CELE Hydraulics & Fluid Mechanics — Hydrodynamics and Fluid MachineryRevision Notes
Final-week revision notes for Hydrodynamics and Fluid Machinery. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Hydraulics & Fluid Mechanics subtest.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Hydraulics & Fluid Mechanics subtest is marked as "Core" in the official pattern, and Hydrodynamics and Fluid Machinery appears in position 9th of 10 in the CELE Hydraulics & Fluid Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Hydrodynamics and Fluid Machinery - Revision Notes
Hydrodynamics deals with the forces exerted by moving fluids, and fluid machinery covers the devices (pumps and turbines) that exchange energy with the flow. This chapter is consistently tested in the PRC Civil Engineer Licensure Examination under Hydraulics & Fluid Mechanics. You will apply the linear momentum equation to jets striking vanes and plates, compute anchoring forces on pipe bends, and calculate the input/output power and efficiency of pumps and turbines. Mastery of the four affinity laws and the distinction between pump and turbine efficiency expressions are essential for board-exam success.
Sections
Formulas
Example
Water flows at Q = 0.10 m³/s; inlet velocity = 20 m/s (x-dir), outlet velocity = 0 (plate stops flow). F = 1000(0.10)(0 − 20) = −2000 N. The plate exerts 2000 N on the fluid in the −x direction; by Newton's 3rd law, jet pushes plate with 2000 N in +x direction.
Formula
ΣF = ρQ(v₂ − v₁)
Variables
ρ = fluid density (kg/m³), Q = discharge (m³/s), v₁ = inlet velocity (m/s), v₂ = outlet velocity (m/s)
Application
General momentum equation for a steady, incompressible flow control volume in one direction.
Exam Tips
- Always draw the free-body diagram of the fluid in the CV, showing all forces (reaction, pressure, weight).
- State the positive direction before writing ΣF = ρQ(v₂ − v₁) for each axis.
- Check: does the fluid turn? If yes, resolve outlet velocity into components. If no (straight pipe), only pressure and area changes matter.
- For quick checks: units of ρQv are (kg/m³)(m³/s)(m/s) = kg·m/s² = N. ✓
Key Points
- Newton's 2nd Law applied to a control volume: ΣF = ρQ(v_out − v_in) in each coordinate direction.
- Forces include: reaction from the vane/bend (what we solve for), pressure forces at control-volume boundaries, and weight (often neglected for horizontal flows).
- For a stationary control volume, use absolute velocities. For a moving vane, transform to the relative frame first.
- The sign convention must be consistent: positive direction chosen, inlet velocity negative contribution, outlet velocity positive.
- Momentum is a vector equation — solve x and y components separately, then combine for the resultant.
- ρ (density) = 1000 kg/m³ for fresh water; γ (specific weight) = 9810 N/m³ = 9.81 kN/m³.
Definitions
Term
Control Volume (CV)
Definition
A fixed or moving region in space through which fluid flows, used to apply conservation laws without tracking individual particles.
Importance
Every momentum and energy problem is solved by drawing a proper CV — examiners check if pressure and momentum terms are correctly identified.
Term
Steady Flow
Definition
Flow conditions (velocity, pressure, density) at every point within the CV do not change with time.
Importance
The simplified momentum equation ΣF = ρQ(v_out − v_in) is valid only for steady flow — the standard assumption in board problems.
Section Title
1. Linear Momentum Equation — The Foundation
Common Mistakes
- Forgetting to include pressure forces (pA) at the inlet and outlet of a pipe bend control volume.
- Using absolute velocity instead of relative velocity (v − u) for a moving vane problem.
- Treating the momentum equation as scalar — failing to resolve into x and y components for non-collinear flows.
- Mixing up ρ (for momentum) and γ (for power) — ρ is in kg/m³; γ is in N/m³.
Formulas
Example
Jet: d = 50 mm → A = π(0.05)²/4 = 0.001963 m², v = 20 m/s. F = 1000(0.001963)(20²) = 785 N.
Formula
F = ρAv² = ρQv
Variables
ρ = 1000 kg/m³, A = jet cross-sectional area (m²), v = jet velocity (m/s), Q = Av (m³/s)
Application
Force of a jet on a stationary flat plate, perpendicular impact.
