General Engineering: Materials, Applied Mechanics, Machines and Design, Thermal and Fluids
1. Engineering materials: structure, properties and the five families
Most engineering metals crystallise in one of three structures. Body-centred cubic (BCC) — iron below 912 °C, chromium, tungsten, molybdenum — has 2 atoms per cell, coordination number 8 and atomic packing factor 0.68. Face-centred cubic (FCC) — aluminium, copper, nickel, austenitic iron — has 4 atoms per cell, coordination number 12 and packing factor 0.74. Hexagonal close-packed (HCP) — zinc, magnesium, titanium at room temperature — also packs to 0.74 with coordination 12. The packing factor is not trivia: FCC metals have many slip systems on close-packed planes and are ductile down to very low temperatures, BCC metals show a ductile-to-brittle transition as temperature falls, and HCP metals, with few active slip systems, are the hardest of the three to cold work.
| Family | Bonding and structure | Characteristic properties | Typical applications |
|---|---|---|---|
| Metals and alloys | Metallic bond, crystalline | Ductile, tough, good thermal and electrical conductors | Structures, shafts, machine frames, tools |
| Semiconductors | Covalent (Si, Ge), band gap about 1 eV | Conductivity rises with temperature and with doping | Diodes, transistors, sensors, solar cells |
| Ceramics | Ionic and covalent, oxides, carbides, nitrides | Hard, refractory, brittle, strong in compression, poor in tension | Cutting tool inserts, refractories, insulators |
| Polymers | Covalent chains; thermoplastics (linear) and thermosets (cross-linked) | Light, corrosion-resistant, low stiffness, creep at room temperature | Housings, pipes, films, gears for light duty |
| Composites (metal, polymer, ceramic matrix) | Reinforcement (fibre or particle) in a matrix | Tailored stiffness-to-weight; properties depend on direction | GFRP and CFRP panels (PMC), Al-SiC brake parts (MMC), SiC-SiC turbine parts (CMC) |
For a fibre composite loaded along the fibres, the rule of mixtures gives the longitudinal modulus as the volume-weighted average, E_c = E_f V_f + E_m V_m (the iso-strain case). Loaded across the fibres the phases carry the same stress instead, and 1/E_c = V_f/E_f + V_m/E_m (the iso-stress case), which is always the lower of the two. That gap is why a unidirectional laminate is stiff along its fibres and weak across them, and why real parts stack plies at several angles.
2. The iron-carbon diagram, heat treatment and the stress-strain curve
The iron-iron carbide diagram has three invariant reactions, and GATE asks the two lower ones most. The eutectoid at 727 °C and about 0.76 % carbon turns austenite into pearlite, a lamellar mixture of ferrite and cementite: γ → α + Fe3C. The eutectic at about 1147 °C and 4.3 % carbon turns liquid into ledeburite: L → γ + Fe3C. The peritectic near 1493 °C (δ + L → γ) sits at very low carbon. Austenite dissolves up to about 2.1 % carbon, which is the conventional boundary between steels and cast irons; ferrite dissolves only about 0.022 %. Below 0.76 % a steel is hypo-eutectoid (proeutectoid ferrite plus pearlite), above it hyper-eutectoid (proeutectoid cementite plus pearlite).
| Treatment | Cycle | Resulting structure | Effect on properties |
|---|---|---|---|
| Full annealing | Austenitise, furnace cool | Coarse pearlite | Softest, most ductile, stresses relieved |
| Normalising | Austenitise, cool in still air | Fine pearlite, refined grain | Stronger and harder than annealed, uniform |
| Hardening | Austenitise, quench in water or oil | Martensite (supersaturated body-centred tetragonal) | Very hard and brittle, high residual stress |
| Tempering | Reheat hardened steel below 727 °C | Tempered martensite | Trades some hardness for toughness |
| Austempering | Quench to above the martensite start, hold isothermally | Bainite | Tough and hard with little distortion |
| Case hardening (carburising, nitriding, induction, flame) | Enrich or heat only the surface, then quench | Hard martensitic case on a tough core | Wear resistance with shock resistance, as in gears and cams |
On the engineering stress-strain curve of a ductile metal, stress is proportional to strain up to the proportional limit (slope E), the material yields (a sharp upper and lower yield point in mild steel, or a 0.2 % offset yield elsewhere), strain-hardens to the ultimate tensile strength, then necks and the engineering stress falls to fracture. True stress σ_T = σ(1 + ε) and true strain ε_T = ln(1 + ε) are valid up to necking; the true curve keeps rising. The area under the elastic part per unit volume is the resilience, σ_y²/(2E); the area under the whole curve is the toughness. Percentage elongation and reduction in area measure ductility.
