📚 Year 12 CIE Engineering: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练
In CIE Engineering, examination questions often require you to integrate knowledge from multiple disciplines such as mechanics, materials science, electronics and energy systems. This article presents a series of interdisciplinary problem-solving exercises that mimic the style of real CIE exam questions. Working through these will sharpen your ability to connect concepts and apply them to complex engineering scenarios.
在CIE工程中，考题常常要求你综合运用力学、材料科学、电子学和能量系统等多个学科的知识。本文提供一系列模拟真实CIE考题风格的跨学科综合训练题，通过这些练习，可以增强你连接概念并应用于复杂工程情境的能力。
1. Static Equilibrium & Material Selection in a Cantilever | 悬臂梁的静力学平衡与材料选择
A cantilever beam of length 1.5 m supports a point load of 2 kN at its free end. The beam has a rectangular cross‑section; width is fixed at 50 mm. Two candidate materials are steel (allowable bending stress 250 MPa, Young’s modulus 210 GPa, density 7850 kg/m³) and aluminium alloy (allowable stress 140 MPa, modulus 70 GPa, density 2700 kg/m³). Determine the minimum depth required for each material to satisfy strength, and select the lighter beam. The deflection must also be kept below L⁄200. Assess which material meets both criteria with the smallest mass.
一根长度为 1.5 m 的悬臂梁在其自由端承受 2 kN 的集中载荷。梁的截面为矩形，宽度固定为 50 mm。候选材料有两种：钢（许用弯曲应力 250 MPa，杨氏模量 210 GPa，密度 7850 kg/m³）与铝合金（许用应力 140 MPa，模量 70 GPa，密度 2700 kg/m³）。试求每种材料满足强度所需的最小深度，并选择质量较轻的梁；同时挠度须限制在 L⁄200 以下。评断哪一种材料在满足两项准则的前提下质量最小。
Begin with strength: maximum bending moment M = F × L. Section modulus for a rectangle Z = b d² / 6. For bending, σ = M / Z ≤ σallow. Rearranging gives d ≥ √(6 F L / (b σallow)). Substituting values, for steel: d ≥ √(6 × 2000 N × 1.5 m / (0.05 m × 250 × 10⁶ Pa)) ≈ 0.0379 m = 37.9 mm. For aluminium: d ≥ √(6 × 2000 × 1.5 / (0.05 × 140 × 10⁶)) ≈ 0.0507 m = 50.7 mm. Now check deflection: δ = F L³ / (3 E I) with I = b d³ / 12. The limit is L/200 = 0.0075 m. Compute for the strength‑sized beams.
先从强度入手：最大弯矩 M = F × L。矩形截面模量 Z = b d² / 6。弯曲应力 σ = M / Z ≤ σallow。整理得 d ≥ √(6 F L / (b σallow))。代入数值，钢的场合：d ≥ √(6 × 2000 N × 1.5 m / (0.05 m × 250 × 10⁶ Pa)) ≈ 0.0379 m = 37.9 mm。铝合金的场合：d ≥ √(6 × 2000 × 1.5 / (0.05 × 140 × 10⁶)) ≈ 0.0507 m = 50.7 mm。接下来校核挠度：δ = F L³ / (3 E I)，其中 I = b d³ / 12。界限为 L/200 = 0.0075 m。针对按强度选定的截面计算。
δ = F L³ ÷ (3 E I)
Steel with d = 0.0379 m gives I = (0.05)(0.0379)³ / 12 = 2.27 × 10⁻⁷ m⁴, δ = 2000 × (1.5)³ / (3 × 210 × 10⁹ × 2.27 × 10⁻⁷) ≈ 0.047 m, which exceeds the limit. Hence depth must be increased. For aluminium, I = (0.05)(0.0507)³ / 12 = 5.43 × 10⁻⁷ m⁴, δ = 2000 × (1.5)³ / (3 × 70 × 10⁹ × 5.43 × 10⁻⁷) ≈ 0.059 m, also over limit. To meet deflection, set δ = 0.0075 m and solve d³ = 4 F L³ / (E b × 0.0075). For steel, required d ≈ 0.064 m; for aluminium, d ≈ 0.089 m. The mass per unit length is ρ b d. Compare: steel mass ∝ 7850 × 0.064 = 502, aluminium ∝ 2700 × 0.089 = 240. Aluminium wins, giving a lighter beam despite larger depth.
