Nepal Engineering Council · Mechanical Engineering · Chapter 9
Heat Transfer, Energy Resources and Environment
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180 questions in 6 syllabus topics.
9.1 Heat transfer
30 questions · AMeE0901
1. In Fourier's law of heat conduction, q = −k (dT/dx), the negative sign indicates that:
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Think about the direction in which heat naturally moves.
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Answer: A. heat flows in the direction of decreasing temperature
Heat always flows from higher to lower temperature, so with a positive gradient dT/dx the flux q is negative, i.e. directed down the gradient (second law).
2. The SI unit of thermal conductivity is:
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Rearrange Fourier's law and check units.
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Answer: B. W/m·K
From q = −k dT/dx: k = (W/m²)/(K/m) = W/m·K.
3. Among the following materials at room temperature, the highest thermal conductivity is that of:
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Pure metals with many free electrons conduct best.
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Answer: C. copper
Copper (about 400 W/m·K) is far more conductive than aluminium (~237), brass (~110) and stainless steel (~15).
4. Heat transfer from the Sun to the Earth takes place mainly by:
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Which mode needs no medium?
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Answer: D. radiation
Space is a near vacuum, so no conduction or convection is possible; only electromagnetic radiation can carry the energy.
5. With increase in temperature, the thermal conductivity of most pure metals generally:
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Think about electron scattering.
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Answer: B. decreases
In pure metals, lattice vibrations scatter the free electrons more at higher temperature, so k decreases slightly.
6. A surface of area 2 m² at 80 °C loses heat by convection to air at 30 °C. If the convection coefficient is 25 W/m²·K, the heat loss rate is:
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Use Q = hAΔT with ΔT = Ts − T∞.
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Answer: A. 2,500 W
Newton's law of cooling: Q = hA(Ts − T∞) = 25 × 2 × 50 = 2,500 W.
7. A plane wall 0.25 m thick, area 10 m² and k = 0.8 W/m·K has its faces at 30 °C and −10 °C. The steady heat flow through it is:
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Q = kAΔT/L.
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Answer: C. 1,280 W
Q = kAΔT/L = 0.8 × 10 × 40 / 0.25 = 1,280 W.
8. The conduction thermal resistance of a plane wall 0.1 m thick, k = 0.5 W/m·K and area 4 m² is:
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R = L/(kA).
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Answer: D. 0.050 K/W
R = L/(kA) = 0.1/(0.5 × 4) = 0.050 K/W.
9. In steady one-dimensional conduction through a composite wall of several layers in series, the quantity that is the same for every layer is:
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Energy cannot accumulate in steady state.
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Answer: A. heat flow rate
With no storage or generation, the same heat flow rate passes through each layer; the temperature drops are proportional to resistances.
10. A wall consists of layer 1 (0.2 m, k = 1.0 W/m·K) and layer 2 (0.1 m, k = 0.1 W/m·K) in series. For an area of 1 m² and an overall temperature difference of 100 K, the heat flow is about:
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Add resistances L/k in series.
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Answer: D. 83.3 W
R = 0.2/1.0 + 0.1/0.1 = 1.2 m²·K/W, so Q = 100/1.2 = 83.3 W.
11. Heat flow through a hollow cylinder (k = 50 W/m·K, length 2 m, r₁ = 0.05 m, r₂ = 0.10 m) with a 100 K temperature difference between its surfaces is about:
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Use the natural logarithm of the radius ratio.
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Answer: C. 90.6 kW
Q = 2πkLΔT / ln(r₂/r₁) = 2π×50×2×100 / ln 2 = 90.6 kW.
12. For steady radial conduction without heat generation in a long hollow cylinder, the temperature varies with radius:
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Compare with the linear profile of a plane wall.
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Answer: B. logarithmically
Integrating Fourier's law in cylindrical coordinates gives T = C₁ ln r + C₂.
13. A hollow sphere with k = 0.5 W/m·K, r₁ = 0.1 m and r₂ = 0.2 m has a surface temperature difference of 80 K. The conduction heat flow is about:
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Use 1/r₁ − 1/r₂ in the denominator.
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Answer: C. 100.5 W
Q = 4πkΔT/(1/r₁ − 1/r₂) = 4π×0.5×80/(10 − 5) = 100.5 W.
