orifices, Weirs, Tubes and NozzlesDetailed Explanation
Detailed explanations for CELE Hydraulics & Fluid Mechanics — Orifices, Weirs, Tubes and Nozzles. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Orifices, Weirs, Tubes and Nozzles questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Orifices, Weirs, Tubes and Nozzles is the 8th chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.
Orifices, Weirs, Tubes and Nozzles - Detailed Explanation
Orifices, weirs, tubes, and nozzles are the foundational flow-measurement and flow-control devices encountered in hydraulic engineering practice. Every irrigation canal headwork in the Philippines, every water-supply reservoir outlet, and every stormwater detention basin relies on the principles covered in this chapter. The unifying concept is Torricelli's theorem — that fluid escaping through an opening under a head h attains a theoretical velocity v = √(2gh) — which is then corrected by empirical discharge coefficients to match real-world behavior. For PRC CE board examinees, this chapter consistently appears in the hydraulics portion, carrying questions that range from direct formula substitution to multi-step tank-emptying problems. Mastery of the three coefficients (Cv, Cc, Cd), the distinction between rectangular and triangular weir formulas, and the time-to-empty equation will allow you to solve virtually any board-exam problem in this topic within three to five minutes.
Concepts
Orifice Flow and Discharge Coefficients
An orifice is any opening — circular, rectangular, or otherwise — through which fluid flows under a driving head h measured to the centroid of the opening. Applying Bernoulli's equation between the free surface and the vena contracta (the narrowest section of the emerging jet) gives the ideal velocity v_ideal = √(2gh). Real flow deviates from this ideal because of two physical effects: 1. FRICTION (velocity reduction): The coefficient of velocity Cv = v_actual / v_ideal. For a sharp-edged circular orifice, Cv ≈ 0.97–0.99 (use 0.98 unless stated otherwise). 2. JET CONTRACTION (area reduction): The streamlines converge past the orifice edge, producing a contracted cross-section called the vena contracta where area A_vc < A_orifice. The coefficient of contraction Cc = A_vc / A_orifice. For sharp-edged orifices, Cc ≈ 0.61–0.64 (use 0.62). The combined effect defines the coefficient of discharge: Cd = Cv × Cc For a sharp-edged circular orifice: Cd ≈ 0.97 × 0.62 ≈ 0.60–0.62. Actual discharge: Q = Cd × A × √(2gh) where A is the gross orifice area (not the vena contracta area). This single equation is the workhorse of orifice hydraulics. SUBMERGED ORIFICE: When the orifice discharges into a pool (tailwater), the effective driving head h = h1 − h2, where h1 is the upstream surface elevation and h2 is the downstream surface elevation, both measured from the orifice centerline (or any common datum). The formula Q = Cd A √(2gh) still applies with this differential head. Orifice shapes and Cd values: - Sharp-edged circular: Cd ≈ 0.61 - Short cylindrical tube (L/D = 2–3): Cd ≈ 0.82 (re-entrant flow fills the tube, eliminating contraction) - Re-entrant (Borda's mouthpiece): Cd ≈ 0.51 - Well-rounded entrance: Cd ≈ 0.98 For board exams, the value of Cd (or Cv and Cc separately) is always given in the problem unless you are asked to compute Cd from the other two.
Examples
The sequence — compute A, compute √(2gh), then multiply by Cd and A — is the standard three-step approach for every orifice discharge problem. Always convert diameter to meters before computing area.
Scenario
A sharp-edged circular orifice 100 mm in diameter is installed in the vertical wall of a tank. The water surface is maintained at a constant head of 3.6 m above the orifice center. Given Cd = 0.62, find: (a) the theoretical velocity, (b) the actual velocity, and (c) the actual discharge.
Solution
Given: D = 100 mm = 0.10 m, h = 3.6 m, Cd = 0.62, Cv = 0.98 (standard sharp-edge), g = 9.81 m/s² Orifice area: A = π(0.10)²/4 = 7.854 × 10⁻³ m² (a) Theoretical velocity: v_th = √(2gh) = √(2 × 9.81 × 3.6) = √70.632 = 8.404 m/s (b) Actual velocity: v_act = Cv × v_th = 0.98 × 8.404 = 8.236 m/s (c) Actual discharge: Q = Cd × A × √(2gh) Q = 0.62 × 7.854×10⁻³ × 8.404 Q = 0.62 × 0.06602 Q = 0.04093 m³/s ≈ 40.9 L/s Note: Alternatively, Q = Cv × Cc × A × √(2gh). Using Cc = Cd/Cv = 0.62/0.98 = 0.6327 gives the same result.
For a submerged orifice, the formula is identical to the free-discharge case; only the head changes. Many examinees incorrectly use h1 alone — remember it is always the DIFFERENCE in surface elevations.
Scenario
A submerged orifice has an area of 0.015 m². The upstream water level is 5.0 m and the downstream water level is 2.5 m above the orifice center. If Cd = 0.60, find Q.
Solution
Differential head: h = h1 − h2 = 5.0 − 2.5 = 2.5 m Q = Cd × A × √(2gh) = 0.60 × 0.015 × √(2 × 9.81 × 2.5) = 0.009 × √49.05 = 0.009 × 7.004 = 0.0630 m³/s ≈ 63.0 L/s
Applications
- Outlet works of irrigation dams and reservoirs (National Irrigation Administration projects)
- Calibration of flowmeters using known Cd values
- Drainage of water from construction cofferdams
- Determining jet trajectory for hydraulic model studies
- Water supply tank outlet sizing for MWSS and LGU water districts
Misconceptions
- Using orifice diameter instead of area — always compute A = πD²/4 first.
- Measuring head to the TOP of the orifice instead of its centroid (center).
- Using Cv alone (without Cc) to compute discharge — this gives Q = Cv A √(2gh) which ignores contraction and overestimates Q.
- For submerged orifices, using only the upstream head instead of the differential head.
- Confusing the vena contracta area with the orifice area when substituting into Q formula.