Example
Q = 0.05 m³/s, v = 15 m/s, α = 30°. F_n = 1000(0.05)(15) sin 30° = 750(0.5) = 375 N.
Formula
F_n = ρQv sin α
Variables
α = angle between jet direction and plate surface (degrees or radians)
Application
Normal force on an inclined stationary flat plate. When α = 90°, recovers F = ρQv.
Example
Jet turns 120° (θ = 120°): Q = 0.10 m³/s, v = 20 m/s. F_x = 1000(0.10)(20)(1 − cos 120°) = 2000(1 − (−0.5)) = 2000(1.5) = 3000 N. F_y = 2000 sin 120° = 2000(0.866) = 1732 N. F_R = √(3000² + 1732²) = 3464 N.
Formula
F_x = ρQv(1 − cos θ); F_y = ρQv sin θ
Variables
θ = total deflection angle of the jet by the vane (degrees). Resultant: F_R = √(F_x² + F_y²)
Application
Force components on a stationary curved vane deflecting the jet.
Example
v = 30 m/s, u = 12 m/s, θ = 165°, A = 0.002 m². F_x = 1000(0.002)(18²)(1 − cos 165°) = 648(1 − (−0.966)) = 648(1.966) = 1274 N. Power = 1274 × 12 = 15,288 W ≈ 15.3 kW.
Formula
F_x = ρA(v − u)²(1 − cos θ)
Variables
u = vane speed (m/s), (v − u) = relative velocity of jet with respect to vane
Application
Tangential force on a moving curved vane (e.g., Pelton bucket). Power = F_x · u.
Exam Tips
- Memorize: F_flat = ρQv; F_vane_x = ρQv(1 − cos θ); for moving vanes, replace Q → A(v−u) and v → (v−u).
- For a 180° U-vane: F = 2ρQv (stationary) or 2ρA(v−u)² (moving) — doubles the force.
- Check the problem: 'strikes and does not rebound' implies flat plate (θ = 90° turn, all momentum in original direction is absorbed).
- When θ = 120° appears in a problem, immediately compute: 1 − cos 120° = 1.5 and sin 120° = 0.866.
Key Points
- Stationary flat plate (normal impact): F = ρQv = ρAv². The jet comes to rest relative to the plate; all momentum is absorbed.
- Stationary flat plate (inclined at angle α to the jet axis): Normal force F_n = ρQv sin α. Here α is measured between the jet direction and the plate surface.
- Stationary curved vane turning jet by angle θ (deflection from original direction): F_x = ρQv(1 − cos θ), F_y = ρQv sin θ.
- For θ = 180° (U-turn vane): F_x = ρQv(1 − cos 180°) = 2ρQv — maximum possible force for a given jet.
- Moving flat vane at speed u: replace Q with A(v − u) and v with (v − u) in the momentum equation. The relative velocity is (v − u).
- Moving curved vane (Pelton bucket): F_x = ρA(v − u)²(1 − cos θ). Power delivered = F_x · u.
- Maximum power from a moving vane occurs when u = v/3 (for a flat plate, u = v/3) or u = v/2 (for a curved 180° vane).
- The discharge actually intercepted by a moving vane is Q_rel = A(v − u), not Av.
Definitions
Term
Deflection Angle (θ)
Definition
The total angle through which the jet is turned by the vane, measured from the initial jet direction to the final exit direction.
Importance
θ = 180° gives maximum force (1 − cos 180° = 2). θ = 90° gives F_x = ρQv (1 − 0) = ρQv.
Term
Relative Velocity (v_rel = v − u)
Definition
The velocity of the fluid jet as seen from the reference frame of the moving vane.
Importance
All momentum calculations for moving vanes must use the relative velocity and the relative discharge Q_rel = A(v − u).
Section Title
2. Jet Forces on Plates and Vanes
Common Mistakes
- Using full jet velocity v instead of relative velocity (v − u) for a moving vane — this is the most common board exam error.
- Using the full jet discharge Q = Av instead of the intercepted discharge A(v − u) for moving vanes.
