Fundamentals of Fluid Flow - Cheat Sheet
Your 30-minute exam companion for continuity, energy equations, momentum, and power in hydraulic systems. All formulas, key facts, and common pitfalls for the PRC Civil Engineer Licensure Examination.
CONTINUITY EQUATION (Conservation of Mass)
Formulas
Q = A₁v₁ = A₂v₂
Meaning
Q = discharge (m³/s); A = area (m²); v = velocity (m/s); subscripts 1, 2 = two points
Watch out
Students forget that velocity is INVERSELY proportional to area. Smaller pipe = FASTER flow. Check: (D₁/D₂)² is squared, not linear.
When to use
Whenever flow enters a different area or diameter — velocity MUST change to keep Q constant
v₂ = v₁(D₁/D₂)²
Meaning
v = velocity; D = diameter; direct relationship between velocity and diameter ratio squared
Watch out
Diameter ratio is SQUARED. Many students forget the exponent. D₁/D₂ ≠ v₂/v₁; it is (D₁/D₂)² = v₂/v₁.
When to use
Finding velocity change in a pipe reducer or enlargement — avoid solving for area first
Q = nAv̄
Meaning
n = number of flow paths (parallel pipes); A = cross-sectional area per path; v̄ = average velocity
Watch out
Only add discharges if paths are truly independent. Series paths use continuity, not addition.
When to use
Multiple identical parallel pipes or channels carrying water
Common values
Value
9.81 m/s²
Symbol
g
Quantity
Gravitational acceleration
Value
1000 kg/m³
Symbol
ρ
Quantity
Water density (fresh water, 4°C)
Value
9.81 kN/m³ or 9810 N/m³
Symbol
γ = ρg
Quantity
Specific weight of water
Important facts
- Continuity holds for ANY cross-sectional shape if A is the total area perpendicular to flow.
- For circular pipes: A = πD²/4; use consistent units (D in m, A in m²).
- In a series pipe system, discharge is the same at every section (Q₁ = Q₂ = Q₃ = ... = constant).
- Velocity is fastest where the pipe is narrowest; slowest where it is widest.
- The continuity equation assumes NO fluid accumulation in the control volume (steady state).
Key definitions
Discharge (Q)
Example
A 300 mm pipe at 2 m/s velocity has Q = 0.1414 m³/s.
Volume of fluid passing a section per unit time (m³/s); also called volumetric flow rate.
Steady Flow
Example
Water in a pipe at constant speed; not a dam suddenly releasing.
Flow properties at any point do not change with time; Q, v, p are independent of time.
Incompressible Flow
Example
Water at ρ = 1000 kg/m³ assumed constant throughout a pipe system.
Fluid density ρ is constant; used for liquids (water, oil) at normal pressures.
Stream Tube
Example
Flow path in a convergent nozzle acts as a single stream tube.
A bundle of streamlines forming an imaginary tube; fluid entering one end must exit the other (continuity).
Diagrams to know
- Convergent nozzle: wide pipe → narrow pipe; velocity arrow increases; area arrow decreases.
- Divergent diffuser: narrow pipe → wide pipe; velocity arrow decreases; area arrow increases.
- Velocity profile across pipe diameter: parabolic (laminar) or nearly flat (turbulent).
BERNOULLI EQUATION (Energy Equation)
Formulas
p₁/γ + v₁²/(2g) + z₁ + hₐ = p₂/γ + v₂²/(2g) + z₂ + hₑ + hₗ
Meaning
p/γ = pressure head (m); v²/(2g) = velocity head (m); z = elevation head (m); hₐ = pump head added (m); hₑ = turbine head extracted (m); hₗ = head lost to friction (m)
Watch out
SIGN convention: pump head hₐ is POSITIVE (adds energy); turbine head hₑ is POSITIVE but SUBTRACTED (removes energy). Head loss hₗ is always SUBTRACTED. Pressure must be in the same units (kPa or Pa); convert if needed.
When to use
Compare energy at two points in a flow system; rearrange to solve for unknown pressure, velocity, elevation, or pump/turbine head
Total Head (H_total) = p/γ + v²/(2g) + z
Meaning
Sum of pressure, velocity, and elevation heads; represents total energy per unit weight of fluid
Watch out
Total head DECREASES in the direction of flow due to friction loss. Do not confuse total head with pump head or turbine head.
