📚 Natural Resource Issues – Energy: Mathematical Modelling | 自然资源问题——能源：数学建模
Energy is one of the most fundamental natural resources, but its supply, demand, and environmental impact are rarely understood without quantitative analysis. In A-Level Mathematics, energy problems provide a rich context for applying exponential functions, logarithms, differentiation, integration, and statistical methods to real-world resource issues such as fossil fuel depletion, renewable capacity, and carbon emissions.
能源是最基本的自然资源之一，但若不进行定量分析，人们很难真正理解其供应、需求和环境影响。在 A-Level 数学中，能源问题为指数函数、对数、微分、积分和统计方法提供了丰富的实际应用背景，涉及化石燃料枯竭、可再生能源容量和碳排放等资源议题。
1. Energy Units and Conversion | 能源单位与换算
Before modelling any energy resource issue, we must be comfortable converting between energy units. Common units include the joule (J), watt-hour (Wh), tonne of oil equivalent (toe), and British thermal unit (Btu). The standard conversions are 1 kWh = 3.6 × 10⁶ J, 1 toe = 4.1868 × 10¹⁰ J, and 1 Btu = 1055 J.
在对任何能源资源问题进行建模之前，我们必须熟悉能量单位之间的换算。常见单位包括焦耳 (J)、瓦时 (Wh)、吨油当量 (toe) 和英热单位 (Btu)。标准换算关系为：1 kWh = 3.6 × 10⁶ J，1 toe = 4.1868 × 10¹⁰ J，1 Btu = 1055 J。
For example, if a power station generates 500 MW for 24 hours, the total energy output in joules is 500 × 10⁶ W × 24 × 3600 s = 4.32 × 10¹³ J. Converting this to tonnes of oil equivalent gives 4.32 × 10¹³ ÷ 4.1868 × 10¹⁰ ≈ 1032 toe.
例如，若一座发电站以 500 MW 的功率运行 24 小时，则总输出能量为 500 × 10⁶ W × 24 × 3600 s = 4.32 × 10¹³ J。将其转换为吨油当量，得到 4.32 × 10¹³ ÷ 4.1868 × 10¹⁰ ≈ 1032 toe。
These conversions are essential because energy reserves, consumption, and emissions are often reported in different units across countries and industries. Using one consistent unit avoids serious errors in resource estimation.
这些换算至关重要，因为不同国家和行业在报告能源储量、消耗和排放时常常使用不同单位。使用一致的单位可以避免资源估算中的严重错误。
2. Exponential Growth in Energy Demand | 能源需求的指数增长
Global energy demand has grown approximately exponentially over the past century. If the annual growth rate is r, and the initial demand is E₀, then the demand after t years is given by:
过去一个世纪，全球能源需求大致呈指数增长。如果年增长率为 r，初始需求为 E₀，则 t 年后的需求可表示为：
E(t) = E₀ × e^(rt)
This continuous model is equivalent to E(t) = E₀ × (1 + r)ᵗ when using annual discrete compounding. The continuous form is preferred in calculus because it differentiates cleanly.
该连续模型等同于离散年复利下的 E(t) = E₀ × (1 + r)ᵗ。连续形式在微积分中更受欢迎，因为其求导更为简洁。
For instance, if world energy demand is 600 EJ (exajoules) per year and grows at 2.3% per year, the demand after 10 years is 600 × e^(0.023 × 10) ≈ 600 × e^0.23 ≈ 600 × 1.2586 ≈ 755 EJ.
例如，若世界能源需求为每年 600 EJ（艾焦），年增长率为 2.3%，则 10 年后的需求为 600 × e^(0.023 × 10) ≈ 600 × e^0.23 ≈ 600 × 1.2586 ≈ 755 EJ。
This exponential behaviour means that even a small continuous growth rate leads to a doubling time of about ln 2 / r. With r = 0.023, the doubling time is approximately 30 years, which has major implications for resource depletion and investment planning.