- Confusing the deflection angle θ with the vane geometry angle — always identify the angle from inlet to outlet direction.
- Forgetting that cos(120°) = −0.5, so (1 − cos 120°) = 1.5, not 0.5.
- Not computing both F_x and F_y when asked for the resultant force on a vane.
Formulas
Example
90° horizontal bend, D = 200 mm (A = 0.03142 m²), v = 3 m/s, p₁ = 120 kPa, p₂ = 100 kPa. Q = 0.03142(3) = 0.09425 m³/s. x-dir: +p₁A₁ + R_x − 0 = ρQ(0 − v₁). R_x = ρQ(−v) − p₁A₁ = 1000(0.09425)(−3) − 120000(0.03142) = −282.8 − 3770 = −4053 N. y-dir: R_y − p₂A₂ = ρQ(v₂ − 0). R_y = ρQv + p₂A₂ = 282.8 + 3142 = 3425 N. F_R = √(4053² + 3425²) = 5308 N ≈ 5.31 kN.
Formula
F_x = ρQ(v₂ₓ − v₁ₓ) − p₁A₁ + p₂A₂ cos β (sign depends on geometry)
Variables
p = gauge pressure at section (Pa = N/m²), A = pipe cross-section area (m²), v_x = x-component of velocity (m/s), β = angle of outlet pipe with x-axis
Application
x-component of force balance on fluid in a pipe bend CV. Similarly for y.
Exam Tips
- Step 1: Draw CV with all forces labeled (R_x, R_y, p₁A₁ pointing inward, p₂A₂ pointing inward at outlet).
- Step 2: Write ΣF_x = ρQ(v₂ₓ − v₁ₓ) and ΣF_y = ρQ(v₂y − v₁y) — solve for R_x and R_y.
- Step 3: The anchor force on the bend = −R (Newton's 3rd law). Compute F_R = √(R_x² + R_y²).
- Typical board problem: 90° bend, given D, v (or Q), and gauge pressures at inlet/outlet. If pressures are not given, use Bernoulli to find them.
Key Points
- A pipe bend changes the direction (and possibly magnitude) of flow momentum — an anchoring force is required to hold it in place.
- Apply the momentum equation in x and y separately, including BOTH pressure forces (p₁A₁, p₂A₂) AND momentum flux terms.
- The control volume is the fluid inside the bend. Forces on the fluid: anchor reaction (R_x, R_y), pressure at inlet (p₁A₁), pressure at outlet (p₂A₂).
- For a 90° bend with equal pipe diameters: v₁ = v₂ = v (continuity). The x-direction loses all momentum; y-direction gains full momentum.
- By Newton's 3rd Law, the force the fluid exerts on the bend is equal and opposite to the force the bend exerts on the fluid.
- The resultant anchoring force: F_R = √(F_x² + F_y²); angle with x-axis: tan φ = F_y / F_x.
- Gauge pressures are used because atmospheric pressure acts uniformly and cancels out.
Definitions
Term
Anchoring Force
Definition
The external force (provided by pipe supports, concrete thrust blocks, or flanges) required to keep a pipe fitting or bend in static equilibrium against the net fluid force.
Importance
In practice, thrust blocks are designed from this force. In board exams, you solve for the magnitude and direction of this resultant anchoring force.
Term
Gauge Pressure
Definition
Pressure measured relative to atmospheric pressure. Gauge pressure = Absolute pressure − Atmospheric pressure.
Importance
Use gauge pressure in bend momentum calculations because atmospheric pressure acts on all external surfaces and cancels out of the equilibrium equation.
Section Title
3. Forces on Pipe Bends
Common Mistakes
- Omitting the pressure-force terms (p₁A₁ and p₂A₂) — this is the single most common error in pipe-bend problems.
- Incorrect sign for pressure forces — the pressure at the inlet pushes the fluid in the positive flow direction (same sign as flow), while at the outlet it pushes opposite to the assumed positive direction.
- Using absolute pressure instead of gauge pressure in the force calculation.
- Forgetting that for a 90° bend, the x-component of outlet velocity is zero and the y-component of inlet velocity is zero.
