Bernoulli’s Equation
Energy Conservation in Flowing Fluids — Formula, Derivation, Applications & Solved Problems
Last Updated: March 2026
📌 Key Takeaways
- Bernoulli’s equation: P + ½ρv² + ρgh = constant along a streamline.
- It expresses conservation of energy for flowing fluids — the total of pressure energy, kinetic energy, and potential energy is constant.
- Key insight: When velocity increases, pressure decreases (and vice versa).
- Assumptions: Steady, incompressible, inviscid (frictionless) flow along a streamline.
- Used with the continuity equation to solve most fluid flow problems.
- Applications: Venturi meters, pitot tubes, orifice meters, siphons, flow over weirs, aerofoil lift.
1. The Equation — Three Forms
Pressure Form (most common)
P₁ + ½ρV₁² + ρgz₁ = P₂ + ½ρV₂² + ρgz₂
Each term has units of pressure (Pa). This form is used most in engineering problems.
Head Form (hydraulics)
P/(ρg) + V²/(2g) + z = constant = H (total head)
Each term has units of length (metres of fluid). Widely used in civil and hydraulic engineering.
- P/(ρg) = pressure head
- V²/(2g) = velocity head
- z = elevation head (datum head)
- H = total head
Energy Form (per unit volume)
P + ½ρV² + ρgz = constant
Each term has units of energy per unit volume (J/m³ = Pa).
2. Physical Meaning — Energy Conservation
Bernoulli’s equation is the First Law of Thermodynamics applied to a flowing fluid element. It states that the total mechanical energy of a fluid particle remains constant as it moves along a streamline (in the absence of friction and external work).
The three forms of energy are:
| Energy Type | Term | Physical Meaning |
|---|---|---|
| Pressure energy | P | Energy stored due to fluid pressure — the work the fluid can do by expanding |
| Kinetic energy | ½ρV² | Energy due to fluid motion — faster fluid has more kinetic energy |
| Potential energy | ρgz | Energy due to elevation — higher fluid has more gravitational potential energy |
The fundamental trade-off: when one form of energy increases, at least one other must decrease to keep the total constant. This is why:
- When flow speeds up (higher V), pressure drops (lower P) — explains aerofoil lift, Venturi effect.
- When fluid rises (higher z), velocity and/or pressure decrease — explains why pumped water slows as it goes uphill.
- When a pipe narrows, velocity increases and pressure decreases — fundamental to flow measurement.
3. Assumptions & Limitations
| Assumption | What It Means | When It Fails |
|---|---|---|
| Steady flow | Properties at any point don’t change with time | Pulsating flow, water hammer, wave motion |
| Incompressible | Constant density throughout | High-speed gas flows (Mach > 0.3), shock waves |
| Inviscid (frictionless) | No energy lost to viscous friction | Long pipes, viscous fluids, boundary layers, turbulence |
| Along a streamline | Equation applies between two points on the same streamline | Cannot compare points on different streamlines (unless irrotational) |
| No shaft work | No pump or turbine between the two points | Systems with pumps, fans, or turbines (use modified Bernoulli) |
Exam tip: Many GATE questions ask you to identify which assumption is violated in a given scenario. Always check whether the problem involves viscous effects, compressibility, or external work — these require modifications to the basic equation.
4. Understanding the Three Pressure Terms
Stagnation Pressure (Total Pressure)
P₀ = P + ½ρV² (for horizontal flow, z = constant)
P₀ = static pressure + dynamic pressure
| Pressure | Symbol | What It Measures | How to Measure |
|---|---|---|---|
| Static pressure | P | Pressure of the fluid at rest relative to the flow | Wall tap (hole flush with pipe wall) |
| Dynamic pressure | ½ρV² | Pressure equivalent of the fluid’s kinetic energy | Cannot be measured directly — calculated from P₀ − P |
| Stagnation (total) pressure | P₀ | Pressure if the fluid were brought to rest isentropically | Pitot tube facing the flow |
5. Application: Venturi Meter
A Venturi meter measures flow rate by exploiting the pressure drop that occurs when fluid accelerates through a constriction. It consists of a converging section, a narrow throat, and a diverging section.
Applying Bernoulli between the inlet (1) and throat (2), and using continuity (A₁V₁ = A₂V₂):
Venturi Meter — Flow Rate
Q = Cd × A₂ × √[2(P₁ − P₂) / (ρ(1 − (A₂/A₁)²))]
Or using pressure head difference Δh:
Q = Cd × A₂ × √[2gΔh / (1 − (A₂/A₁)²)]
Cd = coefficient of discharge (typically 0.95–0.98 for Venturi meters)
The Venturi meter is the most accurate flow measurement device, with very low pressure loss (most of the pressure is recovered in the diverging section).
6. Application: Pitot Tube
A pitot tube measures flow velocity by bringing a small stream of fluid to rest (stagnation) and measuring the resulting pressure increase.
At the stagnation point: V = 0, so all kinetic energy converts to pressure. Bernoulli gives:
Pitot Tube — Velocity
V = √[2(P₀ − P) / ρ]
Where P₀ = stagnation pressure (measured by pitot tube), P = static pressure (measured by wall tap)
A pitot-static tube combines both measurements in a single instrument — the forward-facing hole measures P₀ and side holes measure P. The difference gives the dynamic pressure, from which velocity is calculated.
