📚 Year 12 SQA Engineering: Unit Test Mock Exam Walkthrough | Year 12 SQA 工程：单元测试模拟卷解析
This mock unit test is designed to mirror the style and depth of a typical SQA Higher Engineering Science assessment. It covers core topics in mechanics, thermodynamics, electrical principles, systems, and energy, providing a rigorous workout for students. The following walkthrough explains each question in detail, highlighting the reasoning and calculations needed to secure top marks.
这份模拟单元测试卷仿照 SQA 高级工程科学典型评估的模式和深度设计，涵盖力学、热力学、电学原理、系统及能量等核心主题，为学生提供一次严格的训练。以下解析详细讲解每道题目，突出解题思路和计算过程，帮助获得高分。
1. Overview of the Mock Exam | 模拟卷概览
The paper consists of seven structured questions, each targeting a different area of the Engineering Science specification. Marks are awarded for correct method, accurate numerical answers with appropriate units, and clear justifications where required. Questions range from direct calculations to extended analysis of engineering systems. Time management, careful unit conversion, and systematic working are essential for success.
本试卷由七道结构化的题目组成，每道题针对工程科学大纲的不同领域。评分依据正确的方法、准确的数值结果及单位，以及必要的清晰解释。题目从直接计算延伸到工程系统的扩展分析。时间管理、仔细的单位换算和系统的解题步骤是成功的关键。
2. Question 1: Axial Stress and Strain | 轴向应力与应变
A steel tie rod of circular cross‑section with diameter 20 mm is subjected to an axial tensile load of 50 kN. The Young’s modulus for steel is 200 GPa. Calculate the tensile stress in the rod and the longitudinal strain.
一根直径为20 mm的圆形截面钢拉杆受到50 kN的轴向拉伸载荷。钢材的杨氏模量为200 GPa。计算杆中的拉应力和纵向应变。
First, convert all quantities to consistent SI units: diameter d = 0.020 m, force F = 50 × 10³ N. The cross‑sectional area A = πd²/4 = π × (0.020)² / 4 = 3.1416 × 10⁻⁴ m². Tensile stress σ is given by the ratio of force to area.
首先将所有量转换为一致的SI单位：直径 d = 0.020 m，力 F = 50 × 10³ N。截面积 A = πd²/4 = π × (0.020)² / 4 = 3.1416 × 10⁻⁴ m²。拉应力 σ 等于力与面积的比值。
σ = F / A = 50 × 10³ N / 3.1416 × 10⁻⁴ m² = 159.2 × 10⁶ Pa = 159.2 MPa
Stress is well within the typical yield strength of structural steel. Strain ε is found from Hooke’s law: σ = E ε, so ε = σ / E.
该应力远低于结构钢的典型屈服强度。应变 ε 由胡克定律求得：σ = E ε，故 ε = σ / E。
ε = 159.2 × 10⁶ Pa / 200 × 10⁹ Pa = 7.96 × 10⁻⁴ (dimensionless)
This means the rod stretches by approximately 0.08% of its original length. Students often forget to convert mm² to m² or mix GPa and MPa; always use base SI units to avoid scale errors.
这意味着杆的伸长量约为原始长度的0.08%。学生常忘记将mm²换算为m²或混淆GPa与MPa；务必使用基本SI单位以避免数量级错误。
3. Question 2: Simply Supported Beam – Shear Force and Bending Moment | 简支梁 – 剪力与弯矩
A simply supported beam of length 4 m carries a point load of 8 kN at its mid‑span. Neglect the weight of the beam. Determine the reactions at the supports and sketch the shear force and bending moment diagrams, giving principal values.
一根长4 m的简支梁在中点承受8 kN的集中载荷。忽略梁的自重。求支座反力，并画出剪力图和弯矩图，标出主要数值。
By symmetry, the upward reactions at each support are equal: Rₐ = R₆ = F/2 = 4 kN. With sign conventions: shear force is positive when causing a clockwise rotation to the right of a section. Moving from left to right, shear force jumps to +4 kN at A, remains constant, then drops by 8 kN at mid‑span to -4 kN, and returns to zero at B.