钢梁取 d = 0.0379 m 时 I = (0.05)(0.0379)³ / 12 = 2.27 × 10⁻⁷ m⁴，挠度 δ = 2000 × (1.5)³ / (3 × 210 × 10⁹ × 2.27 × 10⁻⁷) ≈ 0.047 m，超过限制。因而须增大深度。铝合金梁 I = (0.05)(0.0507)³ / 12 = 5.43 × 10⁻⁷ m⁴，δ ≈ 0.059 m，同样超限。为满足挠度条件，令 δ = 0.0075 m，解出 d³ = 4 F L³ / (E b × 0.0075)。钢所需 d ≈ 0.064 m；铝合金所需 d ≈ 0.089 m。单位长度质量 ∝ ρ b d。比较得钢质量因子 7850 × 0.064 = 502，铝 2700 × 0.089 = 240。铝虽深度较大，却更轻，成为最佳选择。
2. DC Motor Efficiency with Mechanical Load | 带机械负载的直流电机效率
A 24 V DC motor lifts a 25 kg mass at a steady speed of 0.8 m/s using a light pulley. Current drawn is 12 A. The armature resistance is 0.5 Ω. Determine (a) the mechanical output power, (b) electrical input power, (c) total losses, (d) motor efficiency, and (e) the energy lost as heat in the armature per minute. Also comment on where the remaining losses occur.
一台 24 V 直流电机通过轻型滑轮组以 0.8 m/s 的恒速提升 25 kg 的重物，工作电流为 12 A，电枢电阻为 0.5 Ω。求：(a) 机械输出功率；(b) 电输入功率；(c) 总损耗；(d) 电机效率；(e) 每分钟在电枢中以热量形式损耗的能量。并说明其余损耗发生在何处。
Mechanical power: Pout = force × velocity = m g v = 25 kg × 9.81 m/s² × 0.8 m/s ≈ 196.2 W. Electrical input: Pin = V × I = 24 V × 12 A = 288 W. Total losses = Pin – Pout = 91.8 W. Efficiency η = Pout / Pin × 100% = 196.2 / 288 × 100% ≈ 68.1%.
机械功率：Pout = 力 × 速度 = m g v = 25 kg × 9.81 m/s² × 0.8 m/s ≈ 196.2 W。电输入功率：Pin = V × I = 24 V × 12 A = 288 W。总损耗 = 288 – 196.2 = 91.8 W。效率 η = Pout / Pin × 100% ≈ 68.1%。
Copper loss (I²R) in armature: PCu = I² × R = (12 A)² × 0.5 Ω = 72 W. In one minute, energy dissipated as heat = 72 J/s × 60 s = 4320 J. The difference, 91.8 W – 72 W = 19.8 W, represents iron losses, friction and windage. Cross‑discipline insight: the wasted heat must be removed to avoid insulation damage, linking to thermal management.
电枢铜损 (I²R)：PCu = (12 A)² × 0.5 Ω = 72 W。每分钟散热能量 = 72 J/s × 60 s = 4320 J。差额 91.8 W – 72 W = 19.8 W 即为铁耗、摩擦和风阻损耗。跨学科启示：多余的热量必须被带走以免损坏绝缘，这关联到热管理。
3. Solar‑Powered Pumping System | 太阳能水泵系统
A photovoltaic panel with an area of 2.0 m² receives solar irradiance of 800 W/m². Its conversion efficiency is 18%. The panel drives a water pump that lifts water 15 m through a pipe. The pump and motor combined efficiency is 70%. Calculate the maximum volume of water that can be lifted per hour. If a storage battery of 12 V, 100 Ah is used to power the pump during cloudy intervals, for how many minutes can the system operate at half the rated flow without sunlight?
一块面积为 2.0 m² 的光伏板接收 800 W/m² 的太阳辐照，转换效率为 18%。该板驱动一台水泵，将水提升 15 m，泵与电机综合效率为 70%。计算每小时最多可提升多少体积的水。若使用一块 12 V、100 Ah 的蓄电池在多云间隙为水泵供电，在无阳光时系统以额定流量的一半能运行多少分钟？
Solar power captured: Psolar = irradiance × area × ηpv = 800 W/m² × 2.0 m² × 0.18 = 288 W. Mechanical power available at pump: Ppump = 288 W × 0.70 = 201.6 W. The power required to lift water: P = ρ g h Q, where ρ = 1000 kg/m³, g = 9.81, h = 15 m. So volume flow rate Q = Ppump / (ρ g h) = 201.6 / (1000 × 9.81 × 15) ≈ 1.37 × 10⁻³ m³/s. Per hour, volume = 1.37 × 10⁻³ × 3600 ≈ 4.93 m³.