3. Equilibrium, stress-strain relations, failure theories and Mohr's circle
Any system of forces on a rigid body reduces to an equivalent force-couple system at a chosen point: a single resultant force R = ΣF and a couple M = Σ(r × F). A body is in equilibrium when both vanish, which in a plane gives three scalar equations, ΣFx = 0, ΣFy = 0 and ΣM = 0 — so a planar free-body diagram can yield at most three unknowns. Draw the free body by isolating it completely and replacing every support by the reactions it can supply: a roller one force normal to its surface, a pin two force components, a fixed support two forces and a moment. A two-force member carries its load along the line joining the two points.
Within the elastic range Hooke's law links stress and strain, σ = Eε and τ = Gγ, and a stretched bar contracts laterally by Poisson's ratio ν = −ε_lateral/ε_axial. The three elastic constants of an isotropic material are not independent: E = 2G(1 + ν) = 3K(1 − 2ν). Since ν cannot exceed 0.5 without K going negative, ν = 0.5 is the incompressible limit. In three dimensions ε_x = [σ_x − ν(σ_y + σ_z)]/E, and the volumetric strain of a body is ε_v = ε_x + ε_y + ε_z.
Mohr's circle for plane stress: plot (σx, τxy) and (σy, −τxy); the circle through them has centre C = (σx + σy)/2 and radius R = √[((σx − σy)/2)² + τxy²]. The principal stresses are σ1,2 = C ± R, the maximum in-plane shear stress is R, the principal planes are at tan 2θp = 2τxy/(σx − σy), and a rotation θ of the element is 2θ on the circle. For σx = 80 MPa, σy = 20 MPa, τxy = 40 MPa: C = 50, R = √(30² + 40²) = 50, so σ1 = 100 MPa, σ2 = 0 and the maximum in-plane shear is 50 MPa.
| Theory | Failure when | Shear yield it predicts | Suited to |
|---|---|---|---|
| Maximum principal stress (Rankine) | σ1 = σ_ut | τ_y = σ_y | Brittle materials |
| Maximum shear stress (Tresca, Guest) | (σ1 − σ3)/2 = σ_y/2 | τ_y = 0.5 σ_y | Ductile materials, conservative |
| Maximum distortion energy (von Mises, Hencky) | σ1² + σ2² − σ1σ2 = σ_y² (plane stress) | τ_y = σ_y/√3 = 0.577 σ_y | Ductile materials, closest to test data |
| Maximum principal strain (St. Venant) | σ1 − ν(σ2 + σ3) = σ_y | τ_y = σ_y/(1 + ν) | Rarely used now |
4. Bending and shear stresses, beam deflection, thin and thick cylinders, torsion
The flexure formula M/I = σ/y = E/R gives the bending stress σ = My/I, zero at the neutral axis and largest at the outer fibre, so σ_max = M/Z with section modulus Z = I/y_max (bd²/6 for a rectangle, πd³/32 for a solid circle). The transverse shear stress is τ = VQ/(Ib), where Q is the first moment of the area beyond the fibre considered; it is zero at the outer fibre and peaks at the neutral axis, at 1.5 times the average for a rectangle and 4/3 times for a solid circle.
| Case | Maximum deflection | Maximum slope |
|---|---|---|
| Cantilever, point load P at free end | PL³/(3EI) | PL²/(2EI) |
| Cantilever, UDL w over length L | wL⁴/(8EI) | wL³/(6EI) |
| Simply supported, central point load P | PL³/(48EI) | PL²/(16EI) |
| Simply supported, UDL w | 5wL⁴/(384EI) | wL³/(24EI) |
A thin cylinder (t < d/20) under internal pressure p carries a hoop stress σ_h = pd/(2t) and a longitudinal stress σ_l = pd/(4t), half the hoop value — so a thin boiler shell splits along its length first. A thin sphere carries pd/(4t) in every direction. In a thick cylinder the stresses vary through the wall and follow Lamé's equations, σ_r = A − B/r² and σ_θ = A + B/r², with A and B fixed by the pressures at the two radii; the hoop stress is greatest at the inner surface. For internal pressure only, σ_θ at the bore is p(r_o² + r_i²)/(r_o² − r_i²).