由对称性，两端支座的向上反力相等：Rₐ = R₆ = F/2 = 4 kN。按照规定符号约定：使截面右边发生顺时针转动的剪力为正。从左向右移动，在A处剪力跃升至+4 kN，保持恒定，到跨中时突降8 kN至-4 kN，最后在B处回归零。
The bending moment at a section x from the left support for 0 ≤ x ≤ 2 m is M = 4x (kN·m). It increases linearly from 0 to a maximum at mid‑span. At x = 2 m, Mₘₐₓ = 4 × 2 = 8 kN·m. For the right half, M decreases linearly: M = 4(4 – x) = 16 – 4x (kN·m). The moment diagram is triangular.
距左支座x处（0 ≤ x ≤ 2 m）的弯矩为M = 4x (kN·m)，从0线性增加到跨中最大值。在x = 2 m处，Mₘₐₓ = 4 × 2 = 8 kN·m。右半段弯矩线性递减：M = 4(4 – x) = 16 – 4x (kN·m)。弯矩图为三角形。
In examinations, clearly label diagrams with values at key points: reaction forces, location of zero shear, and maximum moment. Also state that the point of maximum bending moment occurs where the shear force changes sign.
考试中应当在图上清晰标注关键点的数值：反力、剪力为零的位置及最大弯矩。同时指明最大弯矩出现在剪力改变符号的位置。
4. Question 3: Ideal Gas Law and Polytropic Process | 理想气体定律与多方过程
Air is compressed in a cylinder from an initial state of 1 bar and 300 K to a final pressure of 8 bar. The process follows a polytropic relation pVⁿ = constant, with n = 1.3. Find the final temperature and the work done during compression per kg of air. For air, R = 0.287 kJ/kg·K.
空气在气缸中从初态1 bar、300 K压缩到终压8 bar，过程遵循多方关系 pVⁿ = constant，指数 n = 1.3。求终态温度及每千克空气的压缩功。空气气体常数 R = 0.287 kJ/kg·K。
For a polytropic process, T₂ / T₁ = (p₂ / p₁)^((n-1)/n). Substituting: T₂ = 300 K × (8/1)^((1.3-1)/1.3) = 300 × (8)^(0.3/1.3). 0.3/1.3 ≈ 0.2308, 8^0.2308 ≈ 1.617. So T₂ ≈ 300 × 1.617 = 485 K.
对于多方过程，T₂ / T₁ = (p₂ / p₁)^((n-1)/n)。代入数值：T₂ = 300 K × (8/1)^((1.3-1)/1.3) = 300 × (8)^(0.3/1.3)。0.3/1.3 ≈ 0.2308，8^0.2308 ≈ 1.617。因此 T₂ ≈ 300 × 1.617 = 485 K。
Work done per kg is w = (p₂v₂ – p₁v₁) / (1 – n) or equivalently w = R (T₂ – T₁) / (1 – n). Using this form: w = 0.287 × (485 – 300) / (1 – 1.3) = 0.287 × 185 / (-0.3) = -177 kJ/kg. The negative sign indicates work done on the gas (compression).
每千克气体作功量 w = R (T₂ – T₁) / (1 – n)。代入：w = 0.287 × (485 – 300) / (1 – 1.3) = 0.287 × 185 / (-0.3) = -177 kJ/kg。负号表示对气体做功（压缩）。
Common pitfalls: forgetting to convert bar to absolute pressure (since it is a ratio, units cancel), misusing the polytropic index exponent, or omitting the sign for work. Always specify if work is on or by the system.
常见错误：忘记将bar转换为绝对压力（此处为比值，单位可消去），错误使用多方指数，或遗漏功的符号。务必指明系统是对外做功还是接受功。
5. Question 4: DC Circuit Analysis – Kirchhoff’s Laws | 直流电路分析 – 基尔霍夫定律
A network consists of a 12 V battery with internal resistance 1 Ω, connected in series with a 3 Ω resistor and a parallel combination of 6 Ω and 4 Ω. Calculate the current through the 4 Ω resistor using Kirchhoff’s laws and verify by calculating total resistance.
