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Bernoulli's Equation Solver

Solve Bernoulli's fluid dynamics equation for unknown pressure, velocity, or elevation at any point in a pipe or flow system. Includes continuity equation for area changes.

Solve for unknown fluid dynamics across varying pipe diameters, pressures, and elevations using aerodynamic energy balance equations.

kg/m³

Location 1

Pa
m/s
m

Location 2

Pa
m/s
m

Solved Output

v₂ (Velocity at Point 2)

Calculation Result

10.1980
m/s
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What is The Conservation of Fluid Energy?

Bernoulli's principle states that for an inviscid flow of a non-conducting fluid, an increase in the speed of the fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. It is essentially the law of conservation of energy applied to a steadily flowing fluid.

Mathematical Foundation

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2=ConstantP_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2 = \text{Constant}
P1,P2P_1, P_2= Static Pressure energy exerted by the fluid.
12ρv2\frac{1}{2}\rho v^2= Dynamic Pressure (Kinetic Energy per unit volume).
ρgh\rho g h= Hydrostatic Pressure (Potential Energy via gravity).

Laws & Principles

  • Incompressible fluids only: The standard Bernoulli equation is mathematically invalid for compressible gases (like high-speed air or shockwaves) unless the Mach number is less than 0.3. It strictly relies on the density $\rho$ remaining completely constant throughout the pipe.
  • No Friction/Viscosity: The equation assumes the fluid slides perfectly smooth without generating heat through wall friction. In real plumbing, pressure will steadily drop due to viscous losses over long distances, requiring the complex Darcy-Weisbach addition.
  • Airfoil Lift Illusion: While Bernoulli's equation is often used to explain airplane wings (faster air over the top = lower pressure), this explanation is highly incomplete. Airplanes fly primarily by pushing huge masses of air downward (Newton's 3rd Law of action/reaction).

Step-by-Step Example Walkthrough

" Water (ρ = 1000 kg/m³) flows steadily at 2 m/s through a wide pipe at 150,000 Pa. The pipe narrows and goes up a 3-meter hill, dropping pressure to 110,000 Pa. What is the new velocity? "

  1. Identify P1=150k, v1=2, h1=0. Identify P2=110k, h2=3. Solve for v2.
  2. Calculate P1 Side (C1): 150,000 + 0.5(1000)(2^2) + 0 = 152,000 Joules/m³.
  3. Set P2 side equal to C1: 110,000 + 0.5(1000)v2^2 + 1000(9.81)(3) = 152,000.
  4. Rearrange: 110,000 + 500(v2^2) + 29,430 = 152,000.
  5. Solve: 500(v2^2) = 12,570. v2^2 = 25.14.
Final Result: Taking the square root, v2 = 5.014 m/s. The water accelerated heavily to compensate for the narrower pipe and pressure differential.

What is Bernoulli's Equation & Fluid Energy Balance?

Bernoulli's Equation is the energy conservation law for steady, incompressible, inviscid fluid flow along a streamline. It states that the total mechanical energy per unit volume (pressure energy + kinetic energy + potential energy) remains constant. It is the foundation of Venturi meters, Pitot tubes, nozzle design, airfoil lift theory, and pipe flow analysis across every engineering discipline.

Mathematical Foundation

Bernoulli's Equation
P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2
PP= Static pressure at point 1 or 2 (Pa or psi). The force per unit area exerted by the fluid perpendicular to a surface.
ρ\rho= Fluid density (kg/m³). For water: 1000 kg/m³; for air at sea level: 1.225 kg/m³.
vv= Fluid velocity at point 1 or 2 (m/s). The mean flow velocity at that cross-section.
gg= Gravitational acceleration (9.81 m/s² standard).
hh= Elevation of the point above a reference datum (m). Higher elevation = higher potential energy = lower pressure.
Continuity Equation (Incompressible Flow)
A1v1=A2v2⇒v2=v1⋅A1A2A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad v_2 = v_1 \cdot \frac{A_1}{A_2}
AA= Cross-sectional area of the pipe at point 1 or 2 (m²). For a circular pipe: A = πr².
vv= Mean velocity at each cross-section. Velocity increases as cross-section decreases (the Venturi effect).

Laws & Principles

  • Bernoulli's equation applies only to: (1) steady flow (no time-varying conditions), (2) incompressible flow (density constant, valid for liquids and low-speed gases below Mach 0.3), (3) inviscid flow (frictionless), (4) flow along a single streamline.
  • Cavitation occurs when static pressure P drops below the fluid's vapor pressure (2,337 Pa for water at 20°C). Venturi meters, pump inlets, and control valve orifices are the most common cavitation sites. Always verify P2 > Pvapor after solving.
  • The Venturi principle: where flow is constricted (A decreases), velocity increases and pressure decreases. This is the operating principle of carburetors, Venturi meters, aspirators, and aircraft wings.
  • For real pipe flow, add a head loss term: P1 + ½pv1² + pgh1 = P2 + ½pv2² + pgh2 + hL, where hL = Darcy-Weisbach friction losses (f × L/D × ½v²/g). Bernoulli alone overestimates outlet pressure in real systems.