Formulas
Example
Q = 0.05 m³/s, H = 30 m. P_w = 9.81(0.05)(30) = 14.715 kW.
Formula
P_w = γQH
Variables
γ = 9.81 kN/m³ (or 9810 N/m³), Q = discharge (m³/s), H = total pump head (m). Result in kW if γ in kN/m³.
Application
Water (hydraulic) power — the power actually transferred to the fluid.
Example
P_w = 14.715 kW, η = 0.75. P_input = 14.715 / 0.75 = 19.62 kW.
Formula
P_input = γQH / η
Variables
η = overall pump efficiency (decimal, e.g., 0.75 for 75%)
Application
Shaft/brake power required to drive the pump — what the motor must supply.
Example
Pump at N₁ = 1450 rpm gives H₁ = 25 m. At N₂ = 1750 rpm: H₂ = 25(1750/1450)² = 25(1.2069)² = 25(1.457) = 36.4 m.
Formula
Q₂/Q₁ = N₂/N₁; H₂/H₁ = (N₂/N₁)²; P₂/P₁ = (N₂/N₁)³
Variables
N = rotational speed (rpm), Q = discharge (m³/s), H = head (m), P = power (kW)
Application
Affinity Laws — predict pump performance when speed changes for the same pump.
Example
p_a/γ = 10.33 m, h_vapor = 0.24 m (at 20°C), h_s = 4 m (suction lift), h_f,s = 0.5 m. NPSH_a = 10.33 − 0.24 − 4 − 0.5 = 5.59 m.
Formula
NPSH_a = p_a/γ − h_vapor − h_s − h_f,s
Variables
p_a = atmospheric pressure (N/m²), h_vapor = vapor pressure head (m), h_s = suction head (m, positive = above pump), h_f,s = friction loss in suction pipe (m)
Application
Available NPSH — must be greater than NPSH_required (given by manufacturer) to avoid cavitation.
Exam Tips
- Memory aid: PUMP = Power In > Power Out → P_input = γQH / η (divide — bigger number goes in).
- For affinity laws: write the ratios in a table: Q ratio = N ratio; H ratio = (N ratio)²; P ratio = (N ratio)³.
- Check units: γQH → (kN/m³)(m³/s)(m) = kN·m/s = kW. ✓
- If a problem gives motor efficiency AND pump efficiency, total efficiency = η_pump × η_motor; P_motor = γQH / (η_pump × η_motor).
- NPSH: 'available must exceed required' — if NPSH_a < NPSH_r, cavitation occurs → pump must be lowered or suction pipe enlarged.
Key Points
- A pump adds energy (head) to the fluid. The useful (water/hydraulic) power is P_w = γQH.
- Input (shaft/brake) power is always GREATER than water power because of losses: P_input = γQH / η.
- Overall efficiency η = P_w / P_input = γQH / P_input. Board problems may also give motor efficiency separately.
- Total head H = H_static + h_f + h_minor (static lift + friction losses + minor losses).
- Affinity Laws (for the same pump, changing speed from N₁ to N₂): Q₂/Q₁ = N₂/N₁; H₂/H₁ = (N₂/N₁)²; P₂/P₁ = (N₂/N₁)³.
- Specific speed N_s = N√Q / H^(3/4) — classifies pump type: centrifugal (low N_s), mixed-flow (medium), axial/propeller (high N_s).
- NPSH (Net Positive Suction Head): available NPSH must exceed required NPSH to prevent cavitation on the suction side.
- NPSH_available = (p_atm/γ) − h_vapor − h_s − h_f,suction where h_s = suction lift.
- Cavitation causes pitting, noise, vibration, and loss of pump performance — avoid by limiting suction lift.
- Series pumps: heads add, same Q. Parallel pumps: flows add, same H.
Definitions
Term
Total Pump Head (H)
Definition
The energy per unit weight added by the pump to the fluid, expressed in metres. H = (p₂−p₁)/γ + (v₂²−v₁²)/2g + (z₂−z₁) + h_L.
Importance
H is the key parameter linking power and discharge. It must account for all losses in the system, not just the static lift.
