Unit 8.3 | Fluid Flow and the Continuity Equation
- 13 minutes
- 3 formulas
- 11 questions
- 3 videos
Squeeze a hose and the water shoots out faster. The continuity equation, A₁v₁ = A₂v₂, explains why, and solves every flow rate problem on the exam.
Unit 8 Breakdown
In this lesson:
- Volume flow rate, [katex] Q = Av [/katex], and mass flow rate, [katex] \rho A v [/katex]
- The continuity equation, [katex] A_1 v_1 = A_2 v_2 [/katex]: what flows in must flow out
- Why halving a pipe’s diameter makes the fluid four times faster
- The 4 problem types: pipes that narrow or widen, flow rate and filling time, branching pipes, and falling streams
Introduction
This lesson covers fluid flow and the continuity equation. It is the first of two lessons on moving fluids, and it is conservation of mass written for a pipe.
In Unit 8.1 and Unit 8.2 the fluid sat still. Now it moves. Put your thumb over the end of a garden hose and the water shoots out faster. Nothing pushed harder; the same amount of water just has less room to get through.
Flow Rate
The volume flow rate [katex] Q [/katex] is the volume of fluid passing a point each second. In one second the fluid moves a distance [katex] v [/katex], so a slice of volume [katex] Av [/katex] goes by:
[katex] Q = \frac{V}{t} = Av [/katex]
Its unit is m³/s. Multiply by the density and you get the mass flow rate, [katex] \rho A v [/katex], in kg/s. Watch the letters: here [katex] V [/katex] is a volume and [katex] v [/katex] is a speed.
▶ Watch: what changes when a fluid starts to flow (starts at 16:26).
The Continuity Equation
An ideal fluid is incompressible, so it cannot pile up anywhere in the pipe. Every kilogram that enters one end must leave the other, every second:
- Mass in per second = mass out per second: [katex] \rho A_1 v_1 = \rho A_2 v_2 [/katex].
- The density is the same everywhere (incompressible), so it cancels.
[katex] A_1 v_1 = A_2 v_2 [/katex]
Where the pipe is narrow, the fluid is fast. Where it is wide, the fluid is slow. For a round pipe [katex] A = \pi r^2 [/katex], so the speed goes as one over the radius squared:
[katex] \frac{v_2}{v_1} = \frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 [/katex]
▶ Watch: the continuity equation, worked from scratch (starts at 18:13).
Every continuity question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Pipes that narrow or widen | two diameters or radii, a nozzle, an artery | [katex] A_1 v_1 = A_2 v_2 [/katex] with [katex] A = \pi r^2 [/katex] |
| 2. Flow rate and filling time | “how long to fill”, liters per second, “volume per second” | [katex] Q = Av = \frac{V}{t} [/katex] |
| 3. Branching pipes | one pipe splits into two or more | [katex] Av = \sum A_i v_i [/katex] |
| 4. Falling streams | a faucet stream that gets thinner as it falls | Kinematics for [katex] v [/katex], then [katex] A_1 v_1 = A_2 v_2 [/katex] |
Problem Type 1: Pipes that narrow or widen
How to spot it: A pipe, hose or blood vessel changes size, given by diameters, radii or areas: “what is the speed in the narrow section”.
Setup: Write [katex] A_1 v_1 = A_2 v_2 [/katex]. With radii or diameters, use the ratio squared: [katex] v_2 = v_1\left(\frac{d_1}{d_2}\right)^2 [/katex].
Water flows at [katex] 1.5\,\text{m/s} [/katex] through a garden hose [katex] 2.0\,\text{cm} [/katex] across. The nozzle is [katex] 0.50\,\text{cm} [/katex] across. How fast does the water leave the nozzle?
The diameter shrinks by a factor of 4, so the area shrinks by [katex] 4^2 = 16 [/katex]. [katex] v_2 = (1.5)\left(\frac{2.0}{0.50}\right)^2 = (1.5)(16) = 24 \text{ m/s} [/katex]. The units of the diameters cancel, so you do not even need to convert to meters.
