In Flow Rate and Continuity water sped up in a narrow pipe. But speeding up takes a net force. Something must push the water from behind harder than the water ahead pushes back, so the pressure must drop where the water is fast. Bernoulli's equation makes this exact, using conservation of energy.
Why it works
Follow a small parcel of water of volume $\Delta V$ (mass $\rho\,\Delta V$) from point 1 to point 2 in steady flow. The water behind pushes it forward and does work $P_1 \Delta V$; the water ahead pushes back and takes away $P_2 \Delta V$. With no friction, this net work changes the parcel's kinetic and gravitational energy: $$(P_1 - P_2)\Delta V = \tfrac{1}{2}\rho \Delta V (v_2^2 - v_1^2) + \rho \Delta V g (h_2 - h_1).$$ Divide by $\Delta V$ and gather each point's terms on its own side. For still water ($v_1 = v_2 = 0$) it gives back $P_2 = P_1 + \rho g (h_1 - h_2)$, the rule from Pressure and Depth.
Key idea
Bernoulli's equation: along a streamline of steady, frictionless flow of a liquid, $$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}.$$ Each term is energy per cubic meter. At the same height, faster fluid has lower pressure; at the same speed, higher fluid has lower pressure.
Worked examples
Worked example: A narrowing level pipe
Water flows at $2.0\,\text{m/s}$ and $200\,\text{kPa}$ in a level pipe that narrows to half its area. What is the pressure in the narrow part?
- Continuity: $v_2 = 2v_1 = 4.0\,\text{m/s}$.
- Level, so $$P_2 = P_1 - \tfrac{1}{2}\rho(v_2^2 - v_1^2) = 200{,}000 - \tfrac{1}{2}(1000)(16 - 4)\,\text{Pa}$$
- $\tfrac{1}{2}(1000)(12) = 6000\,\text{Pa}$, so $P_2 = 194{,}000\,\text{Pa} = 194\,\text{kPa}$.
Answer: $194\,\text{kPa}$
Worked example: A leak in a tank (Torricelli)
A large open tank has a small hole $1.8\,\text{m}$ below the water surface. How fast does the water squirt out?
- Point 1 is the surface, point 2 the hole. Both are open to the air, so $P_1 = P_2$; the surface barely moves, so $v_1 \approx 0$.
- Bernoulli leaves $\rho g h = \tfrac{1}{2}\rho v_2^2$, so $v_2 = \sqrt{2gh}$.
- $v_2 = \sqrt{2(9.8)(1.8)} = \sqrt{35.3} = 5.94\,\text{m/s}$, as fast as a stone dropped $1.8\,\text{m}$.
Answer: $5.94\,\text{m/s}$
Worked example: Pressure upstairs
Water at $300\,\text{kPa}$ enters a house through a pipe of constant diameter. What is the pressure $10\,\text{m}$ higher, on the third floor, while it flows?
- Same diameter means the same speed, so the $\tfrac{1}{2}\rho v^2$ terms cancel.
- $$P_2 = P_1 - \rho g \Delta h = 300\,\text{kPa} - (1000)(9.8)(10)\,\text{Pa} = 300 - 98 = 202\,\text{kPa}$$
Answer: $202\,\text{kPa}$
Common mistake
Thinking faster fluid pushes harder, so its pressure must be higher. A fast jet does push hard on whatever stops it, but that is the force of stopping its momentum. Inside the moving stream, sideways pressure is lower where it moves faster. Check with forces: water speeds up only if the pressure behind it is higher than the pressure ahead.
Bernoulli's equation leaves out friction (viscosity) and assumes the liquid's density stays fixed. Real pipes lose some pressure to friction, so it gives an upper limit for speeds and a lower limit for pressure drops.