Memory AnchorsCELE · Hydraulics & Fluid MechanicsReal content

CELE Hydraulics & Fluid Mechanics — Orifices, Weirs, Tubes and NozzlesMemory Anchors

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

Memory anchors for Orifices, Weirs, Tubes and Nozzles — mnemonic devices, acronyms, and tricks that make the CELE Hydraulics & Fluid Mechanics syllabus stick. Use these when a concept just will not stay in your head.

Exam context

For the Civil Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Civil Engineering tests Hydraulics & Fluid Mechanics under a "Core" label, with Orifices, Weirs, Tubes and Nozzles in the 8th slot across 10 chapters. CELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Hydraulics & Fluid Mechanics questions. Date to watch: May and November 2026.

Orifices, Weirs, Tubes and Nozzles - Memory Anchors

Memory techniques can increase long-term recall by up to 400% compared to passive re-reading. The human brain is wired to remember stories, vivid images, emotions, and absurd situations — not abstract equations. By anchoring the dry formulas of orifice flow, weir discharge, and tank emptying to memorable mental hooks — a jeepney driver, a rice terraces analogy, or a dramatic barangay fiesta — you create neural pathways that survive exam-day stress. This chapter's anchors use mnemonics for formula structures, analogies drawn from Filipino daily life, micro-stories that encode sequences, and visual association maps. When you see 'Q = Cd·A·√(2gh)' on a board exam, you will automatically hear the 'CD bargain story.' When you see a V-notch weir, you will picture the Victory sign. These are not tricks — they are cognitive shortcuts built on how memory actually works. Use these anchors actively: recite them aloud, sketch them, and test yourself with the Revision Game at the end.

Anchors

Topic

Orifice Discharge Coefficients

Concept

The three discharge coefficients: Cv (velocity), Cc (contraction), Cd (discharge) and their relationship Cd = Cv × Cc

Anchor Id

Difficulty

easy

Memory Aid

Remember: 'V comes before C, and D is their BABY.' Cv is the parent of velocity (V = velocity), Cc is the parent of contraction (C = constriction), and Cd is their baby — the product of the two parents. Cv × Cc = Cd. Think of it as a family tree: Mom (Cv ≈ 0.98) × Dad (Cc ≈ 0.62) = Baby Cd ≈ 0.61. Baby Cd is always smaller than either parent alone.

Anchor Type

acronym

Why It Works

Personifying abstract coefficients as a family leverages our social brain's strength in remembering relationships. The 'parent-child' structure encodes the multiplication relationship and approximate values simultaneously.

Example Usage

Board exam asks: 'What is Cd if Cv = 0.97 and Cc = 0.64?' Recall VCD family: Cd = 0.97 × 0.64 = 0.621.

Recall Trigger

Think 'VCD family' — same as a VCD disc at a tiangge (market stall).

Topic

Orifice Discharge

Concept

Orifice discharge formula: Q = Cd · A · √(2gh)

Anchor Id

Difficulty

easy

Memory Aid

Picture a sari-sari store owner named 'Kuya Cedric Angas' (Cd·A) who always yells '√2g·h!' when opening a barrel of buko juice. He ALWAYS multiplies his 'discount factor Cd' by the 'hole area A' and then applies 'the energy speed √(2gh).' Kuya Cedric's barrel is the orifice, his discount is Cd, the hole is A, and the splash speed is √(2gh). You can hear him shouting: 'Cd! A! Root-two-gee-aitch!'

Anchor Type

micro_story

Why It Works

Micro-stories with named characters create episodic memories, which are more durable than semantic memory. The phonetic similarity of 'Cd·A' to 'Cedric Angas' activates a vivid mental movie.

Example Usage

When given orifice area, head, and Cd, immediately recall Cedric's barrel scene: Q = Cd × A × √(2 × 9.81 × h).

Recall Trigger

Think of Kuya Cedric opening a buko juice barrel at the sari-sari store.

