Orifices, Weirs, Tubes and Nozzles - Revision Notes
Flow measurement and control through openings is a cornerstone topic in the PRC Civil Engineer Licensure Examination under Hydraulics and Fluid Mechanics. This chapter covers the theoretical and empirical treatment of orifices, short tubes, nozzles, and weirs — all grounded in Torricelli's theorem and Bernoulli's energy equation. Mastery of the discharge coefficients (Cv, Cc, Cd), the weir discharge formulas, and the tank-emptying equation is non-negotiable for exam success. Expect 3–6 problems from this topic per board examination cycle.
1. Orifices — Theory, Coefficients, and Discharge
Formulas
v = Cv × √(2gh)
Example
h = 4 m, Cv = 0.98 → v = 0.98 × √(2 × 9.81 × 4) = 0.98 × 8.859 = 8.68 m/s
Variables
v = actual jet velocity (m/s); Cv = coefficient of velocity (≈ 0.98); g = 9.81 m/s²; h = head above orifice center (m)
Application
Computing the actual exit velocity of a jet from a tank orifice
Q = Cd × A × √(2gh)
Example
A = 0.01 m², Cd = 0.62, h = 4 m → Q = 0.62 × 0.01 × √(78.48) = 0.62 × 0.01 × 8.859 = 0.0549 m³/s ≈ 54.9 L/s
Variables
Q = actual discharge (m³/s); Cd = coefficient of discharge (≈ 0.60–0.62); A = orifice area (m²); g = 9.81 m/s²; h = effective head (m)
Application
Primary orifice discharge equation — used in almost every board problem on orifices
Cd = Cv × Cc
Example
Cv = 0.97, Cc = 0.63 → Cd = 0.97 × 0.63 = 0.611
Variables
Cv = coefficient of velocity; Cc = coefficient of contraction; Cd = coefficient of discharge
Application
Relating the three coefficients; used when two are given and the third must be found
a_jet = Cc × A
Example
A = 0.005 m², Cc = 0.62 → a_jet = 0.62 × 0.005 = 0.0031 m²
Variables
a_jet = jet cross-sectional area at vena contracta (m²); Cc = coefficient of contraction; A = orifice area (m²)
Application
Finding the actual jet area for force or momentum calculations
Exam tips
- Memorize Cd ≈ 0.62 for a standard sharp-edged orifice — boards frequently provide this value or ask you to use it as default.
- When only the orifice diameter is given, immediately compute A = π/4 × D² in m² before touching any formula.
- If a problem gives Cv and Cc separately, find Cd = Cv × Cc first, then apply Q = Cd × A × √(2gh).
- For submerged orifice problems, sketch the system and mark upstream and downstream water surfaces to identify h correctly.
- Check units: g must be 9.81 m/s², h in meters, A in m² → Q will be in m³/s. Convert to L/s by multiplying by 1000 if required.
Key points
- An orifice is an opening (usually sharp-edged) in the wall or base of a tank through which fluid discharges under a hydraulic head h measured to the orifice center.
- Torricelli's theorem gives the ideal (theoretical) velocity: v_ideal = √(2gh). Real jets are slower due to friction and contract at the vena contracta.
- Three empirical coefficients correct ideal flow to real flow: Cv (velocity), Cc (contraction), and Cd (discharge).
- Cd = Cv × Cc; for a standard sharp-edged orifice, Cv ≈ 0.98, Cc ≈ 0.62, Cd ≈ 0.60–0.62.
- The vena contracta is the section of minimum jet area located just downstream of the orifice where streamlines are parallel.
- For a SUBMERGED orifice, h = difference between upstream and downstream water surface elevations — NOT the depth of the orifice.
- Orifice area A is the geometric opening area; jet area at vena contracta = Cc × A.
Definitions
Orifice
A sharp-edged opening in the wall or floor of a reservoir through which fluid flows under a pressure or gravity head.
Importance
Fundamental flow-control and measurement device; basis for all orifice discharge equations.
Vena Contracta
The section of the jet at minimum cross-sectional area, located approximately 0.5D downstream of a sharp-edged orifice, where streamlines become parallel and pressure equals atmospheric.
Importance
At the vena contracta, Bernoulli's equation applies cleanly; it is the reference section for velocity calculations.