Related Concepts
- Bernoulli's equation and energy heads
- Continuity equation (Q = Av)
- Torricelli's theorem
- Hydraulic grade line and energy grade line
- Velocity of approach correction for large orifices
Common Exam Questions
Example
A 75-mm diameter orifice (Cd = 0.61) under 5 m head. Q = 0.61 × π(0.075)²/4 × √(2×9.81×5) = 0.61 × 4.418×10⁻³ × 9.905 = 0.02666 m³/s ≈ 26.7 L/s.
Approach
Identify head h (to centroid), orifice area A, and Cd; apply Q = Cd A √(2gh) directly.
Question Type
Direct discharge computation
Example
Given Q = 0.05 m³/s, Cd = 0.62, D = 100 mm: h = (0.05)²/[2×9.81×(0.62)²×(7.854×10⁻³)²] = 0.0025/[19.62×0.3844×6.168×10⁻⁵] = 5.42 m.
Approach
Rearrange Q = Cd A √(2gh): h = Q²/(2g Cd² A²) or A = Q/(Cd √(2gh)).
Question Type
Back-calculation of head or area from given Q
Example
Cd = 0.62, Cv = 0.98: Cc = 0.62/0.98 = 0.6327. For 100-mm orifice: d_jet = 100√0.6327 = 79.5 mm.
Approach
Cc = Cd / Cv. If jet diameter is asked, A_jet = Cc × A_orifice; d_jet = D_orifice × √Cc.
Question Type
Finding Cc from Cd and Cv
Key Points To Remember
- Q = Cd × A × √(2gh) — commit this to memory first.
- Head h is measured to the CENTROID (center) of the orifice, not the top or bottom.
- Cd = Cv × Cc; typical sharp-edge values: Cv ≈ 0.98, Cc ≈ 0.62, Cd ≈ 0.61.
- Submerged orifice: replace h with (h_upstream − h_downstream).
- A in the formula is the gross orifice area, not the vena contracta area.
- Ideal velocity = √(2gh); actual velocity = Cv × √(2gh).
- Vena contracta area = Cc × A (use this if asked for jet diameter or cross-sectional area of jet).
Tubes and Nozzles
A tube is a short conduit attached to an opening that modifies the flow pattern and hence the discharge coefficient. A nozzle is a converging tube designed to maximize exit velocity by converting pressure head into kinetic energy. TYPES OF TUBES AND THEIR Cd VALUES: 1. Standard short tube (L/D ≈ 2.5, square entrance): Cd ≈ 0.82. The tube runs full; the vena contracta occurs inside, then the flow re-expands to fill the tube. Because Cc ≈ 1.0 (no contraction at exit), Cd = Cv ≈ 0.82. 2. Re-entrant tube (Borda's mouthpiece, projecting inward): Cd ≈ 0.51 (running full), or Cd ≈ 0.61 (running partially). For the fully running case, Cc = 0.5 and Cv = 1.0 theoretically. 3. Converging tube (nozzle): Cd approaches 0.95–0.99 because contraction is minimized and friction is low. 4. Diverging tube (Venturi-type exit): Can increase pressure recovery but discharge is less predictable. NOZZLES: A nozzle attached to a pipe under pressure converts internal pressure to velocity. Given the head H (total energy head at the nozzle inlet referred to the nozzle tip), the exit velocity: v = Cv × √(2gH) Discharge: Q = Cd × A_nozzle × √(2gH) where A_nozzle is the exit (throat) area. For a well-designed nozzle, Cv ≈ 0.96–0.99 and Cc ≈ 1.0, so Cd ≈ Cv. FORCE ON A NOZZLE (supplementary board topic): Applying the momentum equation: F = ρQv_exit − ρQv_pipe. The nozzle must be bolted or restrained against this thrust force, which is relevant in firefighting hose design. VENTURI METER (related device): Uses converging-diverging geometry. Discharge: Q = Cd × A_throat × √(2g Δh / [1 − (A_throat/A_pipe)²]) Board exams occasionally combine the Venturi with orifice concepts.
Examples
The negative pressure at the internal vena contracta is the key physical insight for tubes. If this vacuum exceeds about 7.5 m of water (atmospheric limit), the tube will cavitate and the flow becomes unstable — a practical design constraint.
Scenario
A standard short tube (Cd = 0.82) has a 75-mm diameter and is installed in a tank wall under a constant head of 4.5 m. Find (a) Q and (b) the pressure at the vena contracta inside the tube, if the vena contracta area ratio Cc_internal = 0.62.
Solution
A_tube = π(0.075)²/4 = 4.418 × 10⁻³ m² (a) Q = Cd × A × √(2gh) = 0.82 × 4.418×10⁻³ × √(2×9.81×4.5) = 0.82 × 4.418×10⁻³ × 9.396 = 0.03403 m³/s ≈ 34.0 L/s (b) Pressure at internal vena contracta: At vena contracta, area = 0.62 × 4.418×10⁻³ = 2.739×10⁻³ m² Velocity at vc: v_vc = Q / A_vc = 0.03403 / 2.739×10⁻³ = 12.42 m/s Apply Bernoulli from free surface to internal vc (datum at tube centerline): h + 0 + 0 = 0 + v_vc²/(2g) + p_vc/γ 4.5 = (12.42)²/(2×9.81) + p_vc/γ 4.5 = 7.869 + p_vc/γ p_vc/γ = 4.5 − 7.869 = −3.369 m (vacuum/negative gauge pressure) This sub-atmospheric pressure at the internal vena contracta explains why standard tubes have higher discharge than orifices — the pressure drop 'pulls' more flow through.
Applications
- Fire hose nozzles — high-velocity jets for firefighting
- Irrigation sprinkler nozzle sizing
- Hydraulic turbine penstock outlet nozzles (Pelton wheel)
- Culvert pipe ends acting as short tubes for road drainage
- Industrial process pipe nozzles for chemical injection
Misconceptions
- Applying the sharp-edge orifice Cd = 0.62 to a tube — tubes have higher Cd due to full-bore flow.
- Assuming Cc = 0.62 for a nozzle — nozzles have Cc ≈ 1.0 because the exit jet fills the nozzle throat.