Step-by-Step Example Walkthrough

" A water pipe narrows from 4-inch diameter (Point 1) to 2-inch diameter (Point 2) at the same elevation. Water enters at 5 ft/s and 30 psi. What is the pressure at Point 2? "

  1. Convert: D1 = 0.333 ft, D2 = 0.167 ft, v1 = 5 ft/s, P1 = 30 psi = 4,320 lb/ft²
  2. Continuity: A1/A2 = (D1/D2)² = (0.333/0.167)² = 4.0
  3. v2 = v1 × (A1/A2) = 5 × 4.0 = 20 ft/s
  4. Bernoulli (h1 = h2, same elevation): P2 = P1 + ½ρ(v1² - v2²)
  5. P2 = 4,320 + ½(1.94)(25 - 400) = 4,320 + 0.97(-375) = 4,320 - 364 = 3,956 lb/ft²
Final Result: P2 = 3,956 lb/ft² = 27.5 psi. The 4:1 area reduction increased velocity from 5 to 20 ft/s and dropped pressure by 2.5 psi. This pressure drop is the basis of Venturi flow metering: measure ΔP, use Bernoulli + continuity to back-calculate Q.

Quick Answer: How do you use Bernoulli's Equation to solve for unknown pressure or velocity?

Bernoulli's Equation: P1 + ½ρv1² + ρgh1 = P2 + ½ρv2² + ρgh2. The equation states that the sum of pressure energy, kinetic energy, and potential energy is constant along a streamline. To use it: know 5 of the 6 variables (P1, v1, h1, P2, v2, h2) and solve for the 6th. Worked example — water pipe narrows from 4″ to 2″ diameter at the same elevation, entering at 5 ft/s and 30 psi: Continuity gives v2 = 5 × (4/1)² = 20 ft/s. Bernoulli gives P2 = 30 psi − ½ρ(v2²−v1²) = 27.5 psi. The 4:1 area ratio quadrupled velocity and dropped pressure by 2.5 psi — this pressure drop is exactly how Venturi meters measure flow rate.

Common Fluid Densities & Vapor Pressures for Bernoulli Calculations

Density (ρ) is the most critical input after pressure and velocity. Use actual process temperature — water density drops from 1000 to 983 kg/m³ between 4°C and 60°C, shifting your kinetic energy term by 1.7%. Vapor pressure matters for cavitation analysis at pump inlets and Venturi throats.

Fluid Density (kg/m³) Density (slug/ft³) Vapor Pressure at 20°C
Water (20°C)998 kg/m³1.936 slug/ft³2,337 Pa (0.339 psi)
Water (60°C)983 kg/m³1.907 slug/ft³19,940 Pa (2.89 psi) — cavitation risk rises sharply
Air (sea level, 20°C)1.204 kg/m³0.00234 slug/ft³N/A (gas, no vapor pressure)
Air (2,000 m altitude)1.007 kg/m³0.00195 slug/ft³N/A (16% less dense than sea level)
Gasoline720–750 kg/m³1.40–1.45 slug/ft³~55,000 Pa (8 psi) — very high cavitation risk
Hydraulic Oil (ISO 32)860–880 kg/m³1.67–1.71 slug/ft³< 1 Pa (essentially zero)
For air and compressible gases at velocities above Mach 0.3 (~102 m/s or 335 ft/s), Bernoulli's incompressible form introduces significant error. Use the compressible isentropic flow equations instead. Below Mach 0.3, the incompressible Bernoulli error is less than 0.5% and can be safely ignored.

Real-World Bernoulli Applications

Application Which Terms Dominate Key Relationship
Venturi Flow MeterP and v — same elevationQ = A2√(2ΔP / ρ(1−(A2/A1)²)) — measure ΔP, calculate volumetric flow
Pitot Tube (Airspeed)P and v — same elevationv = √(2ΔP/ρ) — stagnation pressure minus static pressure gives dynamic pressure
Water Tank Drainh and v — P = atm both endsv = √(2gh) (Torricelli's theorem) — drain velocity depends only on head height
Pump Pressure RiseP and h — v often equals at suction/dischargeP2 = P1 + ρgHpump − ρgΔh — add head loss term for real systems
Airfoil Lift (simplified)P and v — same elevation top/bottomvupper > vlower → Pupper < Plower → net upward lift force
Note: the airfoil lift calculation using Bernoulli alone (circulation not included) gives the correct pressure distribution but requires the Kutta-Joukowski theorem to correctly sum total lift force. Bernoulli describes why faster-moving air has lower pressure — but the reason the upper surface has faster air is due to circulation, not simply the longer path length (the “longer path” explanation is a common misconception in textbooks).