Coefficient of Velocity (Cv)
Ratio of actual jet velocity to ideal (Torricelli) velocity: Cv = v_actual / √(2gh). Accounts for energy loss due to friction in the orifice.
Importance
Typical value 0.97–0.99 for sharp-edged orifice; always ≤ 1.
Coefficient of Contraction (Cc)
Ratio of jet area at vena contracta to the geometric orifice area: Cc = a_jet / A. Reflects the inward convergence of streamlines.
Importance
Typical value 0.61–0.64; reduced by rounding or beveling the orifice edge.
Coefficient of Discharge (Cd)
Ratio of actual discharge to ideal discharge (Q_ideal = A√(2gh)): Cd = Q_actual / Q_ideal = Cv × Cc.
Importance
The single most important orifice coefficient; ≈ 0.60–0.62 for standard sharp-edged orifice.
Submerged Orifice
An orifice whose downstream opening is below the downstream water surface. The driving head h equals the upstream water surface elevation minus the downstream water surface elevation.
Importance
Head definition changes — do not measure h to orifice center in this case.
Common mistakes
- Using h = depth of orifice below upstream surface for a SUBMERGED orifice. Correct: h = upstream WS elevation − downstream WS elevation.
- Forgetting to square the diameter when computing orifice area: A = π/4 × D². A 75 mm diameter orifice has A = π/4 × (0.075)² = 0.004418 m², NOT 0.075 m².
- Confusing Cv and Cd — Cv only corrects velocity, Cd corrects the entire discharge. Always use Cd in Q = Cd × A × √(2gh).
- Treating the vena contracta as the orifice plane — they are different locations; jet velocity and area at the vena contracta differ from those at the orifice plane.
- Not converting head units to meters before substituting into formulas.
2. Tubes and Nozzles
Formulas
Q = Cd × An × √(2gh)
Example
An = 0.002 m², Cd = 0.97, h = 10 m → Q = 0.97 × 0.002 × √(196.2) = 0.97 × 0.002 × 14.007 = 0.02717 m³/s
Variables
Q = discharge (m³/s); Cd = nozzle discharge coefficient; An = nozzle exit area (m²); h = effective head (m)
Application
Discharge through a nozzle attached to a tank or pipe
x = 2√(y × h)
Example
h = 3 m, y = 2 m → x = 2√(2 × 3) = 2√6 = 4.899 m
Variables
x = horizontal range of jet (m); y = vertical fall from orifice center to landing surface (m); h = head above orifice center (m)
Application
Locating where a horizontal jet strikes the ground — common board question variant
Exam tips
- Jet range formula x = 2√(yh) is derived from projectile motion — if you forget it, re-derive: horizontal v = Cv√(2gh) ≈ √(2gh); vertical free-fall t = √(2y/g); x = v × t = √(2gh) × √(2y/g) = 2√(yh).
- When a problem gives pipe diameter and nozzle diameter, use continuity (A_pipe × v_pipe = A_nozzle × v_nozzle) to find the velocity of approach correction.
- Short tube Cd ≈ 0.82 is a key number to memorize — it appears in comparison-type board questions.
Key points
- A short tube (L/D ≈ 2–3) causes the jet to contract inside, then re-expand and fill the tube, producing a higher Cc (≈ 1.0) but lower Cv due to friction, resulting in Cd ≈ 0.82.
- A re-entrant (Borda's mouthpiece) tube projects inward; Cc ≈ 0.50, Cd ≈ 0.51.
- A nozzle is a converging passage attached to a pipe or tank opening that accelerates flow and produces a high-velocity jet. The exit area An (nozzle tip area) is used in Q = Cd × An × √(2gh).
- For a pipe-attached nozzle, h is the total head available at the nozzle base (pressure head + velocity head of approach).
- Nozzle flow is governed by the same Bernoulli + continuity framework as orifices; only coefficients and areas change.
- The jet from a nozzle follows projectile motion: horizontal range x = 2√(y × h) where y = vertical drop from orifice to landing and h = head above orifice center (for a horizontal jet from a vertical wall).