Topic

Torricelli's Theorem

Concept

Torricelli's theorem: ideal exit velocity v = √(2gh)

Anchor Id

Difficulty

easy

Memory Aid

Imagine dropping a bato (stone) from height h. It hits the ground at v = √(2gh). Now imagine the water itself is a stone falling that same height h — it exits the orifice at exactly the same speed as the falling stone. Torricelli's genius was recognizing that water through a hole 'falls' the same way a stone does. Water = falling stone. Height = head above orifice.

Anchor Type

analogy

Why It Works

Mapping an unfamiliar concept (water velocity through orifice) onto a familiar one (free-fall of an object) exploits existing neural pathways. Students have intuitive feel for falling objects from Physics 1.

Example Usage

For h = 5 m: v_ideal = √(2 × 9.81 × 5) = √98.1 = 9.9 m/s — same as a stone falling 5 m.

Recall Trigger

A stone thrown off a cliff of height h hits ground at √(2gh). Your orifice water does the same.

Topic

Vena Contracta

Concept

Vena contracta — the jet contracts to minimum area after exiting the orifice

Anchor Id

Difficulty

medium

Memory Aid

Visualize a dramatic TV telenovela moment: after the heroine escapes through a narrow door (the orifice), she SQUEEZES even further in the hallway outside — her waist pinches to minimum size (vena = vein, contracta = contracted). The tightest point of the telenovela actress's silhouette in that hallway IS the vena contracta. The coefficient of contraction Cc = Avc / Aorifice ≈ 0.62 is the ratio of her 'telenovela waist' to the door width.

Anchor Type

visual_association

Why It Works

Vivid visual imagery with an emotional/dramatic scene (telenovela) creates a memorable mental picture. Associating the abstract 'vena contracta' with a physical squeeze makes it concrete.

Example Usage

If board exam asks about where maximum velocity occurs in an orifice jet — recall: maximum velocity (and minimum area) is at the vena contracta, downstream of the opening.

Recall Trigger

The telenovela actress squeezing through the hallway after the door.

Topic

Rectangular Weir

Concept

Rectangular weir formula: Q = (2/3) · Cd · √(2g) · L · H^(3/2)

Anchor Id

Difficulty

medium

Memory Aid

Phrase: 'Two-thirds of a CD playing Long Hits to the power 3.' Break it down: Two-thirds = 2/3, CD = Cd, √(2g) = playing (the 'speed constant'), Long = L (weir length), Hits = H (head), power 3 = H^(3/2). Say it rhythmically: 'Two-thirds CD-playing Long-Hits-cubed-half.' Also remember the exponent 3/2 by the number of 'rails' on a rectangular weir sketch — a rectangle has 3 top-and-side visible lines when you draw it flat on paper.

Anchor Type

mnemonic

Why It Works

Rhythm and rhyme exploit the phonological loop in working memory. Breaking the formula into a short sentence with meaningful word-substitutions reduces cognitive load during recall.

Example Usage

Q_rect = (2/3)(0.62)(√19.62)(L)(H^1.5). The 2/3 and H^(3/2) are the signature identifiers of a rectangular weir.

Recall Trigger

Say 'Two-thirds CD playing Long Hits 3/2' whenever you see a rectangular weir problem.

Topic

Triangular V-Notch Weir

Concept

Triangular (V-notch) weir formula: Q = (8/15) · Cd · √(2g) · tan(θ/2) · H^(5/2)

Anchor Id

Difficulty

medium

Memory Aid

The V-notch has FIVE sides if you trace it and its water triangle: two notch sides, two water sides, one water surface — and so Q is proportional to H^(5/2). Mnemonic phrase: 'Eight-fifteenths CD-playing TAN-half-theta HITS-five.' The Victory sign (V-notch = V for Victory) has 2 fingers — halve the angle: tan(θ/2). And 8/15 = 'APAT divided by LABING-LIMA' — remember 8 and 15 by: an octopus (8 legs) delivering 15-peso fish balls from a triangular cart (V-notch shape).

