📚 Pre-U CIE Engineering: Unit Test Mock Paper Analysis | Pre-U CIE 工程：单元测试模拟卷解析
This article provides a comprehensive walkthrough of a representative Pre-U CIE Engineering unit test mock paper. We will dissect key questions spanning mechanics, thermofluids, electrical circuits, materials selection and control systems. By understanding the detailed solutions and methodology, you can reinforce core concepts and sharpen your exam technique.
本文对一份有代表性的 Pre-U CIE 工程单元测试模拟卷进行了全面解析。我们将剖析涵盖力学、热流体、电路、材料选择和控制系统等关键题目。通过理解详尽的解题过程和方法，你可以巩固核心概念，提升应试技巧。
1. Mock Paper Structure | 模拟卷结构
The mock paper replicates the actual unit test format. It consists of Section A with 20 multiple-choice questions worth 40 marks, and Section B containing 5 structured questions totalling 60 marks. Topics are distributed across Mechanics, Thermofluids, Electrical & Electronic Engineering, Materials, and Systems & Control.
本模拟卷复制了真实单元测试的格式。试卷包含 Section A 的 20 道选择题（40 分）以及 Section B 的 5 道结构化大题（60 分）。专题分布在力学、热流体、电气与电子工程、材料以及系统与控制。
The multiple-choice items assess breadth of knowledge and quick problem-solving, while structured questions demand detailed calculations, explanations and application of principles. The total time allowed is 1 hour 30 minutes, requiring effective time management.
选择题评估知识的广度和快速解题能力，而结构化大题则要求详细的计算、解释以及原理应用。总考试时间为 1 小时 30 分钟，需要有效的时间管理。
2. Core Concepts Assessed | 评估的核心概念
The table below summarises the major themes and specific concepts tested in the mock paper. Familiarity with these topics is essential for success in Pre-U Engineering.
下表总结了模拟卷中考察的主要主题和具体概念。熟悉这些专题对在 Pre-U 工程中取得成功至关重要。
| Topic Area | 专题领域 | Key Concepts | 核心概念 |
|---|---|
| Mechanics of Materials | 材料力学 | Stress σ, strain ε, Young's modulus E, yield strength, UTS, stress-strain curves |
| Thermofluids | 热流体 | Ideal Rankine cycle, enthalpy, entropy, dryness fraction, thermal efficiency, Bernoulli equation, continuity |
| Electrical Circuits | 电路 | Ohm's law, Kirchhoff's current/voltage laws (KCL/KVL), mesh analysis, power dissipation |
| Materials Selection | 材料选择 | Ashby charts, performance indices (M₁ = E1/2/ρ, M₂ = σ_y2/3/ρ), material properties |
| Systems & Control | 系统与控制 | Block diagram reduction, transfer functions, feedback loops, Mason's rule basics |
A solid grasp of these fundamentals enables confident tackling of both familiar and novel problems. Each question in the mock paper is designed to probe at least two of these areas, often in combination.
扎实掌握这些基础知识能让你自信地应对熟悉和新颖的问题。模拟卷中的每道题目都被设计为考察至少两个领域的结合。
3. Question 1: Tensile Test and Stress-Strain Analysis | 题目 1：拉伸试验与应力-应变分析
s 0.1 mm; at yield onset the load is 38 kN with extension 0.25 mm; the maximum load is 55 kN at an extension of 5 mm; and fracture occurs at 45 kN, extension 7 mm. Assume constant cross-section until necking begins. Determine (a) engineering stress σ and strain ε at each point, (b) Young's modulus E, (c) yield strength σ_y, and (d) ultimate tensile strength UTS.
一根原始标距 L₀ = 50 mm、直径 d = 10 mm 的钢试样进行拉伸。记录数据为：载荷 20 kN 时延伸量 0.1 mm；屈服开始载荷 38 kN，延伸量 0.25 mm；最大载荷 55 kN，延伸量 5 mm；断裂发生在 45 kN，延伸量 7 mm。假设颈缩前截面积不变。求 (a) 各点的工程应力 σ 和应变 ε，(b) 杨氏模量 E，(c) 屈服强度 σ_y，(d) 抗拉强度 UTS。
³)² / 4 = 7.854 × 10⁻⁵ m². This area will be used for all engineering stress calculations.