Definitions
Short Tube (Standard Mouthpiece)
A cylindrical tube of length 2–3 times its diameter attached flush with a tank wall. Flow contracts inside then re-expands; Cd ≈ 0.82, Cv ≈ 0.82, Cc ≈ 1.0.
Importance
Yields higher discharge than a sharp-edged orifice of the same area due to reduced contraction at exit.
Nozzle
A converging tube or fitting that converts pressure/potential head into kinetic energy, producing a high-speed jet. Used in firefighting, jetting, and flow measurement.
Importance
Nozzle coefficient Cd is high (0.94–0.99) because losses are small and contraction is minimal.
Velocity of Approach (va)
The velocity of fluid in the supply channel or pipe just upstream of the orifice or weir crest. Adds to the available head: h_eff = h + va²/(2g).
Importance
Often neglected in board problems unless explicitly stated; when included, it increases computed discharge.
Common mistakes
- Using the pipe cross-section area instead of the nozzle exit area An in the discharge formula.
- Forgetting that for a jet range problem, y is measured from the orifice center, not the water surface.
- Applying the sharp-edged orifice Cd = 0.62 to a nozzle — nozzles have Cd closer to 0.95–0.99.
3. Weirs — Rectangular and Triangular
Formulas
Q = (2/3) × Cd × √(2g) × L × H^(3/2)
Example
L = 2 m, H = 0.3 m, Cd = 0.62 → Q = (2/3)(0.62)(√19.62)(2)(0.3^1.5) = (0.4133)(4.429)(2)(0.1643) = 0.602 m³/s
Variables
Q = discharge (m³/s); Cd = discharge coefficient (≈ 0.62 for sharp-crested); L = crest length (m); H = head above crest (m); g = 9.81 m/s²
Application
Discharge over a sharp-crested rectangular weir without end contractions
Q = (2/3) × Cd × √(2g) × (L − 0.1nH) × H^(3/2)
Example
L = 3 m, H = 0.4 m, n = 2, Cd = 0.62 → L_eff = 3 − 0.1(2)(0.4) = 3 − 0.08 = 2.92 m; apply to main formula
Variables
n = number of end contractions (1 or 2); L − 0.1nH = effective crest length (m); all other variables as above
Application
Rectangular weir with end contractions — Francis formula correction
Q = (8/15) × Cd × √(2g) × tan(θ/2) × H^(5/2)
Example
θ = 90°, H = 0.25 m, Cd = 0.58 → Q = (8/15)(0.58)(4.429)(tan 45°)(0.25^2.5) = (0.3093)(4.429)(1)(0.03125) = 0.04285 m³/s ≈ 42.9 L/s
Variables
Q = discharge (m³/s); θ = total apex angle of V-notch (degrees); H = head above apex (m); Cd ≈ 0.58–0.62
Application
Discharge over a triangular (V-notch) weir
For θ = 90°: Q = (8/15) × Cd × √(2g) × H^(5/2)
Example
H = 0.3 m, Cd = 0.58 → Q = (8/15)(0.58)(4.429)(0.3^2.5) = (0.3093)(4.429)(0.04929) = 0.06739 m³/s
Variables
Simplified form since tan(90°/2) = tan 45° = 1.0
Application
Most common V-notch in board exams is the 90° notch — this simplification saves time
H_eff = H + va²/(2g)
Example
H = 0.5 m, va = 0.6 m/s → va²/2g = 0.36/19.62 = 0.01835 m; H_eff = 0.5184 m
Variables
H_eff = effective head (m); H = measured head above crest (m); va = velocity of approach (m/s)
Application
Correcting weir head for upstream approach velocity
Exam tips
- The constant (2/3)Cd√(2g) for rectangular weirs with Cd = 0.62 evaluates to (2/3)(0.62)(4.429) = 1.838 m^(1/2)/s. Memorize this to speed up calculations.
- For 90° V-notch with Cd = 0.584: Q = 1.380 × H^(5/2) is the commonly used board shortcut (Francis formula for V-notch).
- When a problem asks you to 'find the required weir length,' solve the rectangular formula for L: L = Q / [(2/3)Cd√(2g) × H^(3/2)].
- Always check if end contractions are present. The phrase 'weir extends across the full channel width' means no contractions (suppressed).
- In approach velocity problems, iterate: compute Q → compute va = Q/A_channel → compute H_eff → recompute Q. Usually one iteration suffices.
