Hydrodynamics and Fluid Machinery - Memory Anchors
Instead of re-reading formulas repeatedly, your brain stores them as vivid mental images or emotionally charged stories that are retrieved instantly under exam pressure. For the PRC Civil Engineer Licensure Examination, where you must solve Hydraulics problems quickly and accurately, these anchors act as mental shortcuts: one trigger word or image unlocks an entire formula, process, or concept. Use these anchors actively — say them aloud, draw them, and test yourself — and you will find that Hydrodynamics and Fluid Machinery becomes one of your strongest exam topics rather than a source of anxiety.
Memory anchors
Force of a jet on a stationary flat plate: F = ρQv = ρAv²
Memory aid
Imagine you are holding an umbrella against a Habagat typhoon. The harder the rain hits (higher v) and the more rain that falls per second (higher Q), the harder the umbrella is pushed back. The force is simply the mass of water arriving per second (ρQ) times how fast it is moving (v). You feel it directly — no angles, no curves, just straight-on impact. That is F = ρQv for a flat plate hit normally.
Why it works
Connecting to the universally familiar Filipino experience of monsoon rain and umbrellas makes the formula emotionally vivid and physically intuitive, anchoring the equation to a sensory memory.
Example usage
In an exam: 'A jet hits a flat plate normally' → picture the Habagat umbrella → F = ρQv. Substitute ρ=1000, Q=Av, solve.
Recall trigger
Typhoon umbrella push
Force on a curved vane (stationary): Fx = ρQv(1 − cosθ)
Memory aid
Mang Boy operates a buko (coconut) shaving machine shaped like a curved blade. When the jet of water hits the curved blade and turns by angle θ, Mang Boy notices: 'If the blade is flat (θ=0°), the water just slides along and pushes nothing — cosθ=1, so (1−cos0°)=0, no force! If the blade curves the water a full 180°, it pushes back hardest — cos180°=−1, so (1−(−1))=2, maximum force!' Mang Boy's rule: the more the water is turned, the bigger the force.
Why it works
The micro-story with Mang Boy shows the physical meaning of the cosθ term through extreme cases (θ=0° and θ=180°), making the formula self-explaining rather than arbitrary.
Example usage
Vane turns jet by 120°: Fx = ρQv(1−cos120°) = ρQv(1−(−0.5)) = ρQv(1.5). If θ=90°: Fx=ρQv(1−0)=ρQv.
Recall trigger
Mang Boy's curved buko blade
Moving vane: replace v with relative velocity (v − u)
Memory aid
Think of riding a jeepney at speed u while someone runs alongside at speed v throwing a ball at you. The ball hits you with the relative speed (v − u), not v. If the jeepney moves as fast as the ball, no impact at all! Moving vanes are the same — only the relative velocity (v − u) matters for the momentum exchange.
Why it works
The jeepney-runner analogy is a culturally familiar Filipino scenario that makes the concept of relative velocity concrete and intuitive.
Example usage
Moving vane at u=8 m/s, jet at v=20 m/s: use (v−u)=12 m/s in all momentum formulas instead of 20 m/s.
Recall trigger
Jeepney runner ball
Pump input power formula: P_input = γQH / η
Memory aid
Remember the phrase: 'Give Queen Her Energy' → G=γ, Q=discharge, H=head, E=efficiency (η in denominator). For a pump, you GIVE more energy than useful work (divide by η < 1 makes it bigger). The pump is hungry — it needs MORE than what it delivers.
Why it works
The phrase 'Give Queen Her Energy' encodes the variables in order, and the concept 'pump is hungry' reinforces why we divide (input > output) rather than multiply.
Example usage
Exam: pump, Q=0.05 m³/s, H=30 m, η=75% → P = γQH/η = 9.81×0.05×30/0.75 = 19.62 kW
Recall trigger
Give Queen Her Energy (pump is hungry)
Turbine output power formula: P_output = ηγQH
Memory aid
Turbine is 'Efficient Queen Gets Help' → η, γ, Q, H — all multiplied. A turbine GIVES LESS than it receives (multiply by η < 1 makes it smaller). The turbine is generous — it gives away some energy as losses.