Term
Cavitation
Definition
Formation and collapse of vapor bubbles in the fluid when local pressure drops to the vapor pressure. Causes damage to impeller blades, noise, and performance degradation.
Importance
Board exams test NPSH calculations. Cavitation is prevented by keeping suction lift low and minimizing friction losses in the suction line.
Term
Specific Speed (N_s)
Definition
N_s = N√Q / H^(3/4) — a dimensionless (or dimensional) index that characterizes the shape and type of pump impeller at best efficiency point.
Importance
Distinguishes centrifugal (N_s < 2000 in SI rpm units), mixed-flow (2000–5000), and axial-flow (> 5000) pumps. Appears in selection-type board questions.
Term
Pump Efficiency (η)
Definition
η = Water Power Output / Shaft Power Input = γQH / P_input. Accounts for hydraulic, mechanical, and volumetric losses.
Importance
For pumps, efficiency is the ratio of useful output to input — meaning you DIVIDE by η to get input power. This is the reverse of the turbine convention.
Section Title
4. Pumps — Power, Efficiency, and Affinity Laws
Common Mistakes
- Dividing instead of multiplying by efficiency for pump input power (or vice versa for turbines).
- Using γ = 9.81 N/m³ instead of 9810 N/m³ or 9.81 kN/m³ — causing a factor-of-1000 error.
- Applying affinity laws with speed ratio inverted (N₁/N₂ instead of N₂/N₁).
- Forgetting that P ∝ N³ (cube, not square) in the affinity laws.
- Confusing series (heads add) vs parallel (flows add) pump configurations.
Formulas
Example
η = 0.85, Q = 2 m³/s, H = 15 m. P = 0.85 × 9.81 × 2 × 15 = 0.85 × 294.3 = 250.2 kW.
Formula
P_output = η · γ · Q · H
Variables
η = turbine efficiency (decimal), γ = 9.81 kN/m³, Q = m³/s, H = net head (m). Result in kW.
Application
Power delivered by a turbine to the shaft (mechanical output).
Example
P = 500 kW, η = 0.88, H = 120 m. Q = 500 / (0.88 × 9.81 × 120) = 500 / 1034.5 = 0.4835 m³/s.
Formula
Q = P_output / (η · γ · H)
Variables
Rearrangement of the turbine power formula to solve for discharge.
Application
Given rated power, efficiency, and head — find the required discharge.
Example
N = 300 rpm, P = 500 kW, H = 120 m. N_s = 300√500 / 120^(5/4) = 300(22.36) / 277.1 = 6707 / 277.1 = 24.2 (low — Pelton range).
Formula
N_s = N√P / H^(5/4)
Variables
N = speed (rpm), P = power (kW or metric hp, depending on system), H = net head (m). N_s is dimensional.
Application
Turbine specific speed — identifies turbine type. Pelton: low N_s; Francis: medium; Kaplan: high N_s.
Exam Tips
- Memory aid: TURBINE = Power Out < Power In → P_output = η × γQH (multiply — smaller number comes out).
- Quick type identification: High head → Pelton (impulse); Medium head → Francis (reaction); Low head → Kaplan (reaction, adjustable blades).
- For Pelton problems: bucket velocity u = φ√(2gH) where φ ≈ 0.46; jet velocity v_j = C_v√(2gH) where C_v ≈ 0.98.
- If asked for discharge from a Pelton turbine given power and head: Q = P / (η·γ·H). Straightforward rearrangement.
Key Points
- A turbine extracts energy from the fluid. Output power = η × γQH (multiply by efficiency).
- Net head H = gross head − head losses in penstock (the pipe bringing water to the turbine).
- Impulse turbines (Pelton wheel): convert all available head to kinetic energy in a nozzle; water jet strikes buckets. Best for high head (> 200 m), low flow.
- Reaction turbines (Francis, Kaplan): runner is fully submerged; both pressure and velocity energy drive the runner. Best for low-to-medium head, high flow.
- Francis turbine: mixed-flow (radial-to-axial); medium head (40–600 m).
- Kaplan turbine: axial-flow, adjustable blades; low head (< 40 m), high discharge.