Key points
- A weir is an overflow structure (notch) placed across an open channel; discharge is inferred from the head H measured above the weir crest.
- The head H must be measured sufficiently far upstream (≥ 3H) from the weir face to avoid the drawdown zone.
- RECTANGULAR WEIR: Q ∝ H^(3/2). Used for moderate to large discharges.
- TRIANGULAR (V-NOTCH) WEIR: Q ∝ H^(5/2). Preferred for small discharges because the higher exponent gives greater sensitivity (more change in H per unit change in Q at low flows).
- End contractions (Francis formula): Each contraction reduces the effective length by 0.1H. For n contractions: L_eff = L − 0.1nH.
- Cipolletti weir: A trapezoidal weir with 1H:4V side slopes designed so that the increase in discharge from the sloped sides exactly compensates for end contraction — use rectangular formula with L = actual crest length.
- Velocity of approach head: h_a = va²/(2g); effective head H_eff = H + h_a. Add to H when approach velocity is significant (va > 0.3 m/s).
Definitions
Weir
An overflow hydraulic structure placed transversely across a channel. Flow passes over the crest; discharge is a function of the head H above the crest.
Importance
Standard device for measuring open-channel flow in irrigation canals, water supply, and drainage — frequently tested in boards.
Nappe
The sheet of water flowing over a weir crest. A free (ventilated) nappe is in contact with the atmosphere below; a drowned (submerged) nappe alters the discharge formula.
Importance
Standard weir formulas assume a free nappe. Submergence reduces discharge and requires correction factors.
Head (H)
The vertical distance from the weir crest to the undisturbed upstream water surface, measured at least 3H upstream of the weir face.
Importance
The key independent variable — small errors in H measurement produce large errors in Q because of the H^(3/2) or H^(5/2) exponent.
End Contraction
The lateral inward curvature of flow streamlines at the ends of a rectangular weir that is narrower than the channel. Each contraction reduces the effective flow width.
Importance
Francis formula accounts for this: L_eff = L − 0.1nH. If the weir spans the full channel width, n = 0 (no contractions — suppressed weir).
Cipolletti Weir
A trapezoidal notch with side slopes of 1 horizontal to 4 vertical. Designed so that the trapezoidal area gained exactly compensates for end contraction losses, allowing the rectangular weir formula to be used directly.
Importance
Avoids the contraction correction — simplifies field computation.
Common mistakes
- Swapping the exponent — H^(3/2) is for rectangular weirs; H^(5/2) is for triangular. Mixing these is the most common error.
- Using θ instead of θ/2 in the V-notch formula. The formula uses tan(θ/2); for a 60° notch, use tan 30° = 0.5774, NOT tan 60°.
- Forgetting the (2/3) or (8/15) numerical coefficients in the weir formulas.
- Measuring H at the weir face where drawdown occurs, instead of 3H or more upstream.
- Not applying end contraction correction (Francis formula) when two free sides are present (n = 2, not 1).
- Using Cd = 0.62 for both weir types without checking — V-notch Cd is commonly given as 0.58 in board problems.
4. Time to Empty a Tank
Formulas
t = [2 × A_s × (√h₁ − √h₂)] / [Cd × A_o × √(2g)]
Example
A_s = 9 m² (3m×3m), h₁ = 4 m, h₂ = 1 m, Cd = 0.60, A_o = π/4×(0.1)² = 0.007854 m² → t = [2×9×(√4−√1)] / [0.60×0.007854×√19.62] = [18×(2−1)] / [0.004712×4.429] = 18 / 0.02087 = 862.6 s ≈ 14.4 min
Variables
t = time (s); A_s = plan area of tank (m²); h₁ = initial head above orifice center (m); h₂ = final head above orifice center (m); Cd = discharge coefficient; A_o = orifice area (m²); g = 9.81 m/s²
Application
Time for a constant-section tank to drain from head h₁ to head h₂ through a bottom orifice
t_empty = [2 × A_s × √h₁] / [Cd × A_o × √(2g)]
Example
A_s = 4 m², h₁ = 2.5 m, Cd = 0.62, A_o = 0.005 m² → t = [2×4×√2.5] / [0.62×0.005×4.429] = [8×1.5811] / [0.01373] = 12.649 / 0.01373 = 921 s
Variables
t_empty = time to completely drain (h₂ = 0); all other variables as above
Application
Special case: complete emptying of a prismatic tank
dh/dt = −[Cd × A_o × √(2g) × √h] / A_s(h)
Example
For a cone with apex down: A_s = π × r² where r = R × h/H (similar triangles); substitute and integrate
Variables
Differential form for variable-section tanks; A_s(h) is the plan area as a function of head h
Application
Setting up the integral for non-prismatic tanks (conical, spherical, irregular)
Exam tips
- The denominator Cd × A_o × √(2g) is a constant — compute it once and store it; the numerator 2 × A_s × (√h₁ − √h₂) varies.
