Urban Drainage: A-Level Maths Modelling & Calculus | 城市排水：A-Level数学建模与微积分

📚 Urban Drainage: A-Level Maths Modelling & Calculus | 城市排水：A-Level数学建模与微积分

Urban drainage is a rich applied context for A-Level Mathematics. It brings together differential equations, integration, related rates, numerical methods, optimisation and statistical modelling. This article works through the key ideas you need for Edexcel-style questions.

城市排水是 A-Level 数学中一个丰富的应用情境。它把微分方程、积分、相关变化率、数值方法、优化和统计建模结合在一起。本文梳理 Edexcel 风格题目所需的核心概念。

1. Understanding the problem: inflows and outflows | 理解问题：流入与流出

In an urban drainage problem, we usually model a storage basin, drain, or flood tank. The main variable is the volume of water V(t) in the system at time t. The rate of change of volume is the difference between inflow rate and outflow rate.

在城市排水问题中，我们通常模拟蓄水池、排水管或洪水调蓄池。主要变量是系统中 t 时刻的水体积 V(t)。体积的变化率等于流入速率与流出速率之差。

dV/dt = R_in(t) − R_out(t)

Inflow usually comes from rainfall or surface runoff, so R_in(t) may be a constant, a linear function, or an exponential peak after a storm. Outflow depends on drainage capacity, often modelled as a constant pump rate or as a function of water depth.

流入通常来自降雨或地表径流，因此 R_in(t) 可能是常数、线性函数或暴雨后的指数峰值。流出取决于排水能力，通常模拟为恒定泵送速率或水深的函数。

2. Differential equation for water volume | 水体积的微分方程

If inflow is a constant a and outflow is proportional to the volume already stored, say kV, we obtain a first-order linear differential equation:

如果流入为常数 a，流出与已储存体积成正比，即 kV，我们得到一个一阶线性微分方程：

dV/dt = a − kV

This equation can be solved by separating variables or by using an integrating factor. The general solution approaches the equilibrium volume a/k as t becomes large, which represents the long-term storage balance.

该方程可通过分离变量法或积分因子法求解。当 t 增大时，通解趋近平衡体积 a/k，代表长期储水平衡。

For example, with a = 50 m³/min, k = 0.1 min⁻¹, and V(0) = 100 m³, the solution is V(t) = 500 − 400e^(−0.1t).

例如，若 a = 50 m³/min，k = 0.1 min⁻¹，且 V(0) = 100 m³，则解为 V(t) = 500 − 400e^(−0.1t)。

3. Separable differential equations in drainage | 排水中的可分离变量微分方程

Many real drainage systems have outflow that depends on √h, where h is water depth. Since V = A h for a tank of constant cross-sectional area A, we can write:

许多实际排水系统的出流速率与 √h 有关，其中 h 为水深。对于横截面积恒定的水池，V = A h，因此可以写成：

A dh/dt = −k√h

This is separable: bring all h terms to one side and all t terms to the other. Integrating both sides gives an implicit relationship between h and t, which can then be solved for emptying time.

这是可分离变量方程：将所有含 h 的项移到一边，所有含 t 的项移到另一边。两边积分后可得到 h 与 t 的隐式关系，进而求解排空时间。

In an exam, always show the separation step clearly because method marks are awarded for correct rearrangement before integration.

考试中务必清晰展示分离变量步骤，因为正确移项后再积分可以获得方法分。

4. Integration to find total volume | 积分求总排水量

If the outflow rate Q(t) is known as a function of time, the total volume drained over the interval [a,b] is the definite integral:

如果出流速率 Q(t) 是时间的已知函数，则在区间 [a,b] 内排出的总水量为定积分：

Total volume = ∫ₐᵇ Q(t) dt

For example, if Q(t) = 20e^(−0.5t) m³/min, the volume drained in the first 10 minutes is ∫₀¹⁰ 20e^(−0.5t) dt = 40(1 − e^(−5)) m³.

例如，若 Q(t) = 20e^(−0.5t) m³/min，则前 10 分钟排出的水量为 ∫₀¹⁰ 20e^(−0.5t) dt = 40(1 − e^(−5)) m³。

Using a calculator in Edexcel papers, you should be able to evaluate definite integrals to 3 significant figures unless the question states otherwise.

在 Edexcel 考试中，除非题目另有说明，应能使用计算器将定积分保留 3 位有效数字。

5. Related rates: water depth and outflow | 相关变化率：水深与出流

Related rates problems connect two or more changing quantities through differentiation with respect to time. In drainage, we often relate the water depth h to the outflow rate.

相关变化率问题通过对时间求导将两个或多个变化的量联系起来。在排水问题中，我们常常需要联系水深 h 与出流速率。

If the volume is V = A h with constant cross-sectional area A, then differentiating gives dV/dt = A dh/dt. If we know dV/dt from the flow balance, we can find dh/dt.

如果体积为 V = A h，且横截面积 A 恒定，则求导得到 dV/dt = A dh/dt。若已知流量平衡给出的 dV/dt，就可以求出 dh/dt。

dh/dt = (R_in(t) − R_out(t)) / A

For example, if R_in − R_out = −15 m³/min and A = 30 m², then dh/dt = −0.5 m/min, meaning the water level is falling at half a metre per minute.

例如，若 R_in − R_out = −15 m³/min，A = 30 m²，则 dh/dt = −0.5 m/min，表示水位以每分钟 0.5 米的速度下降。

6. Torricelli’s law and square-root outflow | 托里拆利定律与平方根出流

Torricelli’s law states that water exits a small hole at speed v = √(2gh), where g is acceleration due to gravity and h is the height of water above the hole.

托里拆利定律指出，水从小孔流出时的速度为 v = √(2gh)，其中 g 为重力加速度，h 为孔口以上的水深。

If the hole has area a, the volume outflow rate is Q = a√(2gh). For a tank of cross-sectional area A, conservation of volume gives:

若孔口面积为 a，则体积出流率为 Q = a√(2gh)。对于横截面积为 A 的水池，体积守恒给出：

A dh/dt = −a√(2gh)

This is a separable differential equation. Integrating gives the time to drain from initial depth H to final depth 0:

这是一个可分离变量微分方程。积分可得到从初始深度 H 排空到深度 0 所需的时间：

T = (2A / a) √(H / (2g))

Notice the square-root relationship: doubling the initial depth increases the draining time by a factor of √2, not 2.

注意平方根关系：初始深度加倍只会使排空时间增加 √2 倍，而不是 2 倍。

7. Numerical methods: Euler’s method for drainage | 数值方法：欧拉法模拟排水

When a differential equation cannot be solved analytically, Edexcel papers may ask for Euler’s method. The iterative formula is:

当微分方程无法解析求解时，Edexcel 试卷可能要求使用欧拉法。其迭代公式为：

y_{n+1} = y_n + h f(x_n, y_n)

Here h is the step size. In a drainage model, if dh/dt = f(t,h), we approximate the water depth after each step. Smaller step sizes

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Urban Drainage: A-Level Maths Modelling & Calculus | 城市排水：A-Level数学建模与微积分