某电路包含一个12 V电池（内阻1 Ω），串联一个3 Ω电阻，再与6 Ω和4 Ω的并联组合串联。用基尔霍夫定律计算通过4 Ω电阻的电流，并通过计算总电阻进行验证。
Label the parallel branch currents: let I₁ be current through 6 Ω, I₂ through 4 Ω, so I₁ + I₂ = I (total current from battery). Apply KVL to the left loop containing battery, internal resistance, 3 Ω, and the parallel pair where the p.d. across the parallel section is Vₚ = 6 I₁ = 4 I₂.
标记并联支路电流：设I₁为流过6 Ω的电流，I₂为流过4 Ω的电流，则I₁ + I₂ = I（电池总电流）。对包含电池、内阻、3 Ω电阻及并联部分的左回路应用基尔霍夫电压定律，并联段电压 Vₚ = 6 I₁ = 4 I₂。
From 6 I₁ = 4 I₂, we get I₁ = (2/3) I₂. Then total current I = I₁ + I₂ = (2/3) I₂ + I₂ = (5/3) I₂. The total loop equation: 12 V = I(1 Ω + 3 Ω) + Vₚ = I × 4 + 4 I₂. Substitute I = (5/3) I₂: 12 = 4 × (5/3) I₂ + 4 I₂ = (20/3) I₂ + (12/3) I₂ = (32/3) I₂. Thus I₂ = 12 × 3 / 32 = 36/32 = 1.125 A.
由6 I₁ = 4 I₂ 得 I₁ = (2/3) I₂。总电流 I = I₁ + I₂ = (2/3) I₂ + I₂ = (5/3) I₂。全回路方程：12 V = I(1 Ω + 3 Ω) + Vₚ = I × 4 + 4 I₂。代入I = (5/3) I₂：12 = 4 × (5/3) I₂ + 4 I₂ = (20/3) I₂ + (12/3) I₂ = (32/3) I₂。于是 I₂ = 12 × 3 / 32 = 1.125 A。
Verification by total resistance: 6 Ω // 4 Ω = (6×4)/(6+4) = 2.4 Ω. Total circuit resistance = 1 + 3 + 2.4 = 6.4 Ω. Total current I = 12/6.4 = 1.875 A. Then I₂ = I × (6/(6+4)) = 1.875 × 0.6 = 1.125 A, confirming the result. Always check consistency.
用总电阻验证：6 Ω // 4 Ω = 2.4 Ω，电路总电阻 = 1 + 3 + 2.4 = 6.4 Ω，总电流 I = 12/6.4 = 1.875 A。则 I₂ = I × (6/(6+4)) = 1.875 × 0.6 = 1.125 A，结果一致。务必检查一致性。
6. Question 5: Operational Amplifier Circuits | 运算放大器电路
An inverting amplifier uses an ideal op‑amp with input resistor Rᵢ = 2 kΩ and feedback resistor Rₓ = 10 kΩ. The input voltage vᵢ is a 0.5 V peak sinusoidal signal. Determine the voltage gain, the output voltage waveform, and the input impedance seen by the signal source.
一个反向放大器采用理想运放，输入电阻 Rᵢ = 2 kΩ，反馈电阻 Rₓ = 10 kΩ。输入电压 vᵢ 为峰值0.5 V的正弦信号。求电压增益、输出电压波形以及信号源看到的输入阻抗。
For an ideal inverting amplifier, the closed‑loop voltage gain A = -Rₓ / Rᵢ = -10 kΩ / 2 kΩ = -5. The negative sign indicates a 180° phase inversion. Therefore, the output amplitude is 5 × 0.5 V = 2.5 V peak, and the output is an inverted sine wave.
对于理想反向放大器，闭环电压增益 A = -Rₓ / Rᵢ = -10 kΩ / 2 kΩ = -5。负号表示180°相位反转。因此输出幅值为5 × 0.5 V = 2.5 V峰值，输出为反相正弦波。
Due to the virtual earth at the inverting input, the input impedance is simply Zᵢₙ = Rᵢ = 2 kΩ. This is because no current flows into the op‑amp inputs, and the potential at the inverting terminal is virtually zero; all input current flows through Rᵢ to ground potential.