Torsion of a circular shaft: T/J = τ/r = Gθ/L, with polar moment J = πd⁴/32 (solid) or π(D⁴ − d⁴)/32 (hollow). The surface shear stress of a solid shaft is 16T/(πd³). Torsional stiffness is GJ/L, and shafts in series add angles of twist while shafts in parallel share torque in proportion to their stiffness. Power transmitted is P = 2πNT/60 with N in rpm.
5. Planar mechanisms, cams and followers, governors and flywheels
The mobility of a planar mechanism follows Gruebler's criterion, F = 3(n − 1) − 2j − h, where n counts links including the frame, j lower pairs (turning or sliding, one degree of freedom each) and h higher pairs. A four-bar linkage has n = 4, j = 4, so F = 1. A Grashof four-bar (s + l ≤ p + q, shortest plus longest not more than the other two) has at least one link that can rotate fully: fix the link next to the shortest for a crank-rocker, the shortest itself for a double-crank, and the link opposite the shortest for a double-rocker. The slider-crank, with its inversions, gives the reciprocating engine, the Whitworth and the crank-and-slotted-lever quick-return motions.
Velocity analysis uses instantaneous centres: a mechanism of n links has n(n − 1)/2 of them, and Kennedy's theorem says the three instantaneous centres of any three links lie on a straight line. The velocity of any point is ω times its distance from the instantaneous centre of its link with the frame. Acceleration of a point on a link rotating at ω has a centripetal part ω²r towards the centre and a tangential part αr; a slider on a rotating link adds the Coriolis component 2vω.
A cam drives a follower (knife-edge, roller, flat-faced) through rise, dwell and return. Uniform velocity motion has infinite acceleration at the ends and is used only with modifications; simple harmonic motion has finite acceleration but a jump at the ends; uniform acceleration and retardation (parabolic) gives the least maximum acceleration for a given rise; cycloidal motion has zero acceleration at both ends and suits high speeds. The pressure angle — between the follower's direction of motion and the common normal — should stay small (about 30° is a common limit for a translating roller follower) to keep side thrust down; a larger base circle reduces it.
A governor controls mean speed over many cycles by regulating fuel; a flywheel controls fluctuation within one cycle by storing energy, and neither does the other's job. For a Watt governor the height is h = g/ω² (h ≈ 895/N² metres with N in rpm), so height falls as the square of speed. Porter (a loaded central sleeve) and Hartnell (spring-loaded) governors are more powerful. Terms: sensitive — a small speed change moves the sleeve a lot; stable — radius of rotation increases with speed; isochronous — one equilibrium speed for all radii, infinitely sensitive and prone to hunting. For a flywheel, the maximum fluctuation of energy is ΔE = Iω²C_s, where ω is the mean speed and the coefficient of fluctuation of speed is C_s = (ω_max − ω_min)/ω.
6. Design of joints, fits, shafts, keys, couplings, gears, belts, brakes and clutches
- Bolted joints: the bolt is designed on its tensile stress area; an initial preload keeps the joint closed so the bolt sees only a fraction of the external load, the share set by the stiffness of bolt against members.
- Riveted joints fail by tearing of the plate between rivets, P_t = (p − d)tσ_t, shearing of the rivets, P_s = n(π/4)d²τ per shear plane, or crushing, P_c = n d t σ_c. Joint efficiency = the least of the three divided by the strength of the unpunched plate, p t σ_t.
- Welded joints: a fillet weld of leg s fails across its throat, t = s cos 45° = 0.707s, so a parallel fillet of length L carries P = 0.707 s L τ per weld; a butt weld is designed on the plate thickness.