- Turbine specific speed: N_s = N√P / H^(5/4) — different formula from pump specific speed.
- Pelton wheel: bucket speed u ≈ 0.46v_jet at maximum efficiency (u/v ≈ 0.46).
- Draft tube in reaction turbines: recovers kinetic energy at exit and allows setting turbine above tailwater without losing head.
- Affinity laws also apply to turbines (same form as pumps).
Definitions
Term
Net Head (H_net)
Definition
The effective head available at the turbine inlet after subtracting all hydraulic losses from the gross head (total elevation difference between headwater and tailwater).
Importance
Always use net head in turbine power calculations. H_net = H_gross − h_penstock losses.
Term
Pelton Wheel
Definition
An impulse turbine where one or more nozzles produce high-velocity jets that strike cup-shaped buckets on the runner periphery. All energy conversion from pressure to kinetic occurs in the nozzle.
Importance
Identified by: high head (> 200 m), low Q, jet striking buckets, operates in air (not submerged). Common PRC board question: find Q given P, η, and H.
Term
Draft Tube
Definition
A diverging tube connected from the reaction turbine exit to the tailwater. It reduces exit pressure below atmospheric, recovering kinetic energy and effectively increasing the net head.
Importance
Allows the turbine to be installed above tailwater level without head loss. The negative pressure in the draft tube enables this — but must not drop below vapor pressure (cavitation risk).
Term
Turbine Efficiency (η)
Definition
η = Shaft Power Output / Water Power Input = P_shaft / (γQH). Accounts for hydraulic (flow losses), mechanical (bearing friction), and volumetric (leakage) losses.
Importance
For turbines, efficiency multiplies (reduces) the available power to give what you actually get. MULTIPLY by η — the reverse of the pump formula.
Section Title
5. Turbines — Power, Efficiency, and Types
Common Mistakes
- Using gross head instead of net head (penstock losses reduce the available head).
- Applying P = γQH without multiplying by η for a turbine (forgetting efficiency).
- Confusing pump and turbine efficiency directions: Pump → divide; Turbine → multiply.
- Using the pump specific speed formula N_s = N√Q/H^(3/4) for a turbine instead of N_s = N√P/H^(5/4).
Formulas
Example
A pump: γQH = 20 kW, P_input = 25 kW → η = 20/25 = 0.80 = 80%. A turbine: γQH = 300 kW, P_output = 255 kW → η = 255/300 = 0.85 = 85%.
Formula
η_pump = P_w / P_input = γQH / P_input; η_turbine = P_output / P_water = P_output / (γQH)
Variables
P_w = water power (kW), P_input = shaft input power (kW), P_output = shaft output power (kW)
Application
Unified efficiency expressions — rearrange to find the unknown power or efficiency.
Exam Tips
- Create a one-line equation table during the exam: write P_pump_in = γQH/η and P_turbine_out = ηγQH to avoid confusion.
- Affinity law shortcut: if speed increases by 20% (ratio = 1.2), then Q → ×1.2, H → ×1.44, P → ×1.728.
- Board exam format: most problems give you 3 of the 4 variables (P, Q, H, η) and ask for the 4th — plug into the correct formula.
Key Points
- The SAME affinity laws apply to both pumps and turbines: Q∝N, H∝N², P∝N³.
- Efficiency direction: Pump = divide by η; Turbine = multiply by η.
- Momentum uses ρ (kg/m³); power uses γ (kN/m³ or N/m³).
- Jet on stationary flat plate: F = ρAv² (maximum force, all momentum absorbed).
- Jet on 180° curved vane: F = 2ρQv (maximum force for curved vane geometry).
- Series pumps: ΣH, same Q; Parallel pumps: ΣQ, same H.
- Prevent cavitation: maximize NPSH_available by minimizing suction lift and suction-line losses.
- For pipe bends: always include both momentum flux AND pressure force (pA) terms.
Definitions
Term
Water Power (P_w)
Definition
P_w = γQH — the rate of energy transferred to (pump) or from (turbine) the water. Also called hydraulic power.
Importance
This is the reference power for computing efficiency. It appears in both pump and turbine efficiency expressions.