Why it works
Pairing with A4 (pump) using a contrasting analogy (hungry vs generous) lets students quickly distinguish the two formulas by remembering the opposite nature of pumps and turbines.
Example usage
Turbine: η=0.85, Q=2 m³/s, H=15 m → P = 0.85×9.81×2×15 = 250.2 kW
Recall trigger
Turbine is generous (multiply by η)
Pump vs Turbine efficiency direction
Memory aid
Visualize a PUMP as a person PUSHING water uphill carrying a heavy basket (input is more than output — divide by η). Visualize a TURBINE as a water slide GIVING you a ride (the water gives you output — multiply by η). Pump DIVIDES (you pay more), Turbine MULTIPLIES (you get less than 100%).
Why it works
Spatial mental images of pushing uphill (pump) and receiving a ride (turbine) encode the mathematical operation (divide vs multiply) through physical effort direction.
Example usage
Exam asks for motor power driving a pump → automatically divide γQH by η. Exam asks for turbine shaft output → multiply ηγQH.
Recall trigger
Pushing basket uphill (pump) vs water slide ride (turbine)
Affinity Laws: Q∝N, H∝N², P∝N³
Memory aid
The exponents go 1, 2, 3 — the simplest sequence possible! Think: '1-2-3 of the pump's ABCs': Q is the most basic (exponent 1), H is second (exponent 2), P is the most powerful (exponent 3). Say it: 'Q-one, H-two, P-three' or simply '1, 2, 3 — Q, H, P!'
Why it works
Chunking the exponents as the trivial sequence 1-2-3 makes them impossible to confuse. Linking Q (flow) as simplest and P (power, most dependent on N) as the cube root association reinforces the logic.
Example usage
At N₁=1450 rpm, H₁=25 m. At N₂=1750 rpm: H₂=H₁(N₂/N₁)²=25(1750/1450)²=36.42 m
Recall trigger
1-2-3: Q, H, P
Pipe bend momentum equation includes both ρQ(Δv) AND pressure force pA terms
Memory aid
Engineer Nene is designing anchor blocks for a water main bend in Metro Manila. Her supervisor reminds her: 'Nene, dalawa ang kalaban mo!' (You have two enemies!) Enemy 1: the MOMENTUM of the flowing water changing direction. Enemy 2: the PRESSURE pushing on the pipe walls at each end. Forget either one and the anchor block fails, the pipe breaks, and it's your license on the line (RA 544 covers that!). Always write: F = ρQ(Δv) + pA terms.
Why it works
The cautionary professional story with a Filipino name and a reference to RA 544 (Engineer's Law) adds emotional weight, making students remember both terms. The 'dalawa ang kalaban' phrase is culturally resonant.
Example usage
Pipe bend: Fx = ρQ(v₂ₓ − v₁ₓ) + p₁A₁cosα₁ − p₂A₂cosα₂. Never forget the pA terms!
Recall trigger
Nene's two enemies on a pipe bend
ρ (density) is used in momentum/force; γ (specific weight) is used in power
Memory aid
Remember: 'Rho (ρ) = Racing (momentum/force); Gamma (γ) = Go Power!' Momentum is about mass flying through space — use the raw mass density ρ. Power is about energy stored in the weight of water lifted — use specific weight γ. ρ RACES (kinetic, force), γ GIVES POWER (potential energy). Also recall: γ = ρg, so γ is always bigger than ρ.
Why it works
Assigning an action word (Racing, Power) to each Greek letter creates a direct semantic link between the symbol and its physical role.
Example usage
Force on jet: F = ρQv (use 1000 kg/m³). Pump power: P = γQH/η (use 9810 N/m³ or 9.81 kN/m³).
Recall trigger
Rho Races, Gamma Gives Power
Cavitation and NPSH (Net Positive Suction Head)
Memory aid
NPSH is like your cellphone battery warning — it tells you how much suction margin you have before the pump 'dies' (cavitates). If your available NPSH drops below the required NPSH, bubbles form on the impeller like a phone that overheats and shuts down. Rule: NPSH_available > NPSH_required, or your pump will eat itself alive (impeller pitting, vibration, noise). Always check the suction side!