When to use
Plotting the Energy Grade Line (EGL); comparing energy levels at different sections
Ideal Bernoulli: p₁/γ + v₁²/(2g) + z₁ = p₂/γ + v₂²/(2g) + z₂
Meaning
No pump, turbine, or friction loss; energy at point 1 equals energy at point 2
Watch out
Real flows ALWAYS have losses. Using ideal Bernoulli without hₗ gives unrealistically high pressure or velocity. This is the #1 exam mistake.
When to use
Short smooth pipes, or as a starting approximation before accounting for losses
Energy Grade Line (EGL): elevation = p/γ + v²/(2g) + z
Meaning
Graphical plot of total head along a pipe; represents the total mechanical energy line
Watch out
EGL always slopes downward in direction of flow (due to hₗ). EGL is ABOVE the Hydraulic Grade Line (HGL) by exactly v²/(2g).
When to use
Visualizing energy distribution in a pipe system; identifying where losses occur
Hydraulic Grade Line (HGL): elevation = p/γ + z
Meaning
Plot of pressure head plus elevation; EGL minus velocity head
Watch out
HGL is where piezometer readings plot. If HGL drops BELOW the pipe centerline, pressure becomes negative (cavitation risk). HGL is always BELOW EGL.
When to use
Determining if cavitation risk exists (when HGL falls below pipe invert) or surface elevation of water in open channels
Common values
Value
101.3 kPa
Symbol
p_atm
Quantity
Atmospheric pressure (sea level)
Value
9.81 m/s²
Symbol
g
Quantity
Standard gravity
Value
9.81 kN/m³
Symbol
γ
Quantity
Water specific weight
Important facts
- Bernoulli is a form of conservation of energy: total energy at one point = total energy at another ± work done by/on fluid ± losses.
- In ideal flow (no losses, no machines), H_total is CONSTANT: EGL is horizontal.
- Pressure head is GAUGE pressure if using p (in kPa or Pa); use absolute pressure for cavitation checks.
- Velocity head v²/(2g) is NEVER negative; fast flow → high velocity head.
- In a horizontal pipe with no machines, if diameter decreases (v increases), pressure MUST decrease to conserve energy.
- At the outlet of a nozzle, pressure typically equals atmospheric; use this as boundary condition.
- EGL and HGL are shown on the SAME graph; EGL is always above HGL by v²/(2g).
Key definitions
Pressure Head (p/γ)
Example
100 kPa pressure = 100/(9.81) ≈ 10.2 m of water head.
Height of water column equivalent to pressure p; expressed in meters of water.
Velocity Head (v²/2g)
Example
Velocity 2 m/s gives v²/(2g) = 4/19.62 ≈ 0.204 m.
Height of water column equivalent to kinetic energy per unit weight; increases with velocity squared.
Elevation Head (z)
Example
Point 5 m above datum has z = +5 m; point 3 m below has z = −3 m.
Vertical distance above a reference datum (usually ground or sea level); positive upward.
Pump Head (hₐ)
Example
Pump raising water 20 m adds hₐ = 20 m head.
Energy per unit weight added to fluid by a pump; always positive in Bernoulli equation.
Turbine Head (hₑ)
Example
Turbine extracting 25 m of head: subtract hₑ = 25 m.
Energy per unit weight extracted from fluid by a turbine; subtracted as positive value in Bernoulli.
Head Loss (hₗ)
Example
Long rough pipe at high velocity: hₗ = 8 m (energy wasted).
Energy per unit weight dissipated by friction; always positive and subtracted in Bernoulli.
Diagrams to know
- EGL vs HGL diagram: two lines, EGL above HGL, both sloping downward; vertical distance = v²/(2g).
- Pressure variation in a convergent nozzle: p₁ (high) → p₂ (low); velocity v₁ (low) → v₂ (high).
- Siphon system: total head line showing how siphon maintains flow even when discharge is above source.
MOMENTUM EQUATION (Conservation of Momentum)
Formulas
ΣF = ρQ(v₂ − v₁)
Meaning
ΣF = sum of forces on control volume (N); ρ = fluid density (kg/m³); Q = discharge (m³/s); v₁, v₂ = velocities (m/s) at inlet and outlet
Watch out
This is a VECTOR equation: apply COMPONENT-WISE (x-direction, y-direction, z-direction). Velocity change v₂ − v₁ is signed; reversing direction makes it negative. Always include pressure forces on the control surface.
