Bernoulli's principle · Physics Class 11–12 | StemPy Lab
Course 9.2·Lesson 2 of 3
Bernoulli's principle
Bernoulli's principle: along a streamline of an ideal fluid in steady flow, p+21ρv2+ρgh
Lesson 2 of 3Show all lessonsHide
Try the code and the simulations beside the lesson.
Energy in a moving fluid
As water speeds up into a narrow pipe, something must push it forward. That push comes from a pressure difference: the pressure behind is higher than in front. So where a fluid moves faster, its pressure is lower.
Daniel Bernoulli put this as a law of energy. For an ideal fluid (no viscosity, no change of volume) in steady flow, along one streamline:
p+21ρv2+ρgh=the same everywhere
Each term is an energy per unit volume, in J/m3, which is the pascal:
p, the work the pressure can do;
21ρv2, the kinetic energy;
ρgh, the potential energy.
A pipe that is narrow and low on the left, and rises to a wider section on the right. On the left, the height is h one, the pressure is p one and the speed is v one. On the right, the height is h two, the pressure is p two and the speed is v two. The sum of p plus one half rho v squared plus rho g h is the same at both places, along a streamline.
You should find that both speeding up and climbing lower the pressure, and that each metre of height costs about 10kPa.
Two special cases
A level pipe (h the same): p1−p2=21ρ(v22−v12).
A fluid at rest (v=0): p1−p2=ρg(h2−h1), the pressure-depth law of the first course.
Watch out. Bernoulli's equation needs the difference of the squares, v22−v12, never (v2−v1)2.
A worked example
Water in a main is at 2.8×105Pa, moving at 1.2m/s at ground level. The main rises 6.0m and its diameter halves. Find the pressure up there.
Speed. Half the diameter is a quarter of the area: v2=4×1.2m/s=4.8m/s.