这种指数行为意味着即使是很小的持续增长率，其翻倍时间也约为 ln 2 / r。当 r = 0.023 时，翻倍时间约为 30 年，这对资源枯竭和投资规划具有重大影响。
3. Logistic Growth and the Hubbert Peak Model | 逻辑斯蒂增长与哈伯特峰值模型
Resource extraction cannot grow exponentially forever. The Hubbert peak model uses a logistic curve to describe cumulative production of a finite resource. The annual production rate P(t) often follows a bell-shaped curve:
资源开采不可能永远指数增长。哈伯特峰值模型使用逻辑斯蒂曲线描述有限资源的累计产量。年产量 P(t) 通常呈现钟形曲线：
P(t) = K / [1 + e^(–b(t – t₀))]
Here K is the peak production rate, t₀ is the time of peak output, and b controls the steepness of the rise and decline. This model is widely used for oil, coal, and natural gas.
其中 K 是峰值产量，t₀ 是产量达到峰值的时间，b 控制产量上升和下降的陡峭程度。该模型广泛用于石油、煤炭和天然气。
For example, if a region’s oil production follows P(t) = 4 / [1 + e^(–0.15(t – 20))] million barrels per day, then when t = 20, P = 4 / (1 + 1) = 2 million barrels per day. The maximum rate is 4 million barrels per day as t becomes large before the decline, but the actual peak occurs only if the curve is centred correctly.
例如，若某地区的石油产量遵循 P(t) = 4 / [1 + e^(–0.15(t – 20))] 百万桶/日，则当 t = 20 时，P = 4 / (1 + 1) = 2 百万桶/日。随着 t 增大，最大产量趋近于 4 百万桶/日，但实际峰值只有在曲线正确居中时才会出现。
The derivative dP/dt = bP(1 – P/K) shows that the growth rate is largest when P = K/2. This is a classic logistic differential equation, and integrating it gives the cumulative production curve, which approaches a finite total resource.
导数 dP/dt = bP(1 – P/K) 表明，当 P = K/2 时增长率最大。这是经典的逻辑斯蒂微分方程，对其积分可得到累计产量曲线，该曲线趋近于有限的总资源量。
4. Reserves-to-Production Ratio | 储采比
The reserves-to-production ratio (R/P ratio) is a simple indicator of how long remaining reserves will last at current production rates. If R is the remaining reserve and P is the annual production, then:
储采比 (R/P 比) 是一个简单指标，表示在当前产量水平下剩余储量还能开采多少年。如果 R 为剩余储量，P 为年产量，则：
R/P ratio = R / P
This ratio can be misleading because production rates may change, new reserves may be discovered, and demand may fall due to substitution or efficiency. Nevertheless, it is a useful first approximation for resource life.
该比值可能产生误导，因为产量可能发生变化，新储量可能被发现，需求也可能因替代或效率提升而下降。然而，它仍是资源寿命的一个有用初步近似。
For example, if a country has 250 billion tonnes of coal reserves and produces 5 billion tonnes per year, the R/P ratio is 250 / 5 = 50 years. If production grows at 3% per year, exponential growth will shorten the actual lifetime, and we can solve 250 = ∫₀ᵀ P₀e^(0.03t) dt to find the true depletion time.
例如，若某国拥有 2500 亿吨煤炭储量，年产量为 50 亿吨，则储采比为 250 / 5 = 50 年。如果产量以每年 3% 的速度增长，指数增长将缩短实际开采寿命，我们可以通过求解 250 = ∫₀ᵀ P₀e^(0.03t) dt 来找到真实的枯竭时间。
Integrating gives 250 = P₀(e^(0.03T) – 1) / 0.03. With P₀ = 5, this becomes 250 = (5 / 0.03)(e^(0.03T) – 1), so e^(0.03T) = 1 + 1.5 = 2.5, hence T = ln 2.5 / 0.03 ≈ 30.5 years. This is much shorter than the static R/P ratio suggests.
