Year 13 WJEC Engineering Unit Test Mock Paper Analysis | WJEC 工程 Year 13 单元测试模拟卷解析

📚 Year 13 WJEC Engineering Unit Test Mock Paper Analysis | WJEC 工程 Year 13 单元测试模拟卷解析

This mock paper analysis is designed to support Year 13 students following the WJEC Engineering specification, covering the typical depth and breadth required for Unit 3 (Engineering Applications) and reinforcing key principles from AS material. By working through each problem and its detailed solution, you can sharpen your analytical skills, improve your use of formulas, and understand how marks are awarded in real examinations.

本模拟卷解析专为学习 WJEC 工程课程的 Year 13 学生设计，覆盖了单元三（工程应用）所需的典型深度与广度，同时强化了 AS 阶段的核心原理。通过仔细研究每一道问题及其详细解答，你可以提高分析能力、熟练掌握公式的运用，并理解真实考试中的评分方式。

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1. Overview of the Mock Examination | 模拟试卷概览

The paper contains eight structured questions drawn from mechanics, electronics, materials, manufacturing, thermodynamics, design analysis, and quality control. A total of 90 marks is available, to be attempted in 1 hour 45 minutes. Each question includes synoptic elements, requiring you to apply knowledge across different areas of the course.

本试卷包含八个结构性问题，涵盖力学、电子学、材料学、制造工艺、热力学、设计分析以及质量控制等内容。满分为 90 分，考试时间 1 小时 45 分钟。每道题都包含综合性考察点，要求你跨课程模块运用知识。

Question	Topic	Marks
1	Bending Stress in Cantilever Beam	12
2	Transistor Amplifier Biasing	12
3	Al-Cu Phase Diagram & Age Hardening	10
4	CNC Machining Tolerance & Cp/Cpk	10
5	Diesel Cycle Efficiency	10
6	FMEA of a Pump Assembly	12
7	SPC Charts & Out-of-Control Rules	14
8	Material Selection via Ashby Chart	10

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2. Problem 1: Bending Stress in a Cantilever Beam | 问题 1：悬臂梁的弯曲应力

A cantilever beam of length 1.2 m supports a concentrated vertical load of 800 N at its free end. The beam has a solid rectangular cross-section with width b = 50 mm and depth d = 80 mm. Determine the maximum bending stress in the beam and state whether a steel with yield strength 250 MPa would be safe under a safety factor of 1.6.

一根长 1.2 m 的悬臂梁在其自由端承受 800 N 的集中垂直载荷。梁截面为实心矩形，宽度 b = 50 mm，深度 d = 80 mm。试计算梁内的最大弯曲应力，并判断屈服强度为 250 MPa 的钢材在安全系数 1.6 的情况下是否安全。

Step 1 – Maximum bending moment. For a cantilever with end point load, the peak moment occurs at the fixed support: M_max = F × L = 800 N × 1.2 m = 960 N·m.

步骤 1 – 最大弯矩。对于端部受集中力的悬臂梁，峰值弯矩出现在固定端：M_max = F × L = 800 N × 1.2 m = 960 N·m。

Step 2 – Section modulus. For a rectangular section, Z = (b × d²)/6. Convert dimensions to metres: b = 0.05 m, d = 0.08 m. Z = (0.05 × 0.08²)/6 = (0.05 × 0.0064)/6 = 0.00032/6 = 5.333 × 10⁻⁵ m³.

步骤 2 – 截面模量。对于矩形截面，Z = (b × d²)/6。将尺寸转换为米：b = 0.05 m, d = 0.08 m。Z = (0.05 × 0.08²)/6 = (0.05 × 0.0064)/6 = 0.00032/6 = 5.333 × 10⁻⁵ m³。

Step 3 – Bending stress. σ_b = M / Z = 960 N·m / 5.333 × 10⁻⁵ m³ = 18.0 × 10⁶ Pa = 18.0 MPa.

步骤 3 – 弯曲应力。σ_b = M / Z = 960 N·m / 5.333 × 10⁻⁵ m³ = 18.0 × 10⁶ Pa = 18.0 MPa。

Step 4 – Safety check. Allowable stress = yield strength / safety factor = 250 MPa / 1.6 = 156.25 MPa. Since 18.0 MPa ≪ 156.25 MPa, the beam is safe with a large margin.

步骤 4 – 安全性校验。许用应力 = 屈服强度 / 安全系数 = 250 MPa / 1.6 = 156.25 MPa。因 18.0 MPa 远小于 156.25 MPa，梁安全且裕度充足。

σ_b = 18.0 MPa, σ_allow = 156.25 MPa → design is safe.