- Using pipe cross-section area instead of nozzle throat area for nozzle discharge.
Related Concepts
- Orifice discharge coefficients
- Cavitation and minimum pressures in conduits
- Momentum equation for jet force
- Venturi meter principle
- Hydraulic turbine nozzle design
Common Exam Questions
Example
Same opening (A = 0.01 m²), same head (h = 3 m): Q_orifice = 0.62×0.01×7.672 = 0.0476 m³/s; Q_tube = 0.82×0.01×7.672 = 0.0629 m³/s. Increase = 32%.
Approach
Use Q = Cd A √(2gh) for each; ratio Q_tube/Q_orifice = Cd_tube/Cd_orifice since A and h are the same.
Question Type
Comparison of orifice vs. tube discharge
Example
Nozzle exit D = 50 mm, H = 10 m, Cv = 0.97: v = 0.97×√(196.2) = 0.97×14.007 = 13.59 m/s; Q = 13.59 × π(0.05)²/4 = 13.59 × 1.963×10⁻³ = 0.02668 m³/s.
Approach
Identify H (total head at nozzle), Cv, exit area A; then v = Cv√(2gH), Q = v × A (since Cc = 1 for nozzle).
Question Type
Nozzle exit velocity and discharge
Key Points To Remember
- Tube Cd > Orifice Cd because the tube runs full (no vena contracta at exit): standard tube Cd ≈ 0.82 vs. sharp-edge orifice Cd ≈ 0.61.
- Nozzle: Cc ≈ 1.0 (jet fills the exit area), so Cd ≈ Cv ≈ 0.96–0.99.
- Re-entrant (Borda's) tube: Cd ≈ 0.51 — lowest of all standard configurations.
- The governing equation Q = Cd A √(2gH) applies to ALL configurations; only Cd and A change.
- For nozzle problems, H is the total head at the nozzle inlet, not just elevation head.
- Jet power: P = γQH (useful for checking nozzle/turbine problems).
Weirs — Rectangular and Triangular
A weir is a flow-measurement structure consisting of an overflow notch (crest) built across an open channel. Discharge is determined from the head H measured above the crest at a point upstream unaffected by drawdown (typically 3–5× H upstream of the weir face). FUNDAMENTAL DERIVATION (for context): Consider a thin horizontal strip of area dA = L dh at height h above the crest. Applying orifice theory: dQ = Cd L dh √(2gh). Integrating from h = 0 to h = H: Q = Cd L √(2g) ∫₀ᴴ h^(1/2) dh = Cd L √(2g) × [2h^(3/2)/3]₀ᴴ Q_rect = (2/3) Cd √(2g) L H^(3/2) RECTANGULAR WEIR: Q = (2/3) Cd √(2g) L H^(3/2) Numerical constant: (2/3)√(2×9.81) = (2/3)(4.429) = 2.953 → Q = 2.953 Cd L H^(3/2) [SI, m³/s, m] FRANCIS FORMULA (end contractions): If the weir does not span the full channel width, streamlines contract at the ends. Using n = number of end contractions (0, 1, or 2): L_eff = L − 0.1 n H Q = (2/3) Cd √(2g) × (L − 0.1nH) × H^(3/2) VELOCITY OF APPROACH CORRECTION: If approach channel velocity va is significant, replace H with (H + va²/2g): Q = (2/3) Cd √(2g) L (H + va²/2g)^(3/2) TRIANGULAR (V-NOTCH) WEIR: For a V-notch of apex angle θ, the width of the notch at height h above the apex is 2h tan(θ/2). Following the same strip integration: dQ = Cd [2h tan(θ/2)] dh √(2gh) = 2Cd tan(θ/2) √(2g) h^(3/2) dh Q = 2Cd tan(θ/2) √(2g) ∫₀ᴴ h^(3/2) dh = 2Cd tan(θ/2) √(2g) × [2h^(5/2)/5]₀ᴴ Q_tri = (8/15) Cd √(2g) tan(θ/2) H^(5/2) For a 90° V-notch: tan(45°) = 1, so: Q_90° = (8/15) Cd √(2g) H^(5/2) Numerical: (8/15)(4.429) = 2.362 → Q = 2.362 Cd H^(5/2) [SI] For a 60° V-notch: tan(30°) = 0.5774: Q = (8/15) Cd √(2g) × 0.5774 × H^(5/2) = 1.364 Cd H^(5/2) WHY V-NOTCH IS PREFERRED FOR SMALL FLOWS: The H^(5/2) dependence means Q changes steeply with H — a small change in H corresponds to a proportionally larger change in Q at low flows, giving better measurement sensitivity. For large flows, the rectangular weir is preferred because the linear dimension L is adjustable.
Examples
Always apply the end-contraction correction before substituting into the weir formula. The 0.1nH reduction accounts for the lateral contraction of streamlines at the weir ends. Forgetting this step is a common board-exam error.
Scenario
A sharp-crested rectangular weir L = 2.5 m, Cd = 0.62, carries a head H = 0.35 m with two end contractions. Find Q.
Solution
Step 1: Apply Francis end-contraction correction (n = 2): L_eff = L − 0.1(n)(H) = 2.5 − 0.1(2)(0.35) = 2.5 − 0.07 = 2.43 m Step 2: Compute H^(3/2): H^(3/2) = (0.35)^1.5 = 0.35 × √0.35 = 0.35 × 0.5916 = 0.2071 m^(3/2) Step 3: Discharge: Q = (2/3) Cd √(2g) × L_eff × H^(3/2) = (2/3)(0.62)(4.429)(2.43)(0.2071) = (0.4133)(4.429)(2.43)(0.2071) = (0.4133)(4.429) = 1.8306 × 2.43 = 4.448 × 0.2071 = 0.9212 m³/s ≈ 0.921 m³/s Verification using 2.953Cd L H^(3/2): Q = 2.953 × 0.62 × 2.43 × 0.2071 = 1.831 × 2.43 × 0.2071 = 0.921 m³/s ✓
The H^(5/2) power is the step where most computational errors occur. Compute (H)^2 then multiply by √H, or use logarithms: log(0.28^2.5) = 2.5 × log(0.28) = 2.5 × (−0.5528) = −1.382 → 10^(−1.382) = 0.04149 ✓
Scenario
A 90° triangular weir discharges under a head H = 0.28 m. Given Cd = 0.58, find Q.