Anchor Type

mnemonic

Why It Works

Multiple overlapping cues (V for Victory, 5 sides, octopus story) create redundant memory traces for the same formula, making retrieval more robust under exam pressure.

Example Usage

For 90° V-notch: tan(45°) = 1, so Q = (8/15)(Cd)(√19.62)(1)(H^2.5). The tan(θ/2) simplifies beautifully for 90°.

Recall Trigger

Flash the V/Victory sign — V-notch, H^(5/2), tan(θ/2), 8/15.

Topic

Weir Exponent Comparison

Concept

Rectangular weir exponent H^(3/2) vs. V-notch exponent H^(5/2)

Anchor Id

Difficulty

easy

Memory Aid

Rhyme it: 'A rectangle stands on THREE legs (3/2), but a triangle points to FIVE stars (5/2).' Picture a rectangular table with 3 visible legs — that's your H^(3/2). Now picture a triangular pyramid pointing up to 5 stars in the sky — that's your H^(5/2). Every time you sketch a weir on scratch paper, draw three legs under a rectangle and five stars above a triangle.

Anchor Type

rhyme

Why It Works

Rhymes activate the phonological loop; visual sketches activate the visuospatial sketchpad. Using both encoding pathways simultaneously strengthens memory consolidation.

Example Usage

If the problem shows a rectangular notch — write H^(3/2). If it shows a V-notch — write H^(5/2). Never mix them up again.

Recall Trigger

3-legged rectangle (H^3/2), 5-star triangle (H^5/2).

Topic

Weir Selection

Concept

V-notch weir is more accurate for small flows

Anchor Id

Difficulty

easy

Memory Aid

Think of measuring rice in a tabo (dipper) vs. a pail. For a tiny amount of rice, you use the small tabo (V-notch) — tiny changes in rice volume cause a big change in the height. For large amounts, use the pail (rectangular weir). The V-notch's H^(5/2) sensitivity means a small change in H causes a BIG change in Q — perfect for detecting small flows. 'For small laundry, use a tabo.'

Anchor Type

analogy

Why It Works

Everyday Filipino objects (tabo, pail) make the abstract sensitivity argument concrete and culturally resonant. The measurement analogy directly maps to the mathematical sensitivity of the H^(5/2) exponent.

Example Usage

Board exam states: 'A small stream must be accurately gauged.' Answer: Use a triangular (V-notch) weir — it is more sensitive to small flows due to the H^(5/2) relationship.

Recall Trigger

Small flow = tabo = V-notch. Large flow = pail = rectangular weir.

Topic

Time to Empty Tank

Concept

Time to empty a tank: t = 2·As·(√h1 − √h2) / (Cd·Ao·√(2g))

Anchor Id

Difficulty

hard

Memory Aid

Story: Engineer Aisa (As = tank plan area) waits as a tank drains from head h1 down to h2. She measures time t. She stands at the top and counts: 'Two times my area (2·As) multiplied by how much the square-root of the water dropped (√h1 − √h2), divided by the orifice team downstairs (Cd·Ao·√2g).' Remember: the NUMERATOR has the TANK (big, on top — 2·As and the √h difference), the DENOMINATOR has the ORIFICE (small, at the bottom — Cd·Ao·√2g).

Anchor Type

micro_story

Why It Works

A named character (Engineer Aisa) embodying the variable 'As' creates a personal story. The physical position of numerator/denominator maps to the physical position of tank (top) and orifice (bottom).

Example Usage

t = [2 × As × (√h1 − √h2)] / [Cd × Ao × √(2×9.81)]. Plug in As = tank plan area, Ao = orifice area, h1 and h2 in meters.

Recall Trigger

Engineer Aisa on top (numerator), orifice team at bottom (denominator).