Why it works
Cellphone battery anxiety is universally understood by today's engineering students, making the 'margin before failure' concept emotionally vivid and the consequence (pump destruction) memorable.
Example usage
Board exam: pump at high elevation or hot fluid — check if NPSH_a > NPSH_r to prevent cavitation.
Recall trigger
Low battery warning = NPSH warning
Pelton wheel (impulse turbine): high head, low flow
Memory aid
Picture a Pelton wheel as a BASKETBALL player — tall (high head), but handles only one ball at a time (low flow). It gets hit by a single jet of water like a player catching one strong pass. Impulse means the jet does all the work in the open air — no pressure housing needed. High in the mountains, where water falls from great height — think NPC Pantabangan Dam feeding a Pelton wheel.
Why it works
The basketball player analogy (tall but handles one ball) encodes both characteristics (high head, low flow) simultaneously, with a Philippine NPC reference adding local context.
Example usage
Exam: 500 kW, 120 m head → Pelton turbine. P = ηγQH → Q = P/(ηγH) = 500/(0.88×9.81×120) = 0.484 m³/s
Recall trigger
Tall basketball player = Pelton (high head, low flow)
Francis and Kaplan turbines (reaction): low-to-medium head, high flow
Memory aid
Francis and Kaplan are like KALABAW (carabao) — low to the ground (low head) but handle huge volumes of work (high flow). Reaction turbines are always submerged in water under pressure, like a carabao wading through rice paddy floodwaters. Francis is the radial type (like a snail shell); Kaplan has adjustable blades like a propeller fan. Think of Caliraya Reservoir feeding a Francis turbine.
Why it works
The Kalabaw analogy — familiar to Filipino agricultural life — encodes low-head and high-flow characteristics in a single image. Submerged in water reinforces the reaction (pressure) nature.
Example usage
Exam: large river with 10 m head, 50 m³/s → Francis or Kaplan turbine, not Pelton.
Recall trigger
Kalabaw wading in paddy = Francis/Kaplan (low head, high flow)
Continuity equation for jets: Q = Av (also used when computing jet momentum)
Memory aid
Short rhyme to memorize: 'Q is what flows, A times v shows — Bigger the pipe, bigger the dose, Squeeze it down, the speed rose!' Q = Av is the bridge between the geometry of a jet (area A) and its momentum (v). Always compute Q first, then use F = ρQv.
Why it works
Rhymes exploit phonological loop memory — the rhythm makes the formula recall automatic, and the 'squeeze it down, the speed rose' phrase also encodes the continuity-velocity inverse relationship.
Example usage
Jet: D=50 mm → A=π(0.05)²/4=0.001963 m². v=25 m/s → Q=0.001963×25=0.04909 m³/s → F=ρQv=1000×0.04909×25=1227 N
Recall trigger
Q flows, A times v shows
Momentum equation sign convention on pipe bends — include both inlet and outlet
Memory aid
Walk mentally through a pipe bend from INLET to OUTLET. At the FRONT DOOR (inlet), water pushes IN with momentum ρQv₁ and pressure p₁A₁ — both acting in the original flow direction. At the BACK DOOR (outlet), water pushes OUT with ρQv₂ and p₂A₂ — in the new direction. The anchoring force on the bend is the reaction to the net push. Place yourself at the bend: front door (inlet), back door (outlet). Draw all forces, then sum each axis.
Why it works
Method of loci — mentally walking through a familiar house structure (front door, back door) — provides spatial organization to a problem with multiple force components that students commonly mix up.
Example usage
90° bend: x-direction gets inlet momentum; y-direction gets outlet momentum. Draw the free body diagram, label both doors.
Recall trigger
Front door (inlet) and back door (outlet) of the bend
Water power (hydraulic power): P_water = γQH
Memory aid
Remember: 'Gamma Quickly Handles' — γ × Q × H = water power. This is the IDEAL power — what water can theoretically deliver or what a pump ideally needs. No efficiency yet. Think of it as the 'sticker price' of water power before efficiency discounts (for turbine) or surcharges (for pump).