Solution
For 90° V-notch, tan(θ/2) = tan(45°) = 1.0 Q = (8/15) Cd √(2g) × tan(θ/2) × H^(5/2) = (8/15)(0.58)(4.429)(1.0)(0.28)^(2.5) Compute H^(5/2) = (0.28)^2.5: (0.28)² = 0.0784 (0.28)^0.5 = 0.5292 (0.28)^2.5 = 0.0784 × 0.5292 = 0.04149 m^(5/2) Q = (0.5333)(0.58)(4.429)(1.0)(0.04149) = (0.5333 × 0.58) = 0.3093 × 4.429 = 1.370 × 0.04149 = 0.05684 m³/s ≈ 56.8 L/s
Back-calculation for weir length is common in design-type exam questions. Rearrange the formula algebraically before substituting numbers to avoid algebraic errors.
Scenario
A rectangular weir must pass Q = 1.8 m³/s at H = 0.4 m with Cd = 0.62, no end contractions. Find the required weir length L.
Solution
Rearrange Q = (2/3) Cd √(2g) L H^(3/2) for L: L = Q / [(2/3) Cd √(2g) H^(3/2)] H^(3/2) = (0.4)^1.5 = 0.4 × √0.4 = 0.4 × 0.6325 = 0.2530 m^(3/2) (2/3) Cd √(2g) = 2.953 × 0.62 = 1.831 L = 1.8 / (1.831 × 0.2530) = 1.8 / 0.4632 = 3.885 m ≈ 3.89 m
Applications
- Stream gauging stations of PAGASA and DPWH-BMO for flood monitoring
- Irrigation distribution canal flow measurement (NIA headworks)
- Settling basin overflow weirs in water treatment plants (MWSS, MCWD)
- Stormwater detention pond outlet structures under DPWH design standards
- Hydropower intake control structures
Misconceptions
- Mixing up the exponents: rectangular is H^(3/2), triangular is H^(5/2). Never swap them.
- Measuring H from the channel bottom instead of from the weir crest elevation.
- Using the full notch angle θ instead of θ/2 in tan(θ/2) for V-notch formula.
- Forgetting that Cd for V-notch is typically slightly different from rectangular (0.58–0.62 vs 0.61–0.62).
- Applying end-contraction correction to a suppressed weir (weir spans full channel width) where n = 0.
Related Concepts
- Orifice strip integration derivation
- Open-channel flow measurement methods
- Critical flow and Froude number
- Broad-crested weir (Cd ≈ 0.848)
- Spillway design — Ogee crest weir
Common Exam Questions
Example
60° V-notch, H = 0.25 m, Cd = 0.60: Q = (8/15)(0.60)(4.429)(tan 30°)(0.25^2.5) = (0.3200)(4.429)(0.5774)(0.03125) = 0.02571 m³/s.
Approach
Identify weir type (rectangular or triangular), apply correct formula, compute H^(3/2) or H^(5/2) carefully, include end contractions if stated.
Question Type
Find Q given weir dimensions, Cd, and H
Example
Rectangular weir: H = (Q / (2.953 Cd L))^(2/3).
Approach
Rearrange formula: H = [Q / ((2/3)Cd√(2g)L)]^(2/3) for rectangular, or H = [Q / ((8/15)Cd√(2g)tan(θ/2))]^(2/5) for triangular.
Question Type
Find H given Q and weir geometry
Example
For Q < 0.1 m³/s, use V-notch; for Q > 1 m³/s, use rectangular — justification based on sensitivity of dQ/dH.
Approach
V-notch Q ∝ H^(5/2) gives better resolution at low H; rectangular Q ∝ H^(3/2) handles large Q with wider crest.
Question Type
Choose between rectangular and V-notch for accuracy
Key Points To Remember
- Rectangular weir: Q = (2/3) Cd √(2g) L H^(3/2) — exponent is 3/2 = 1.5.
- Triangular weir: Q = (8/15) Cd √(2g) tan(θ/2) H^(5/2) — exponent is 5/2 = 2.5.
- Head H is measured ABOVE THE CREST, not above the channel bottom.
- 90° V-notch: tan(45°) = 1 simplifies the formula significantly.
- Francis end-contraction correction: L_eff = L − 0.1nH (n = 1 for one end, n = 2 for both ends free).
- V-notch is better for small flows; rectangular is better for large flows.
- Cd for weirs: typically 0.60–0.62 for sharp-crested rectangular; 0.58–0.62 for V-notch.
- Memorize (2/3)√(2g) = 2.953 and (8/15)√(2g) = 2.362 for quick SI computation.
Time to Empty (Drain) a Tank
When a tank drains through an orifice and the head decreases as the tank empties, the discharge Q is not constant — it diminishes as h falls. This requires integration to find the time for the head to drop from h1 to h2. DERIVATION: Let As = plan (horizontal cross-sectional) area of the tank (constant for a prismatic tank), Ao = orifice area, Cd = discharge coefficient. At any instant, Q = Cd Ao √(2gh). The continuity equation for the tank (volume leaving = volume decrease): −As dh = Q dt = Cd Ao √(2gh) dt Rearranging: dt = −As dh / (Cd Ao √(2gh)) dt = −(As / (Cd Ao √(2g))) × h^(−1/2) dh Integrate from t = 0 (head = h1) to t = T (head = h2): T = (As / (Cd Ao √(2g))) × ∫_{h1}^{h2} (−h^(−1/2)) dh T = (As / (Cd Ao √(2g))) × [2h^(1/2)]_{h2}^{h1} T = (As / (Cd Ao √(2g))) × 2(√h1 − √h2) FINAL FORMULA: T = 2As(√h1 − √h2) / (Cd Ao √(2g)) To drain COMPLETELY (h2 = 0): T_empty = 2As√h1 / (Cd Ao √(2g)) Note: T_empty = 2 × [time to drain if Q were constant at the initial rate Q1]. This makes physical sense — the average rate during drainage is half the initial rate. NON-PRISMATIC TANK: For a tank where plan area As varies with depth (e.g., conical tank), replace As with As(h) and integrate accordingly. Board exams occasionally present a hemispherical tank where As(h) = π(2Rh − h²) where R = sphere radius. The integral becomes more complex but the approach is identical. SUBMERGED OUTLET (tailwater present): If the tank drains into a space with a constant tailwater level h_tw, then h in the formula is always the differential head. However, if the tank drains to atmosphere (zero tailwater), use the formula as derived. PUMPED EMPTYING: If a pump assists drainage, a pump term appears in the continuity equation. This is an advanced topic sometimes appearing in PRC exams: −As dh = (Q_orifice + Q_pump) dt.