积分后得到 250 = P₀(e^(0.03T) – 1) / 0.03。代入 P₀ = 5，得到 250 = (5 / 0.03)(e^(0.03T) – 1)，因此 e^(0.03T) = 1 + 1.5 = 2.5，从而 T = ln 2.5 / 0.03 ≈ 30.5 年。这比静态储采比所显示的时间短得多。
5. Linear and Quadratic Cost Models for Energy Supply | 能源供应的线性与二次成本模型
Energy production costs often depend on output level. A linear cost model has the form C(x) = mx + c, where x is the amount of energy produced, m is the marginal cost, and c is the fixed cost. For example, a gas plant might have a fixed cost of £50 million and a marginal cost of £20 per MWh.
能源生产成本通常取决于产出水平。线性成本模型的形式为 C(x) = mx + c，其中 x 为生产的能源量，m 为边际成本，c 为固定成本。例如，一座天然气电厂可能有 5000 万英镑的固定成本和每 MWh 20 英镑的边际成本。
In many real situations, costs rise more steeply as output approaches capacity, so a quadratic model C(x) = ax² + bx + c is more appropriate. The marginal cost is then dC/dx = 2ax + b, which increases with output.
在许多实际情况中，随着产出接近产能，成本上升得更快，因此二次模型 C(x) = ax² + bx + c 更为合适。此时边际成本为 dC/dx = 2ax + b，随产出增加而增加。
For instance, if C(x) = 0.002x² + 25x + 100, where x is in MWh, then the marginal cost at x = 1000 MWh is 2(0.002)(1000) + 25 = 4 + 25 = £29 per MWh. The average cost is C(x)/x = 0.002x + 25 + 100/x.
例如，若 C(x) = 0.002x² + 25x + 100，其中 x 以 MWh 计，则当 x = 1000 MWh 时，边际成本为 2(0.002)(1000) + 25 = 4 + 25 = 29 英镑/MWh。平均成本为 C(x)/x = 0.002x + 25 + 100/x。
Minimising average cost by setting its derivative to zero gives d/dx (0.002x + 25 + 100/x) = 0.002 – 100/x² = 0, so x² = 50000, and x = √50000 ≈ 223.6 MWh. This is the output level that minimises the average cost per unit of energy.
通过将平均成本导数设为零来最小化平均成本，得到 d/dx (0.002x + 25 + 100/x) = 0.002 – 100/x² = 0，因此 x² = 50000，x = √50000 ≈ 223.6 MWh。这就是使单位能源平均成本最低的产出水平。
6. Renewable Energy Capacity Factor Calculations | 可再生能源容量因子计算
Renewable sources such as wind and solar do not produce at full capacity all the time. The capacity factor is the ratio of actual energy output over a period to the maximum possible output if the plant operated at rated capacity for the same period.
风能和太阳能等可再生能源并不会始终满负荷运行。容量因子是指在一段时间内实际发电量与同一时期按额定容量满负荷运行所能产生的最大发电量之比。
Capacity factor = Actual energy output / (Rated capacity × Time)
For example, a 2 MW wind turbine operating for 8760 hours in a year produces 5,256 MWh. Its capacity factor is 5256 / (2 × 8760) = 5256 / 17520 = 0.3, or 30%.
例如，一台 2 MW 的风力发电机一年运行 8760 小时，发电量为 5256 MWh。其容量因子为 5256 / (2 × 8760) = 5256 / 17520 = 0.3，即 30%。
Capacity factors vary greatly: offshore wind may achieve 40–50%, solar PV around 10–20%, and nuclear above 90%. When comparing energy sources, using capacity factor is mathematically essential to avoid overestimating renewable output.
容量因子差异很大：海上风电可达 40–50%，太阳能光伏约为 10–20%，核电则超过 90%。在比较能源时，使用容量因子在数学上至关重要，可以避免高估可再生能源的产出。
If we want to match the annual output of a 1 GW nuclear plant with a 90% capacity factor using solar panels with a 15% capacity factor, the required solar capacity is (1 × 0.9) / 0.15 = 6 GW of installed panels.