由于反相输入端的虚地特性，输入阻抗即为 Zᵢₙ = Rᵢ = 2 kΩ。这是因为没有电流流入运放输入端，且反相端电位虚为零；所有输入电流经 Rᵢ 流入地电位。
In an exam answer, always state assumptions: ideal op‑amp (infinite open‑loop gain, infinite input impedance, zero output impedance). Show the formula, substitute values, and comment on the significance of the sign. A small sketch can help illustrate the waveform inversion.
考试答题时，务必说明假设：理想运放（无限开环增益、无限输入阻抗、零输出阻抗）。写出公式，代入数值，并说明负号的意义。一幅小草图有助于表明波形的反相。
7. Question 6: Control Systems – Transfer Function and Stability | 控制系统 – 传递函数与稳定性
A position control system has an open‑loop transfer function G(s) = 20 / (s(s + 4)(s + 5)). Determine the closed‑loop transfer function for unity feedback and, using the Routh‑Hurwitz criterion, establish whether the closed‑loop system is stable.
一个位置控制系统的开环传递函数为 G(s) = 20 / (s(s + 4)(s + 5))。求单位反馈下的闭环传递函数，并利用劳斯‑赫尔维茨判据判断闭环系统的稳定性。
For unity feedback, the closed‑loop transfer function is T(s) = G(s) / (1 + G(s)). The characteristic equation is 1 + G(s) = 0 ⇒ s(s+4)(s+5) + 20 = 0. Expand the denominator: s(s² + 9s + 20) = s³ + 9s² + 20s. Adding 20 gives s³ + 9s² + 20s + 20 = 0.
单位反馈下，闭环传递函数 T(s) = G(s) / (1 + G(s))。特征方程为 1 + G(s) = 0，即 s(s+4)(s+5) + 20 = 0。展开分母：s³ + 9s² + 20s + 20 = 0。
Construct the Routh array:
| s³ | 1 | 20 |
| s² | 9 | 20 |
| s¹ | (9×20 – 1×20)/9 = 160/9 ≈ 17.78 | 0 |
| s⁰ | 20 |
All elements in the first column (1, 9, 160/9, 20) are positive, indicating no sign changes. Therefore, the closed‑loop system is stable. Had any element been negative, the number of sign changes would equal the number of right‑half‑plane poles.
劳斯表第一列所有元素（1, 9, 160/9, 20）均为正，表明无符号变化。因此闭环系统是稳定的。若出现负元素，符号变化的次数即等于右半平面极点个数。
Students should check arithmetic carefully, particularly when computing the s¹ row. They should also note that a zero in the first column requires special treatment (epsilon method), though not needed here. The Routh‑Hurwitz criterion is a key tool for assessing stability without solving for roots.
学生应仔细检查计算，尤其是在计算 s¹ 行时。还需注意若第一列出现零需要特殊处理（ε方法），不过此处不需要。劳斯‑赫尔维茨判据是评价稳定性而不需求根的重要工具。
8. Question 7: Energy and Efficiency in a System | 系统能量与效率
An electric motor drives a pump that lifts water through a height of 12 m. The motor draws 2.5 kW from the supply and operates at 85% efficiency. The pump mechanism has an efficiency of 70%. Determine the mass flow rate of water if gravitational acceleration g = 9.81 m/s².
一台电动机驱动水泵将水提升12 m高度。电动机从电源吸取2.5 kW，运行效率为85%。泵机构的效率为70%。求水的质量流量，取重力加速度 g = 9.81 m/s²。
First, find the useful mechanical power delivered to the pump shaft: Pₛₕₐₓₜ = 2.5 kW × 0.85 = 2.125 kW. Then, the hydraulic power actually transferred to the water is Pₕ = Pₛₕₐₓₜ × pump efficiency = 2.125 kW × 0.70 = 1.4875 kW = 1487.5 W.