σ_b = 18.0 MPa，σ_allow = 156.25 MPa → 设计安全。

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3. Problem 2: Transistor Amplifier Circuit Analysis | 问题 2：晶体管放大电路分析

A common-emitter amplifier uses a silicon NPN transistor with β = 100. The supply voltage V_CC = 12 V, collector resistor R_C = 2 kΩ, base resistor R_B = 200 kΩ, and the emitter is grounded. Calculate the quiescent collector current I_CQ and the collector-emitter voltage V_CEQ. Assume V_BE = 0.7 V and negligible base current effect from the collector load.

一个共射极放大器采用硅 NPN 晶体管，β = 100。电源电压 V_CC = 12 V，集电极电阻 R_C = 2 kΩ，基极电阻 R_B = 200 kΩ，发射极接地。计算静态集电极电流 I_CQ 和集电极-发射极电压 V_CEQ。假设 V_BE = 0.7 V，且忽略集电极负载对基极电流的微小影响。

Base current: I_B = (V_CC – V_BE) / R_B = (12 V – 0.7 V) / 200 kΩ = 11.3 V / 200,000 Ω = 56.5 µA.

基极电流：I_B = (V_CC – V_BE) / R_B = (12 V – 0.7 V) / 200 kΩ = 11.3 V / 200,000 Ω = 56.5 µA。

Collector current: I_CQ ≈ β × I_B = 100 × 56.5 µA = 5.65 mA.

集电极电流：I_CQ ≈ β × I_B = 100 × 56.5 µA = 5.65 mA。

Collector-emitter voltage: V_CEQ = V_CC – I_CQ × R_C = 12 V – (5.65 mA × 2 kΩ) = 12 V – 11.3 V = 0.7 V.

集电极-发射极电压：V_CEQ = V_CC – I_CQ × R_C = 12 V – (5.65 mA × 2 kΩ) = 12 V – 11.3 V = 0.7 V。

The quiescent point sits near saturation, giving very little output swing. To obtain a linear amplifier, the bias point should be adjusted, e.g. by adding an emitter resistor or using a voltage-divider bias.

静态工作点接近饱和区，输出摆幅极小。若要获得线性放大器，应当调整偏置点，例如增加发射极电阻或采用分压式偏置。

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4. Problem 3: Al-Cu Phase Diagram and Precipitation Hardening | 问题 3：铝铜相图与沉淀硬化

An Al-4% Cu alloy is solution treated at 540 °C, water quenched, and aged at 190 °C. Explain the microstructural changes at each stage and state why the aged alloy becomes stronger. Refer to GP zones, θ” and θ’ phases, and the equilibrium θ phase (Al₂Cu).

一种含 4% Cu 的铝合金经过 540 °C 固溶处理、水淬，并在 190 °C 时效。请解释各阶段的微观组织变化，并说明时效态合金为何强度更高。需提及 GP 区、θ” 相和 θ’ 相，以及平衡 θ 相 (Al₂Cu)。

Solution treatment dissolves the θ phase into a single α phase (FCC aluminium solid solution). Rapid quenching traps a supersaturated solid solution at room temperature.

固溶处理使 θ 相溶解形成单相 α 固溶体（FCC 铝固溶体）。快速淬火在室温下形成过饱和固溶体。

Ageing allows copper atoms to cluster into GP zones (coherent, disc-shaped solute-rich regions). These zones distort the lattice and impede dislocation motion, increasing strength. With further time, GP zones transform into semi-coherent θ” and then θ’ phases. Eventually, incoherent equilibrium θ (Al₂Cu) precipitates form, causing over-ageing and a drop in strength.

时效使铜原子聚集形成 GP 区（共格、圆盘状的富溶质区域）。这些区域使晶格畸变并阻碍位错运动，从而提高强度。随着时间延长，GP 区转变为半共格的 θ” 相，再转变为 θ’ 相。最终，非共格的平衡 θ 相 (Al₂Cu) 析出，导致过时效，强度下降。

Peak strength → fine distribution of GP zones and θ”

峰值强度 → 弥散分布的 GP 区和 θ” 相

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5. Problem 4: CNC Machining Tolerances and Process Capability | 问题 4：数控加工公差与过程能力

A CNC lathe produces shafts with a nominal diameter of 25.00 mm. The specification limits are 25.00 ± 0.04 mm. A sample of 50 shafts gives a mean of 25.01 mm and a standard deviation of 0.012 mm. Calculate the process capability indices Cp and Cpk, and comment on the process capability.

一台数控车床加工标称直径为 25.00 mm 的轴。规格限为 25.00 ± 0.04 mm。从 50 根轴中取样得到均值 25.01 mm，标准差 0.012 mm。计算过程能力指数 Cp 和 Cpk，并评价过程能力。

USL = 25.04 mm, LSL = 24.96 mm. Cp = (USL – LSL) / (6σ) = (0.08 mm) / (6 × 0.012 mm) = 0.08 / 0.072 ≈ 1.11. Cpk = min[(USL – μ)/3σ, (μ – LSL)/3σ] = min[(0.03)/0.036, (0.05)/0.036] = min[0.83, 1.39] = 0.83.

