Urban Drainage Modelling | 城市排水系统建模

📚 Urban Drainage Modelling | 城市排水系统建模

Urban drainage systems are vital for preventing flooding and managing stormwater in cities. In A-Level Mathematics, particularly in the Edexcel specification, the flow of water through pipes and channels can be modelled using differential equations. This article demonstrates how core calculus techniques, such as separation of variables and integrating factors, are applied to predict water depth, discharge rates, and tank emptying times, connecting pure mathematics with real-world civil engineering problems.

城市排水系统对于防止城市内涝和管理雨水至关重要。在A-Level数学，尤其是Edexcel考试大纲中，水流经管道和渠道的过程可以用微分方程来建模。本文展示了如何将分离变量法和积分因子等核心微积分技术，应用于预测水深、流量和水箱排空时间，把纯数学与现实世界的土木工程问题联系起来。

1. The Urban Drainage Problem | 城市排水问题

In an urban catchment, rainfall runs off impermeable surfaces and enters a network of gullies, pipes, and storage tanks. Engineers must design the system so that water is conveyed safely without surcharging. The rate of change of water volume in a tank or a pipe section depends on the inflow and outflow rates, leading naturally to first-order differential equations.

在城市集水区，雨水从不透水表面流走，进入由雨水口、管道和蓄水池组成的网络。工程师必须设计这样的系统，使水能安全输送而不发生溢流。水箱或管段中水体积的变化率取决于流入和流出的速率，这自然引出一阶微分方程。

2. Modelling a Simple Storage Tank | 简单储水箱建模

Consider a tank with a constant inflow rate Q_in sectional area A (m²), the volume V = A × h. The net rate of change of volume is dV/dt = Q_in - k√h, where k is a discharge coefficient. This is a differential equation connecting h and t.

考虑一个具有恒定流入速率 Q_in (m³/s) 的水箱，其流出速率与水深 h (m) 成正比。如果水箱具有均匀的横截面积 A (m²)，则体积 V = A × h。体积的净变化率为 dV/dt = Q_in - k√h，其中 k 是流量系数。这是一个联系 h 和 t 的微分方程。

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3. Formulating the Differential Equation | 建立微分方程

Using the chain rule: dV/dt = A × dh/dt. Equating this to the net flow gives A dh/dt = Q_in - k√h. This is a first-order ordinary differential equation (ODE) that describes how the water depth evolves over time. For a tank without inflow after a storm, Q_in might be zero, and the water level drops purely due to drainage: dh/dt = -(k/A)√h.

运用链式法则：dV/dt = A × dh/dt。将其与净流量相等，得到 A dh/dt = Q_in - k√h。这是一个一阶常微分方程(ODE)，描述了水深随时间的变化。对于暴雨过后没有流入的水箱，Q_in 可能为零，水位仅因排水而下降：dh/dt = -(k/A)√h。

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4. Solving by Separation of Variables | 用分离变量法求解

For the pure drainage case dh/dt = -c√h, where c = k/A, we can separate the variables: dh/√h = -c dt. Integrating both sides gives 2√h -ct + D, where D is the constant of integration. If initially h = h₀ at t = 0, then D = 2√h₀, leading to the solution h = (√h₀ - (c/2)t)².

对于纯排水情形 dh/dt = -c√h，其中 c = k/A，我们可以分离变量：dh/√h = -c dt。两边积分得到 2√h = -ct + D，其中 D 是积分常数。如果在 t = 0 时初始水深为 h₀，则 D = 2√h₀，得到解 h = (√h₀ - (c/2)t)²。

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5. Time to Empty the Tank | 水箱排空时间

The tank empties when h = 0. From the solution, this occurs when √h₀ - (c/2)T = 0, so T = 2√h₀ / c. This simple formula is powerful: doubling the initial depth does not double the emptying time—it increases by a factor of √2. This nonlinear behaviour is typical in drainage systems where resistance to flow changes with depth.

水箱在 h = 0 时排空。由解得出，当 √h₀ - (c/2)T = 0 时发生，因此 T = 2√h₀ / c。这个简单的公式非常有用：初始水深加倍并不会使排空时间加倍——它只增加 √2 倍。这种非线性行为在排水系统中很典型，因为流动阻力随水深而变化。

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6. Incorporating Constant Inflow | 纳入恒定流入

When Q_in is constant and non-zero, the equation becomes dh/dt = (Q_in/A) - c√h. This is still separable but requires a substitution. Let u = √h, then h = u², dh/dt = 2u du/dt. Substituting yields 2u du/dt = (Q_in/A) - c u. Rearranging into standard form, we get du/dt + (c/2)u = Q_in/(2A u). This is no longer linear due to the u in the denominator, but can be tackled by further separation or an integrating factor after appropriate manipulation.

当 Q_in 为恒定且非零时，方程变为 dh/dt = (Q_in/A) - c√h。这仍然是可分离的，但需要代换。令 u = √h，则 h = u²，dh/dt = 2u du/dt。代入得到 2u du/dt = (Q_in/A) - c u。重新排列成标准形式，得 du/dt + (c/2)u = Q_in/(2A u)。由于分母中有 u，这个方程不再是线性的，但可以通过进一步分离变量或在适当处理后采用积分因子来求解。

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7. Linear First-Order ODE for Pipe Flow | 管流的一阶线性常微分方程

with laminar resistance, the outflow rate might be b h, where b is a constant. The model then becomes dh/dt = (Q_in/A) - (b/A) h. Writing this as dh/dt + p h = q, with p = b/A and q = Q_in/A, we have a linear first-order ODE that can be solved using an integrating factor.