Topic

Submerged Orifice

Concept

For a submerged orifice, h = difference between upstream and downstream water surfaces

Anchor Id

A10

Difficulty

medium

Memory Aid

Imagine two rice paddies (palayan) at different elevations connected by a pipe with a hole. The water doesn't care about how deep the hole is — it only cares about the DIFFERENCE in water levels between the two paddies. That elevation difference IS your effective head h. Submerged = two paddies. Free jet = one paddy with a hole and air on the other side.

Anchor Type

analogy

Why It Works

The rice paddy irrigation analogy is directly relevant to Philippine agriculture and makes the 'submerged vs. free' distinction memorable through a familiar context.

Example Usage

If upstream water is at elevation 10 m and downstream at 6 m, h = 10 − 6 = 4 m for submerged orifice discharge calculation.

Recall Trigger

Two rice paddies at different levels — h = difference in surface elevations.

Topic

Rectangular Weir End Contractions

Concept

End contractions reduce the effective weir length: L' = L − 0.1nH (Francis formula)

Anchor Id

A11

Difficulty

medium

Memory Aid

FRANCIS says: 'Friends, Remove A Notch Cutting In Side.' Each end contraction REMOVES 0.1H from the effective length. Francis formula: L' = L − 0.1nH where n = number of end contractions (0, 1, or 2). Remember 0.1 by: 'One in ten — each contraction steals one-tenth of H from the length.' Or think: Francis Fontaine (the engineer) is 10% shy at each end.

Anchor Type

mnemonic

Why It Works

Personifying Francis as a named engineer creates an episodic hook. The '10% shy' metaphor makes the 0.1 coefficient intuitive and memorable.

Example Usage

A weir L = 3 m with two end contractions (n = 2), H = 0.4 m: L' = 3 − 0.1(2)(0.4) = 3 − 0.08 = 2.92 m. Use L' in the rectangular weir formula.

Recall Trigger

Francis is 10% shy at each end — L' = L − 0.1nH.

Topic

Standard Coefficient Values

Concept

Standard Cd values: orifice Cd ≈ 0.61, Cv ≈ 0.98, Cc ≈ 0.62; weir Cd ≈ 0.62

Anchor Id

A12

Difficulty

easy

Memory Aid

Chunk: '0.98 — 0.62 — 0.61 — 0.62.' Think of it as a phone number: (098) 062-0162. Cv = 0.98 (nearly 1, almost no friction loss). Cc = 0.62 (the jet squeezes to 62% of the hole area). Cd = 0.61 (the combined product, slightly less than Cc). Weir Cd = 0.62 (same as Cc, easy to remember). Store the 'phone number' in your mental contacts as 'Hydraulics Hotline.'

Anchor Type

chunking

Why It Works

Phone number chunking exploits the brain's natural chunking ability for digit sequences. The 'Hydraulics Hotline' label adds a humorous hook that makes the chunk stand out.

Example Usage

When Cd is not given, use Cd ≈ 0.61 for sharp-edged orifice, Cd ≈ 0.62 for weirs. If Cv and Cc are given instead, compute Cd = Cv × Cc.

Recall Trigger

Dial the Hydraulics Hotline: 098-062-0162.

Topic

Head Measurement for Orifice

Concept

Head h in orifice formula is measured to the CENTER of the orifice, not to the top or bottom

Anchor Id

A13

Difficulty

medium

Memory Aid

Draw a circle (the orifice) on a wall. Now draw a horizontal water surface above it. The dimension h goes from the water surface straight down to the BULLSEYE (center) of the circle. It's like a dartboard game — you always aim for the CENTER. If you measure to the top or bottom of the orifice, you lose points. In board exams, always locate the centroid of the orifice and measure head to that point.

Anchor Type

visual_association

Why It Works

The dartboard/bullseye image is a vivid spatial anchor. It corrects the common exam mistake of measuring to the wrong point by associating the correct behavior with a familiar game.