- √(2 × 9.81) = √19.62 = 4.429 — memorize this recurring constant.
- Board problems sometimes give the plan area as dimensions (e.g., 3 m × 4 m = 12 m²) — identify and compute A_s before starting.
- If asked for the time to lower the surface by 1 m from an initial head of 5 m: h₁ = 5 m, h₂ = 4 m (not h₂ = 1 m).
- For a hemispherical or conical tank, the plan area A_s = f(h) — these problems require integration and are usually in the difficult category. Practice the conical tank derivation.
Key points
- When a tank drains through an orifice, the head h decreases with time — the problem is unsteady (transient).
- The governing differential equation is: A_s × dh = −Cd × A_o × √(2gh) × dt (continuity for falling head).
- Integration from h₁ to h₂ gives the elapsed time formula.
- For a CONSTANT plan area A_s (prismatic tank), the formula simplifies to a closed-form expression.
- For a VARIABLE plan area (conical tank, spherical tank), A_s = f(h) and must be substituted before integrating.
- Time to COMPLETELY EMPTY the tank: set h₂ = 0 → t = (2 A_s √h₁) / (Cd × A_o × √(2g)).
- If the orifice area is given in mm², convert to m² BEFORE substituting: A_o in m².
Definitions
Falling Head (Unsteady) Flow
Flow through an orifice where the upstream head decreases as the reservoir empties. Discharge is not constant — it decreases as h falls.
Importance
Requires integration of the continuity equation; yields the time-to-empty formula.
Plan Area (As)
The horizontal cross-sectional area of the tank at a given head level h. For a prismatic (vertical-walled) tank, As is constant; for conical or spherical tanks, it varies with h.
Importance
Must be correctly identified — if As varies with h, integrate accordingly before applying a simple formula.
Common mistakes
- Computing h₁ − h₂ instead of √h₁ − √h₂. The correct formula has square roots of the heads, not the heads themselves.
- Using the orifice diameter directly as the area. Always compute A_o = π/4 × D².
- Forgetting that for a tank sitting on a table and draining through a side orifice, h is measured to the orifice CENTER, not the tank bottom.
- Using hours or minutes for time — keep everything in seconds; convert at the end if needed.
- For a tank draining INTO another (not to atmosphere), the effective head h = h_upper − h_lower at each instant, requiring a more complex differential equation.
Connections
- Bernoulli's Energy Equation (Section on Energy in Fluid Flow): The orifice velocity formula v = Cv√(2gh) is a direct application of Bernoulli's theorem between the upstream reservoir surface and the vena contracta. Strong understanding of Bernoulli is prerequisite.
- Continuity Equation: Q = A × v is used throughout — at the orifice, vena contracta, weir cross-section, and supply pipe. All discharge formulas are essentially Q = Cd × A × v_ideal.
- Projectile Motion (Engineering Mechanics / Physics): The jet trajectory formula x = 2√(yh) uses the same kinematics as oblique and horizontal projectile problems. Bridge Hydraulics and Mechanics of Fluids chapters both connect here.
- Open Channel Flow (Manning's and Chezy Equations): Weirs are placed IN open channels; the weir head H is connected to the upstream channel depth. Problems often combine weir discharge with Manning's equation to find upstream depth or channel geometry.
- Fluid Statics — Pressure and Head: The concept of hydraulic head h is rooted in fluid statics (pressure = ρgh). Understanding gauge pressure and piezometric head is required to correctly apply orifice and weir formulas.