在某些排水模型中，出流量与水头成正比，从而得到一个线性常微分方程。例如，如果水通过具有层流阻力的管道从水箱流出，出流速率可能为 b h，其中 b 是常数。此时模型变为 dh/dt = (Q_in/A) - (b/A) h。将其写为 dh/dt + p h = q，其中 p = b/A，q = Q_in/A，我们得到一个一阶线性常微分方程，可用积分因子求解。

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8. Using an Integrating Factor | 使用积分因子

The integrating factor is μ(t) = e^{∫ p dt} = e^pt. Multiplying the ODE by μ(t) gives d/dt (h e^pt) = q e^pt. Integrating both sides yields h e^pt = (q/p) e^pt + C, so h = q/p + C e^-pt. If h(0) = h₀, then C = h₀ - q/p. The steady-state depth is h_∞ = q/p = Q_in/b, approached exponentially.

积分因子为 μ(t) = e^{∫ p dt} = e^pt。将常微分方程乘以 μ(t) 得到 d/dt (h e^pt) = q e^pt。两边积分给出 h e^pt = (q/p) e^pt + C，因此 h = q/p + C e^-pt。如果 h(0) = h₀，则 C = h₀ - q/p。稳态水深为 h_∞ = q/p = Q_in/b，以指数方式趋近。

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9. Modelling Stormwater Runoff | 模拟雨水径流

During a storm, the inflow rate Q_in may be a function of time, such as Q_in(t) = R t e^-αt uick rise and gradual decay of rainfall intensity. The ODE becomes dh/dt + p h = (R/A) t e^-αt. The integrating factor method still works, but the integral on the right-hand side requires integration by parts. This models a detention basin's water level during a storm and tests students’ integration skills.

在暴雨期间，流入速率 Q_in 可能是时间的函数，例如 Q_in(t) = R t e^-αt，代表降雨强度的快速上升和逐渐衰减。常微分方程变为 dh/dt + p h = (R/A) t e^-αt。积分因子法仍然有效，但右侧的积分需要使用分部积分法。这模拟了暴雨期间调节池的水位，并考验学生的积分技巧。

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10. Numerical Methods: Euler’s Method | 数值方法：欧拉法

an be solved analytically. For complex outflow functions, A-Level students can apply Euler' to approximate h at discrete time steps. Given dh/dt = f(t, h), the iterative formula is h_n+1 = h_n + f(t_n, h_n) × Δt. For the tank drainage equation dh/dt = -c√h, Euler's method provides a way to compute the depth curve step by step, highlighting the trade-off between step size and accuracy.

并非所有的排水常微分方程都能解析求解。对于复杂的出流函数，A-Level 学生可以应用欧拉法在离散的时间步长上近似求解 h。根据 dh/dt = f(t, h)，迭代公式为 h_n+1 = h_n + f(t_n, h_n) × Δt。对于水箱排水方程 dh/dt = -c√h，欧拉法提供了逐步计算水深曲线的方法，突显了步长与精度之间的权衡。

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11. Interpreting Solutions in Context | 结合情境解读解

omain of h(t) is restricted to t such that h(t) ≥ 0. Furthermore, coefficients like k and b are determined empirically, and sensitivity analysis can show how small changes in parameters affect flooding risks—an excellent crossover with statistical sampling.

数学解必须对照物理约束进行验证。水深不能为负，且当水箱排空时模型失效。h(t) 的定义域限定于使 h(t) ≥ 0 的 t 值。此外，k 和 b 等系数是通过经验确定的，敏感性分析可以显示参数的微小变化如何影响内涝风险——这与统计抽样是一个极好的交叉领域。

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12. Exam-Style Problem | 考试风格问题

Problem: -sectional area A = 200 m². Water drains through an outlet so that the outflow rate is 0.5√h m³/s. After a storm, the initial depth is 4 m, and no further inflow occurs. Find the time taken for the depth to drop to 1 m, and the time to empty completely.

Question: 一个调蓄池塘的横截面积为 A = 200 m²。水通过一个出水口排出，出流速度为 0.5√h m³/s。暴雨过后，初始水深为 4 m，且没有进一步的流入。求水深降至 1 m 所需的时间，以及完全排空所需的时间。

The ODE is 200 dh/dt = -0.5√h, so dh/dt = -0.0025√h. Separating: dh/√h = -0.0025 dt, integrating: 2√h = -0.0025 t + C. Using h(0)=4 gives C=4, so √h = 2 - 0.00125 t. For h=1, √h 2 - 0.00125 t, t = 800 s (≈13.3 min). For h=0, 0 = 2 - 0.00125 t, t = 1600 s (≈26.7 min).

常微分方程为 200 dh/dt = -0.5√h，所以 dh/dt = -0.0025√h。分离变量：dh/√h = -0.0025 dt，积分得：2√h = -0.0025 t + C。利用 h(0)=4 得到 C=4，因此 √h = 2 - 0.00125 t。当 h=1，√h=1，故 1 = 2 - 0.00125 t，t = 800 秒（约13.3分钟）。当 h=0，0 = 2 - 0.00125 t，t = 1600 秒（约26.7分钟）。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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