Section Title
6. Summary of Key Equations and Relationships
Common Mistakes
- Applying series pump head-addition rule when pumps are actually in parallel.
- Forgetting to cube the speed ratio when computing power change using affinity laws.
- Not converting units consistently: mixing kPa with Pa, or kN with N.
Connections
- Bernoulli's equation provides the velocity and pressure values used in momentum calculations for jets and bends — these two principles work together in every pipe-flow problem.
- Continuity equation (Q = A₁v₁ = A₂v₂) is always applied first to find velocities at each section before using the momentum equation.
- The head-loss equations (Darcy-Weisbach, Hazen-Williams) determine friction losses that reduce the net head available to a turbine or increase the head a pump must overcome.
- Energy equation (Bernoulli + head loss + pump/turbine head) connects total pump or turbine head H to the system conditions — H_pump = z₂ − z₁ + (p₂−p₁)/γ + (v₂²−v₁²)/2g + h_L.
- Dimensional analysis (Buckingham Pi theorem) is the theoretical basis for the affinity laws and specific speed — understanding dimensions prevents unit errors in speed-change problems.
- Open-channel flow and pipe flow share the concept of specific energy, which is analogous to the total head concept used in turbine and pump calculations.
- Structural design (reinforced concrete thrust blocks) relies on the fluid force calculations from this chapter — connecting hydraulics to structural engineering in practice.
- Thermodynamics: turbine and pump efficiency concepts mirror those of heat engines and refrigeration cycles — the direction of energy flow determines whether efficiency multiplies or divides.
- Hydrostatics (pressure at a depth) provides the gauge pressure values at pipe-bend sections used in the pressure-force terms of the momentum equation.
- Philippine practice: NPC (National Power Corporation) and DPWH irrigation systems operate turbines and pumps governed by these principles — RA 6395 (NPC Charter) and the Philippine Water Code (PD 1067) provide the legal framework.
Exam Strategy
In the PRC board exam, Hydrodynamics and Fluid Machinery problems appear consistently — typically 3–6 items per exam. Prioritize the following: (1) Jet-on-vane problems: memorize F = ρQv for flat plate and F_x = ρQv(1−cosθ) for curved vane; immediately check if the vane is moving (use relative velocity v−u). (2) Pump/turbine power: write the two formulas P_pump = γQH/η and P_turbine = ηγQH at the top of your scratch paper — avoid the most common error of mixing them up. (3) Affinity laws: set up the ratio table Q∝N, H∝N², P∝N³ — board problems typically give the old and new speed and ask for new H or P. (4) Pipe bends: draw the control volume, label ALL forces (reaction + pressure at both ends), then write ΣF_x and ΣF_y. Never omit the pA terms. (5) Units: use γ = 9.81 kN/m³ consistently so that P = γQH gives kW directly with Q in m³/s and H in m. Check: (kN/m³)(m³/s)(m) = kN·m/s = kW ✓. Allocate 3–4 minutes per computational problem; if you recognize the formula type within 30 seconds, you are on pace. Review Examples 1, 2, and 3 from the reference material — they represent the three most common problem archetypes.
Quick Review Questions
A water jet with velocity 20 m/s and cross-sectional area 0.005 m² strikes a stationary flat plate normally. What is the force exerted on the plate?
Q = Av = 0.005 × 20 = 0.10 m³/s. F = ρQv = 1000 × 0.10 × 20 = 2000 N. Alternatively, F = ρAv² = 1000 × 0.005 × 400 = 2000 N. The jet's entire x-momentum is absorbed by the plate.
A 50 mm diameter jet at 25 m/s strikes a stationary curved vane that deflects the jet by 120°. Find the resultant force on the vane.
A = π(0.05)²/4 = 0.001963 m², Q = 0.001963 × 25 = 0.04909 m³/s. F_x = ρQv(1−cos120°) = 1000(0.04909)(25)(1−(−0.5)) = 1000(0.04909)(25)(1.5) = 1841 N. F_y = ρQv sin120° = 1000(0.04909)(25)(0.866) = 1063 N. F_R = √(1841² + 1063²) = √(3389281 + 1130169) = √4519450 ≈ 2126 N. Note: recalculate with exact Q = 0.04909 m³/s giving F_R ≈ 2126 N. (Rounded: ≈ 2.13 kN.)