Examples
The numerator is 2×As×(difference of square roots). The denominator is Cd × Ao × √(2g). Compute the denominator as one block — a common error is to compute Cd×Ao first and then forget to multiply by √(2g).
Scenario
A rectangular tank 4 m × 3 m in plan drains through a sharp-edged circular orifice of diameter 120 mm at its bottom. Initial head = 3.6 m, final head = 0.9 m. Given Cd = 0.62, find the time of drainage.
Solution
Given: As = 4 × 3 = 12 m² D_o = 120 mm = 0.12 m Ao = π(0.12)²/4 = π(0.0144)/4 = 0.01131 m² h1 = 3.6 m, h2 = 0.9 m Cd = 0.62, g = 9.81 m/s² Compute √(2g) = √(19.62) = 4.429 m^(1/2)/s Compute (√h1 − √h2): √3.6 = 1.8974 m^(1/2) √0.9 = 0.9487 m^(1/2) Difference = 1.8974 − 0.9487 = 0.9487 m^(1/2) T = 2As(√h1 − √h2) / (Cd Ao √(2g)) = 2(12)(0.9487) / (0.62 × 0.01131 × 4.429) = 22.769 / (0.62 × 0.01131 × 4.429) = 22.769 / 0.031041 = 733.5 s ≈ 733 s ≈ 12.2 minutes
This ratio property — equal times for equal increments of √h — is a favorite board-exam concept question. The bottom portion (small h, low flow rate) takes just as long as the upper portion (large h, high flow rate) because the interval happens to correspond to the same √h increment.
Scenario
How long does it take to completely empty the tank in the preceding example if h2 = 0?
Solution
For complete emptying (h2 = 0, √h2 = 0): T = 2As√h1 / (Cd Ao √(2g)) = 2(12)(√3.6) / (0.62 × 0.01131 × 4.429) = 2(12)(1.8974) / 0.031041 = 45.538 / 0.031041 = 1467 s ≈ 24.4 minutes Observation: The time to drain the lower half (h: 0.9 → 0) is 1467 − 733 = 734 s ≈ 12.2 min, almost equal to the time for the upper portion. This confirms the √h behavior: equal time intervals correspond to equal increments of √h, not equal increments of h.
When multiple orifices discharge simultaneously, their areas add directly in the denominator, effectively increasing the total Cd×Ao product.
Scenario
A cylindrical tank (diameter = 2 m, height = 5 m) is full. It drains through two identical orifices, each 80 mm diameter (Cd = 0.60) at the bottom. Find the time to empty half the tank (top 2.5 m drains).
Solution
As = π(2)²/4 = π m² = 3.1416 m² Two orifices: A_o_total = 2 × π(0.08)²/4 = 2 × 5.027×10⁻³ = 0.010053 m² h1 = 5.0 m (full tank head) h2 = 2.5 m (half-drained, water level at mid-height) √h1 = √5.0 = 2.2361 √h2 = √2.5 = 1.5811 Difference = 2.2361 − 1.5811 = 0.6550 m^(1/2) T = 2(3.1416)(0.6550) / (0.60 × 0.010053 × 4.429) = 4.1137 / (0.60 × 0.010053 × 4.429) = 4.1137 / 0.026720 = 153.9 s ≈ 154 s ≈ 2.57 minutes
Applications
- Designing retention pond drawdown time for DPWH flood control projects
- Estimating fire suppression tank drainage time for BFAD/BFP compliance
- Construction dewatering — time to drain an excavation sump
- Agricultural reservoir management for NIA irrigation scheduling
- Septic tank effluent discharge timing
Misconceptions
- Writing √h2 − √h1 (reversed) giving negative time — always write (√h1 − √h2) with h1 > h2.
- Using tank diameter instead of plan area As — remember As = π D²/4 for a circular tank.
- Using the instantaneous Q at h1 as a constant discharge (ignoring the time variation of head).
- Forgetting to include √(2g) = 4.429 in the denominator — this is the most frequent arithmetic error.
- Confusing orifice area Ao with tank plan area As.
Related Concepts
- Orifice discharge equation
- Continuity equation for unsteady flow
- Integration of variable-head flow
- Tank-filling time (pump + orifice equation)
- Flood routing in detention basins
Common Exam Questions
Example
3m×2m tank, 100mm orifice, Cd=0.61, h1=4m, h2=1m: T = 2(6)(√4−√1)/(0.61×7.854×10⁻³×4.429) = 2(6)(2−1)/0.02122 = 12/0.02122 = 566 s.
Approach
Identify As, Ao, Cd, h1, h2; compute √(2g) = 4.429; substitute into T = 2As(√h1−√h2)/(Cd Ao √(2g)).
Question Type
Direct time computation (constant As)
Example
Tank 5m×5m, h1=3m, h2=0, T = 600 s, Cd=0.62: Ao = 2(25)(√3)/(0.62×600×4.429) = 86.60/(1647.6) = 0.05256 m²; D = √(4×0.05256/π) = 0.2588 m ≈ 259 mm.
Approach
Rearrange for Ao: Ao = 2As(√h1−√h2)/(Cd × T × √(2g)).
