Bangladesh Inland Water Transport Authority (BIWTA)- 2026

Solution: Combustion in a CI engine starts by self-ignition of the injected fuel in hot compressed air. The factors that govern it are:

• High compression temperature and pressure (compression ratio 14–22)
• Fuel injection pressure, injection timing and rate of injection
• Atomisation, penetration and spray pattern of the fuel
• Air–fuel ratio and air motion (swirl/turbulence) in the combustion chamber
• Ignition delay period, which depends on the cetane number of the fuel
• Temperature of the cylinder walls and the intake air (supercharging)

Solution: Limits: The two extreme permissible sizes of a shaft — the upper (maximum) limit and the lower (minimum) limit. Example: a shaft of 50.00/49.95 mm.

Tolerance: The difference between the upper and lower limits of size, i.e. the permissible variation in size. Tolerance = Upper limit − Lower limit = 50.00 − 49.95 = 0.05 mm.

Deviation: The algebraic difference between a size (actual or limit size) and the corresponding basic size. Upper deviation = Upper limit − Basic size; Lower deviation = Lower limit − Basic size.

Solution: COP: The ratio of the refrigerating effect (heat removed from the cold space) to the work input, COP = Q_L/W.

Ton of refrigeration (TR): The rate of heat removal required to freeze one short ton (907 kg) of water at 0°C into ice at 0°C in 24 hours. 1 TR = 211 kJ/min ≈ 3.5 kW.

Given, COP = 1.2, W = 60 kJ/min = 1 kW.

$$Q_L=COP\times W=1.2\times 60=72\text{kJ/min}=1.2\text{kW}=\frac{72}{211}=0.341\text{TR}$$

$$Q_H=Q_L+W=72+60=132\text{kJ/min}=2.2\text{kW}$$

∴ Refrigeration effect = 72 kJ/min (0.341 TR); Heat rejected = 132 kJ/min (Answer)

Solution:

Solution: Fuel contamination: The presence of unwanted substances — water, dirt, dust, rust, sludge or microbial growth — in the fuel.

Reasons: water entering the storage tank (condensation, rain), dirty containers and poor handling, corrosion inside tanks and pipes, and poor or damaged filters.

Oil dilution: Mixing of fuel with the lubricating oil in the crankcase, which reduces the viscosity and lubricating ability of the oil.

Reasons: leaking or dribbling injectors, over-fuelling and poor atomisation, frequent cold starting and long idling, incomplete combustion, and worn piston rings or liners allowing fuel to pass into the crankcase.

Solution: (a) Beam with a central point load P (span L, supports at both ends):

R_A = R_B = P/2; shear force = +P/2 from A to the load and −P/2 from the load to B; maximum bending moment at the centre, M_max = PL/4.

If both ends are fixed, the end moments are −PL/8 and the central moment is +PL/8.

(b) Cantilever with a point load P at the free end:

Shear force = P (constant) throughout; bending moment varies linearly from zero at the free end to M_A = −PL at the fixed end.

Solution: Given, D = 0.25 m, L = 0.3 m, N = 60 rpm, Q_act = 16.5 L/s.

$$A=\frac{\pi }{4}(0.25{)}^2=0.04909{\text{m}}^2$$

$$Q_{th}=\frac{ALN}{60}=\frac{0.04909\times 0.3\times 60}{60}=0.01473{\text{m}}^3/\text{s}=14.73\text{L/s}$$

$$C_d=\frac{Q_{act}}{Q_{th}}=\frac{16.5}{14.73}=1.12$$

$$\text{Slip}=Q_{th}-Q_{act}=14.73-16.5=-1.77\text{L/s},\%\text{Slip}=\frac{-1.77}{14.73}\times 100=-12.0\%$$

∴ Q_th = 14.73 L/s; C_d = 1.12; Slip = −1.77 L/s (−12.0%) (Answer)

Since the actual discharge exceeds the theoretical discharge, the slip is negative. Negative slip occurs when the delivery valve opens before the end of the suction stroke, as in pumps with long suction pipes and short delivery pipes running at high speed.

Solution: A Stirling engine is an external-combustion, closed-cycle regenerative heat engine. A fixed mass of gas (air, helium or hydrogen) is alternately heated and cooled; heat is supplied from an outside source (fuel, solar, biomass or waste heat). The cycle consists of two isothermal and two constant-volume processes, with a regenerator that stores heat during constant-volume cooling and returns it during constant-volume heating. With perfect regeneration its efficiency equals the Carnot efficiency.

1–2: isothermal compression (heat rejected at T_L); 2–3: constant-volume heat addition from the regenerator; 3–4: isothermal expansion (heat supplied at T_H); 4–1: constant-volume heat rejection to the regenerator.

Solution: An injector must be tested when:

• The engine is hard to start, misfires or runs unevenly.
• There is loss of power or abnormally high fuel consumption.
• Excessive black/white smoke appears in the exhaust or knocking is heard.
• Fuel dribbling, leakage or a poor spray pattern is observed.
• The opening (injection) pressure is suspected to be incorrect.
• After overhauling, repair or replacement of the injector, and at the scheduled maintenance interval.

Solution: Parallel flow: Both fluids enter at the same end and flow in the same direction. The temperature difference is large at the inlet and small at the outlet; the cold fluid outlet temperature can never exceed the hot fluid outlet temperature.

Counter flow: The fluids enter at opposite ends and flow in opposite directions. The temperature difference is more uniform along the length, the cold fluid can leave hotter than the hot fluid outlet, and for the same duty it needs the least surface area.