如果我们希望用容量因子为 15% 的太阳能电池板来匹配一座容量因子为 90% 的 1 GW 核电站的年发电量，则所需的太阳能装机容量为 (1 × 0.9) / 0.15 = 6 GW。
7. Carbon Intensity and Logarithmic Relationships | 碳强度与对数关系
Carbon intensity is the amount of CO₂ emitted per unit of energy produced, usually measured in gCO₂/kWh. Different energy sources have very different values: coal about 900–1000 gCO₂/kWh, gas about 400–500, solar about 40–50, and wind about 10–20.
碳强度是指生产单位能量所排放的 CO₂ 量，通常以 gCO₂/kWh 计量。不同能源的碳强度差异很大：煤约为 900–1000 gCO₂/kWh，天然气约为 400–500，太阳能约为 40–50，风能约为 10–20。
The relationship between CO₂ concentration in the atmosphere and cumulative emissions is approximately logarithmic. If C is the atmospheric CO₂ concentration and E is the cumulative emissions, then a simplified model is:
大气中 CO₂ 浓度与累计排放量之间的关系近似呈对数关系。如果 C 为大气 CO₂ 浓度，E 为累计排放量，则简化模型为：
C = C₀ + α × ln(E / E₀)
This logarithmic form means that each additional tonne of CO₂ has a smaller marginal effect on concentration, but it also makes emission reduction targets harder to express linearly. A-Level logarithmic skills are directly applicable.
这种对数形式意味着每额外排放一吨 CO₂ 对浓度的边际影响较小，但也使得减排目标更难以用线性方式表达。A-Level 对数技能在此可直接应用。
For instance, if C₀ = 280 ppm, E₀ = 600 GtC, and α = 20, then when cumulative emissions reach 1200 GtC, the concentration is C = 280 + 20 × ln(1200 / 600) = 280 + 20 × ln 2 ≈ 280 + 13.9 = 293.9 ppm.
例如，若 C₀ = 280 ppm，E₀ = 600 GtC，α = 20，则当累计排放量达到 1200 GtC 时，浓度为 C = 280 + 20 × ln(1200 / 600) = 280 + 20 × ln 2 ≈ 280 + 13.9 = 293.9 ppm。
Solving for E when C is known requires exponentiating: E = E₀ × e^((C – C₀)/α). This is a common exam-style manipulation.
已知 C 求解 E 时需要进行指数运算：E = E₀ × e^((C – C₀)/α)。这是常见的考试题型。
8. Time Series Analysis of Energy Data | 能源数据的时间序列分析
Energy statistics such as monthly electricity demand or oil prices are time series. We can smooth these data using moving averages to reveal underlying trends. A 12-point moving average is useful for annual seasonal patterns.
电力月需求或石油价格等能源统计数据属于时间序列。我们可以使用移动平均来平滑这些数据，以揭示潜在趋势。12 点移动平均对于年度季节性模式非常有用。
If monthly electricity demand values are D₁, D₂, …, D₁₂, then the 12-point moving average for the first complete year is (D₁ + D₂ + … + D₁₂) / 12. We can also use exponential smoothing: Sₜ = αDₜ + (1 – α)Sₜ₋₁, where α is between 0 and 1.
如果月度电力需求值为 D₁, D₂, …, D₁₂，则第一个完整年度的 12 点移动平均为 (D₁ + D₂ + … + D₁₂) / 12。我们还可以使用指数平滑：Sₜ = αDₜ + (1 – α)Sₜ₋₁，其中 α 介于 0 和 1 之间。
For example, if S₀ = 100, D₁ = 110, and α = 0.3, then S₁ = 0.3 × 110 + 0.7 × 100 = 33 + 70 = 103. This smoothing method is simple but powerful for identifying energy demand trends.
例如，若 S₀ = 100，D₁ = 110，α = 0.3，则 S₁ = 0.3 × 110 + 0.7 × 100 = 33 + 70 = 103。这种平滑方法简单但功能强大，可用于识别能源需求趋势。
Moving averages and exponential smoothing are part of the Edexcel A-Level statistics content, and energy data provide realistic contexts for these techniques.