USL = 25.04 mm, LSL = 24.96 mm。Cp = (USL – LSL) / (6σ) = (0.08 mm) / (6 × 0.012 mm) = 0.08 / 0.072 ≈ 1.11。Cpk = min[(USL – μ)/3σ, (μ – LSL)/3σ] = min[(0.03)/0.036, (0.05)/0.036] = min[0.83, 1.39] = 0.83。

A Cpk of 0.83 is less than 1.0, indicating the process is not capable as the mean is too close to the upper specification limit. To achieve a capable process (Cpk ≥ 1.33), tool offset adjustment and variation reduction are needed.

Cpk 为 0.83，小于 1.0，表明过程能力不足，因为过程均值过于靠近上规格限。要使过程能力达标（Cpk ≥ 1.33），需调整刀具偏置并减少变差。

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6. Problem 5: Diesel Cycle Thermal Efficiency | 问题 5：狄塞尔循环热效率

An ideal Diesel cycle has a compression ratio r = 18 and a cut-off ratio β = 2.0. The working fluid is air with γ = 1.4. Derive the thermal efficiency and calculate its numerical value.

一个理想狄塞尔循环的压缩比 r = 18，切断比 β = 2.0，工质为空气，比热容比 γ = 1.4。推导热效率并计算其数值。

The Diesel efficiency formula: η = 1 – (1 / r^γ-1) × [(β^γ – 1) / (γ (β – 1))].

狄塞尔效率公式：η = 1 – (1 / r^γ-1) × [(β^γ – 1) / (γ (β – 1))]。

First, r^γ-1 = 18^0.4. Estimate: 18^0.4 = exp(0.4 × ln18) ≈ exp(0.4 × 2.890) = exp(1.156) ≈ 3.178. Next, β^γ = 2^1.4 ≈ exp(1.4 × ln2) = exp(1.4 × 0.693) = exp(0.970) ≈ 2.639. Denominator γ(β-1) = 1.4 × 1 = 1.4. So the factor = (2.639 – 1)/1.4 = 1.639/1.4 ≈ 1.171.

首先，r^γ-1 = 18^0.4。估算：18^0.4 = exp(0.4 × ln18) ≈ exp(0.4 × 2.890) = exp(1.156) ≈ 3.178。其次，β^γ = 2^1.4 ≈ exp(1.4 × ln2) = exp(1.4 × 0.693) = exp(0.970) ≈ 2.639。分母 γ(β-1) = 1.4 × 1 = 1.4。因此，因子 = (2.639 – 1)/1.4 = 1.639/1.4 ≈ 1.171。

η = 1 – (1/3.178) × 1.171 = 1 – 0.3685 = 0.6315 ≈ 63.2 %.

η = 1 – (1/3.178) × 1.171 = 1 – 0.3685 = 0.6315 ≈ 63.2 %。

Compare with Otto cycle with same compression ratio: η_Otto = 1 – 1/r^γ-1 = 1 – 1/3.178 = 68.5 %. The Diesel cycle is slightly less efficient at the same r, but it can operate at higher compression ratios without knock.

与相同压缩比的奥托循环比较：η_Otto = 1 – 1/r^γ-1 = 1 – 1/3.178 = 68.5 %。在相同压缩比下，狄塞尔效率略低，但它可在不爆震的条件下采用更高的压缩比。

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7. Problem 6: Failure Mode and Effects Analysis (FMEA) | 问题 6：失效模式与影响分析 (FMEA)

A centrifugal pump is being designed for chemical transfer. Identify two potential failure modes, assign severity (S), occurrence (O), and detection (D) ratings on a 1–10 scale, and calculate the risk priority number (RPN). Suggest corrective actions to reduce the highest RPN.

一台用于化学品输送的离心泵正在设计中。请识别两种可能的失效模式，按 1–10 分制分配严重度 (S)、发生率 (O) 和检测度 (D)，计算风险优先级数 (RPN)，并针对最高 RPN 项提出纠正措施。

Failure Mode	Effect	S	O	D	RPN
Mechanical seal leakage	Chemical spill, safety hazard	9	4	5	180
Impeller cavitation damage	Loss of flow, vibration	7	6	4	168

RPN = S × O × D. The highest RPN (180) is for seal leakage. Corrective action: use a double mechanical seal with a flush system, and add a leakage sensor for early detection. This reduces O (more robust design) and D (sensor lowers detection rating).

RPN = S × O × D。最高 RPN (180) 来自密封泄漏。纠正措施：采用带冲洗系统的双端面机械密封，并增加泄漏传感器以实现早期检测，从而降低 O（更稳健的设计）和 D（传感器可降低检测度评分）。

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