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Continuity Equation and Flow Rate

Volume flow rate, mass flow rate, and the continuity equation for incompressible fluids

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🌊 Continuity Equation

For an incompressible fluid in steady flow, the same mass passes every cross-section per second.

Volume Flow Rate

Q=AvQ = A v

  • QQ = volume flow rate (m³/s)
  • AA = cross-sectional area (m²)
  • vv = average flow speed (m/s)

Continuity Equation

A1v1=A2v2A_1 v_1 = A_2 v_2

💡 Key Idea: Where the pipe is narrow, the fluid moves faster. Where the pipe is wide, it moves slower. (Think of squeezing a hose.)

For compressible flow: ρ1A1v1=ρ2A2v2\rho_1 A_1 v_1 = \rho_2 A_2 v_2 (mass flow rate). AP Physics 1 only deals with the incompressible case.


Where Continuity Comes From

It's just conservation of mass for an incompressible fluid: in a time Δt\Delta t, the volume entering at section 1 (A1v1ΔtA_1 v_1 \Delta t) must equal the volume leaving at section 2 (A2v2ΔtA_2 v_2 \Delta t).


⚠️ Common Mistakes

  1. Using diameter instead of area. Area scales as the square of radius (or diameter), so halving the diameter quadruples the speed.
  2. Adding speeds at a junction. When two pipes merge, flow rates (AvAv) add, not speeds.
  3. Confusing flow speed with mass flow. Speed depends on cross-section; mass flow rate is constant along a streamline.

Key Formulas

QuantityFormulaUnits
Volume flow rateQ=AvQ = Avm³/s
Continuity (incompressible)A1v1=A2v2A_1 v_1 = A_2 v_2—
Mass flow ratem˙=ρAv\dot m = \rho A vkg/s

📚 Practice Problems

1Problem 1easy

❓ Question:

Water flows at 2.0 m/s through a horizontal pipe of cross-sectional area 4.0×10−44.0\times10^{-4} m². The pipe narrows to area 1.0×10−41.0\times10^{-4} m². Find the speed in the narrow section.

💡 Show Solution

Continuity: A1v1=A2v2A_1 v_1 = A_2 v_2.

v2=v1A1A2=2.0⋅4.0×10−41.0×10−4=8.0 m/s.v_2 = v_1 \frac{A_1}{A_2} = 2.0 \cdot \frac{4.0\times10^{-4}}{1.0\times10^{-4}} = 8.0\text{ m/s}.

2Problem 2medium

❓ Question:

A garden hose with inner diameter 2.0 cm carries water at 1.5 m/s. What is the volume flow rate (in L/s)?

💡 Show Solution

Radius: r=0.010r = 0.010 m → area A=πr2=π(0.010)2≈3.14×10−4A = \pi r^2 = \pi(0.010)^2 \approx 3.14\times10^{-4} m².

Q=Av=3.14×10−4⋅1.5≈4.71×10−4 m3/s=0.47 L/s.Q = Av = 3.14\times10^{-4} \cdot 1.5 \approx 4.71\times10^{-4}\text{ m}^3/\text{s} = 0.47\text{ L/s}.

3Problem 3medium

❓ Question:

A pipe of radius 6 cm narrows to radius 2 cm. By what factor does the fluid speed change?

💡 Show Solution

A∝r2A \propto r^2, so A1A2=(6/2)2=9\frac{A_1}{A_2} = (6/2)^2 = 9.

By continuity, v2=v1⋅9v_2 = v_1 \cdot 9 — the fluid speeds up by a factor of 9.

Explain using:

📌 Related Topics in Fluids

❓ Frequently Asked Questions

What is Continuity Equation and Flow Rate?▾
Volume flow rate, mass flow rate, and the continuity equation for incompressible fluids
How can I study Continuity Equation and Flow Rate effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Continuity Equation and Flow Rate study guide free?▾
Yes — all study notes, flashcards, and practice problems for Continuity Equation and Flow Rate on Study Mondo are free to access. No account is needed.
What course covers Continuity Equation and Flow Rate?▾
Continuity Equation and Flow Rate is part of the AP Physics 1 course on Study Mondo, specifically in the Fluids section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Continuity Equation and Flow Rate?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.