Question Type
Find orifice size for required drainage time
Example
Time to drain top half vs. bottom half of full tank (h1=H, h2=H/2 vs h2=H/2, h3=0): T_top/T_bottom = (√H−√(H/2))/(√(H/2)−0) = (1−1/√2)/(1/√2) = (√2−1)/1 = 0.4142 — top half drains in 41.4% of the time of the bottom half.
Approach
Form ratio T2/T1 = (√h1a−√h2a)/(√h1b−√h2b) if all other parameters are the same.
Question Type
Ratio of drainage times
Key Points To Remember
- T = 2As(√h1 − √h2) / (Cd Ao √(2g)) — memorize this completely.
- The numerator has √h1 MINUS √h2 (larger minus smaller = positive time).
- For complete emptying, set h2 = 0: T = 2As√h1 / (Cd Ao √(2g)).
- Ao is the orifice area, not the tank area.
- √(2g) = √(2×9.81) = √19.62 = 4.429 — compute this constant once.
- Units check: As [m²] × √h [m^(1/2)] / (Ao [m²] × √(2g) [m^(1/2)/s]) = seconds ✓
- For the formula to apply, the tank must be prismatic (constant As); otherwise, integrate with As as a function of h.
- Time to drop from h1 to h1/4 equals time to drop from h1/4 to 0 — useful for ratio-type problems.
Practice Problems
Parts (b) and (c) require decomposing Cd into its components Cv and Cc. The product check Q = A_vc × v_actual confirms the calculation. Note that Cv = Cd/Cc only when you know both Cd and Cc; if Cv is given directly, use v = Cv√(2gh).
Problem
PROBLEM 1 (Orifice — Direct): A sharp-edged orifice of diameter 75 mm is installed in the bottom of a large tank. The water level is maintained at a constant height of 5.0 m above the orifice. Given Cd = 0.61, find: (a) the discharge Q in m³/s, (b) the actual jet velocity in m/s, and (c) the area of the jet at the vena contracta if Cc = 0.62.
Solution
(a) Orifice area: A = π(0.075)²/4 = π(0.005625)/4 = 4.418×10⁻³ m² √(2gh) = √(2×9.81×5.0) = √98.1 = 9.905 m/s Q = Cd × A × √(2gh) = 0.61 × 4.418×10⁻³ × 9.905 = 0.61 × 4.374×10⁻² = 2.668×10⁻² m³/s ≈ 26.7 L/s (b) Actual jet velocity: Cv = Cd/Cc = 0.61/0.62 = 0.9839 v_actual = Cv × √(2gh) = 0.9839 × 9.905 = 9.746 m/s Alternatively: v_actual = Q/A_vc (c) Vena contracta area: A_vc = Cc × A = 0.62 × 4.418×10⁻³ = 2.739×10⁻³ m² Verification: Q = A_vc × v_actual = 2.739×10⁻³ × 9.746 = 2.669×10⁻² m³/s ✓
For submerged orifices, the differential head is simply the difference between the two free surface elevations — the centroid location cancels out. This is a useful shortcut. However, if only one side is submerged, you must measure h to the orifice centroid.
Problem
PROBLEM 2 (Submerged Orifice): A sluice gate opening acts as a submerged orifice with area 0.5 m (wide) × 0.4 m (tall) = 0.20 m². The upstream water level is 6.0 m above the channel bottom and the downstream level is 4.5 m above the channel bottom. The sill of the opening is 1.0 m above the channel bottom. Given Cd = 0.61, find the discharge.
Solution
Centroid of orifice: 1.0 + 0.4/2 = 1.2 m above channel bottom. Upstream head above centroid: h1 = 6.0 − 1.2 = 4.8 m Downstream head above centroid: h2 = 4.5 − 1.2 = 3.3 m Differential head: Δh = h1 − h2 = 4.8 − 3.3 = 1.5 m Note: For submerged orifice, the differential head = upstream surface − downstream surface (centroid cancels): Δh = 6.0 − 4.5 = 1.5 m ← simpler approach, same answer. Q = Cd × A × √(2g Δh) = 0.61 × 0.20 × √(2×9.81×1.5) = 0.122 × √29.43 = 0.122 × 5.425 = 0.6618 m³/s ≈ 0.662 m³/s
The upstream depth is always the weir height (P) plus the head (H). This straightforward part (b) is often asked to verify understanding of the weir geometry. The weir crest height P ≥ 2H is the standard recommendation to minimize velocity of approach effects.
Problem
PROBLEM 3 (Rectangular Weir): A suppressed (no end contractions) rectangular weir spans a 3.0 m wide channel. The weir crest is 1.2 m above the channel bottom. The measured head above the crest is H = 0.45 m. Cd = 0.62. Neglect velocity of approach. Find (a) the discharge Q and (b) the depth of flow upstream.
Solution
(a) Suppressed weir (no end contractions), L = 3.0 m: H^(3/2) = (0.45)^1.5 = 0.45 × √0.45 = 0.45 × 0.6708 = 0.30186 m^(3/2) Q = (2/3) Cd √(2g) L H^(3/2) = (2/3)(0.62)(4.429)(3.0)(0.30186) = (0.4133)(4.429)(3.0)(0.30186) = (1.8306)(3.0)(0.30186) = (5.4919)(0.30186) = 1.658 m³/s (b) Depth of flow upstream: The water surface upstream is at the weir crest plus the head H: Depth = weir crest height + H = 1.2 + 0.45 = 1.65 m
The key insight: at small Q, the H^(5/2) relationship of the V-notch gives a larger H than the H^(3/2) relationship of the rectangular weir for the same discharge. A larger H means smaller percentage error in head measurement, hence better accuracy. Board exams often ask for this comparative analysis.
Problem
PROBLEM 4 (V-Notch Weir + Comparison): Two weirs are to be used to measure the same discharge of Q = 0.080 m³/s. Weir A is a 90° V-notch (Cd = 0.58) and Weir B is a rectangular weir with L = 1.5 m (Cd = 0.62, no end contractions). Find the head H required over each weir and state which gives more accurate measurement.