先求传递到泵轴的有用机械功率：Pₛₕₐₓₜ = 2.5 kW × 0.85 = 2.125 kW。再求出实际传递给水的液压功率：Pₕ = Pₛₕₐₓₜ × 泵效率 = 2.125 kW × 0.70 = 1.4875 kW = 1487.5 W。
Hydraulic power for lifting water is Pₕ = ṁ g h, where ṁ is mass flow rate (kg/s). Rearranged: ṁ = Pₕ / (g h) = 1487.5 / (9.81 × 12) = 1487.5 / 117.72 ≈ 12.64 kg/s. In litres per second (density 1000 kg/m³), that is about 12.64 L/s.
提升水的液压功率为 Pₕ = ṁ g h，其中 ṁ 为质量流量（kg/s）。整理得：ṁ = Pₕ / (g h) = 1487.5 / (9.81 × 12) = 1487.5 / 117.72 ≈ 12.64 kg/s。换算为升每秒（密度1000 kg/m³），约为12.64 L/s。
When cascading efficiencies, multiply them sequentially. Watch the conversion between kW and W, and always include the correct unit in the final answer. In open‑ended engineering problems, a quick sanity check on the order of magnitude helps avoid mistakes.
当效率级联时，需顺序相乘。注意千瓦与瓦的换算，并在最终答案中使用正确单位。在开放式工程问题中，对数量级进行快速合理性检查有助于避免错误。
9. Common Mistakes and Tips | 常见错误与技巧
Many candidates lose marks by neglecting unit conversions (e.g. mm to m, bar to Pa), misapplying sign conventions in beam analysis, or forgetting to state assumptions for ideal components. Always write out the governing equation before substituting numbers, and double‑check the order of operations when using calculators for exponents. In electrical questions, clearly define current directions at the start.
许多考生因忽视单位换算（如mm转m，bar转Pa）、在梁分析中搞错符号约定或忘记陈述理想元件的假设而扣分。务必先写出控制方程再代入数值，并用计算器计算指数时仔细核对运算顺序。在电学题中，一开始就明确规定电流方向。
For thermodynamics, distinguish between Celsius and Kelvin; always use absolute temperature in gas law calculations. When drawing diagrams, label axes and key values – unlabelled sketches receive no credit. Lastly, show all working step by step so that even if the final answer is wrong, method marks can be awarded.
热力学中，注意区分摄氏度和开尔文温标；气体定律计算中必须使用绝对温度。绘制图表时要标注坐标轴和关键数值——未标注的草图不给分。最后，展示所有解题步骤，即使最终答案错误也可获得方法分。
10. Final Review and Key Equations | 最终复习与关键方程
Stress and Strain: σ = F/A, ε = δL/L, E = σ/ε. Beam bending: M/I = σ/y = E/R. Gases: pV = mRT, pVⁿ = const, w = (p₂V₂ – p₁V₁)/(1-n). Circuits: V = IR, ΣI = 0, ΣV = 0. Op‑amp: A = -Rₓ/Rᵢ (inverting). Control: T(s) = G(s)/(1+G(s)H(s)). Energy: P = ṁgh, η = Pₒᵤₜ/Pᵢₙ. Frequent practise with mixed‑unit problems and past paper questions is the best preparation.
应力与应变：σ = F/A，ε = δL/L，E = σ/ε。梁弯曲：M/I = σ/y = E/R。气体：pV = mRT，pVⁿ = const，w = (p₂V₂ – p₁V₁)/(1-n)。电路：V = IR，ΣI = 0，ΣV = 0。运放：A = -Rₓ/Rᵢ（反向）。控制：T(s) = G(s)/(1+G(s)H(s))。能量：P = ṁgh，η = Pₒᵤₜ/Pᵢₙ。经常练习混合单位问题及历年真题是最佳准备方式。
This mock exam reinforces the interconnected nature of engineering science. A deep understanding of fundamental principles, rather than memorisation, allows you to tackle unfamiliar contexts with confidence. Review each step, ask “why”, and always connect the mathematics to the physical situation.
这份模拟卷强化了工程科学各部分的相互联系。对基本原理的深入理解，而非死记硬背，能让你从容应对不熟悉的情境。复习每一步，多问“为什么”，并始终将数学与物理情境联系起来。
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