When to use
Calculate force on pipe bends, nozzles, vanes, or any control volume where flow direction or speed changes
For a 90° bend (horizontal pipe): Fₓ = ρQv₁, Fᵧ = ρQv₂ (if v₁ = v₂ from continuity)
Meaning
Horizontal and vertical force components on a 90° elbow when inlet and outlet velocities are equal
Watch out
If areas are different (v₁ ≠ v₂), use actual velocities. If outlet is to atmosphere, pressure force on control volume outlet = 0. Always check direction: force on FLUID by pipe (then negate for force on pipe by fluid).
When to use
Determining anchor forces on pipe bends (common exam question)
For a nozzle jet hitting a plate: F = ρQ(v − 0) = ρQv (assuming jet stops on plate)
Meaning
Force exerted by an impact jet on a stationary flat plate
Watch out
Momentum BEFORE = ρQv; AFTER = 0 (jet stops). The change is ρQv, so force is F = ρQv. If plate moves, adjust final velocity accordingly.
When to use
Jet propulsion problems, turbine bucket design, or hydraulic ram analysis
Common values
Value
1000 kg/m³
Symbol
ρ
Quantity
Water density
Important facts
- Momentum equation is a direct application of Newton's second law: F = ma = d(mv)/dt.
- For a control volume, ΣF includes ALL external forces: pressure forces, weight, anchor forces, friction.
- In most exam problems, weight and friction are ignored unless stated; focus on pressure and momentum change.
- Force on the pipe from the fluid is OPPOSITE to force on the fluid from the pipe (Newton's 3rd law).
- Direction matters: velocity into the control volume is POSITIVE; velocity out is also positive in its direction. Change v₂ − v₁ is a vector subtraction.
- Momentum equation is independent of friction and internal energy losses; it depends only on velocities (and areas via continuity).
Key definitions
Control Volume
Example
The water inside a pipe elbow between inlet and outlet sections.
Fixed imaginary region in space through which fluid flows; momentum equation applied to mass inside it.
Momentum
Example
Water jet at 1 m³/s and 10 m/s has momentum rate = 1000 × 1 × 10 = 10,000 N (or 10 kN).
Product of mass and velocity: m·v; for a fluid stream, rate of momentum = ρQ·v.
Anchor Force
Example
Elbow must be anchored with a force equal and opposite to the momentum change of the water.
Reaction force required to hold a pipe or nozzle in place against fluid momentum and pressure.
Diagrams to know
- Control volume around pipe bend: inlet arrow, outlet arrow, force vectors on pipe.
- Free body diagram of a jet hitting a flat plate: incoming momentum ρQv, reaction force F.
- Pressure forces on a control volume: act normal to surface, inward on all sides.
POWER OF FLOWING FLUID
Formulas
P = γQH
Meaning
P = power (W); γ = specific weight of fluid (N/m³ = 9.81 kN/m³ for water); Q = discharge (m³/s); H = head (m)
Watch out
Units are critical: if γ is in N/m³ (9810), result is in Watts (W). If γ is in kN/m³ (9.81), result is in kilowatts (kW). Most exams use γ = 9.81 kN/m³ for simplicity. ALWAYS convert to desired power units.
When to use
Calculate power of water flowing under a given head; the MOST COMMON power formula in exam
P_pump = γQH / η
Meaning
Power input to pump; η = pump efficiency (0 to 1); H = total head pump must raise
Watch out
Efficiency is LESS than 100%; actual power input is LARGER than theoretical power γQH. If η = 0.85, then P_input = γQH / 0.85 ≈ 1.176 × γQH.
When to use
Sizing pump motor or calculating electrical input power required
P_turbine = η × γQH
Meaning
Useful power output from turbine; η = turbine efficiency (0 to 1)
Watch out
Efficiency multiplies power (output is smaller than input). If η = 0.88 and γQH = 100 kW, output is only 88 kW.
When to use
Calculating electrical output or mechanical power delivered by a turbine
H = (p/γ + v²/2g + z)₁ − (p/γ + v²/2g + z)₂
Meaning
Net head available between two points; difference in total heads
Watch out
H can include elevation change, pressure drop, and velocity change. Always account for ALL three components; don't drop the velocity head just because the problem doesn't emphasize it.
When to use
Determining available head for turbine or required head for pump in complex systems
Common values
Value
9.81 kN/m³
Symbol
γ
Quantity
Water specific weight
Value
75%–90%
Symbol
η_pump
Quantity
Typical pump efficiency
Value
80%–95%
Symbol
η_turbine
Quantity
Typical turbine efficiency
Important facts
- Power scales linearly with discharge Q: double Q → double power (if H constant).
- Power scales linearly with head H: double H → double power (if Q constant).
- Specific weight γ = ρg = 1000 kg/m³ × 9.81 m/s² = 9810 N/m³ ≈ 9.81 kN/m³.