A pump delivers Q = 0.08 m³/s against a total head of 45 m at 70% efficiency. What motor input power is required?
P_water = γQH = 9.81 × 0.08 × 45 = 35.316 kW. P_input = P_water / η = 35.316 / 0.70 = 50.45 kW ≈ 50.2–50.5 kW. The motor must supply more power than the water receives due to pump losses.
A Pelton turbine develops 500 kW under a net head of 120 m at 88% efficiency. Find the discharge.
P_output = η·γ·Q·H → Q = P_output / (η·γ·H) = 500 / (0.88 × 9.81 × 120) = 500 / 1034.496 ≈ 0.4835 m³/s ≈ 0.484 m³/s.
A pump runs at 1450 rpm and produces H = 25 m and Q = 0.06 m³/s. If speed is increased to 1750 rpm, find the new head and new discharge.
Speed ratio = 1750/1450 = 1.2069. Q₂ = Q₁ × 1.2069 = 0.06 × 1.2069 = 0.07241 m³/s. H₂ = H₁ × (1.2069)² = 25 × 1.4566 = 36.41 m. Affinity law: Q ∝ N (linear), H ∝ N² (quadratic).
For a pipe bend, why must both momentum flux AND pressure force terms be included in the force calculation?
The momentum equation ΣF = ρQ(v_out − v_in) represents Newton's 2nd law for the fluid. All forces on the fluid must appear on the left side: (1) the anchor reaction force, (2) pressure forces at the inlet pushing fluid forward, and (3) pressure forces at the outlet resisting flow. In pressurized pipe systems, p·A terms often exceed the momentum terms in magnitude. Neglecting them is the most common board-exam error in bend problems.
What is the difference in the efficiency formula for a pump versus a turbine, and why?
Efficiency accounts for losses. A pump must consume MORE power than it delivers to the fluid (input > water power), so we divide by η. A turbine produces LESS power than the fluid supplies to it (output < water power), so we multiply by η. Both efficiencies are less than 1.0 (or less than 100%), which means dividing makes the result larger and multiplying makes it smaller — consistent with physical reality.
A moving flat vane travels at u = 8 m/s and a jet at v = 20 m/s strikes it. The jet area is 0.003 m². Find the force on the vane.
Relative velocity = v − u = 20 − 8 = 12 m/s. Relative discharge = A(v − u) = 0.003 × 12 = 0.036 m³/s. Force (normal flat vane) = ρ × A(v−u) × (v−u) = ρA(v−u)² = 1000 × 0.003 × 144 = 432 N. Alternatively, F = ρ·Q_rel·(v−u) = 1000 × 0.036 × 12 = 432 N. Power = F × u = 432 × 8 = 3456 W. (Note: if the problem says 'stationary vane' F = ρAv² = 1000×0.003×400 = 1200 N — much higher.)
Which turbine type is best suited for a site with a 350 m head and low discharge? Justify.
Pelton wheels are designed for high head (> 200 m) and relatively low discharge. At high head, converting the pressure to a high-velocity jet (via nozzle) is energy-efficient. The jet strikes buckets at atmospheric conditions (impulse — no pressure differential across the runner). Francis and Kaplan turbines require the runner to be submerged and work best at lower heads (Francis: 40–600 m but typically medium; Kaplan: < 40 m). At 350 m, Pelton is the standard choice.
State the three affinity laws for a centrifugal pump and derive the power ratio for a 10% increase in speed.
Affinity (similarity) laws: Q₂/Q₁ = N₂/N₁; H₂/H₁ = (N₂/N₁)²; P₂/P₁ = (N₂/N₁)³. These follow from dimensional analysis of pump performance. For N₂/N₁ = 1.10: Q ratio = 1.10 (10% more); H ratio = 1.21 (21% more); P ratio = 1.331 (33.1% more). Key insight: a small speed increase causes a large power increase — important for motor sizing. This cubic relationship is why variable-speed drives save substantial energy when flow is reduced.
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