- Interference (shrink) fit: a diametral interference δ between a solid shaft and a hub of the same material (outer diameter D) creates a contact pressure p = (Eδ/2d)(1 − d²/D²); the torque it can transmit by friction is T = μpπd²L/2.
- Shafts under combined bending M and torque T: by maximum shear stress theory the equivalent torque is T_e = √(M² + T²) and d³ = 16T_e/(πτ); by maximum normal stress the equivalent moment is M_e = [M + √(M² + T²)]/2 and d³ = 32M_e/(πσ).
- Keys transmit torque by shear across the width, τ = 2T/(w l d), and by crushing on the side face, σ_c = 4T/(h l d) for a key half-sunk in the shaft. Couplings: rigid (sleeve, flange) for aligned shafts; flexible (bush-pin, Oldham, universal joint) where misalignment exists.
- Spur gears: module m = d/z, circular pitch πm; the pitch-line velocity ratio is ω1/ω2 = z2/z1. The Lewis equation gives the beam strength of a tooth, F_t = σ_b b m Y, with face width b and form factor Y; interference is avoided by a minimum number of pinion teeth for the pressure angle used (larger for 14.5° than for 20°).
- Belt drives: for a flat belt T1/T2 = eμθ (θ the angle of lap on the smaller pulley, in radians), for a V-belt eμθ/sin β with β the half groove angle. Power = (T1 − T2)v. Centrifugal tension T_c = mv² adds to both sides; power is greatest when T_c = T_max/3.
- Brakes: a band brake obeys the same eμθ tension ratio, braking torque (T1 − T2)r; a block brake gives friction μN at the drum. Clutches (single plate, per friction surface): uniform wear, T = μW(R1 + R2)/2; uniform pressure, T = (2/3)μW(R1³ − R2³)/(R1² − R2²). A new clutch is closer to uniform pressure, a worn one to uniform wear, which is the safer (lower) design value.
Friction and lubrication: dry (Coulomb) friction gives F = μN independent of the apparent area. A journal bearing moves, as speed rises, from boundary lubrication (asperities in contact, high μ) through mixed to hydrodynamic lubrication, where a full oil film carries the load and μ rises again only slowly with speed. Petroff's equation for a lightly loaded concentric journal, f = 2π²(μN/P)(r/c), shows friction growing with viscosity μ and speed N and falling with bearing pressure P; the dimensionless group μN/P (the Sommerfeld variable) is what the Stribeck curve is plotted against.
7. Fluid statics, Bernoulli, pipe flow, continuity and momentum, capillarity, dimensional analysis
In a fluid at rest pressure rises linearly with depth, p = p0 + ρgh, and acts equally in all directions. The resultant force on a plane submerged surface is F = ρg h̄ A (h̄ the depth of the centroid), acting at the centre of pressure, which is always below the centroid by I_G/(A h̄) for a vertical surface. Continuity for incompressible flow is A1V1 = A2V2. Along a streamline of steady, inviscid, incompressible flow, Bernoulli's equation p/ρg + V²/2g + z = constant; with losses, a head-loss term h_L is added on the downstream side. The momentum equation for a control volume, ΣF = ρQ(V_out − V_in), gives forces on bends, nozzles and vanes.
Pipe flow is laminar below a Reynolds number Re = ρVD/μ of about 2000 and turbulent well above it. The Darcy-Weisbach equation, h_f = fLV²/(2gD), gives the friction head loss for both; in laminar flow f = 64/Re, which reproduces the Hagen-Poiseuille result Δp = 32μVL/D², and the velocity profile is parabolic with the maximum twice the mean. In turbulent flow f depends on Re and relative roughness (the Moody chart) and the profile is much flatter. Minor losses at bends, valves and sudden expansions are written K V²/(2g).
Capillary action: in a tube of diameter d, surface tension σ lifts (or depresses) liquid to a height h = 4σ cos θ/(ρgd). Water wets clean glass (θ ≈ 0) and rises; mercury (θ > 90°) is depressed. The rise is inversely proportional to diameter, so halving the bore doubles it. Dimensional analysis by Buckingham's π theorem: n variables in m fundamental dimensions form n − m independent dimensionless groups. Reynolds (inertia/viscous), Froude (inertia/gravity), Weber (inertia/surface tension), Mach (inertia/compressibility) and Euler (pressure/inertia) numbers are the ones models are matched on.