光伏发电量：Psolar = 辐照度 × 面积 × ηpv = 800 W/m² × 2.0 m² × 0.18 = 288 W。传输至水泵的有效机械功率：Ppump = 288 W × 0.70 = 201.6 W。提升水所需功率：P = ρ g h Q，其中 ρ = 1000 kg/m³，g = 9.81，h = 15 m。故体积流量 Q = 201.6 / (1000 × 9.81 × 15) ≈ 1.37 × 10⁻³ m³/s。每小时可提升水量：1.37 × 10⁻³ × 3600 ≈ 4.93 m³。
Battery energy: E = V × Ah = 12 V × 100 Ah = 1200 Wh = 4.32 × 10⁶ J. At half flow, required hydraulic power = 0.5 × 201.6 W = 100.8 W, but pump still operates at 70% efficiency, so electrical power drawn = 100.8 / 0.70 = 144 W. Operating time = battery energy / power = 1200 Wh / 144 W = 8.33 hours = 500 minutes (assuming full battery use). Realistically, usable capacity might be 80%, but the calculation illustrates the integration of energy storage.
蓄电池能量：E = V × Ah = 12 V × 100 Ah = 1200 Wh = 4.32 × 10⁶ J。在半流量下，所需水力功率 = 0.5 × 201.6 W = 100.8 W，但泵仍以 70% 效率运行，故消耗电功率 = 100.8 / 0.70 = 144 W。运行时间 = 电池能量 / 功率 = 1200 Wh / 144 W = 8.33 小时 = 500 分钟（假设完全放电）。实际可用容量可能为 80%，但此计算示出了储能跨学科集成。
4. Truss Analysis & Energy Absorption | 桁架分析与能量吸收
A simply‑supported truss bridge carries a central load of 50 kN. One diagonal member is made of mild steel with a cross‑sectional area of 500 mm² and length 2.5 m. Under load, the force in the member is found to be 40 kN in tension. Calculate the tensile stress, elastic extension, and strain energy stored in the member. Steel’s Young’s modulus is 210 GPa. If the member were replaced by a carbon‑fibre composite of the same cross‑section but E = 140 GPa and yield strength 600 MPa, compare the extension and energy under the same force.
一座简支桁架桥梁承受中心载荷 50 kN。一根斜杆采用低碳钢，截面积为 500 mm²，长度为 2.5 m。承受载荷后，该杆受拉力 40 kN。计算其拉应力、弹性伸长量及储存的应变能。钢材杨氏模量取 210 GPa。若将此杆更换为同截面的碳纤维复合材料，其 E = 140 GPa，屈服强度 600 MPa，试比较相同力作用下的伸长量与能量。
Stress σ = F / A = 40 000 N / (500 × 10⁻⁶ m²) = 80 MPa. Extension ΔL = F L / (A E) = (40 000 N × 2.5 m) / (500 × 10⁻⁶ m² × 210 × 10⁹ Pa) = 0.952 mm. Strain energy U = (1/2) F ΔL = 0.5 × 40 000 N × 0.952 × 10⁻³ m = 19.05 J.
应力 σ = F / A = 40 000 N / (500 × 10⁻⁶ m²) = 80 MPa。伸长 ΔL = F L / (A E) = (40 000 × 2.5) / (500 × 10⁻⁶ × 210 × 10⁹) = 0.952 mm。应变能 U = ½ F ΔL = 0.5 × 40 000 × 0.952 × 10⁻³ = 19.05 J。
For CFRP: ΔL = (40 000 × 2.5) / (500 × 10⁻⁶ × 140 × 10⁹) = 1.428 mm. U = 0.5 × 40 000 × 1.428 × 10⁻³ = 28.56 J. CFRP has lower modulus, so larger extension and absorbs more energy for same load, which can be beneficial for impact but must not exceed strength – here σ = 80 MPa < 600 MPa, safe.
对于碳纤维：ΔL = (40 000 × 2.5) / (500 × 10⁻⁶ × 140 × 10⁹) = 1.428 mm。应变能 U = 0.5 × 40 000 × 1.428 × 10⁻³ = 28.56 J。碳纤维模量更低，故伸长量更大，吸收更多能量，对冲击有利，但须确保不超强度——此处应力 80 MPa 远低于 600 MPa，安全。
5. Thermodynamics & Electrical Power Generation | 热力学与发电
A gas turbine receives air at 15 °C and compresses it to a pressure ratio of 12. The fuel combustion adds 800 kJ of thermal energy per kilogram of air. The turbine drives an alternator with 95% efficiency. If the mass flow of air is 0.8 kg/s, calculate the net shaft power available and the electrical output. Assume the cycle’s thermal efficiency is 36%. Explain how waste heat could be used to raise the overall efficiency.