Applications: aircraft airspeed indicators, wind tunnel measurements, river flow measurement.
7. Application: Orifice & Free Jets
Orifice Meter
An orifice plate is a thin plate with a hole, placed in a pipe. The flow contracts through the orifice, creating a pressure drop proportional to the flow rate. Cheaper than a Venturi but causes more permanent pressure loss.
Q = Cd × Aorifice × √[2ΔP / (ρ(1 − β⁴))]
Where β = dorifice/Dpipe, Cd ≈ 0.6–0.65 for sharp-edged orifice
Torricelli’s Theorem — Free Jet from a Tank
For a liquid draining through a small hole at depth h below the free surface:
Torricelli’s Theorem
V = √(2gh)
The velocity of efflux equals the velocity a body would attain falling freely from height h.
With coefficient of velocity: Vactual = Cv√(2gh), where Cv ≈ 0.97–0.99
8. Modified Bernoulli — With Losses & Pumps
Real systems have friction losses and may include pumps or turbines. The modified Bernoulli equation accounts for these:
Modified Bernoulli Equation (Head Form)
P₁/(ρg) + V₁²/(2g) + z₁ + hpump = P₂/(ρg) + V₂²/(2g) + z₂ + hL + hturbine
Where:
- hpump = head added by a pump (m)
- hturbine = head extracted by a turbine (m)
- hL = total head loss due to friction and fittings (m)
This form connects Bernoulli’s equation to pipe flow analysis (Darcy-Weisbach equation for hL) and is the most practically useful version for real engineering systems.
9. Worked Numerical Examples
Example 1: Horizontal Pipe Narrowing
Problem: Water flows through a horizontal pipe that narrows from 300 mm to 150 mm diameter. Pressure at the wider section is 250 kPa and velocity is 2 m/s. Find the pressure at the narrower section. (ρ = 1000 kg/m³)
Solution
Step 1 — Continuity: A₁V₁ = A₂V₂
V₂ = V₁ × (D₁/D₂)² = 2 × (300/150)² = 2 × 4 = 8 m/s
Step 2 — Bernoulli (horizontal, z₁ = z₂):
P₁ + ½ρV₁² = P₂ + ½ρV₂²
250,000 + ½(1000)(2²) = P₂ + ½(1000)(8²)
250,000 + 2,000 = P₂ + 32,000
P₂ = 220,000 Pa = 220 kPa
Pressure drops by 30 kPa as velocity increases from 2 to 8 m/s.
Example 2: Torricelli — Tank Drainage
Problem: A large open tank contains water to a depth of 5 m. A small orifice is located at the bottom. Find the velocity of efflux and the flow rate if the orifice diameter is 50 mm and Cd = 0.62.
Solution
V = √(2gh) = √(2 × 9.81 × 5) = √(98.1) = 9.9 m/s
A = π(0.05)²/4 = 1.963 × 10⁻³ m²
Q = Cd × A × V = 0.62 × 1.963 × 10⁻³ × 9.9
Q = 0.01205 m³/s = 12.05 L/s
Example 3: Pitot Tube
Problem: A pitot-static tube in an air duct reads a stagnation pressure of 1,050 Pa and a static pressure of 1,000 Pa. Air density is 1.2 kg/m³. Find the air velocity.
Solution
V = √[2(P₀ − P)/ρ] = √[2(1050 − 1000)/1.2] = √[2 × 50/1.2] = √(83.33)
V = 9.13 m/s
10. Common Mistakes Students Make
- Applying Bernoulli without continuity: Most problems require BOTH equations. Bernoulli has two unknowns (P₂ and V₂); continuity provides the second equation. Always use them together.
- Forgetting the assumptions: Applying Bernoulli to viscous pipe flow without accounting for friction gives incorrect pressures. Use the modified Bernoulli equation (with hL) for real pipe systems.
- Using gauge pressure inconsistently: You can use either gauge or absolute pressure in Bernoulli’s equation, but you must be consistent on both sides. Mixing gauge on one side and absolute on the other gives wrong answers.
- Not squaring the velocity ratio: When using continuity with diameters, V₂/V₁ = (D₁/D₂)². Students often forget to square the diameter ratio.
- Ignoring elevation difference: For vertical or inclined flow, the ρgz terms matter. For horizontal flow, they cancel (z₁ = z₂). Omitting them in non-horizontal problems causes significant errors.
11. Frequently Asked Questions
What is Bernoulli’s equation?
Bernoulli’s equation states that for steady, incompressible, inviscid flow along a streamline, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant: P + ½ρV² + ρgz = constant. It is the most important equation in fluid mechanics and forms the basis for analysing flow measurement devices, aerofoil lift, pipe flow, and open channel flow.
What are the assumptions of Bernoulli’s equation?
The five key assumptions are: steady flow, incompressible fluid, inviscid (no friction), flow along a single streamline, and no shaft work (no pumps or turbines). When any of these are violated, the modified Bernoulli equation (with loss terms, pump head, or turbine head) must be used instead.
Why does pressure decrease when velocity increases?
Because total mechanical energy is conserved. When fluid accelerates (higher kinetic energy ½ρV²), the energy must come from somewhere — it comes from the pressure energy P. So pressure drops. This pressure-velocity trade-off is the core of the Venturi effect, aerofoil lift, and many flow measurement techniques. It is not intuitive at first, but it follows directly from energy conservation.