Example Usage

A circular orifice of diameter 0.1 m has its top at 3.05 m below the surface. Center is at 3.05 + 0.05 = 3.10 m below surface. Use h = 3.10 m in Q = Cd·A·√(2gh).

Recall Trigger

Orifice = dartboard. Head = distance to the bullseye (center).

Topic

Head Measurement for Weir

Concept

Head H for a weir is measured ABOVE the weir crest (not the total water depth)

Anchor Id

A14

Difficulty

medium

Memory Aid

A weir crest is like the TOP RAIL of a fence at Luneta Park. People can stroll up to the fence freely (that's the water up to the crest). But only those ABOVE the fence rail (head H) actually OVERFLOW. The park security only counts the people climbing OVER (above crest). H = height of water ABOVE the fence top (weir crest). Total depth includes the crest itself — don't confuse total depth with H.

Anchor Type

analogy

Why It Works

The Luneta Park fence analogy uses a familiar Manila landmark. The people-over-the-fence metaphor makes the distinction between total depth and overflow head (H) vivid and intuitive.

Example Usage

Total water depth = 1.5 m, weir crest height = 1.0 m. Then H = 1.5 − 1.0 = 0.5 m. Use H = 0.5 m in the weir formula, NOT 1.5 m.

Recall Trigger

H = water height above the fence rail (weir crest), not the total depth.

Topic

Numerical Constants

Concept

The √(2g) constant ≈ 4.429 m^(1/2)/s — a frequently needed numerical constant

Anchor Id

A15

Difficulty

easy

Memory Aid

√(2 × 9.81) = √19.62 = 4.429. Remember 4.429 as: 'FORTY-FOUR point TWENTY-NINE' — or chunk it as '4-4-2-9.' Think of a basketball jersey number 44 with a dash and 29 — like a PBA player 'number 44-29.' Or: 4.43 rounded (less than 4.5, greater than 4.4). In exam computations, pre-compute: (2/3)×4.429 = 2.953 for rectangular weirs, and (8/15)×4.429 = 2.362 for V-notch weirs — write these on your scratch paper immediately.

Anchor Type

chunking

Why It Works

Pre-computing constants and chunking numerical values reduces working memory load during multi-step exam problems, freeing cognitive resources for the actual setup.

Example Usage

Rectangular weir: Q = (2/3)(0.62)(4.429)(L)(H^1.5) = (0.4133)(4.429)(L)(H^1.5) = 1.831·L·H^1.5 for Cd = 0.62.

Recall Trigger

PBA jersey 44-29. √(2g) ≈ 4.429.

Topic

Nozzles

Concept

Nozzle converts pressure head to kinetic energy — high exit velocity, small area

Anchor Id

A16

Difficulty

easy

Memory Aid

A nozzle is like the thumb-over-a-garden-hose trick every Filipino kid has done. Cover most of the hose opening with your thumb (reduce area), and the water shoots farther — faster jet, smaller area. The pressure (head) you had in the hose is traded for speed. Energy is conserved: high pressure + low velocity BECOMES low pressure + high velocity. Your thumb IS the nozzle.

Anchor Type

analogy

Why It Works

Near-universal Filipino childhood experience (playing with garden hoses) creates an immediate, embodied memory. The physical sensation of covering the hose encodes the concept viscerally.

Example Usage

Nozzle discharge: Q = Cd·A_exit·√(2gh), where h is the total head driving the jet and A_exit is the nozzle exit area (smaller than the pipe area).

Recall Trigger

Thumb over garden hose = nozzle. Pressure traded for speed.