一台燃气轮机将 15 °C 的空气压缩至压比为 12，燃烧过程每千克空气加入 800 kJ 热能。涡轮驱动一台效率为 95% 的发电机。空气质量流量为 0.8 kg/s，试计算净轴功率和电力输出。假设循环热效率为 36%，并说明如何利用废热提升总效率。
Rate of heat input: Q̇in = mass flow × heat added per kg = 0.8 kg/s × 800 kJ/kg = 640 kJ/s = 640 kW. Net shaft power (mechanical) = Q̇in × ηth = 640 kW × 0.36 = 230.4 kW. Electrical output = shaft power × ηgen = 230.4 × 0.95 = 218.88 kW.
热量输入率：Q̇in = 质量流量 × 单位加热量 = 0.8 kg/s × 800 kJ/kg = 640 kW。净轴功率（机械）= 640 kW × 0.36 = 230.4 kW。电输出 = 轴功率 × 发电机效率 = 230.4 × 0.95 = 218.88 kW。
Approximately 640 – 230.4 = 409.6 kW of heat is rejected. This waste heat could be recovered via a heat exchanger for district heating or industrial processes (combined heat and power, CHP), raising total utilisation to over 80%. Such systems require civil and mechanical engineering integration.
大约有 640 – 230.4 = 409.6 kW 的热量被排弃。这些废热可通过换热器回收，用于区域供暖或工业流程（热电联产，CHP），使整体能量利用率超过 80%。该类系统需要土木与机械工程的集成。
6. Gear Train Design & Strength Calculation | 齿轮传动设计与强度计算
An electric motor delivers torque of 18 N m at 1500 rpm. A two‑stage reduction gearbox reduces the speed to 150 rpm. The first stage uses a 20‑tooth pinion driving a 60‑tooth gear. The second stage uses a 20‑tooth pinion and unknown gear to achieve the overall ratio. Determine the number of teeth on the second gear and the output torque, assuming 95% efficiency per stage. Also, calculate the bending stress at the root of the first stage pinion if module = 3 mm, face width = 30 mm, and the Lewis form factor Y = 0.32.
一台电动机在 1500 r/min 时输出 18 N m 的扭矩。两级减速箱将转速降至 150 r/min。第一级由 20 齿小齿轮驱动 60 齿大齿轮；第二级用 20 齿小齿轮与未知齿轮实现总减速比。求第二级大齿轮的齿数和输出扭矩，假设每级效率 95%。并计算第一级小齿轮齿根弯曲应力，设模数 m = 3 mm，齿宽 30 mm，齿形系数 Y = 0.32。
Overall ratio = 1500 / 150 = 10. First stage ratio = 60/20 = 3. Hence second stage ratio must be 10/3 ≈ 3.333. With 20‑tooth pinion, second gear teeth N₂ = 20 × 3.333 = 66.67, so choose 67 teeth (giving actual ratio 67/20 = 3.35, overall 3 × 3.35 = 10.05, close enough). Torque after first stage: T₁ = 18 N m × 3 × 0.95 = 51.3 N m. After second stage: T₂ = 51.3 × 3.35 × 0.95 ≈ 163 N m.
总减速比 = 1500 / 150 = 10。第一级比 = 60/20 = 3。故第二级比 = 10/3 ≈ 3.333。用 20 齿小齿轮，所需大齿轮齿数 N₂ = 20 × 3.333 = 66.67，取 67 齿（实际比 3.35，总减速比 3 × 3.35 = 10.05，误差很小）。第一级后扭矩 = 18 N m × 3 × 0.95 = 51.3 N m。第二级后扭矩 = 51.3 × 3.35 × 0.95 ≈ 163 N m。
Bending stress in pinion: tangential force Ft = 2 T / dp. Pitch diameter dp = m × N = 3 mm × 20 = 60 mm = 0.06 m. T on pinion = 18 N m (motor torque). So Ft = 2 × 18 / 0.06 = 600 N. Bending stress σb = Ft / (Y m b) = 600 / (0.32 × 3 mm × 30 mm) = 600 / 28.8 ≈ 20.8 MPa. This is very low against typical steel strengths, highlighting that pinion is safe.
小齿轮弯曲应力：圆周力 Ft = 2 T / dp。节圆直径 dp = m × N = 3 mm × 20 = 60 mm = 0.06 m。小齿轮承受扭矩为电机扭矩 18 N m，故 Ft = 2 × 18 / 0.06 = 600 N。弯曲应力 σb = Ft / (Y m b) = 600 / (0.32 × 3 mm × 30 mm) ≈ 20.8 MPa。该值远低于典型钢材强度，表明小齿轮安全。
7. Fluid Mechanics & Hydraulic Press | 流体力学与液压机
A hydraulic press uses a plunger of diameter 40 mm to move a ram of diameter 200 mm. The lever operating the plunger has a mechanical advantage of 15.
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