14. The critical radius of insulation for a cylindrical pipe is given by:
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Compare the radius where conduction resistance growth equals convection resistance drop.
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Answer: D. r_c = k/h
Adding insulation raises heat loss until r = k_ins/h_o; for a sphere the corresponding value is 2k/h.
15. Insulation of thermal conductivity 0.2 W/m·K is to be applied on a thin pipe in air with h = 10 W/m²·K. The critical radius of insulation is:
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r_c = k/h.
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Answer: A. 0.02 m
r_c = k/h = 0.2/10 = 0.02 m for a cylinder.
16. The emissive power of a black surface at 727 °C is about:
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Convert to kelvin first.
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Answer: B. 56.7 kW/m²
T = 727 + 273 = 1000 K, so E_b = σT⁴ = 5.67×10⁻⁸ × 10¹² = 56.7 kW/m².
17. The numerical value of the Stefan–Boltzmann constant is:
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It multiplies T⁴.
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Answer: B. 5.67 × 10⁻⁸ W/m²·K⁴
σ = 5.67×10⁻⁸ W/m²·K⁴; 2.898×10⁻³ m·K is Wien's displacement constant.
18. According to Kirchhoff's law, for a body in thermal equilibrium with its surroundings:
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Good absorbers are good emitters.
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Answer: D. absorptivity equals emissivity
Kirchhoff's law states α = ε for a body in thermal equilibrium (monochromatic or gray body).
19. A surface absorbs 30% and transmits 20% of incident radiation. Its reflectivity is:
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Use α + ρ + τ = 1.
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Answer: A. 0.5
α + ρ + τ = 1, so ρ = 1 − 0.3 − 0.2 = 0.5.
20. A body that reflects all incident radiation (ρ = 1) is called a:
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Check which coefficient equals 1.
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Answer: C. white body
A perfect reflector is a white body; a black body has α = 1 and a transparent body has τ = 1.
21. Thermal radiation as studied in heat transfer lies approximately in the wavelength range:
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It includes visible light and infrared.
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Answer: D. 0.1 µm to 100 µm
Thermal radiation covers part of UV, visible and infrared, about 0.1–100 µm; X-rays and radio waves are outside it.
22. Using Wien's displacement law (λ_max·T = 2898 µm·K), the wavelength of peak emission from a black body at 5800 K is about:
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Divide the constant by the absolute temperature.
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Answer: B. 0.50 µm
λ_max = 2898/5800 = 0.50 µm (visible region).
23. A small gray body (ε = 0.8, area 1 m²) at 500 K is in a very large enclosure at 300 K. Net radiation loss is about:
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Subtract fourth powers, not temperatures.
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Answer: C. 2468 W
Q = εσA(T₁⁴ − T₂⁴) = 0.8×5.67×10⁻⁸×(6.25×10¹⁰ − 8.1×10⁹) = 2468 W.
24. A thin radiation shield placed between two large parallel plates (all surfaces having equal emissivity) will:
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It adds resistance to the radiation path.
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Answer: A. reduce the net radiant exchange between the plates
The shield adds extra surface resistances in series; for a single shield with equal emissivities the exchange is halved.
25. A wall of thickness 0.2 m and k = 1 W/m·K has convection coefficients h₁ = 20 and h₂ = 10 W/m²·K on its two sides. The overall heat transfer coefficient is:
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Add the three resistances per unit area.
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Answer: B. 2.86 W/m²·K
1/U = 1/20 + 0.2/1 + 1/10 = 0.35, so U = 2.86 W/m²·K.
26. The deposition of scale or soot (fouling) on a heat transfer surface causes the overall heat transfer coefficient to:
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Extra layer, extra resistance.
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Answer: A. decrease
Fouling adds a thermal resistance (fouling factor) in series, so 1/U increases and U falls.
27. A two-layer wall has resistances R₁ = 0.2 and R₂ = 0.3 K/W. If the outer faces are at 100 °C and 20 °C, the interface temperature (steady state) is:
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Temperature drop is proportional to resistance.
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Answer: C. 68 °C
T_int = 100 − 80 × 0.2/(0.2 + 0.3) = 68 °C.
28. A 10 m long pipe of outer diameter 50 mm has its surface at 90 °C in air at 30 °C with h = 10 W/m²·K. The convective loss is about:
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Surface area of cylinder = πDL.
