Mechanical Engineering Question Bank

Bangladesh Ordnance Factories (BOF)- 2026

Post: Assistant Engineer (Mechanical)

Exam Date: 30.01.2026

Total Marks: 100 (Departmental 40, Non-Departmental 60)

Departmental:

1) An engine works on the Otto cycle. The bore is 150 mm, the stroke is 225 mm and the clearance volume is 1.23 × 10⁻³ m³. Find the compression ratio and the air-standard efficiency.

Solution: $V_{s} = \frac{π}{4} d^{2} L = \frac{π}{4} (0.15)^{2} (0.225) = 3.976 \times 1 0^{- 3} m^{3}$

$r = \frac{V _{s} + V _{c}}{V _{c}} = \frac{3.976 + 1.23}{1.23} = 4.23, η = 1 - \frac{1}{r ^{γ - 1}} = 1 - \frac{1}{4.23 ^{0.4}} = 0.4385$

∴ r = 4.23; η = 43.85% (Answer)

2) A refrigerator works between 10°C and −10°C and has a cooling load of 7 TR. Find the COP and the power required to drive the compressor.

Solution: T_L = 263 K, T_H = 283 K; Q_L = 7 × 3.517 = 24.62 kW.

$COP = \frac{T _{L}}{T _{H} - T _{L}} = \frac{263}{20} = 13.15, W = \frac{Q _{L}}{COP} = \frac{24.62}{13.15} = 1.87 kW$

∴ COP = 13.15; Compressor power = 1.87 kW (Answer)

3) A power plant has a maximum demand of 75 MW. The connected loads are 45 MW, 30 MW, 20 MW and 10 MW. The load factor is 0.6. Find the average load and the demand factor.

Solution: Q1, QB-131, Bangladesh China Power Company (Pvt.) Limited (BCPCL)-2018.

$Average load = 0.6 \times 75 = 45 MW, Demand factor = \frac{75}{45 + 30 + 20 + 10} = \frac{75}{105} = 0.714$

4) Oil of specific gravity 0.85 flows through a 250 mm diameter pipe at 120 L/s. The pressure at a point 3 m above the datum is 20 kPa. Find the total energy (head) at that point.

Solution: $V = \frac{0.12}{\frac{π}{4} ( 0.25 ) ^{2}} = 2.445 m/s, \frac{p}{ρ g} = \frac{20 \times 1 0 ^{3}}{850 \times 9.81} = 2.399 m, \frac{V ^{2}}{2 g} = 0.305 m$

$H = 2.399 + 0.305 + 3 = 5.70 m of oil$

∴ Total energy = 5.70 m of oil (Answer)

5) A spur gear has a pitch circle diameter of 400 mm and 50 teeth. Find the module, diametral pitch and circular pitch.

Solution: $m = \frac{D}{T} = \frac{400}{50} = 8 mm, P_{d} = \frac{T}{D} = 0.125 teeth/mm (3.175 teeth/inch),$

$p_{c} = πm = 25.13 mm$

6) A solid shaft of 40 mm diameter transmits 40 kW at 200 rpm. Find the maximum torsional shear stress.

Solution: $T = \frac{60 P}{2 π N} = \frac{60 \times 40 , 000}{2 π \times 200} = 1909.9 N \cdotp m, τ_{m a x} = \frac{16 T}{π d ^{3}} = \frac{16 \times 1.91 \times 1 0 ^{6}}{π \times 4 0 ^{3}} = 152 MPa$

∴ Maximum shear stress = 152 MPa (Answer)

7) A thin cylindrical vessel of 400 mm diameter has a wall thickness of 20 mm. If the longitudinal stress is limited to 140 MPa and the circumferential stress to 60 MPa, find the maximum pressure the vessel can withstand.

Solution: $σ_{c} = \frac{p d}{2 t} \Rightarrow p = \frac{2 \times 20 \times 60}{400} = 6 MPa, σ_{l} = \frac{p d}{4 t} \Rightarrow p = \frac{4 \times 20 \times 140}{400} = 28 MPa$

∴ Maximum pressure = 6 MPa (governed by the circumferential stress) (Answer)

8) What is spark ignition? Write the processes of a four-stroke engine.

Solution: In a spark ignition (SI) engine the compressed air–fuel mixture (petrol, CNG or LPG) is ignited by an electric spark from the spark plug near the end of the compression stroke; such engines work on the Otto cycle.

A four-stroke engine completes one cycle in two revolutions of the crankshaft: (1) suction — inlet valve open, piston TDC → BDC, charge drawn in; (2) compression — both valves closed, piston BDC → TDC; (3) power — combustion pushes the piston TDC → BDC; (4) exhaust — exhaust valve open, piston BDC → TDC, burnt gases expelled.

Non-Departmental:

The non-departmental part contained: বাগধারার অর্থসহ বাক্য, এক কথায় প্রকাশ, বিপরীত শব্দ, সন্ধি, English translation, prepositions, sentence making with group verbs, a paragraph on “Referendum 2026”, টীকা and short questions (impact of artificial intelligence; steps of the interim government for environmental protection).

Unnumbered summary of the non-departmental part