8. Laws of thermodynamics, work and heat, air-standard cycles, conduction, convection and radiation
The zeroth law — two bodies each in thermal equilibrium with a third are in equilibrium with each other — is what makes a thermometer meaningful. The first law for a closed system is Q − W = ΔU; for a steady-flow control volume it becomes the steady-flow energy equation, q − w = (h2 − h1) + (V2² − V1²)/2 + g(z2 − z1) per unit mass. The second law in its Kelvin-Planck form forbids a cycle that converts heat from one reservoir wholly into work, and in its Clausius form forbids heat flowing unaided from cold to hot; together they cap every engine at the Carnot efficiency 1 − T_L/T_H and introduce entropy, dS ≥ δQ/T.
| Process | Law | Work done by the gas |
|---|---|---|
| Constant volume | V = const | 0 |
| Constant pressure | p = const | p(V2 − V1) |
| Isothermal | pV = const | p1V1 ln(V2/V1) |
| Polytropic (adiabatic when n = γ) | pV^n = const | (p1V1 − p2V2)/(n − 1) |
Air-standard cycles replace combustion by heat addition to air treated as an ideal gas. Otto (constant-volume heat addition): η = 1 − 1/rγ−1, depending on the compression ratio r alone — 56.5 % for r = 8 and γ = 1.4. Diesel (constant-pressure heat addition): η = 1 − (1/rγ−1)[(ρ^γ − 1)/(γ(ρ − 1))], with cut-off ratio ρ; for the same r the Diesel is less efficient than the Otto, but it runs at much higher r. Dual combines the two. Brayton (the gas turbine): η = 1 − 1/r_p(γ−1)/γ, depending on the pressure ratio r_p.
Conduction follows Fourier's law, Q = −kA dT/dx, so a plane wall conducts Q = kAΔT/L and has a thermal resistance L/(kA); resistances in series add like electrical ones. Convection follows Newton's law of cooling, Q = hAΔT, resistance 1/(hA). Radiation from a black body is σT⁴ per unit area with σ = 5.67 × 10⁻⁸ W/m²K⁴; a grey body emits εσT⁴, and net exchange depends on T1⁴ − T2⁴, never on the difference of temperatures. Insulating a small pipe can increase heat loss until the outer radius passes the critical radius r_c = k/h (2k/h for a sphere).
Key takeaways
- BCC packs to 0.68 and FCC and HCP to 0.74; the eutectoid (727 °C, about 0.76 % C) gives pearlite, the eutectic (about 1147 °C, 4.3 % C) ledeburite, and the lever rule always reads the opposite segment.
- Mohr's circle: centre (σx + σy)/2, radius √[((σx − σy)/2)² + τ²]; Tresca predicts shear yield at 0.5σ_y and von Mises at 0.577σ_y, and Tresca needs σ3 = 0 counted in plane stress.
- Thin cylinder: hoop pd/2t is twice the longitudinal pd/4t; torsion T/J = τ/r = Gθ/L; beam deflection PL³/3EI (cantilever) and PL³/48EI or 5wL⁴/384EI (simply supported).
- Design formulas to carry: T_e = √(M² + T²) for shafts, T1/T2 = eμθ with θ in radians for belts and band brakes, uniform-wear clutch torque μW(R1 + R2)/2, fillet throat 0.707s.
- Laminar pipe flow has f = 64/Re; capillary rise 4σcosθ/(ρgd) falls with diameter; Otto efficiency 1 − 1/rγ−1 and Brayton 1 − 1/r_p(γ−1)/γ; all Carnot and radiation temperatures in kelvin.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the atomic packing factor of a face-centred cubic crystal?
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Answer: C — 0.74
FCC has 4 atoms per cell and a = 4R/√2, so the packing factor is 4 × (4/3)πR³ / a³ = 0.74, the same as HCP. 0.68 is BCC, the usual confusion, and 0.52 is simple cubic.Which reaction occurs at the eutectoid point of the iron-iron carbide diagram?