步骤 1：计算原始横截面积 A₀。A₀ = πd²/4 = π × (10 × 10⁻³)² / 4 = 7.854 × 10⁻⁵ m²。该面积将用于所有工程应力计算。
Engineering stress: σ = F / A₀
Engineering strain: ε = ΔL / L₀
0 = 0.005. At maximum load (55 kN): UTS = 55×10³ / 7.854×10⁻⁵ ≈ 700.3 MPa; ε_UTS = 5/50 = 0.1. At fracture (nominal stress still uses original area): σ_f = 45×10³ / 7.854×10⁻⁵ ≈ 572.8 MPa; ε_f = 7/50 = 0.14.
在 20 kN 点：σ = 20×10³ / 7.854×10⁻⁵ ≈ 254.6 MPa；ε = 0.1/50 = 0.002（2.0×10⁻³）。屈服点（38 kN）：σ_y = 38×10³ / 7.854×10⁻⁵ ≈ 483.7 MPa；ε_y = 0.25/50 = 0.005。最大载荷点（55 kN）：UTS = 55×10³ / 7.854×10⁻⁵ ≈ 700.3 MPa；ε_UTS = 5/50 = 0.1。断裂点（名义应力仍用原始面积）：σ_f = 45×10³ / 7.854×10⁻⁵ ≈ 572.8 MPa；ε_f = 7/50 = 0.14。
Young' cal for steel. The yield strength is 483.7 MPa and the UTS is 700.3 MPa. Always confirm that the strain used for E corresponds to the linear portion.
杨氏模量由线弹性段求得。使用 20 kN 数据点：E = σ/ε = 254.6 MPa / 0.002 = 127.3 × 10⁹ Pa = 127.3 GPa。该值符合钢材典型范围。屈服强度为 483.7 MPa，UTS 为 700.3 MPa。务必确认用于计算 E 的应变落在线性段。
A common mistake is confusing engineering stress with true stress. In this exam, unless specified, use the original area. Also check unit conversions; here mm and kN must be in metres and Newtons to obtain Pascals.
常见错误是混淆工程应力与真实应力。本考试中除非特别说明，一律使用原始面积。同时注意单位换算；此处 mm 和 kN 须化为米和牛顿以得到帕斯卡。
4. Question 2: Ideal Rankine Cycle Analysis | 题目 2：理想朗肯循环分析
superheated vapour at 400°C. Given the following extracted steam table data: at 2 MPa and 400°C, h₃ = 3248 kJ/kg, s₃ = 7.127 kJ/(kg·K); at 10 kPa, saturation data: hf = 191.8 kJ/kg, hg = 2584 kJ/kg, sf = 0.649 kJ/(kg·K), sg ump work. Determine the dryness fraction at turbine exit and the cycle thermal efficiency.
一简单理想朗肯循环运行在锅炉压力 2 MPa（20 bar）和冷凝器压力 10 kPa（0.1 bar）之间。蒸汽以 400°C 过热状态进入透平。给定蒸汽表数据：在 2 MPa、400°C，h₃ = 3248 kJ/kg，s₃ = 7.127 kJ/(kg·K)；在 10 kPa 饱和数据：hf = 191.8 kJ/kg，hg = 2584 kJ/kg，sf = 0.649 kJ/(kg·K)，sg = 8.150 kJ/(kg·K)。忽略泵功。求透平出口干度及循环热效率。
Isentropic expansion: s₄ = s₃
127 kJ/(kg·K). Using the condenser condition, the dryness fraction x₄ is found from s₄ = sf + x₄(sg - sf (8.150 - 0.649) = 0.649 + x₄ × 7.501. Solving gives x₄ = (7.127 - 0.649) / 7.501 ≈ 0.864.
透平出口（状态 4）为等熵过程，因此 s₄ = s₃ = 7.127 kJ/(kg·K)。利用冷凝器条件，干度 x₄ 由 s₄ = sf + x₄(sg - sf) 求得。故 7.127 = 0.649 + x₄ × (8.150 - 0.649)，解得 x₄ = (7.127 - 0.649) / 7.501 ≈ 0.864。
Then the exit enthalpy h₄ = hf + x₄(hg - hf (2584 - 191.8) = 191.8 + 0.864 × 2392.2 ≈ 191.8 + 2067.9 = 2259.7 kJ/kg. The heat input qin = h₃ - h₂. Since pump work is neglected, h₂ ≈ h₁ = hf at condenser pressure = 191.8 kJ/kg. Hence qin = 3248 - 191.8 = 3056.2 kJ/kg. The turbine work output wt = h₃ - h₄ = 3248 - 2259.7 = 988.3 kJ/kg. The thermal efficiency η = wt / qin = 988.3 / 3056.2 ≈ 0.3234 or 32.3%.