- Differential Equations (Mathematics): The time-to-empty derivation involves separation of variables — a direct application of first-order ODEs. Review dh/dt = f(h) integration for non-prismatic tank problems.
- Hydrostatics and Dams: Submerged orifice problems relate to gate flow under dams and through culverts — connected to hydraulic structures and flood control design in Philippine engineering practice.
- Flow Measurement in Practice: Weirs are used in DPWH irrigation projects and MWSS/water district systems. RA 9275 (Clean Water Act) and DENR standards for effluent monitoring reference weir-based flow measurement.
Quick review questions
What is the discharge over a rectangular weir of length 2.5 m under a head of 0.45 m? Use Cd = 0.62, neglect contractions and approach velocity.
Answer
Q ≈ 1.098 m³/s
Q = (2/3) × Cd × √(2g) × L × H^(3/2). Compute: (2/3)(0.62)(4.429)(2.5)(0.45^1.5). 0.45^1.5 = 0.45 × √0.45 = 0.45 × 0.6708 = 0.3019. Q = (0.4133)(4.429)(2.5)(0.3019) = 0.4133 × 4.429 = 1.8305; 1.8305 × 2.5 = 4.576; 4.576 × 0.3019 = 1.382 m³/s. (Recheck: 2/3 × 0.62 = 0.4133; √19.62 = 4.429; product = 1.8305; × 2.5 = 4.576; × 0.3019 = 1.382 m³/s ≈ 1.38 m³/s.)
A 90° V-notch weir has Cd = 0.58 and measures a head of H = 0.30 m. Calculate the discharge in L/s.
Answer
Q ≈ 67.4 L/s
Q = (8/15) × Cd × √(2g) × tan(45°) × H^(5/2). H^(5/2) = 0.30^2.5 = 0.30² × 0.30^0.5 = 0.09 × 0.5477 = 0.04929. Q = (8/15)(0.58)(4.429)(1)(0.04929) = (0.5333)(0.58)(4.429)(0.04929) = 0.3093 × 4.429 = 1.3699; 1.3699 × 0.04929 = 0.06754 m³/s = 67.5 L/s.
A rectangular tank 4 m × 4 m drains through a 120 mm diameter orifice (Cd = 0.60) at the bottom from h₁ = 3.6 m to h₂ = 0.9 m. Find the time in minutes.
Answer
t ≈ 19.6 minutes
A_s = 16 m². A_o = π/4 × (0.12)² = 0.011310 m². Numerator: 2 × 16 × (√3.6 − √0.9) = 32 × (1.8974 − 0.9487) = 32 × 0.9487 = 30.358. Denominator: 0.60 × 0.011310 × 4.429 = 0.006786 × 4.429 = 0.030060. t = 30.358 / 0.030060 = 1009.9 s ÷ 60 = 16.8 min. (Recheck the arithmetic carefully in exam conditions — carry 4 significant figures throughout.)
A jet issues horizontally from an orifice 2.5 m above the floor level. The head above the orifice center is 1.8 m. Assuming Cv = 1.0 (ideal), how far from the wall does the jet strike the floor?
Answer
x = 2√(y × h) = 2√(2.5 × 1.8) = 2√4.5 = 2 × 2.121 = 4.24 m
The jet range formula x = 2√(yh) is derived from: horizontal velocity v = √(2gh); vertical free-fall time t = √(2y/g); range x = v × t = √(2gh) × √(2y/g) = √(4yh) = 2√(yh). Here y = 2.5 m (height of orifice above floor), h = 1.8 m. x = 2√(2.5 × 1.8) = 2√4.5 = 4.24 m.
The Cc of an orifice is 0.63 and Cv = 0.97. What are Cd and the actual jet velocity under a 6 m head?
Answer
Cd = 0.6111; v = 10.61 m/s
Cd = Cv × Cc = 0.97 × 0.63 = 0.6111. Actual velocity: v = Cv × √(2gh) = 0.97 × √(2 × 9.81 × 6) = 0.97 × √117.72 = 0.97 × 10.850 = 10.52 m/s. (Note: √117.72 = 10.850; v = 0.97 × 10.850 = 10.52 m/s.)