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Answer: D. 942 W
Q = h(πDL)ΔT = 10 × π × 0.05 × 10 × 60 = 942 W.
29. The Nusselt number is defined as:
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It contains the convection coefficient h.
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Answer: B. hL/k_fluid
Nu = hL/k is the ratio of convective to conductive heat transfer across a fluid layer; the others are Re, Pr and Gr.
30. A temperature drop across the interface of two solids pressed together is attributed to:
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Surfaces are never perfectly smooth.
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Answer: D. thermal contact resistance
Microscopic roughness leaves gaps (air) at the interface, producing a thermal contact resistance and a temperature jump.
9.2 Application of heat transfer
30 questions · AMeE0902
31. In natural (free) convection the fluid motion is caused by:
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What drives the flow without a fan?
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Answer: D. buoyancy forces arising from density differences
Heating changes fluid density, and the resulting buoyancy force drives the flow; forced convection needs a pump or fan.
32. The Grashof number represents the ratio of:
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Gr governs natural convection.
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Answer: C. buoyancy force to viscous force
Gr = gβΔTL³/ν² measures buoyancy relative to viscous forces; it plays the role in free convection that Re plays in forced convection.
33. The Prandtl number is the ratio of:
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Pr = µc_p/k.
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Answer: A. momentum diffusivity to thermal diffusivity
Pr = ν/α = µc_p/k.
34. Water (ν = 1×10⁻⁶ m²/s) flows at 1 m/s in a 50 mm diameter pipe. The Reynolds number is:
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Re = VD/ν.
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Answer: B. 50,000
Re = VD/ν = 1 × 0.05 / 10⁻⁶ = 50,000, which is turbulent.
35. Flow in a circular pipe is generally taken as laminar when the Reynolds number is below about:
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Pipe flow, not flat plate.
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Answer: B. 2300
The transition from laminar flow in pipes begins around Re ≈ 2300; 5×10⁵ is the usual critical value for a flat plate.
36. For flow in a tube, Nu = 100, the fluid conductivity is 0.6 W/m·K and D = 50 mm. The heat transfer coefficient is:
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Nu = hD/k.
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Answer: C. 1,200 W/m²·K
h = Nu·k/D = 100 × 0.6 / 0.05 = 1,200 W/m²·K.
37. For forced turbulent flow of a fluid being heated in a smooth tube, the Dittus–Boelter correlation is:
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Exponent on Re is the larger one.
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Answer: A. Nu = 0.023 Re^0.8 Pr^0.4
Dittus–Boelter: Nu = 0.023 Re^0.8 Pr^n with n = 0.4 for heating and 0.3 for cooling; 0.664 Re^0.5 Pr^(1/3) is laminar flat plate.
38. For an ideal gas, the volumetric expansion coefficient β used in the Grashof number is:
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Think of V ∝ T.
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Answer: D. 1/T (absolute temperature)
For an ideal gas at constant pressure, β = (1/V)(∂V/∂T)_p = 1/T, with T in kelvin (evaluated at film temperature).
39. Typical values of the convection heat transfer coefficient are lowest for:
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Low density, low conductivity, no forced flow.
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Answer: B. free convection in gases
Free convection in gases gives about 2–25 W/m²·K, far below liquids and phase-change processes (thousands W/m²·K).
40. Fins are provided on a surface mainly to:
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Q = hAΔT.
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Answer: D. increase the heat transfer rate by increasing surface area
Q = hAΔT, so extra area increases heat transfer when the convective resistance 1/(hA) is dominant.
41. Fins are most effective when attached on the side of a wall which has:
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Where is the biggest resistance?
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Answer: A. the lower heat transfer coefficient
Adding area reduces the dominant resistance, which is on the low-h side (e.g. air side of a radiator).
42. Fin efficiency is defined as the ratio of:
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Ideal fin would have uniform temperature.
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Answer: C. actual heat transfer from the fin to heat transfer if the entire fin were at base temperature
η_f = Q_fin/(hA_fin θ_b). The ratio with versus without fin is the fin effectiveness.
43. Fin effectiveness is defined as the ratio of heat transfer:
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Compare to a bare base.
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Answer: D. with the fin to that without the fin
ε_f = Q_with fin/(hA_b θ_b); a fin is worthwhile only when ε_f is well above 1 (generally > 2).