Why it works
The sticker price vs discounts/surcharges metaphor elegantly connects water power to both pump (add efficiency surcharge → divide) and turbine (apply efficiency discount → multiply), unifying three concepts.
Example usage
Water power = 9.81×0.08×45 = 35.32 kW. With 70% pump efficiency → Motor power = 35.32/0.70 = 50.45 kW
Recall trigger
Gamma Quickly Handles = γQH (sticker price of water power)
Force on inclined flat plate: Fn = ρQv sinα (α = angle between jet and plate surface)
Memory aid
Visualize a broom (walis tingting) angled to the floor. If you spray water straight down the broom handle (α=0°), the water just slides along the broom — sin0°=0, zero normal force. If you spray perpendicular to the broom (α=90°), maximum force — sin90°=1. The sine of the jet-to-plate angle tells you how much of the jet's punch is 'wasted' sliding vs. used pushing the plate normally.
Why it works
The broom analogy gives a physical sense of why sinα appears: the component of force perpendicular to the plate is the effective push. α=0° and α=90° extreme cases are self-verifying.
Example usage
Jet at 30° to plate: Fn = ρQv sin30° = ρQv × 0.5. At 90°: Fn = ρQv sin90° = ρQv (normal plate case).
Recall trigger
Spray water on a tilted broom — sinα determines the push
Efficiency formula: η = P_output / P_input (general definition)
Memory aid
Efficiency is your GWA (General Weighted Average) — how much of what went IN (study hours) came OUT as useful grades. η = what you get / what you put in. For a pump: you put in motor power, you get useful hydraulic power — but pump is inefficient, so output < input, η < 1. Always: η = P_out/P_in, rearrange to find what's asked.
Why it works
GWA is a universally understood metric for Filipino engineering students. Connecting efficiency to their personal academic performance creates an emotionally salient and memorable analogy.
Example usage
Given pump efficiency 75%, motor power = P_in. Useful hydraulic power P_out = 0.75 × P_in. Or given P_out and η, find P_in = P_out/η.
Recall trigger
GWA = output/input (efficiency)
Affinity Laws applied to head ratio: H₂/H₁ = (N₂/N₁)²
Memory aid
Direk runs a pumping station in Marikina. He increases pump speed from 1450 to 1750 rpm. 'How much more head?' he asks. His assistant says: 'Direk, head scales as the SQUARE of speed — like exam stress squares when deadlines double!' The speed ratio is 1750/1450=1.207, so H₂=H₁×(1.207)²=H₁×1.457. A 20.7% speed increase gives a 45.7% head increase — the square punishes small speed changes.
Why it works
The relatable exam stress analogy (doubling deadlines squares stress) makes the squared relationship memorable, while the actual board exam numbers are embedded in the story.
Example usage
N₁=1450 rpm, H₁=25 m, N₂=1750 rpm → H₂=25×(1750/1450)²=25×1.457=36.4 m
Recall trigger
Direk's speed increase — exam stress squares
Impulse (Pelton) turbine operates in AIR; Reaction (Francis, Kaplan) turbine operates SUBMERGED
Memory aid
Pelton = STREET FIGHTER standing in air, hit by a targeted jet (like a fire hose aimed at spinning cups). Reaction = UNDERWATER WRESTLER, fully submerged and surrounded by pressurized water on all sides. Pelton converts only kinetic energy; reaction turbines convert both kinetic AND pressure energy. Air vs. Water environment is the key visual distinction.
Why it works
Two contrasting combat images (air fighter vs. underwater wrestler) create a binary memory structure that is visually distinct and easy to recall under pressure.
Example usage
Exam mentions 'partially full casing' or 'jet from nozzle' → Pelton. Exam mentions 'runner submerged' or 'draft tube' → reaction turbine.