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Answer: A — Austenite → ferrite + cementite
The eutectoid at 727 °C is a solid-state reaction, γ → α + Fe3C, whose product is pearlite. Liquid → austenite + cementite is the eutectic (ledeburite), δ + L → γ is the peritectic, and austenite → martensite is a non-equilibrium quench transformation that does not appear on the equilibrium diagram at all.A plain carbon steel with 0.4 wt% C is slowly cooled to just below 727 °C. Taking the eutectoid composition as 0.76 wt% C and the carbon solubility in ferrite as 0.022 wt%, what percentage of the microstructure is pearlite? (Answer to one decimal place.)
Numerical answer — type the value.
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Answer: 51.2
Just below 727 °C all the austenite present at the eutectoid becomes pearlite, so apply the lever rule between ferrite (0.022) and the eutectoid (0.76): pearlite fraction = (0.4 − 0.022)/(0.76 − 0.022) = 0.378/0.738 = 0.512, i.e. 51.2 %. Reading the near segment instead, (0.76 − 0.4)/0.738 = 48.8 %, gives the proeutectoid ferrite, which is the tempting wrong answer.A steel gear must have a hard, wear-resistant surface and a tough core able to take shock. Which treatment achieves this?
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Answer: C — Carburising followed by quenching and tempering
Carburising enriches only the surface with carbon, so the quench turns the case into hard martensite while the low-carbon core stays tough; tempering then relieves the quench stresses. Annealing and normalising soften the whole section, and an untempered through-hardened gear is hard everywhere and brittle in the core, which is the opposite of what shock loading needs.At a point in plane stress σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. What is the maximum in-plane shear stress, in MPa?
Numerical answer — type the value.
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Answer: 50
The radius of Mohr's circle is the maximum in-plane shear: R = √[((80 − 20)/2)² + 40²] = √(900 + 1600) = 50 MPa. The principal stresses are 50 ± 50, i.e. 100 and 0 MPa. Taking (σx − σy)/2 = 30 MPa alone ignores the applied shear, and taking σ1/2 happens to give 50 here only because σ2 = 0.Which of the following statements about failure theories are correct?
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Answer: A — For a ductile material the maximum shear stress (Tresca) theory is more conservative than the distortion energy (von Mises) theory; B — The von Mises theory predicts yield in pure shear at τ = σ_y/√3; D — The Tresca theory predicts yield in pure shear at τ = 0.5σ_y
Tresca's hexagon lies inside the von Mises ellipse, so it predicts yield earlier — more conservative. In pure shear σ1 = τ, σ2 = −τ: von Mises gives 3τ² = σ_y², τ = 0.577σ_y, and Tresca gives 2τ = σ_y, τ = 0.5σ_y. The maximum principal stress theory ignores shear-driven yielding and is used for brittle materials, so that statement is false.A thin cylindrical vessel of internal diameter 1000 mm and wall thickness 10 mm holds a gas at 2 MPa. What is the hoop stress in the wall, in MPa?
Numerical answer — type the value.
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Answer: 100
Hoop stress σ_h = pd/(2t) = 2 × 1000/(2 × 10) = 100 MPa. The longitudinal stress pd/(4t) = 50 MPa is the value candidates give when they mix up the two formulas; the hoop stress is always the larger, which is why a thin shell fails along its length.A solid shaft carries a bending moment of 3 kN·m and a torque of 4 kN·m. Using the maximum shear stress theory with an allowable shear stress of 50 MPa, what is the minimum shaft diameter in mm? (Answer to one decimal place.)
Numerical answer — type the value.
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Answer: 79.9
Equivalent torque T_e = √(3² + 4²) = 5 kN·m = 5 × 10⁶ N·mm. Then d³ = 16T_e/(πτ) = 16 × 5 × 10⁶/(π × 50) = 5.093 × 10⁵ mm³, so d = 79.9 mm. Designing on the torque alone (4 kN·m) gives 74.1 mm, which ignores the bending the shaft also carries.A planar mechanism has 6 links (including the frame) connected by 7 lower pairs and no higher pairs. What is its degree of freedom by Gruebler's criterion?