- For pump: P_input = P_theoretical / η; for turbine: P_output = P_theoretical × η.
- In a pump, higher head requires higher power; in a turbine, higher head produces higher power.
- Power is zero if either Q = 0 or H = 0; both are needed for power generation.
Key definitions
Power
Example
Water at 0.1 m³/s flowing under 20 m head delivers power = 9.81 × 0.1 × 20 = 19.62 kW.
Rate of energy transfer; for fluid flow, P = γQH (energy per unit weight × discharge).
Efficiency (η)
Example
Pump with 85% efficiency needs 1/0.85 = 1.176 times the theoretical power.
Ratio of useful output power to input power (for pump) or input power to output power (for turbine); always 0 < η < 1.
Head (H)
Example
20 m head means the fluid has enough energy to lift 1 kg of water 20 m vertically.
Energy per unit weight of fluid, expressed as an equivalent height of water column (m).
Diagrams to know
- Pump system: inlet at lower elevation, outlet at higher elevation; pump head hₐ added; power = γQ(hₐ) / η.
- Turbine system: inlet at high head, outlet at low; turbine extracts head hₑ; power = η × γQ(hₑ).
- Power vs discharge curve: linear relationship (for constant H).
WORKED EXAMPLE SOLUTIONS (Common Exam Formats)
Important facts
- Example 1 — Continuity in a reducer: D₁ = 300 mm, v₁ = 2 m/s; D₂ = 150 mm → v₂ = 2 × (300/150)² = 8 m/s; Q = π(0.3)²/4 × 2 = 0.1414 m³/s.
- Example 2 — Bernoulli with pressure change: p₁ = 200 kPa, v₁ = 2 m/s, z₁ = 0; v₂ = 8 m/s (from continuity), z₂ = 5 m; neglect loss → p₂/γ = 200/9.81 + 0.204 − 3.262 − 5 ≈ 12.33 m → p₂ ≈ 121 kPa.
- Example 3 — Power of flow: Q = 0.1414 m³/s, H = 20 m → P = 9.81 × 0.1414 × 20 = 27.74 kW.
- Example 4 — Pump head required: water rises z = 10 m, head loss hₗ = 8 m → pump must add hₐ = 10 + 8 = 18 m (from Bernoulli rearranged).
- Example 5 — Nozzle exit pressure: horizontal nozzle, p₁ = 300 kPa, v₁ = 3 m/s, v₂ = 12 m/s → p₂/γ = 300/9.81 + (9 − 144)/19.62 ≈ 30.58 − 6.90 = 23.68 m → p₂ ≈ 232 kPa.
- Example 6 — Turbine power: Q = 5 m³/s, H = 25 m, η = 0.88 → P_out = 0.88 × 9.81 × 5 × 25 ≈ 1080 kW.
- Example 7 — Force on 90° elbow: ρ = 1000 kg/m³, Q = 0.1 m³/s, v = 2 m/s (same in/out from continuity) → F_x = F_y = 1000 × 0.1 × 2 = 200 N each; resultant = 200√2 ≈ 283 N at 45°.
Must remember
- Continuity: Q = A₁v₁ = A₂v₂ — ALWAYS true for incompressible flow in series; smaller area → faster velocity; diameter ratio is SQUARED in velocity relation.
- Bernoulli with machines: pump head hₐ is +; turbine head hₑ is −; head loss hₗ is −; rearrange to solve for unknowns (pressure, velocity, pump size, etc.).
- Velocity head v²/(2g) MUST be included whenever flow velocity changes; it represents kinetic energy per unit weight; cannot be dropped even in short pipes.
- Energy Grade Line (EGL) slopes downward in direction of flow; always above HGL by v²/(2g); horizontal only in ideal flow with no losses.
- Momentum equation ΣF = ρQ(v₂ − v₁) is a VECTOR; apply component-wise for pipe bends; includes pressure forces and anchor reaction forces.
- Power: P = γQH is universal (water specific weight × discharge × head); pump input = γQH/η; turbine output = η×γQH; η < 1 always.
- At pipe exit to atmosphere: assume pressure = atmospheric (usually gauge p = 0); use this boundary condition in Bernoulli.
- Cavitation risk: occurs when HGL drops below pipe invert or when absolute pressure falls below vapor pressure; use absolute pressure, not gauge.
- In a reducer (convergent), pressure DECREASES and velocity INCREASES; in a diffuser (divergent), pressure INCREASES and velocity DECREASES (Bernoulli trade-off).
