流体连续性方程

AP Physics 1· AP物理1课程考试说明（CED）——流体· 14 分钟阅读

1. 核心概念：定义与推导★★☆☆☆⏱ 4 min

流体连续性方程是质量守恒定律应用于不可压缩流体在管道、血管、河道等流道中定常流动的表述。在AP物理1的考察范围内，我们仅考虑定常流动，即任意固定位置处的流体流速与密度不随时间发生变化。

📘 定义

流体连续性方程（不可压缩流动）

$A_{1}, A_{2}$ = cross-sectional areas; $v_{1}, v_{2}$ = average flow speeds; $Q$ = volume flow rate

对于定常不可压缩流动，管道内没有质量累积，因此体积流量为恒定值。由此可以得到横截面积与流速之间的核心反比关系。

例:

When a pipe narrows, flow speed increases because cross-sectional area decreases.

🔬 推导

目标:

从质量守恒定律推导连续性方程

起点:

定常流动下不可压缩流体的质量守恒（密度 $ρ$ 为常数）

1
在管道内的两个截面1和截面2之间定义一个控制体。在 $Δ t$ 时间内，流入的流体移动的距离为 $v_{1} Δ t$ ，因此流入的流体体积为：
2
$V_1 = A_1 v_1 \Delta t$
3
流入的流体质量为 $m_{1} = ρ V_{1}$ 。同理，从截面2流出的流体质量为：
4
$m_2 = \rho A_2 v_2 \Delta t$
5
对于定常流动，流入质量等于流出质量，因此：
6
$\rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$
7
将等式两侧的常数项 $ρ$ 和 $Δ t$ 约去。

结果:

AP物理1要求掌握的核心连续性方程为：

📐 例题

一根水平管道的半径从0.12 m收窄至0.06 m。若宽段的水流速度为1.0 m/s，求窄段的水流速度为多少？

1
将连续性方程变形求解 $v_{2}$ ，回忆圆的面积公式为 $A = π r^{2}$ ：
2
$v_2 = v_1 \frac{A_1}{A_2} = v_1 \frac{\pi r_1^2}{\pi r_2^2}$
3
约去 $π$ 后代入数值：
4
$\frac{r_1^2}{r_2^2} = \frac{(0.12\ \text{m})^2}{(0.06\ \text{m})^2} = 4$
5
计算得到最终流速：
6
$v_2 = 1.0\ \text{m/s} \times 4 = 4.0\ \text{m/s}$

2. 体积流量计算★★☆☆☆⏱ 4 min

体积流量 $Q$ 的定义为单位时间内通过某一位置的流体体积，国际单位为立方米每秒（ $m^{3} / s$ ）。对于不可压缩流动，即使横截面积发生变化，沿任意流动路径的 $Q = A v$ 均为恒定值。该物理量可用于计算一段时间内输送的流体总体积，也可在已知总流量的情况下求解未知流速。

考试中常用的单位换算：1立方米 = 1000升。将升每分钟换算为 $m^{3} / s$ 时，除以60000即可。在计算面积与流量前，务必确认所有长度单位都已转换为米。

📐 例题

厨房水龙头用15秒接满一个12升的锅，水龙头的供水管内径为1.8 cm，求供水管内水流的平均速度为多少？

1
将所有物理量转换为国际单位：12 L = 0.012 m $^{3}$ ，时间 = 15 s，直径 = 1.8 cm = 0.018 m，因此半径 $r = 0.009$ m。
2
计算体积流量 $Q$ ：
3
$Q = \frac{\text{Total Volume}}{\text{Time}} = \frac{0.012\ \text{m}^3}{15\ \text{s}} = 8.0 \times 10^{-4}\ \text{m}^3/\text{s}$
4
计算管道的横截面积：
5
$A = \pi r^2 = \pi (0.009\ \text{m})^2 \approx 2.54 \times 10^{-4}\ \text{m}^2$
6
将 $Q = A v$ 变形求解流速 $v$ ：
7
$v = \frac{Q}{A} = \frac{8.0 \times 10^{-4}\ \text{m}^3/\text{s}}{2.54 \times 10^{-4}\ \text{m}^2} \approx 3.1\ \text{m/s}$

3. 非圆形截面与分支流动★★★☆☆⏱ 3 min

连续性原理适用于任意流道，而不仅限于圆形管道。无论流道形状如何， $A$ 始终为垂直于流动方向的横截面积。例如，矩形的开放式河道的横截面积等于河道宽度乘以平均水深。

对于一根主管道分为多根更小支管的分支流动场景，核心规则是流入系统的总体积流量等于所有支管流出的体积流量之和。两点关系式 $A_{1} v_{1} = A_{2} v_{2}$ 仅适用于单根无分支的流动路径。

📐 例题

横截面积为 $9.0 \times 1 0^{- 4} m^{2}$ 的主管道以1.2 m/s的流速向住宅输水，之后主管道分为三根完全相同的支管。若每根支管内的流速为1.8 m/s，求每根支管的横截面积为多少？