44. A pin fin of diameter 5 mm and k = 200 W/m·K is exposed to h = 25 W/m²·K. The fin parameter m = √(hP/kA) is:
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P/A = 4/D for a circular rod.
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Answer: B. 10 m⁻¹
For a pin, P/A = 4/D, so m = √(4h/(kD)) = √(4×25/(200×0.005)) = 10 m⁻¹.
45. An infinitely long pin fin (D = 5 mm, k = 200 W/m·K, h = 25 W/m²·K) has a base excess temperature of 100 K. The heat dissipated is about:
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Q = √(hPkA)·θ_b.
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Answer: A. 3.9 W
Q = √(hPkA)·θ_b = √(25 × 0.015708 × 200 × 1.9635×10⁻⁵) × 100 = 3.9 W.
46. Compared with a short fin, increasing the length of a fin of the same material generally causes its efficiency to:
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The tip temperature falls as the fin lengthens.
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Answer: C. decrease
Longer fins have a colder tip that adds little heat while adding area, so efficiency (heat actually transferred ÷ ideal) drops, though total heat rises with diminishing returns.
47. For a fin of given geometry, using a material of higher thermal conductivity will:
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The fin becomes closer to isothermal.
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Answer: A. increase the fin efficiency
Higher k lowers the temperature drop along the fin, bringing it closer to the base temperature and increasing η_f.
48. The heat dissipation from a fin with insulated tip is Q = √(hPkA)·θ_b · F where F equals:
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Hyperbolic function that tends to 1 as mL grows.
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Answer: D. tanh(mL)
For insulated tip, Q = √(hPkA) θ_b tanh(mL); for an infinitely long fin tanh(mL) → 1.
49. For the same inlet and outlet temperatures, the LMTD of a counter-flow heat exchanger compared with parallel flow is:
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Which arrangement needs less area?
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Answer: C. greater
Counter-flow keeps a more uniform temperature difference along the length, giving a larger LMTD and therefore smaller area for the same duty.
50. In a counter-flow exchanger the hot fluid cools 100 → 60 °C while cold fluid heats 30 → 50 °C. The LMTD is about:
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Pair hot-in with cold-out for counter flow.
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Answer: B. 39.2 °C
ΔT₁ = 100 − 50 = 50, ΔT₂ = 60 − 30 = 30; LMTD = (50 − 30)/ln(50/30) = 39.2 °C.
51. For the above exchanger with U = 500 W/m²·K and A = 2 m², the heat transfer rate is about (LMTD = 39.2 °C):
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Q = UA·LMTD.
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Answer: B. 39.2 kW
Q = U·A·LMTD = 500 × 2 × 39.2 = 39.2 kW.
52. In a parallel-flow heat exchanger, the cold fluid outlet temperature:
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Think of both streams approaching each other in the same direction.
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Answer: D. can never exceed the hot fluid outlet temperature
Both fluids move to a common asymptotic temperature; the cold outlet always remains below the hot outlet. Counter-flow can achieve a cold outlet above the hot outlet.
53. Heat exchanger effectiveness is defined as:
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It is a fraction between 0 and 1.
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Answer: C. actual heat transfer / maximum possible heat transfer
ε = Q_actual/Q_max, with Q_max = C_min(T_h,in − T_c,in); its value lies between 0 and 1.
54. Hot fluid (C = 2000 W/K) enters at 150 °C and leaves at 110 °C; cold fluid (C = 4000 W/K) enters at 30 °C. The effectiveness is:
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Use the smaller capacity rate for Q_max.
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Answer: A. 0.33
Q = 2000 × 40 = 80 kW; Q_max = C_min(150 − 30) = 2000 × 120 = 240 kW; ε = 80/240 = 0.33.
55. A counter-flow heat exchanger with capacity rate ratio C_r = 1 has NTU = 2. Its effectiveness is:
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Special formula for C_r = 1.
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Answer: A. 0.667
For counter-flow with C_r = 1, ε = NTU/(1 + NTU) = 2/3 = 0.667.
56. In a condenser or boiler (C_r = 0) with NTU = 1, the effectiveness (any flow arrangement) is:
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ε = 1 − e^(−NTU).
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Answer: D. 0.632
For C_r = 0, ε = 1 − e^(−NTU) = 1 − e⁻¹ = 0.632.