则出口焓 h₄ = hf + x₄(hg - hf) = 191.8 + 0.864 × (2584 - 191.8) ≈ 2259.7 kJ/kg。输入热量 qin = h₃ - h₂，忽略泵功有 h₂ ≈ h₁ = 冷凝压力下饱和水焓 = 191.8 kJ/kg。因此 qin = 3248 - 191.8 = 3056.2 kJ/kg。透平做功 wt = h₃ - h₄ = 988.3 kJ/kg。热效率 η = wt / qin ≈ 0.3234 即 32.3%。
Always double-check the isentropic assumption and the correct substitution of saturation data. Many students forget that pump work, though small, affects h₂ slightly, but for this unit test the approximation is accepted. Also note that the back-work ratio is negligible in steam cycles.
务必检查等熵假设以及饱和数据的正确代入。许多学生忘记泵功虽小但会轻微影响 h₂，但在此单元测试中允许近似。同时注意蒸汽循环中回功比可忽略。
5. Question 3: DC Circuit Analysis Using Kirchhoff's Laws | 题目 3：运用基尔霍夫定律分析直流电路
Consider a circuit with two voltage sources and three resistors. A 12 V source is in series with a 2 Ω resistor in the left branch. A 6 V source is in series with a 1 Ω resistor in the right branch. These two branches connect at nodes A and B, and a 4 Ω resistor is placed between nodes A and B. Determine the three branch currents I₁ (left), I₂ (right) and I₃ (middle 4 Ω) using mesh analysis or KCL/KVL. Assume currents flow from A to B through the 4 Ω resistor.
考虑一个含两个电压源和三个电阻的电路。左支路为 12 V 电源串接 2 Ω 电阻，右支路为 6 V 电源串接 1 Ω 电阻。两支路在节点 A 和 B 相连，且 A、B 之间跨接一个 4 Ω 电阻。用网孔分析法或 KCL/KVL 求三个支路电流 I₁（左）、I₂（右）和 I₃（中间 4 Ω）。假设电流从 A 流向 B 经过 4 Ω 电阻。
Ω - 4 Ω): 12 = 2I₁ + 4I₃ (equation 2). For the right mesh (6 V - 1 Ω - 4 Ω): 6 = 1I₂ - 4I₃? s 4 Ω, the drop is +4×I₃ (current I₃ flows from A to B, so the mesh current sees the same direction drop). However I₂ is defined downward through the 1 Ω. Better to use loop currents: Let mesh current i₁ loop left (clockwise) and mesh current i₂ loop right (clockwise). Then the current through the 4 Ω is i₁ - i₂ (downward). Using mesh equations:
在节点 A 应用 KCL：I₁ = I₂ + I₃（方程 1）。左网孔（12 V - 2 Ω - 4 Ω）：12 = 2I₁ + 4I₃（方程 2）。右网孔需仔细确定极性：定义网孔电流 i₁（左回路顺时针）与 i₂（右回路顺时针）。流过 4 Ω 的电流为 i₁ - i₂（方向向下）。网孔方程：
Mesh 1: 12 = 2i₁ + 4(i₁ - i₂)
Mesh 2: -6 = 1i₂ + 4(i₂ - i₁)
₁ = 2.571 A. Then from (1): 12 = 6×2.571 - 4i₂ => 12 = 15.426 - 4i₂ => 4i₂ = 3.426 => i₂ = 0.857 A. The branch currents: I₁ = i₁ = 2.571 A; I₂ = i₂ = 0.857 A; I₃ = i₁ - i₂ = 1.714 A. Check KCL: I₁ = I₂ + I₃ => 2.571 = 0.857 + 1.714, correct.
简化得：(1) 12 = 6i₁ - 4i₂。(2) -6 = -4i₁ + 5i₂。联立求解：5×(1) 加 4×(2) 得 60 - 24 = 30i₁ - 20i₂ -16i₁ +20i₂ => 36 = 14i₁ => i₁ = 2.571 A。代入得 i₂ = 0.857 A。支路电流：I₁ = i₁ = 2.571 A，I₂ = i₂ = 0.857 A，I₃ = i₁ - i₂ = 1.714 A。KCL 验证正确。
<Published by TutorHao | Pre-U 工程 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图