1
对于三根完全相同的支管，总体积流量守恒给出 $Q_{主} = 3 Q_{支管}$ ，因此 $Q_{支管} = \frac{Q _{主}}{3}$ 。
2
Calculate $Q_{main}$ :
3
$Q_{\text{main}} = A_{\text{main}} v_{\text{main}} = (9.0 \times 10^{-4}\ \text{m}^2)(1.2\ \text{m/s}) = 1.08 \times 10^{-3}\ \text{m}^3/\text{s}$
4
求 $Q_{支管}$ ：
5
$Q_{\text{branch}} = \frac{1.08 \times 10^{-3}\ \text{m}^3/\text{s}}{3} = 3.6 \times 10^{-4}\ \text{m}^3/\text{s}$
6
整理公式求解 $A_{支管}$ ：
7
$A_{\text{branch}} = \frac{Q_{\text{branch}}}{v_{\text{branch}}} = \frac{3.6 \times 10^{-4}\ \text{m}^3/\text{s}}{1.8\ \text{m/s}} = 2.0 \times 10^{-4}\ \text{m}^2$

4. AP 风格概念检测★★★☆☆⏱ 3 min

✓ 快速检测

请通过以下AP风格题目检验你的理解程度：

A circular garden hose narrows from diameter 3 cm to diameter 1.5 cm at the nozzle. What is the ratio of the flow speed in the nozzle to the flow speed in the main hose?
- 1:2
- 2:1
- 4:1
- 1:4
显示答案
4:1 —
Area scales with the square of diameter, so $\frac{v _{2}}{v _{1}} = (\frac{d _{1}}{d _{2}})^{2} = 2^{2} = 4$ , so the ratio is 4:1. The most common error is forgetting the square relationship.
A mountain river is 15 m wide and 2.0 m deep on average, with an average flow speed of 1.2 m/s. The river flows through a narrow canyon that is only 3.0 m wide and 1.5 m deep. What is the average flow speed in the canyon?
显示答案
8.0 m/s —
Cross-sectional area of the open section is $15 \times 2 = 30 m^{2}$ , canyon area is $3 \times 1.5 = 4.5 m^{2}$ . $v_{2} = 1.2 \times \frac{30}{4.5} = 8.0 m/s$ .

5. 常见陷阱

错误做法:

Using diameter instead of area, or forgetting to square the diameter when calculating the speed ratio

原因:

Students mix up linear pipe dimension and area, which scales with the square of linear size

正确做法:

Always explicitly write $\frac{A _{1}}{A _{2}} = (\frac{d _{1}}{d _{2}})^{2}$ before plugging in values for circular pipes

错误做法:

Applying $A_{1} v_{1} = A_{2} v_{2}$ to one main pipe and one branch when the main pipe splits into two branches

原因:

Students memorize the two-point continuity equation and forget it only applies to a single unbranched flow path

正确做法:

For any branched system, write total inflow equals the sum of all outflows, adding $Q$ for each branch

错误做法:

Leaving length units in centimeters when calculating area, leading to speed values off by a factor of 10,000

原因:

Problems often give pipe diameter in centimeters for convenience, and students skip unit conversion

正确做法:

Circle all length units in the problem statement, and convert every length to meters before starting calculations

错误做法:

Assuming doubling the diameter of a pipe doubles the flow speed for constant $Q$

原因:

Students confuse linear proportionality with area proportionality

正确做法:

Remember that speed is inversely proportional to the square of diameter, so doubling diameter quarters the speed for constant flow rate

错误做法:

Using the simplified continuity equation for compressible flow and assuming $Q$ is constant

原因:

Students forget the simplified equation only holds for constant density

正确做法:

On AP Physics 1, all fluids tested for continuity are incompressible, so the simplified form always applies unless the problem explicitly states density changes

6. 速查表

Category	Formula	Notes
Core Continuity Equation	$A_{1} v_{1} = A_{2} v_{2}$	Steady, incompressible unbranched flow
Volume Flow Rate	$Q = A v$	Units: $m^{3} / s$ ; constant for incompressible flow
Branched Flow	$Q_{in} = \sum Q_{out}$	Total inflow equals sum of all outflows
Area (Circle)	$A = π r^{2} = \frac{π d ^{2}}{4}$	Area scales with square of linear dimension
Speed-Diameter Proportionality	$v \propto \frac{1}{d ^{2}}$	Inverse square relation for constant $Q$
Total Volume Over Time	$V_{total} = Qt$	Calculate total fluid delivered over time $t$
Non-Circular (River)	$A = width \times depth$	Rectangular cross-section for open channels

下一步

The fluid continuity equation is the foundational prerequisite for Bernoulli's equation, which connects flow speed to fluid pressure and gravitational potential energy in moving fluids. Almost all AP Physics 1 free-response problems on fluids combine continuity and Bernoulli's equation, so you cannot solve these full problems if you cannot correctly apply continuity to find flow speed at different points. Beyond fluids, the core idea of continuity (conservation of a quantity through steady flow) reappears in other AP Physics 1 topics, including conservation of charge in electric circuits. Next, you will use flow speeds from continuity to solve for pressure changes in moving fluids.

→Bernoulli's Principle