Cross flow: The fluids flow perpendicular to each other, as in automobile radiators and air-cooled condensers. Either fluid may be mixed or unmixed.

Solution: Axial force in AB = 4 + 3 = 7 kN (tension).

$$A_{AB}=\frac{\pi }{4}(30{)}^2=706.86{\text{mm}}^2,{\sigma }_{AB}=\frac{7000}{706.86}=9.90{\text{N/mm}}^2$$

$${\delta }_{AB}=\frac{PL}{AE}=\frac{7000\times 300}{706.86\times 200\times 10^3}=0.0149\text{mm}$$

∴ σ_AB = 9.90 MPa (tensile); δ_AB = 0.0149 mm (elongation) (Answer)

Solution: Common causes of abnormal boiler operation (low steam output, falling efficiency, pressure drop or overheating) are: scale on the water side and soot on the gas side, low water level or poor circulation, priming and foaming, faulty burner and wrong air–fuel ratio, air leakage into the furnace, leaking or stuck safety valves and steam traps, and faulty feed pump or water level controls.

Solution: Priming is the carry-over of water particles with the steam leaving the boiler, producing wet steam. It is caused by a very high water level, sudden increase in steam demand, high velocity of steam, and impurities (oil, soap, suspended solids) in the boiler water. It reduces the dryness of steam and can damage turbine blades and engine cylinders. It is prevented by maintaining a proper water level, using steam separators/dry pipes, avoiding sudden steam demand and using clean feed water.

Solution: Diesel has low volatility and a low self-ignition temperature (low octane number). In a petrol engine it does not vaporise properly in the intake, the spark cannot ignite it reliably and the engine misfires, gives heavy smoke and may stall or fail to start. Unburnt diesel fouls the spark plugs, clogs injectors/filters and dilutes the lubricating oil. The tank and fuel lines must be drained and flushed before restarting.

Solution: $|A|=-1(1\cdot 1-1\cdot 1)-0+1(1\cdot 1-1\cdot 0)=0+1=1$

Co-factors:

$$C_{11}=0,C_{12}=-1,C_{13}=1$$

$$C_{21}=1,C_{22}=-1,C_{23}=1$$

$$C_{31}=-1,C_{32}=2,C_{33}=-1$$

$$\text{Adj}(A)={\left[C_{ij}\right]}^T=\left[\begin{array}{ccc}0& 1& -1\\ -1& -1& 2\\ 1& 1& -1\end{array}\right]$$

$$A^{-1}=\frac{1}{|A|}\text{Adj}(A)=\left[\begin{array}{ccc}0& 1& -1\\ -1& -1& 2\\ 1& 1& -1\end{array}\right]$$

Check: A·A⁻¹ = I.

Solution: $$A={\int }_{-2}^2\left(4-x^2\right)dx=2{\int }_0^2\left(4-x^2\right)dx=2{\left[4x-\frac{x^3}{3}\right]}_0^2=2\left(8-\frac{8}{3}\right)=\frac{32}{3}$$

∴ Area = 32/3 = 10.67 square units (Answer)

Solution: $$\frac{dy}{dx}=4x^3+6x^2-6x-4=2(x-1)(2x+1)(x+2)=0\Rightarrow x=-2,-\frac{1}{2},1$$

$$\frac{d^{2}y}{d{x}^2}=12x^2+12x-6:x=-2\to 18\left(\text{min}\right),x=-\frac{1}{2}\to -9\left(\text{max}\right),x=1\to 18\left(\text{min}\right)$$

$$y(-2)=16-16-12+8=-4,y\left(-\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{4}-\frac{3}{4}+2=\frac{17}{16},y(1)=1+2-3-4=-4$$

∴ Increasing in (−2, −1/2) and (1, ∞); decreasing in (−∞, −2) and (−1/2, 1). Maximum value = 17/16 at x = −1/2; minimum value = −4 at x = −2 and x = 1 (Answer)

Solution: Put t = tan(x/2), so cos x = (1 − t²)/(1 + t²) and dx = 2dt/(1 + t²).

$$4+5\cos x=\frac{4\left(1+t^2\right)+5\left(1-t^2\right)}{1+t^2}=\frac{9-t^2}{1+t^2}$$

$$\int \frac{dx}{4+5\cos x}=\int \frac{2dt}{9-t^2}=2\times \frac{1}{2\times 3}\ln \left|\frac{3+t}{3-t}\right|+C=\frac{1}{3}\ln \left|\frac{3+tan\frac{x}{2}}{3-tan\frac{x}{2}}\right|+C$$

Solution: Annalena Baerbock (Germany), President of the 80th session. [Bangladesh’s Foreign Minister Dr. Khalilur Rahman was elected (June 2026) President of the 81st session.]

Answer: International Convention for the Safety of Life at Sea.

Answer: Internet of Things.

Answer: Exclusive Economic Zone; up to 200 nautical miles from the baseline.

Answer: Sustainable Development Goals; 17 goals (169 targets).

উত্তর: দৈনিক আজাদী (চট্টগ্রাম, ১৭ ডিসেম্বর ১৯৭১)।

Answer: The Atlantic Ocean (Caribbean Sea) and the Pacific Ocean.

Answer: Rodri (Spain).

Answer: Five — West Bengal, Assam, Meghalaya, Tripura and Mizoram.

Answer: Finland.

Answer: Saudi Arabia (as recalled by the examinees).

Answer: 11 (Timor-Leste became the 11th member in October 2025).