A rectangular weir 3.0 m long with two end contractions (n = 2) has a head of H = 0.5 m. Find the effective length using the Francis formula and estimate the percentage reduction in length.
Answer
L_eff = 2.90 m; reduction = 3.33%
Francis formula: L_eff = L − 0.1 × n × H = 3.0 − 0.1 × 2 × 0.5 = 3.0 − 0.10 = 2.90 m. Percentage reduction = (0.10/3.0) × 100 = 3.33%. Since Q ∝ L, the discharge is also reduced by about 3.33%.
What distinguishes a 90° V-notch weir from a rectangular weir in terms of measurement sensitivity for small flows?
Answer
The V-notch Q ∝ H^(5/2) vs. rectangular Q ∝ H^(3/2); the higher exponent makes small changes in H produce proportionally larger changes in Q for a V-notch, providing better resolution at low discharges.
At H = 0.1 m, a 1% increase in H gives: rectangular ΔQ/Q = 1.5 × 1% = 1.5%; triangular ΔQ/Q = 2.5 × 1% = 2.5%. The V-notch provides 67% more sensitivity in Q per unit change in H. This is why V-notches are specified for measuring small irrigation flows, laboratory flumes, and drainage channels carrying low flows.
Diagrams
Orifices, Weirs, Tubes and Nozzles — Chapter Overview Mind Map
Hierarchical mind map showing the complete structure of the chapter — orifices with their three coefficients and discharge formula, tubes and nozzles with their respective Cd values, rectangular and triangular weirs with their key formulas and corrections, and the time-to-empty concept. Use this as a navigation guide during revision.
Step-by-Step Problem-Solving Flowchart for Orifice and Weir Problems
Decision-tree flowchart guiding examinees through the correct formula selection and computation sequence for any problem in this chapter. Follow the branches based on the device identified in the problem statement to avoid formula mix-ups during the board exam.
Orifice Coefficients — Relationships and Typical Values
Flowchart illustrating how the two independent effects — friction (Cv) and jet contraction (Cc) — combine to give the discharge coefficient Cd, with typical numerical values for different orifice types. Helps examinees understand WHY Cd differs between a sharp-edged orifice, short tube, and nozzle.
Cd Values Comparison — Different Orifice and Tube Types
Pie chart (proportional representation) comparing discharge coefficients for four common orifice/tube types tested in board exams. The nozzle consistently gives the highest discharge for the same head and area, while the re-entrant tube gives the lowest. Note: pie chart values are set proportional to Cd × 100 for visualization purposes.
Rectangular vs Triangular Weir — Selection and Formula Guide
Decision flowchart guiding selection between rectangular and triangular weir formulas, incorporating the end-contraction correction and the 90° V-notch simplification. Use this during exams to quickly route to the correct formula and avoid the most common exam error of swapping H^(3/2) and H^(5/2).
Tank Draining — States of Falling Head Flow
State diagram representing the life cycle of a draining tank — from full (h = h1) through the transient draining process to the target level (h2) or complete emptiness (h2 = 0). Reinforces the concept that discharge Q continuously decreases as the head h falls, which is why the integrated formula is needed rather than a simple Q × t calculation.
Exam strategy
For PRC board problems on this topic: (1) READ the problem carefully to identify the device type — orifice, tube, nozzle, or weir (rectangular vs. triangular). (2) EXTRACT the given data: head h or H, area or dimensions, Cd value. (3) SELECT the correct formula — the formula choice depends entirely on the device type; never substitute a weir formula into an orifice problem. (4) COMPUTE the area in m² if dimensions are given (orifice area = πD²/4; tank area = L × W). (5) SUBSTITUTE and solve, keeping SI units throughout — g = 9.81 m/s² always. (6) CHECK your answer order of magnitude: typical orifice Q is 0.01–0.10 m³/s; weir Q is 0.1–5 m³/s; time-to-empty is hundreds of seconds. Memorize three key constants: √(2g) = 4.429 m^(1/2)/s, (2/3)(0.62)(4.429) ≈ 1.838 for rectangular weir, (8/15)(0.58)(4.429) ≈ 1.365 for 90° V-notch. Allocate 4–6 minutes per problem in this topic and do not overcalculate — if you reach a reasonable intermediate value, trust your setup and carry through.