AP trap: Scaling the speed with the diameter instead of the area. Half the diameter means a quarter of the area, so four times the speed, not twice.
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Problem Type 2: Flow rate and filling time
How to spot it: A tank, pool, room or bucket fills or drains: “how long”, “liters per second”, “how large must the pipe be”.
Setup: [katex] Q = Av = \frac{V}{t} [/katex]. Convert first: [katex] 1 \text{ L} = 10^{-3} \text{ m}^3 [/katex] and [katex] 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 [/katex].
A garden hose with an inside radius of [katex] 1.0\,\text{cm} [/katex] fills a [katex] 20\,\text{L} [/katex] bucket in [katex] 50\,\text{s} [/katex]. Find the flow rate and the speed of the water in the hose.
[katex] Q = \frac{0.020 \text{ m}^3}{50 \text{ s}} = 4.0 \times 10^{-4} \text{ m}^3\text{/s} [/katex]. The hose area is [katex] \pi(0.010)^2 \approx 3.14 \times 10^{-4} \text{ m}^2 [/katex], so [katex] v = \frac{Q}{A} = \frac{4.0 \times 10^{-4}}{3.14 \times 10^{-4}} \approx 1.3 \text{ m/s} [/katex].
AP trap: Mixing up the volume [katex] V [/katex] with the speed [katex] v [/katex], or leaving liters and cm² unconverted. Most wrong answers here are off by a power of ten.
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Problem Type 3: Branching pipes
How to spot it: One pipe splits into two or more smaller pipes (or several join into one).
Setup: The flow in equals the total flow out: [katex] A v = A_1 v_1 + A_2 v_2 + \dots [/katex]. For identical branches, [katex] Av = n\,A_b v_b [/katex].
A pipe with a cross-sectional area of [katex] 4.0\,\text{cm}^2 [/katex] carries water at [katex] 2.0\,\text{m/s} [/katex]. It splits into two identical branches, each [katex] 1.0\,\text{cm}^2 [/katex]. How fast is the water in each branch?
[katex] (4.0)(2.0) = 2(1.0)\,v_b [/katex], so [katex] v_b = 4.0 \text{ m/s} [/katex]. The two branches together have only half the area of the main pipe, so the water moves twice as fast.
AP trap: Comparing the main pipe to one branch, [katex] A v = A_1 v_1 [/katex]. The flow divides between the branches; compare the main pipe to all of them together.
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Problem Type 4: Falling streams
How to spot it: A stream of water pours from a faucet and gets narrower as it falls.
Setup: The water speeds up as it falls, so the stream must narrow to keep [katex] Av [/katex] the same. Find the speed lower down with kinematics, [katex] v^2 = v_0^2 + 2gh [/katex], then use continuity for the area.
Water leaves a faucet at [katex] 0.50\,\text{m/s} [/katex] through an opening of [katex] 1.0\,\text{cm}^2 [/katex]. What is the area of the stream [katex] 10\,\text{cm} [/katex] below the faucet?
Speed: [katex] v^2 = (0.50)^2 + 2(9.8)(0.10) = 2.21 [/katex], so [katex] v \approx 1.49 \text{ m/s} [/katex]. Area: [katex] A = \frac{(1.0)(0.50)}{1.49} \approx 0.34 \text{ cm}^2 [/katex]. The stream is a third of its starting area.
AP trap: Thinking the stream narrows because of pressure or surface tension. It narrows because it speeds up, and continuity makes the faster flow thinner.
Accounts are free! Log in to try this question, see explanations, save progress, and more!
Lesson 8.4 Preview
When the fluid speeds up in a narrow pipe, its kinetic energy grows. In the final lesson we find where that energy comes from: Bernoulli’s equation, conservation of energy for a flowing fluid.
Accounts are free! Log in to read the rest, try the practice questions, and save your progress.
This article was written by Jason