Recall trigger
Air fighter (Pelton) vs. Underwater wrestler (Francis/Kaplan)
Units check: γ in kN/m³ gives power in kW; γ in N/m³ gives power in W
Memory aid
Remember: 'kilo-gamma = kilowatts' (k goes with k). Use γ=9.81 kN/m³ with Q in m³/s and H in m → get kW directly. Use γ=9810 N/m³ → get Watts. The kilo-family stays together! If an exam gives motor power in kW (most board problems do), always use γ=9.81 kN/m³.
Why it works
The 'kilo-family stays together' rule creates a unit-tracking mnemonic that prevents the most common arithmetic error in power problems (off by factor of 1000).
Example usage
P = 9.81 kN/m³ × 0.05 m³/s × 30 m / 0.75 = 19.62 kW (answer in kW, no conversion needed)
Recall trigger
kilo-gamma = kilowatts (k with k)
Revision game
I am the force a typhoon-strength jet exerts on a flat wall. My formula multiplies density, flow rate, and jet speed. Who am I?
Answer
F = ρQv (jet force on flat plate)
Memory link
A1 — Habagat typhoon umbrella analogy
Mang Boy's curved buko blade turns a jet by 180°. What is the (1 − cosθ) factor for maximum force?
Answer
2 (since cos180° = −1, so 1−(−1) = 2)
Memory link
A2 — Mang Boy's buko blade micro-story
A pump is 'hungry' — it needs MORE power than useful work. What does this tell you about its efficiency formula?
Answer
Divide γQH by η → P_input = γQH/η (dividing by η<1 makes it bigger)
Memory link
A4 — Give Queen Her Energy mnemonic
A turbine is 'generous' — it gives LESS shaft power than the water provides. What is its output power formula?
Answer
P_output = η × γ × Q × H (multiplying by η<1 makes it smaller)
Memory link
A5 — Efficient Queen Gets Help mnemonic
Count 1-2-3 for pump affinity. Q scales as N to the power of ___, H as N to the power of ___, P as N to the power of ___.
Answer
1, 2, 3
Memory link
A7 — 1-2-3 Q-H-P chunking mnemonic
Engineer Nene has TWO enemies when designing anchor blocks for a pipe bend. Name both.
Answer
- Momentum change: ρQ(Δv). 2. Pressure forces: p₁A₁ and p₂A₂ at each end.
Memory link
A8 — Nene's two enemies micro-story
Which Greek letter belongs to momentum and force calculations — ρ or γ? Which belongs to power calculations?
Answer
ρ (density) for momentum/force. γ (specific weight) for power.
Memory link
A9 — Rho Races, Gamma Gives Power mnemonic
Your pump's NPSH_available drops below NPSH_required. What happens? Relate to a Filipino student's cellphone.
Answer
Cavitation occurs — like a phone shutting down on low battery. Bubbles form, impeller erodes, pump fails.
Memory link
A10 — Low battery warning = NPSH warning analogy
Formula mnemonics
F = ρQv = ρAv² (jet on stationary flat plate, normal impact)
Mnemonic
Rho Quickly Vrooms — ρ, Q, v all together = force. Or: 'Rho Arrives Very fast' = ρ × Av (since Q=Av).
When to use
Jet hits a flat plate perpendicularly (normal impact). Also the baseline case before modifying for angles or curves.
What each part means
ρ = fluid density (1000 kg/m³ for water), Q = discharge (m³/s) = Av, v = jet velocity (m/s). F comes out in Newtons.
Fx = ρQv(1 − cosθ) (curved stationary vane, x-component)
Mnemonic
1 minus cosine = how much the curve fights back. θ=0 → no fight. θ=180° → maximum fight (factor 2).
When to use
Stationary curved vane deflecting a jet by angle θ. For moving vane at speed u, replace v with (v−u) and Q with A(v−u).
What each part means
ρQv = jet momentum flux. (1−cosθ) = the turning factor. θ = total angle the jet is deflected by the vane.
P_pump_input = γQH / η
Mnemonic
Give Queen Her Energy — γ, Q, H divided by η. Pump is GREEDY — needs MORE than it gives (divide makes bigger when η<1).
When to use
Finding motor or shaft power needed to drive a pump at known flow, head, and efficiency.