- For pump sizing: required head = elevation rise + friction loss + velocity head change (if any); do not forget any component in Bernoulli analysis.
Last minute tips
- Draw the Bernoulli energy line (EGL and HGL) if given pipe geometry — visualizing slopes and intersections catches 80% of head-loss problems instantly.
- Always check units in power calculations: γ = 9.81 kN/m³ gives power in kW; if using 9810 N/m³, convert Watts to kW after. One wrong unit = wrong answer.
- For any two-point problem (continuity or Bernoulli), define your datum (z = 0) clearly; all elevations must be measured from THE SAME reference; use ground level or pipe centerline consistently.
- In Bernoulli, if the problem mentions 'negligible friction' or 'ideal flow,' set hₗ = 0 and simplify early; if it mentions 'long pipe' or 'rough pipe,' DO NOT neglect hₗ — it will dominate the energy balance.
- Momentum exam questions often hide pressure forces: read carefully — if the problem specifies 'both inlet and outlet exposed to atmosphere,' then both pressure forces cancel, and only momentum change drives the anchor force.
Comparison tables
Bernoulli Terms Comparison
| Head Type | Formula | Units | When It Increases | When It Decreases |
|---|---|---|---|---|
| Pressure Head | p/γ | meters (m) | Higher pressure; smaller area | Lower pressure; larger area |
| Velocity Head | v²/(2g) | meters (m) | Higher velocity; smaller pipe | Lower velocity; larger pipe |
| Elevation Head | z | meters (m) | Higher point; upstream | Lower point; downstream |
| Total Head | p/γ + v²/(2g) + z | meters (m) | Any head component increases | Friction loss; machine extraction |
Pump vs Turbine in Bernoulli Equation
| Parameter | Pump | Turbine |
|---|---|---|
| Direction of hₐ or hₑ | + hₐ (energy added) | − hₑ (energy removed) |
| Effect on total head | Increases total head | Decreases total head |
| Power direction | Input (motor drives pump) | Output (water drives turbine) |
| Efficiency correction | P_input = γQH / η | P_output = η × γQH |
| Exam appearance | Lift water uphill; increase pressure | Drop water downhill; extract power |
Continuity: Area vs Velocity Relationship
| Condition | Diameter Change | Velocity Change | Energy Status | Example |
|---|---|---|---|---|
| Convergent (reducer) | D decreases | v increases | Pressure ↓ | 300 mm → 150 mm: v quadruples |
| Divergent (diffuser) | D increases | v decreases | Pressure ↑ | 150 mm → 300 mm: v reduces to 1/4 |
| Constant diameter | D constant | v constant | p unchanged | Straight pipe: v same throughout |
Common Pitfalls in Bernoulli Exam Problems
| Pitfall | Wrong Approach | Correct Approach | Typical Exam Mark Loss |
|---|---|---|---|
| Neglecting velocity head | Drop v²/(2g) term | Always include; it matters when v changes | 5–10 points |
| Wrong sign on pump head | Subtract hₐ | Add hₐ (energy input) | 5 points |
| Wrong sign on turbine | Add hₑ | Subtract hₑ (energy extracted) | 5 points |
| Mixing gauge and absolute pressure | Use gauge p in cavitation check | Use absolute p | 10 points for cavitation problem |
| Forgetting continuity linkage | Assume v₁ = v₂ | Calculate v₂ from A₁v₁ = A₂v₂ | 10 points |
| Units mismatch in power | γ = 9810 N/m³ but forget to convert W → kW | Be consistent: 9.81 kN/m³ → kW directly | 5 points |
Diagrams
Fundamentals of Fluid Flow — Concept Map
Comprehensive mind map showing the four pillars of fluid flow (continuity, Bernoulli, momentum, power) and their key sub-concepts; serves as a quick hierarchical reference for understanding how topics interconnect.
Bernoulli Problem-Solving Decision Tree
Step-by-step flowchart guiding students through any Bernoulli problem: starting from continuity if needed, identifying machine terms (pump/turbine), accounting for losses, solving, and verifying units. This is the #1 exam roadmap.
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) Logic
Decision and logic flowchart for constructing Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) on a pipe system. Shows how machines and losses affect both lines, and highlights cavitation risk condition. Essential for visual problem-solving in exams.
Pipe Diameter Transition States (Continuity)
State diagram showing how pipe diameter changes trigger velocity changes via continuity; illustrates the cyclical nature of convergent and divergent sections and emphasizes that discharge Q remains constant throughout (key exam insight).