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Answer: B — 1
F = 3(n − 1) − 2j − h = 3 × 5 − 2 × 7 − 0 = 15 − 14 = 1, a constrained mechanism such as the Watt or Stephenson six-bar. Using 3n instead of 3(n − 1) forgets that the frame is fixed and gives 4, and miscounting pairs as links gives the other options.An open flat belt has a tight-side tension of 1000 N, a coefficient of friction of 0.3 and an angle of lap of 180° on the smaller pulley. Neglecting centrifugal tension, what power does it transmit at a belt speed of 10 m/s?
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Answer: B — 6.10 kW
T1/T2 = e0.3π = e0.942 = 2.566, so T2 = 1000/2.566 = 389.7 N. Power = (T1 − T2)v = 610.3 × 10 = 6.10 kW. 3.90 kW is T2 × v, the slack-side tension mistaken for the effective pull, and 10 kW is T1 × v, which forgets that the slack side pulls back.Which of the following statements about governors, flywheels and clutches are correct?
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Answer: A — A flywheel limits the speed fluctuation within one cycle; a governor controls the mean speed as load changes; B — An isochronous governor is infinitely sensitive and tends to hunt; D — For the same axial load and friction radii, the uniform-wear assumption gives a lower clutch torque than the uniform-pressure assumption
The flywheel and governor do different jobs, and an isochronous governor, with one speed for every radius, is infinitely sensitive and hunts. The Watt height is h = g/ω², so doubling the speed QUARTERS the height — that statement is false. Uniform wear gives μW(R1 + R2)/2, which is always slightly below the uniform-pressure value, so it is the safe design assumption.A solid steel shaft of 50 mm diameter is shrunk into a steel hub of 100 mm outer diameter with a diametral interference of 0.05 mm. Taking E = 200 GPa for both, what is the contact pressure at the interface?
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Answer: B — 75 MPa
For a solid shaft in a hub of the same material, p = (Eδ/2d)(1 − d²/D²) = (200 000 × 0.05/100)(1 − 0.25) = 100 × 0.75 = 75 MPa. 100 MPa drops the (1 − d²/D²) factor, treating the hub as infinitely thick, and 150 MPa uses δ/d without the factor of 2 that splits the interference between shaft and hub.In a capillary tube, the height to which water rises is reduced to half its original value. How has the tube diameter changed, other things being equal?
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Answer: B — It has been doubled
Capillary rise h = 4σ cos θ/(ρgd) is inversely proportional to diameter, so halving h means the diameter doubled. Four times would follow only if h went as 1/d², which is the area, not the perimeter the surface tension acts along.For fully developed laminar flow in a circular pipe, how does the Darcy friction factor depend on the Reynolds number?
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Answer: B — f = 64/Re
The Hagen-Poiseuille solution gives Δp = 32μVL/D², which matches h_f = fLV²/(2gD) only when f = 64/Re. 16/Re is the Fanning friction factor, a quarter of Darcy's; 0.316/Re^0.25 is the Blasius correlation for smooth turbulent flow; and independence of Re belongs to fully rough turbulent flow.An air-standard Otto cycle has a compression ratio of 8. Taking γ = 1.4, what is its thermal efficiency in percent? (Answer to one decimal place.)
Numerical answer — type the value.
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Answer: 56.5
η = 1 − 1/rγ−1 = 1 − 1/80.4 = 1 − 1/2.297 = 1 − 0.4353 = 0.5647, i.e. 56.5 %. Using γ instead of γ − 1 in the exponent gives 1 − 8−1.4 = 94.6 %, far above anything a real engine reaches, which is the warning sign of that slip.Heat flows through a furnace wall made of two layers in series, of thermal resistance 0.2 K/W and 0.3 K/W, with a temperature difference of 500 K across the whole wall. What is the heat flow?
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Answer: A — 1000 W
Series thermal resistances add like electrical resistors: R = 0.2 + 0.3 = 0.5 K/W, so Q = ΔT/R = 500/0.5 = 1000 W. 4167 W treats the layers as parallel (R = 0.12 K/W), and 2500 W uses only the first layer.