What each part means
γ = specific weight (9.81 kN/m³), Q = discharge (m³/s), H = total head added by pump (m), η = pump efficiency (decimal). Result in kW.
P_turbine_output = η × γ × Q × H
Mnemonic
Efficient Queen Gets Help — η, γ, Q, H all multiplied. Turbine is GENEROUS — gives LESS than it receives (multiply makes smaller when η<1).
When to use
Finding shaft power output of a turbine at known net head, flow rate, and efficiency.
What each part means
η = turbine efficiency, γ = 9.81 kN/m³, Q = discharge through turbine (m³/s), H = net head across turbine (m). Result in kW.
Q₂/Q₁ = N₂/N₁ (Affinity Law — flow)
Mnemonic
Q rhymes with N: 'Q follows N one-to-one.' Linear, simplest law.
When to use
Predicting new flow rate when pump speed changes, at constant impeller diameter.
What each part means
Q = volumetric flow rate, N = rotational speed (rpm). At same pump/impeller geometry.
H₂/H₁ = (N₂/N₁)² (Affinity Law — head)
Mnemonic
Head is SQUARE of speed: 'H goes N-squared.' Twice the speed = four times the head.
When to use
Predicting new head when pump speed changes. Most common affinity law question on board exams.
What each part means
H = total head (m), N = rotational speed (rpm). Exponent 2 in the 1-2-3 sequence.
P₂/P₁ = (N₂/N₁)³ (Affinity Law — power)
Mnemonic
Power is CUBE of speed: 'P goes N-cubed.' Small speed increase = big power jump. Explains why overspeed is dangerous.
When to use
Predicting power consumption or demand when pump speed is varied. Critical for motor sizing.
What each part means
P = shaft power (kW), N = rotational speed (rpm). Exponent 3, the last in the 1-2-3 sequence.
Fn = ρQv sinα (inclined flat plate, normal component)
Mnemonic
sinα = how 'normal' the hit is. α is jet-to-plate angle. sin90°=1 (direct hit). sin0°=0 (parallel, no hit).
When to use
Jet strikes a flat plate at an angle. Note: α is measured from plate surface, not from the plate normal.
What each part means
Fn = force normal to plate, ρ = density, Q = discharge, v = jet speed, α = angle between jet direction and plate surface.
η = P_output / P_input (general efficiency)
Mnemonic
GWA formula: Efficiency = what you get / what you put in. Always ≤ 1 (or ≤ 100%).
When to use
Any problem asking for efficiency, or rearranged to find unknown power given efficiency and another power value.
What each part means
Dimensionless ratio. For pump: η=γQH/P_motor. For turbine: η=P_shaft/(γQH).
Quick recall chains
Steps to Solve Any Jet Force Problem
Recall test
A jet at 20 m/s hits a vane moving at 8 m/s and deflects 90°. What is Fx? → Use (v−u)=12, Q_rel=A×12, Fx=ρQ_rel(v−u)(1−cos90°)=ρQ_rel×12×1
Memory chain
Remember 'FIVE-QR': Find the type (flat/curved), Identify motion (stationary/moving), Velocity adjustment (v or v−u), Enter Q=Av, Recall correct formula, check Units. 'FIGURE: Flat/curved, If moving, Get Q, Use (v−u), Recall formula, Examine units.'
Items to remember
- 1. Identify: flat plate, inclined plate, or curved vane?
- 2. Identify: stationary or moving vane?
- 3. Compute discharge Q = Av
- 4. If moving: use relative velocity v_rel = (v − u)
- 5. Apply the correct formula: F = ρQv, ρQv sinα, or ρQv(1−cosθ)
- 6. Check units (ρ in kg/m³, F in Newtons)
Pump vs Turbine Problem-Solving Sequence
Recall test
Pump: Q=0.08 m³/s, H=45 m, η=70%. Find motor power. → Tag=9.81×0.08×45=35.32 kW. Pump pays more: 35.32/0.70=50.46 kW.
Memory chain
The shopping analogy: 'Water Power is the Price Tag (γQH). Pump PAYS MORE (divide by η — budget goes up). Turbine EARNS LESS (multiply by η — take-home is less than gross).' Sequence: Tag → Pump pays more / Turbine earns less → Use kN for kW.