Solution
WEIR A — 90° V-notch: Q = (8/15) Cd √(2g) tan(45°) H^(5/2) 0.080 = (8/15)(0.58)(4.429)(1.0) H^(5/2) 0.080 = (0.5333)(0.58)(4.429) H^(5/2) 0.080 = 1.3693 H^(5/2) H^(5/2) = 0.080/1.3693 = 0.05843 m^(5/2) H = (0.05843)^(2/5) = (0.05843)^0.4 log H = 0.4 × log(0.05843) = 0.4 × (−1.2334) = −0.4934 H = 10^(−0.4934) = 0.3209 m ≈ 0.321 m WEIR B — Rectangular: Q = (2/3) Cd √(2g) L H^(3/2) 0.080 = (2/3)(0.62)(4.429)(1.5) H^(3/2) 0.080 = (0.4133)(4.429)(1.5) H^(3/2) 0.080 = 2.747 H^(3/2) H^(3/2) = 0.080/2.747 = 0.02913 m^(3/2) H = (0.02913)^(2/3) log H = (2/3) × log(0.02913) = (2/3)(−1.5357) = −1.0238 H = 10^(−1.0238) = 0.09465 m ≈ 0.0947 m CONCLUSION: - V-notch head H_A = 0.321 m (larger, easier to measure accurately) - Rectangular head H_B = 0.0947 m (small, harder to read precisely) The V-notch provides a larger, more readable head for small flows — confirming that V-notch weirs are more accurate for small discharges.
For non-prismatic tanks, replace As with As(h) and integrate from scratch — do NOT use the standard prismatic formula T = 2As(√h1−√h2)/(Cd Ao √(2g)). The key step is correctly expressing the plan area as a function of h from the geometry of the container.
Problem
PROBLEM 5 (Tank Emptying): A conical tank with the apex at the bottom has its axis vertical. The tank is 2 m in diameter at the top and 3 m tall (full depth). It drains through a 60-mm diameter orifice at the apex (Cd = 0.60). Find the time to completely empty the full tank. (Hint: At depth h from the apex, the tank radius r = h × (1/3), since r_top/h_total = 1/3.)
Solution
Geometry of conical tank (apex at bottom, widens upward): At elevation h above apex: radius r(h) = (1 m / 3 m) × h = h/3 Plan area: As(h) = π r² = π(h/3)² = πh²/9 Orifice area: Ao = π(0.06)²/4 = π(0.0036)/4 = 2.827×10⁻³ m² Continuity equation: −As(h) dh = Cd Ao √(2gh) dt −(πh²/9) dh = Cd Ao √(2g) √h dt dt = −(πh²/9) / (Cd Ao √(2g) √h) dh dt = −π/(9 Cd Ao √(2g)) × h^(3/2) dh Integrate from h = H = 3 m to h = 0: T = π/(9 Cd Ao √(2g)) × ∫₀³ h^(3/2) dh = π/(9 Cd Ao √(2g)) × [2h^(5/2)/5]₀³ = π/(9 Cd Ao √(2g)) × (2/5)(3)^(5/2) (3)^(5/2) = 3² × 3^(1/2) = 9 × 1.7321 = 15.588 Numerator: (π/9) × (2/5) × 15.588 = (0.34907) × (0.4000) × 15.588 = 0.34907 × 6.2354 = 2.177 m^(5/2) × m² / — (include units) Denominator: Cd Ao √(2g) = 0.60 × 2.827×10⁻³ × 4.429 = 0.60 × 0.012524 = 7.514×10⁻³ m^(5/2)/s (in effective units) T = 2.177 / 7.514×10⁻³ = 2.177 / 7.514×10⁻³ = 289.7 s ≈ 290 s ≈ 4.83 minutes Full computation check: T = [π × (2/5) × H^(5/2)] / [9 × Cd × Ao × √(2g)] = [3.1416 × 0.4 × 15.588] / [9 × 0.60 × 2.827×10⁻³ × 4.429] = [19.584] / [9 × 7.514×10⁻³] = 19.584 / 0.067626 = 289.6 s ✓
This combined problem requires setting Q_weir = Q_orifice and solving for the common steady-state head. Trial-and-error iteration is the standard approach. In an exam, you would be given the basin geometry more explicitly, but the method remains identical. This type of problem assesses deep understanding over rote formula application.
Problem
PROBLEM 6 (Combined — Weir and Orifice in Series): A reservoir discharges over a 90° V-notch weir (Cd = 0.60) into a stilling basin, from which the flow exits through a 150-mm diameter submerged orifice (Cd = 0.62) to a downstream channel. The downstream channel maintains a constant level 0.5 m above the orifice center. Under steady-state conditions, Q_weir = Q_orifice. Find the head H over the weir crest and the discharge Q.
Solution
Let H = head over V-notch crest [m], h_basin = head in stilling basin above orifice center [m] Orifice: h_orifice_net = h_basin − 0.5 (differential head) Q_orifice = 0.62 × π(0.15)²/4 × √(2×9.81×(h_basin − 0.5)) Ao = 0.017671 m² Q_orifice = 0.62 × 0.017671 × 4.429 × √(h_basin−0.5) Q_orifice = 0.04849 √(h_basin−0.5) ... (1) V-notch weir (90°, tan 45°=1): Q_weir = (8/15)(0.60)(4.429)(1.0) H^(5/2) Q_weir = 1.4153 H^(5/2) ... (2) For steady state: Q_weir = Q_orifice → 1.4153 H^(5/2) = 0.04849 √(h_basin−0.5) This requires a second relationship between H and h_basin. If the weir crest is at the orifice centerline (simplifying assumption): h_basin = H + weir_crest_height. Let weir crest be 0.8 m above orifice center: h_basin = H + 0.8. Then: 1.4153 H^(5/2) = 0.04849 √(H + 0.8 − 0.5) = 0.04849 √(H + 0.3) Try H = 0.15 m: LHS = 1.4153 × (0.15)^2.5 = 1.4153 × (0.15)² × √0.15 = 1.4153 × 0.0225 × 0.3873 = 0.01234 RHS = 0.04849 × √(0.45) = 0.04849 × 0.6708 = 0.03252 — LHS < RHS Try H = 0.25 m: LHS = 1.4153 × (0.25)^2.5 = 1.4153 × 0.0625 × 0.5 = 1.4153 × 0.03125 = 0.04423 RHS = 0.04849 × √(0.55) = 0.04849 × 0.7416 = 0.03596 — LHS > RHS Linear interpolation between H=0.15 (LHS-RHS = −0.02018) and H=0.25 (LHS-RHS = +0.00827): H = 0.15 + 0.10 × (0.02018)/(0.02018+0.00827) = 0.15 + 0.10 × 0.709 = 0.221 m Check H = 0.221 m: LHS = 1.4153 × (0.221)^2.5 = 1.4153 × (0.04884)(0.4701) = 1.4153 × 0.02296 = 0.03249 RHS = 0.04849 × √(0.521) = 0.04849 × 0.7218 = 0.03500 (close) Final answer: H ≈ 0.23 m, Q ≈ 1.4153 × (0.23)^2.5 ≈ 1.4153 × 0.02534 ≈ 0.0359 m³/s ≈ 35.9 L/s (Exact value requires iteration or numerical solver — the method is what boards test, not the precise decimal.)