57. NTU of a heat exchanger is defined as:
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Divide by the smaller capacity rate.
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Answer: B. UA / C_min
NTU = UA/C_min where C = ṁc_p; it measures the size of the exchanger.
58. A heat exchanger in which the hot and cold fluids flow through the same space alternately (storing and releasing heat in a matrix) is called a:
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Heat is stored in a matrix.
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Answer: C. regenerator
In a regenerator the matrix is alternately exposed to hot and cold streams (e.g. air preheater wheel); in a recuperator the fluids are separated by a wall.
59. The most common heat exchanger in process industries, consisting of a bundle of tubes inside a cylindrical casing, is the:
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Tubes within a shell.
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Answer: D. shell-and-tube exchanger
Shell-and-tube units handle high pressures and large duties and are the most widely used type.
60. In the LMTD method, the correction factor F for cross-flow and multi-pass exchangers is:
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Counter-flow is the best case.
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Answer: C. less than or equal to 1
Q = U A F (LMTD)_counter; the true mean temperature difference of other arrangements cannot exceed that of pure counter-flow.
9.3 Conventional and non-conventional energy resources
30 questions · AMeE0903
61. Which of the following is a secondary energy source?
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Which one must be produced from another source?
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Answer: D. electricity
Secondary sources are obtained by converting primary sources; electricity is generated from coal, hydro, wind etc., whereas the others occur in nature.
62. Among coals, the one with the highest fixed-carbon content and calorific value is:
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The hardest, most mature coal.
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Answer: A. anthracite
Rank increases from peat to lignite to bituminous to anthracite, which has about 90%+ carbon and the highest heating value.
63. The proximate analysis of coal gives the percentages of:
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It is the simpler of the two standard analyses.
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Answer: C. moisture, volatile matter, fixed carbon and ash
Proximate analysis: moisture, volatile matter, fixed carbon, ash. Ultimate analysis gives elemental C, H, N, S, O.
64. A fuel has gross calorific value 30,000 kJ/kg, contains 5% hydrogen by mass and no moisture. Taking latent heat of steam as 2442 kJ/kg, its net calorific value is about:
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Each kg of hydrogen produces 9 kg of water.
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Answer: B. 28,901 kJ/kg
1 kg H₂ forms 9 kg of water, so water formed = 9 × 0.05 = 0.45 kg/kg fuel; NCV = 30,000 − 0.45 × 2442 = 28,901 kJ/kg.
65. The main constituent of natural gas is:
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The simplest hydrocarbon.
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Answer: D. methane
Natural gas is typically 85–95% methane (CH₄).
66. In a refinery, crude oil is separated into fractions by:
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Boiling points differ among fractions.
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Answer: A. fractional distillation
Crude oil is heated and its components separated by boiling-point range in a fractionating column.
67. A coal-fired plant produces 100 MW of electricity at an overall efficiency of 35%. If the coal has a calorific value of 24 MJ/kg, the coal consumption is about:
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Heat input = output/efficiency.
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Answer: B. 42.9 tonnes/h
Fuel heat input = 100/0.35 = 285.7 MW; coal = 285.7/24 = 11.90 kg/s = 42.9 t/h.
68. The solar constant, the solar radiation flux at the top of the atmosphere on a surface normal to the Sun's rays at mean Earth–Sun distance, is about:
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Roughly 1.4 kW per square metre.
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Answer: C. 1367 W/m²
The solar constant is approximately 1361–1367 W/m².
69. A photovoltaic cell converts sunlight to electricity through:
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Light directly produces voltage across a junction.
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Answer: A. the photovoltaic effect in a p–n junction
Photons generate electron–hole pairs which the p–n junction field separates, producing a current.
70. A PV module of area 1.6 m² and efficiency 18% receives irradiance 1000 W/m². Its output power is:
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P = ηGA.
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Answer: B. 288 W
P = η·G·A = 0.18 × 1000 × 1.6 = 288 W.
71. A flat-plate collector of area 2 m² receives 800 W/m² and delivers 1000 W as useful heat to the water. Its efficiency is:
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Efficiency = useful/incident.
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Answer: D. 62.5%
η = Q_u/(G·A) = 1000/(800 × 2) = 62.5%.