Items to remember
- 1. Identify: pump (adds head) or turbine (extracts head)?
- 2. Find γQH = water power (sticker price)
- 3. Pump: divide by η to get motor input power
- 4. Turbine: multiply by η to get shaft output power
- 5. Use γ = 9.81 kN/m³ for kW answers
Affinity Laws in Order
Recall test
Pump at 1200 rpm delivers 30 L/s and uses 5 kW. At 1500 rpm: Q₂=30×(1500/1200)=37.5 L/s; H₂ scales squared; P₂=5×(1500/1200)³=9.77 kW
Memory chain
Chant: 'Q is ONE, H is TWO, P is THREE — 1-2-3, Q-H-P!' Count on your fingers: thumb=Q (one), index=H (two), middle=P (three). Practice it three times and it is permanent.
Items to remember
- Q₂/Q₁ = (N₂/N₁)¹ — Flow is linear
- H₂/H₁ = (N₂/N₁)² — Head is squared
- P₂/P₁ = (N₂/N₁)³ — Power is cubed
Turbine Classification Quick Keys
Recall test
Which turbine for a 200 m head, 0.5 m³/s site? → Pelton (high head, low flow). Which for 8 m head, 100 m³/s? → Kaplan (very low head, very high flow).
Memory chain
PFK order: 'Pelton Flies Kite' (air/high, medium, low head respectively). Or: Pelton=Peak (high head), Francis=Flat (medium), Kaplan=Kayak (low head, high flow rivers).
Items to remember
- Pelton — Impulse — High head — Low flow — Jet in air
- Francis — Reaction — Medium head — Medium flow — Submerged, radial
- Kaplan — Reaction — Low head — High flow — Submerged, axial, adjustable
Pipe Bend Force Free-Body Diagram Steps
Recall test
90° bend, same diameter, find anchoring force. → Inlet in x-direction, outlet in y-direction. Fx = ρQv + p₁A₁ (no x-component at outlet). Fy = ρQv + p₂A₂ (no y-component at inlet). R=√(Fx²+Fy²).
Memory chain
Remember 'CIVIL BD': Control volume, Identify inlet/outlet, Velocities labelled, Include pA terms, Lay out x and y equations, Build resultant. 'CIVIL engineers Build Dams' — and they also solve bend problems!
Items to remember
- 1. Draw the control volume around the bend
- 2. Label inlet (1) and outlet (2) with velocities and pressures
- 3. Write ΣFx = ρQ(v₂ₓ − v₁ₓ) + p₁A₁cosα₁ − p₂A₂cosα₂
- 4. Write ΣFy = ρQ(v₂y − v₁y) + p₁A₁sinα₁ − p₂A₂sinα₂
- 5. Find resultant R = √(Fx² + Fy²)
Diagrams
Hydrodynamics and Fluid Machinery — Master Concept Map
A hierarchical mind map covering all major topics in Hydrodynamics and Fluid Machinery: jet forces (flat, inclined, curved, moving), pipe bend forces, pump power and affinity laws, turbine types, and key unit distinctions. Use this to orient yourself before studying individual topics.
Decision Flowchart: Solving Any Hydrodynamics Board Exam Problem
A step-by-step decision flowchart for classifying any Hydrodynamics board exam problem and selecting the correct formula. Branches cover jet forces (flat/inclined/curved, stationary/moving), pump power, turbine power, affinity laws, and pipe bend analysis. Follow the arrows to arrive at the correct equation every time.
Pump vs Turbine Power: The Efficiency Direction Guide
A side-by-side flowchart showing how efficiency affects the power formula differently for pumps (divide by η, input is larger) versus turbines (multiply by η, output is smaller). The sticker price analogy (γQH) is shown as the starting point for both machine types.
Affinity Laws: Speed Change Effects on Pump Performance
A visual summary of the three affinity laws showing how pump flow, head, and power scale with speed change using exponents 1, 2, and 3 respectively. Includes a worked numerical example from the board exam exercises.