Pro Tips & Common Bernoulli Mistakes

Do This

  • ✓Always apply the continuity equation first to find the unknown velocity before using Bernoulli. If you know pipe diameters at both points and one velocity, the continuity equation (A1v1 = A2v2) gives you the second velocity for free. Then Bernoulli has only one unknown. Skipping continuity and trying to solve Bernoulli with two unknowns requires additional information that may not be available.
  • ✓Check for cavitation after solving: verify P2 > vapor pressure at operating temperature. At a Venturi throat, pump suction, or valve orifice, the calculated static pressure can go below the fluid's vapor pressure — meaning the liquid will flash to vapor. For water at 20°C, Pvapor = 2,337 Pa (absolute). If your Bernoulli result gives a static pressure below this, the flow will cavitate and the actual flow rate will be significantly lower than calculated.

Avoid This

  • ✗Don't apply the ideal Bernoulli equation to long pipe runs without adding head loss. Bernoulli assumes frictionless flow. Real pipes have viscous friction losses (Darcy-Weisbach: hL = f × (L/D) × v²/(2g)) that reduce outlet pressure. For a 100-ft pipe at moderate velocity, friction losses can easily exceed 5–10 psi — more than the static head or kinetic energy terms you're trying to calculate. The extended Bernoulli (sometimes called the energy equation) adds hL to the right-hand side. Always include friction when pipe length is more than ∼20 pipe diameters.
  • ✗Don't confuse static pressure, dynamic pressure, and total pressure. Static pressure (P): what a pressure gauge reads — the force per unit area on a surface parallel to flow. Dynamic pressure (½ρv²): the pressure rise when flow is brought to rest (stagnation). Total (stagnation) pressure: P + ½ρv² — what a Pitot tube reads. Bernoulli says total pressure is constant (for inviscid flow). A pressure gauge on a pipe wall reads static pressure — not total pressure.

Frequently Asked Questions

What are the assumptions of Bernoulli's equation and when does it break down?

The four assumptions are: 1) Steady flow (flow conditions don't change with time at any point). 2) Incompressible flow (density is constant — valid for liquids and gases below Mach 0.3). 3) Inviscid (frictionless) flow (no viscous losses along the streamline). 4) Along a streamline (you cannot apply Bernoulli between two points that are not on the same streamline unless the flow is also irrotational). It breaks down in: long pipe runs with significant friction; turbulent mixing zones; flow through pumps, fans, or turbines (which add/remove energy); high-speed compressible gas flow; and near flow separations where streamlines diverge. For real pipe networks, use the extended Bernoulli equation (energy equation) which adds pump head and subtracts head loss terms.

How does Bernoulli's equation explain the Venturi effect and flow metering?

When a pipe narrows (Venturi throat), the continuity equation forces velocity to increase: v2 = v1(A1/A2). Then Bernoulli says the increased kinetic energy (½ρv2²) must come at the expense of pressure: P2 = P1 − ½ρ(v2²−v1²). The resulting pressure difference ΔP = P1−P2 is measurable with a differential pressure gauge. Solving for volumetric flow rate: Q = A2√(2ΔP / ρ(1−(A2/A1)²)). In practice, multiply by a discharge coefficient Cd (typically 0.95–0.98 for well-designed Venturi meters) to correct for real-fluid friction losses at the throat. This is the most accurate and lowest-pressure-loss flow measurement method available for large pipes — which is why Venturi meters are standard on major water mains and industrial process lines.

What is cavitation and how do I check for it using Bernoulli?

Cavitation occurs when the local static pressure drops below the fluid's vapor pressure, causing the liquid to vaporize and form vapor bubbles. These bubbles collapse violently when they move to higher-pressure zones downstream, causing intense localized pressure spikes (>10,000 psi) that erode metal surfaces within hours. To check: solve Bernoulli for P2 at the minimum-pressure point (usually the throat or pump suction), then compare to Pvapor at operating temperature. For water: Pvapor = 2,337 Pa at 20°C; 19,940 Pa at 60°C; 101,325 Pa at 100°C (boiling). If P2 (absolute) < Pvapor, cavitation will occur. Fixes: increase upstream pressure, reduce flow velocity, increase throat/orifice area, lower fluid temperature, or use a back-pressure valve to raise system pressure. Cavitation is the leading cause of premature pump failure in process industries.

Is Bernoulli's equation valid for air and gas flow?

Yes, for low-speed airflow (below Mach 0.3 — approximately 102 m/s or 335 ft/s at sea level). At these speeds, air compressibility is negligible and the density can be treated as constant. This covers the vast majority of HVAC duct flows, wind tunnel speeds below ∼70 m/s, low-speed aerodynamics (cars, buildings, drones), and most industrial ventilation. For speeds above Mach 0.3, compressibility effects become significant and the standard Bernoulli equation systematically overpredicts dynamic pressure. At Mach 0.5, the error is about 6%; at Mach 1.0, it is completely invalid. For compressible flow, use the isentropic flow relations: P0/P = (1 + (γ−1)/2 × M²)γ/(γ−1) where γ = 1.4 for air and M is the local Mach number.

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