Bernoulli Equation Calculator
The Bernoulli equation is one of the most important results in fluid mechanics. It states that for an ideal, incompressible fluid flowing steadily along a streamline, the sum of static pressure, dynamic pressure, and hydrostatic pressure remains constant. Written out: P1 plus one half rho v1 squared plus rho g h1 equals P2 plus one half rho v2 squared plus rho g h2. The three terms represent static pressure (force per unit area exerted by the fluid at rest), dynamic pressure (the pressure due to fluid motion, proportional to the square of velocity), and hydrostatic pressure (the pressure due to the fluid's weight above that point). The equation is derived from conservation of energy applied to a fluid element and says that where fluid speeds up it must slow down in pressure, which explains how an aircraft wing generates lift, why a Venturi meter works, and why flow accelerates through a narrowing pipe. This calculator takes the conditions at point 1 (upstream pressure P1 in pascals, flow velocity v1, elevation h1) and the conditions at point 2 (flow velocity v2, elevation h2), along with the fluid density, and solves for the unknown pressure P2 at the downstream point. The default values represent water flowing horizontally: density 1000 kg/m³, atmospheric pressure upstream at 101,325 Pa, velocity increasing from 2 m/s to 4 m/s at the same height. Results assume ideal flow without viscous losses; real systems will have a lower P2 due to pipe friction and fittings losses.
Point 1 (upstream)
Point 2 (downstream)
Assumes ideal (inviscid, incompressible, steady) flow along a single streamline. Real systems have additional friction losses. For pipe networks, apply the Darcy-Weisbach equation for head loss.
How it works
The Bernoulli equation rearranged for P2: P2 = P1 + 0.5 * rho * (v1² - v2²) + rho * g * (h1 - h2). Where g = 9.80665 m/s². Dynamic pressure at each point: q = 0.5 * rho * v². Pressure change = P2 - P1. When v2 exceeds v1 and both heights are equal, P2 is less than P1, which is the classic result: faster flow means lower pressure.
Worked example
Water (rho = 1000 kg/m³) flows horizontally (h1 = h2 = 0 m) through a narrowing pipe. At point 1: P1 = 101,325 Pa, v1 = 2 m/s. At point 2: v2 = 4 m/s. P2 = 101325 + 0.5 times 1000 times (4 - 16) + 0 = 101325 + 0.5 times 1000 times (-12) = 101325 - 6000 = 95,325 Pa = 95.33 kPa. Pressure drop = -6,000 Pa. Dynamic pressure at 1 = 0.5 times 1000 times 4 = 2,000 Pa. Dynamic pressure at 2 = 0.5 times 1000 times 16 = 8,000 Pa.
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