Topic

Tank Emptying Assumptions

Concept

Time to empty formula applies only when plan area As is constant (prismatic tank)

Anchor Id

A17

Difficulty

medium

Memory Aid

Story: Engr. Torres uses the emptying formula for a perfect rectangular GI water tank (constant As). But his neighbor has a tapering clay pot (palayok) — the plan area changes with depth! Engr. Torres warns: 'My formula only works for PRISMATIC tanks — same cross-section all the way up.' For non-prismatic tanks, you must integrate. Remember: t = 2As(√h1−√h2)/(Cd·Ao·√2g) is the PRISONER formula — it only works inside the 'prison' of a constant-section tank.

Anchor Type

micro_story

Why It Works

Contrast between a GI tank and a palayok (clay pot) encodes the limitation through a memorable comparison. The 'PRISONER in a prismatic prison' wordplay adds a second hook.

Example Usage

Always check: is the tank prismatic (cylindrical, rectangular box)? If yes, apply t = 2As(√h1−√h2)/(Cd·Ao·√2g) directly. If not, set up the integral dV/dt = −Q.

Recall Trigger

Rectangular GI tank = constant As = use the formula. Tapered palayok = must integrate.

Topic

Coefficient Relationships

Concept

Coefficient of discharge Cd < Cv and Cd < Cc (Cd is always the smallest of the three)

Anchor Id

A18

Difficulty

easy

Memory Aid

Rhyme: 'Velocity is near to one, Contraction cuts it down. Discharge is the smallest one, the humblest in the crown.' Cv ≈ 0.98 (almost 1, barely any loss). Cc ≈ 0.62 (significant squeeze). Cd = Cv × Cc ≈ 0.61 (the product of two numbers less than 1 is always smaller than either). Think: every multiplication makes things SMALLER when both factors < 1.

Anchor Type

rhyme

Why It Works

The rhyme embeds the ordering relationship in a memorable auditory pattern. The 'humblest in the crown' metaphor creates an emotional/hierarchical anchor for Cd's position.

Example Usage

If a problem gives Cv = 0.95 and asks which is largest, answer: Cv > Cc > Cd. If they ask for Cd, compute Cd = Cv × Cc, which will be less than either.

Recall Trigger

Cd is the humblest — always less than both Cv and Cc.

Topic

Unit Consistency

Concept

Units check: Q in m³/s, h in meters, A in m², g = 9.81 m/s²

Anchor Id

A19

Difficulty

easy

Memory Aid

Walk through a building in your mind. At the DOOR (entry) you check units: Q [m³/s] — volume per time — cubic meter per second. At the WINDOW you see Area A [m²]. On the FLOOR you see Head h [m] — it's a floor-to-surface depth. At the CEILING you see g = 9.81 m/s² pulling everything DOWN. Every time you set up a hydraulics problem, take a mental tour: Door(Q) → Window(A) → Floor(h) → Ceiling(g).

Anchor Type

method_of_loci

Why It Works

The Method of Loci (memory palace) is one of the most powerful memory techniques, used by memory champions. Spatial anchoring of units to familiar locations makes them easy to reconstruct during exams.

Example Usage

Before computing: check that A is in m² (not cm²), h in m (not cm), g = 9.81 m/s². Convert all units FIRST, then compute Q = Cd·A·√(2gh) to get m³/s.

Recall Trigger

Mental building tour: Door=Q, Window=A, Floor=h, Ceiling=g.

Topic

Tubes — Types and Coefficients

Concept

Re-entrant (Borda's) tube vs. standard short tube — different Cd values

Anchor Id

A20

Difficulty

hard

Memory Aid

Visualize a standard short tube as a simple straw cut flush with a wall — Cd ≈ 0.82 (higher because the tube re-attaches flow and reduces contraction). A re-entrant tube (Borda's) sticks INWARD like a finger poking into the tank — lower Cd ≈ 0.51 because the re-entrant geometry causes more contraction. Memory image: the INWARD-pointing finger (re-entrant) gives a WORSE (smaller Cd) discharge, like pointing inward blocks the flow. Flush straw (standard) = 0.82. Inward finger (Borda's) = 0.51.