移动平均和指数平滑是 Edexcel A-Level 统计学内容的一部分，能源数据为这些技术提供了真实的应用背景。
9. Optimisation of Energy Mix Using Differentiation | 使用微分优化能源结构
Countries often aim to minimise the total cost or carbon emissions of their energy mix subject to a demand constraint. Suppose energy comes from gas (g) and wind (w), and total emissions are given by E = 500g + 20w, while total output must satisfy g + w = 100 TWh. Substituting w = 100 – g gives:
各国通常力求在满足需求约束的前提下，使能源结构的总成本或碳排放最小化。假设能源来自天然气 (g) 和风能 (w)，总排放量为 E = 500g + 20w，而总产出必须满足 g + w = 100 TWh。代入 w = 100 – g，得到：
E = 500g + 20(100 – g) = 480g + 2000
This is a linear function, so minimising emissions would suggest g = 0 and w = 100, but this ignores cost and reliability constraints. A more realistic model might be E = 500g + 20w + 0.1g² + 0.05w².
这是一个线性函数，因此最小化排放意味着 g = 0 且 w = 100，但这忽略了成本和可靠性约束。更现实的模型可能是 E = 500g + 20w + 0.1g² + 0.05w²。
Using w = 100 – g, we get E = 500g + 20(100 – g) + 0.1g² + 0.05(100 – g)². Expanding and differentiating with respect to g gives dE/dg = 480 + 0.2g – 0.1(100 – g) = 480 + 0.2g – 10 + 0.1g = 470 + 0.3g. Setting this to zero gives a negative g, indicating the minimum occurs at the boundary g = 0 under these coefficients.
利用 w = 100 – g，我们得到 E = 500g + 20(100 – g) + 0.1g² + 0.05(100 – g)²。对 g 求导得 dE/dg = 480 + 0.2g – 0.1(100 – g) = 480 + 0.2g – 10 + 0.1g = 470 + 0.3g。令其为零会得到负的 g，表明在这些系数下最小值出现在边界 g = 0 处。
This type of constrained optimisation is central to energy policy mathematics. In an exam, you would be expected to form the objective function, substitute the constraint, differentiate, and justify boundary solutions.
这种带约束的优化是能源政策数学的核心。在考试中，你需要构建目标函数、代入约束、求导，并论证边界解。
10. Edexcel-Style Worked Example | Edexcel 风格例题解析
A typical exam question might state: “The annual production of a coal mine is modelled by P(t) = 12te^(–0.05t) million tonnes, where t is time in years. Find the time at which production is maximised and the maximum production.”
一道典型考题可能是：“某煤矿的年产量可建模为 P(t) = 12te^(–0.05t) 百万吨，其中 t 为时间（年）。求产量达到最大的时间以及最大产量。”
We differentiate using the product rule: dP/dt = 12e^(–0.05t) + 12t(–0.05)e^(–0.05t) = 12e^(–0.05t)(1 – 0.05t). Setting this to zero gives 1 – 0.05t = 0, so t = 20 years.
我们使用乘积法则求导：dP/dt = 12e^(–0.05t) + 12t(–0.05)e^(–0.05t) = 12e^(–0.05t)(1 – 0.05t)。令其为零，得到 1 – 0.05t = 0，因此 t = 20 年。
Substituting t = 20 into P(t) gives P(20) = 12 × 20 × e^(–1) = 240 / e ≈ 240 / 2.718 ≈ 88.3 million tonnes. The second derivative test confirms this is a maximum because d²P/dt² = 12e^(–0.05t)(–0.05) + derivative of the bracket, and at t = 20 it is negative.
将 t = 20 代入 P(t)，得到 P(20) = 12 × 20 × e^(–1) = 240 / e ≈ 240 / 2.718 ≈ 88.3 百万吨。二阶导数检验确认这是最大值，因为 d²P/dt² = 12e^(–0.05t)(–0.05) 加上括号项的导数，在 t = 20 时为负。
This example combines exponential functions, product rule differentiation, and optimisation, all of which are core Edexcel A-Level Mathematics skills applied to a natural resource energy issue.
这个例子结合了指数函数、乘积法则求导和优化，这些都是 Edexcel A-Level 数学的核心技能，并应用于自然资源的能源问题。
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