AP Physics 1 · Lesson 3

Fluid flow: continuity and Bernoulli’s equation

Know Know it

An ideal fluid cannot be compressed, has no viscosity and flows steadily. These are the assumptions behind both flow equations.

The volume flow rate is Q=AvQ = Av. In one pipe with no leaks, the same volume must pass every cross section each second, so A1v1=A2v2A_1v_1 = A_2v_2 (the continuity equation): the fluid moves faster where the pipe is narrower. For a round pipe, the area goes as the diameter squared.

Bernoulli’s equation is energy conservation for each cubic meter of fluid along the flow: P1+12ρv12+ρgy1=P2+12ρv22+ρgy2P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2. At one height, where the fluid moves faster its pressure is lower.

From a wide section to a narrow section of a level pipe
QuantityIn the narrow section, compared with the wide one
Volume flow ratethe same
Speedgreater
Pressurelower
Kinetic energy per cubic metergreater

Worked example Easy

Blood flows at 0.25 m/s through a section of artery with a cross-sectional area of 0.90 cm². Farther along, a narrowing reduces the area to 0.30 cm². Treat the blood as an ideal fluid. How fast does it move through the narrowed part?

  1. 0.083 m/s
  2. 2.3 m/s
  3. 0.75 m/sAnswer
  4. 0.25 m/s

How to solve it

  1. Continuity: A1v1=A2v2A_1v_1 = A_2v_2.
  2. v2=(0.25)(0.90)/(0.30)=0.75v_2 = (0.25)(0.90)/(0.30) = 0.75 m/s, three times faster through one-third of the area.

Why each choice is right or wrong

  • A. Incorrect. 0.083 m/s multiplies by the area ratio upside down; the blood must speed up, not slow down, in the narrower part.
  • B. Incorrect. 2.3 m/s squares the ratio of 3, as if 0.90 and 0.30 were diameters; they are already areas.
  • C. Correct. v2=v1A1/A2=(0.25)(0.90/0.30)=0.75v_2 = v_1A_1/A_2 = (0.25)(0.90/0.30) = 0.75 m/s.
  • D. Incorrect. 0.25 m/s keeps the same speed, but the same volume each second through a third of the area needs three times the speed.

Trap Narrow means faster and at lower pressure

It feels as if squeezing fluid into a narrow section should raise its pressure. For steady flow in a level pipe it is the opposite: the fluid speeds up as it enters the narrow part, the push that speeds it up comes from higher pressure behind it, and so the pressure in the narrow part is lower.

Explain Explain it

Task verbs: explain, calculate

Work flow problems in two steps: use continuity to find any unknown speed, then put the speeds, heights and the known pressure into Bernoulli’s equation. Keep every term in pascals (which are also joules per cubic meter).

A fluid at rest is a special case: with v = 0 everywhere, Bernoulli’s equation becomes P1−P2=ρg(y2−y1)P_1 - P_2 = \rho g(y_2 - y_1), the same rule as P=P0+ρghP = P_0 + \rho g h. For water leaving a small hole a depth h below the open surface of a wide tank, the same equation gives v=2ghv = \sqrt{2gh}.

Worked example Medium

Water flows through a level pipe. In the wide part it moves at 0.80 m/s at a pressure of 1.80 × 10⁵ Pa. In a narrow part it moves at 2.6 m/s. Treat the water as an ideal fluid with density 1000 kg/m³. What is the pressure in the narrow part?

  1. 1.77 × 10⁵ PaAnswer
  2. 1.83 × 10⁵ Pa
  3. 1.74 × 10⁵ Pa
  4. 1.80 × 10⁵ Pa

How to solve it

  1. Same height, so P1+12ρv12=P2+12ρv22P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2.
  2. 12ρ(v22−v12)=500(2.62−0.802)=500(6.12)=3060\frac{1}{2}\rho(v_2^2 - v_1^2) = 500(2.6^2 - 0.80^2) = 500(6.12) = 3060 Pa.
  3. P2=1.80×105−3060=1.77×105P_2 = 1.80 \times 10^5 - 3060 = 1.77 \times 10^5 Pa.

Why each choice is right or wrong

  • A. Correct. P2=P1−12ρ(v22−v12)=1.80×105−500(6.76−0.64)=1.80×105−3060=1.77×105P_2 = P_1 - \frac{1}{2}\rho(v_2^2 - v_1^2) = 1.80 \times 10^5 - 500(6.76 - 0.64) = 1.80 \times 10^5 - 3060 = 1.77 \times 10^5 Pa.
  • B. Incorrect. 1.83 × 10⁵ Pa adds the 3060 Pa; where the water moves faster, its pressure is lower, not higher.
  • C. Incorrect. 1.74 × 10⁵ Pa leaves out the one-half in 12ρv2\frac{1}{2}\rho v^2, subtracting 6120 Pa instead of 3060 Pa.
  • D. Incorrect. 1.80 × 10⁵ Pa assumes the pressure is the same along a level pipe, but a change in speed means a change in pressure.

Explain Your turn

During a windstorm, air rushes quickly across the flat roof of a closed house, while the air inside the house is still. Using Bernoulli’s principle, explain why the wind can push the roof upward.

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Compare with a model answer

A full answer does this

  • Says the fast-moving air above the roof is at a lower pressure
  • Says the higher pressure of the still air inside pushes up harder than the air above pushes down, giving a net upward force

Model answer

By Bernoulli’s principle, air that moves faster is at lower pressure, so the fast wind across the roof is at a lower pressure than still air. The air inside the closed house is still, so it stays at a higher pressure. It pushes up on the underside of the roof harder than the moving air pushes down on the top, so there is a net upward force that can lift the roof.

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