Anchor Type

visual_association

Why It Works

The gestural association (pointing finger vs. flush straw) creates a kinesthetic memory. The spatial distinction (inward vs. flush) directly maps to the physical geometry that produces the different Cd values.

Example Usage

Problem states 'Borda's mouthpiece' — recall inward finger, use Cd ≈ 0.51. Problem states 'standard short tube' — recall flush straw, use Cd ≈ 0.82.

Recall Trigger

Flush straw (standard) Cd ≈ 0.82. Inward finger (Borda's re-entrant) Cd ≈ 0.51.

Revision Game

Coefficient of Discharge, Cd = Cv × Cc ≈ 0.61

Clue

I am the product of two coefficients less than one. I combine friction loss and jet squeeze into a single multiplier. Who am I?

Memory Link

A1 — VCD Family: Cd is the baby born from Cv-mom and Cc-dad.

Vena Contracta — the minimum cross-section of the jet downstream of the orifice.

Clue

I am the tightest point in the water jet after it escapes an orifice. My name sounds like a contracted vein. What am I?

Memory Link

A4 — Telenovela actress squeezing through the hallway after the door.

Rectangular Weir: Q = (2/3)·Cd·√(2g)·L·H^(3/2)

Clue

My exponent is 3/2. I have a length L and I flow over a flat crest. Which weir am I?

Memory Link

A5 and A7 — Three-legged rectangular table. Two-thirds CD-playing Long Hits 3/2.

Triangular V-Notch Weir: Q = (8/15)·Cd·√(2g)·tan(θ/2)·H^(5/2)

Clue

Flash a V sign with your hand. My exponent is 5/2. I am better at measuring small flows. My formula includes tan(θ/2). What am I?

Memory Link

A6 — Victory sign, five-star pyramid exponent, octopus with fishballs.

t = 2·As·(√h1 − √h2) / (Cd·Ao·√(2g))

Clue

I am an engineer named Aisa. My plan area goes in the top of a fraction. The orifice team lives in my denominator. What equation describes how long it takes for a tank to drain?

Memory Link

A9 — Engineer Aisa on top (numerator), orifice team below (denominator).

Re-entrant Tube (Borda's Mouthpiece), Cd ≈ 0.51

Clue

I stick inward into the tank like a pointing finger. My Cd is only about 0.51 — the worst of all tube types. What am I called?

Memory Link

A20 — Inward-pointing finger vs. flush straw. Inward = worse Cd.

√(2g) = √(2 × 9.81) = √19.62 = 4.429 m^(1/2)/s — the energy constant derived from Torricelli's theorem, common to all orifice and weir discharge equations.

Clue

I am a number. Multiply me by the square root of 2 times 9.81 and you get 4.429. What am I used for, and why do I appear in EVERY weir formula?

Memory Link

A15 — PBA jersey 44-29. Pre-compute (2/3)×4.429=2.953 and (8/15)×4.429=2.362 on scratch paper.

End contraction — L' = L − 0.1nH, where n = 0, 1, or 2 contractions (Francis formula).

Clue

I reduce the effective length of a rectangular weir. Engineer Francis said each one of me subtracts 0.1H from the weir length. What am I, and how many can there be?

Memory Link

A11 — Francis is 10% shy at each end.

L' = effective weir length (m); L = actual crest length (m); n = number of end contractions (0, 1, or 2); H = head over weir (m); 0.1 = empirical Francis constant.

Numerator: 2 · As · (√h1 − √h2)
Denominator: Cd · Ao · √(2g)
As = tank plan area
Ao = orifice area
h1 = initial head, h2 = final head
Valid only for prismatic (constant cross-section) tanks

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CELE Hydraulics & Fluid Mechanics — Orifices, Weirs, Tubes and NozzlesMemory Anchors

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Orifices, Weirs, Tubes and Nozzles - Memory Anchors

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