Exam Preparation Tips
- FORMULA CARD PRIORITY: Write these four formulas on your scratch paper immediately at the start of the exam: (1) Q = Cd A √(2gh), (2) Q_rect = (2/3)Cd√(2g)LH^(3/2), (3) Q_tri = (8/15)Cd√(2g)tan(θ/2)H^(5/2), (4) T = 2As(√h1−√h2)/(Cd Ao√(2g)). With these four, you can solve 95% of board problems in this chapter.
- CONSTANT COMPUTATION SHORTCUT: Memorize √(2×9.81) = √19.62 = 4.429 m^(1/2)/s. Also memorize (2/3)(4.429) = 2.953 and (8/15)(4.429) = 2.362. These reduce the rectangular and triangular weir formulas to Q = 2.953 Cd L H^(3/2) and Q = 2.362 Cd tan(θ/2) H^(5/2).
- UNIT DISCIPLINE: Every formula in this chapter is in SI — meters, seconds, m³/s. Never mix cm or mm without converting. Orifice areas must be in m² (convert D to meters first). Heads must be in meters.
- EXPONENT CHECK: The single most common error on board exams is using H^(3/2) for a V-notch or H^(5/2) for a rectangular weir. Always write the weir type and exponent explicitly before computing.
- POWER COMPUTATION ON CALCULATOR: For H^(5/2) = H^2.5, use the y^x or x^y key: enter H, press y^x, enter 2.5, press =. Alternatively, H^2.5 = H^2 × H^0.5 = H² × √H. Practice both methods.
- HEAD REFERENCE DISCIPLINE: Orifice h = to CENTROID of opening; Weir H = above CREST; Tank emptying h = above ORIFICE. These three are distinct and non-interchangeable.
- SUBMERGED ORIFICE SHORTCUT: For a submerged orifice, the effective head is simply the water surface elevation difference (upstream minus downstream). The orifice centroid location cancels out and is irrelevant.
- END CONTRACTIONS CHECK: If the problem says 'full-width weir' or 'suppressed weir,' n = 0 (no correction). If 'two end contractions' or 'weir narrower than channel,' n = 2, L_eff = L − 0.2H.
- TANK EMPTYING — DENOMINATOR CHECKLIST: The denominator Cd × Ao × √(2g) requires three separate values. Write them out explicitly: Cd = ___, Ao = π D²/4 = ___ m², √(2g) = 4.429. Multiply in sequence to avoid errors.
- TIME ALLOWED: A typical orifice or weir problem should take 3–4 minutes; a tank-emptying problem 4–5 minutes. If you exceed 6 minutes, mark it, move on, and return. Never abandon the rest of the exam for one problem.
- REVIEW COMMON Cd VALUES: Sharp-edge orifice 0.61, short tube 0.82, re-entrant tube 0.51, sharp-crested rectangular weir 0.62, 90° V-notch 0.58–0.62. These are the values used in PRC exam problems unless explicitly stated otherwise.
- CHECK ANSWER REASONABLENESS: Q for a 100-mm orifice under 3–5 m head should be in the range 20–50 L/s. Q for a 1-m wide weir under 0.3–0.5 m head should be 0.3–1.0 m³/s. If your answer is far outside these ranges, recheck your computation.
In summary
Orifices, weirs, tubes, and nozzles represent some of the most formula-intensive yet physically intuitive topics in the PRC CE hydraulics examination. The underlying physics is always the same — Bernoulli's equation applied from a reservoir surface to a discharge point — modified by empirical coefficients (Cv, Cc, Cd) that account for real-fluid behavior. Four master formulas govern the entire chapter: Q = Cd A √(2gh) for orifices and tubes; Q_rect = (2/3)Cd√(2g)LH^(3/2) for rectangular weirs; Q_tri = (8/15)Cd√(2g)tan(θ/2)H^(5/2) for V-notch weirs; and T = 2As(√h1−√h2)/(Cd Ao√(2g)) for tank emptying. The most critical board-exam discipline is correct head reference: orifice head h is measured to the opening centroid, weir head H is measured above the crest, and the tank-emptying formula uses the instantaneous water surface elevation above the orifice. End-contraction corrections (Francis formula), the choice between V-notch and rectangular weirs for different flow ranges, and the non-prismatic tank integration are the higher-order skills that distinguish examinees who merely memorize formulas from those who deeply understand hydraulics. As Filipino civil engineers entering practice in a country with extensive irrigation infrastructure (NIA), water supply networks (MWSS, LWDs), and flood-prone terrain requiring accurate hydrology (DPWH-BMO), this knowledge is not merely academic — it is a fundamental professional competency. Study the derivations to understand where the coefficients enter, practice the worked examples until the four-step procedure becomes reflexive, and you will consistently score in this chapter on the licensure examination.
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