University Grants Commission of Bangladesh (UGC)- 2025

Solution: At steady state, heat generated = heat convected: Q = hA(T_s − T_∞).

$$A=\pi DL=\pi \times 0.1\times 0.2=0.0628{\text{m}}^2,T_s=30+\frac{8}{2\times 0.0628}=30+63.7={93.7}^{\circ }\text{C}$$

∴ Surface temperature = 93.7°C (Answer)

Solution: Engine B (6 cylinders) needs the smaller flywheel. The mass moment of inertia of a flywheel is

$$I=\frac{\Delta E}{{\omega }^2{C}_s}$$

In a four-stroke engine, the interval between power strokes is 720°/4 = 180° for engine A and 720°/6 = 120° for engine B. With more frequent and overlapping power strokes, the turning moment of the 6-cylinder engine is more uniform and the maximum fluctuation of energy ΔE is much smaller. For the same speed and coefficient of fluctuation of speed, a smaller ΔE requires a smaller (lighter) flywheel.

Solution: $$P_{in}=\sqrt{3}{V}_L{I}_L\cos \varphi =\sqrt{3}\times 400\times 17\times 0.9=10.6\text{kW}$$

$$P_{out}=\rho gQH=1000\times 9.81\times 0.03\times 26.8=7.887\text{kW}$$

$$\eta =\frac{7.887}{10.6}=0.744$$

∴ Efficiency = 74.4% (Answer)

Solution: d = D/C = 25/8 = 3.125 mm.

$$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}=\frac{31}{28}+\frac{0.615}{8}=1.184$$

$$\tau =K\frac{8WD}{\pi d^3}\Rightarrow W=\frac{\pi d^3\tau }{8KD}=\frac{\pi \times {3.125}^3\times 100}{8\times 1.184\times 25}=40.5\text{N}$$

Deflection per active coil at this load = 8WC³/(Gd) = 8 × 40.5 × 512/(80,000 × 3.125) = 0.663 mm; the 40 mm deflection limit is satisfied if the number of active coils does not exceed 40/0.663 ≈ 60.

∴ Safe axial load = 40.5 N (governed by shear stress) (Answer)

Solution: $${\eta }_{bth}=\frac{3600}{bsfc\times CV}=\frac{3600}{0.22\times 40,000}=0.409$$

(m_f = 0.22 × 3.6 = 0.792 kg/h; heat input = 8.8 kW; BP = 3.6 kW.)

∴ Brake thermal efficiency = 40.9% (Answer)

Solution: 5 TR = 60,000 Btu/h = 17.585 kW; EER is in Btu/h per W.

$$P=\frac{60,000}{6.5}=9231\text{W}=9.23\text{kW},COP=\frac{17.585}{9.231}=1.905\left(=\frac{EER}{3.412}\right)$$

∴ Power = 9.23 kW; COP = 1.905 (Answer)

Solution: (a) At idling the throttle is almost closed, so the intake pressure is low and the residual exhaust gas dilutes the small fresh charge; fuel vaporisation and mixing are poor. A rich mixture (A/F ≈ 11–12 : 1) is therefore needed for reliable ignition, stable flame propagation and smooth running without misfire or stalling.

(b) Advanced (early) injection: fuel enters cooler, lower-pressure air, so the ignition delay is longer; more fuel accumulates and burns suddenly, causing rapid pressure rise, diesel knock, high peak pressure and temperature, higher NO_x, negative work before TDC and fuel wetting of the cylinder walls.

Retarded (late) injection: combustion continues well into the expansion stroke, so peak pressure falls, power and thermal efficiency drop, fuel consumption and exhaust temperature rise, and smoke, CO and unburnt HC increase; NO_x decreases.

Solution: In the austenitic range the steel transforms completely (or partly) into austenite (FCC), which can dissolve much more carbon, giving a uniform solid solution. Only from this structure can the desired final structures be obtained on cooling — martensite on quenching (hardening), fine pearlite on air cooling (normalising) or soft coarse pearlite on furnace cooling (annealing). Heating and soaking in this range also homogenise the composition and refine the grains, so austenitising is essential to achieve the required mechanical properties.