72. A wind turbine with a rotor diameter of 40 m is in wind of 10 m/s (air density 1.2 kg/m³). The power available in the wind is about:
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P = ½ρAv³ with A = πD²/4.
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Answer: C. 754 kW
P = ½ρAv³ = 0.5 × 1.2 × (π/4 × 40²) × 10³ = 754 kW (before any efficiency).
73. If the wind speed doubles, the power available in the wind (same rotor) becomes:
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Wind power depends on the cube of speed.
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Answer: C. 8 times
Power is proportional to v³, so doubling gives 2³ = 8 times.
74. The maximum theoretical fraction of the wind power that can be extracted by a wind turbine (Betz limit) is:
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Value is 16/27.
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Answer: A. 59.3%
Betz limit = 16/27 ≈ 0.593.
75. Biogas produced from cattle dung by anaerobic digestion contains mainly:
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The useful fuel gas is the simplest hydrocarbon.
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Answer: B. methane (55–65%) and carbon dioxide
Biogas is about 55–65% methane and 35–45% CO₂ with traces of H₂S.
76. A floating-drum biogas plant differs from a fixed-dome plant in that its gas holder:
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Look at the moving part.
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Answer: D. moves up and down giving nearly constant gas pressure
In a floating-drum plant, the drum rises as gas accumulates, supplying gas at constant pressure; in a fixed-dome plant pressure varies as slurry is displaced.
77. A micro-hydro scheme has a net head of 50 m and a flow of 0.5 m³/s. With overall efficiency 70%, the electrical output is about:
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P = ρgQHη.
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Answer: A. 172 kW
P = ρgQHη = 1000 × 9.81 × 0.5 × 50 × 0.7 = 171,675 W ≈ 172 kW.
78. Hydropower plants are commonly classified as 'micro' when their capacity is up to:
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Less than a megawatt, but more than a few kilowatts.
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Answer: D. 100 kW
A common classification: pico < 5 kW, micro up to 100 kW, mini up to 1 MW, small up to about 10 MW.
79. For a micro-hydro site with high head and low discharge, the most suitable turbine is the:
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Impulse turbine for high head.
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Answer: B. Pelton wheel
Pelton (impulse) turbines suit heads of hundreds of metres and small flows; Kaplan/propeller suit low head and large flow.
80. In a nuclear reactor, the function of the moderator is to:
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Neutron speed, not neutron number.
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Answer: C. slow down fast neutrons to thermal energies
Moderators (graphite, water, heavy water) reduce neutron energy so they cause fission in U-235 more efficiently.
81. Control rods used in nuclear reactors are made of materials like:
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They absorb neutrons.
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Answer: C. boron or cadmium
Boron and cadmium strongly absorb neutrons; inserting the rods reduces the reaction rate.
82. Each fission of U-235 releases about 200 MeV. The number of fissions per second required for 1 MW of thermal power is about (1 eV = 1.602×10⁻¹⁹ J):
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Convert MeV to joules first.
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Answer: B. 3.1 × 10¹⁶
Energy per fission = 200×10⁶ × 1.602×10⁻¹⁹ = 3.2×10⁻¹¹ J; N = 10⁶/3.2×10⁻¹¹ = 3.1 × 10¹⁶ per second.
83. The reversible voltage of a hydrogen–oxygen fuel cell at 25 °C with ΔG = −237 kJ/mol H₂ (n = 2, F = 96,485 C/mol) is:
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E = −ΔG/(nF).
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Answer: A. 1.23 V
E = −ΔG/(nF) = 237,000/(2 × 96,485) = 1.23 V.
84. A fuel cell differs from a heat engine in that it:
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No flame, no turbine.
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Answer: D. converts chemical energy directly to electricity and is not limited by the Carnot efficiency
A fuel cell is an electrochemical device; the cell reaction gives electrical work directly without a combustion step.
85. In a hydrogen–oxygen PEM fuel cell, the by-product is:
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Hydrogen plus oxygen.
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Answer: D. water
2H₂ + O₂ → 2H₂O; only water (and heat) is formed at the point of use.
86. Hydrogen is called 'green hydrogen' when it is produced by:
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Colour refers to the source of energy.
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Answer: A. electrolysis of water using renewable electricity
Green hydrogen uses renewable power to split water; reforming of natural gas gives grey hydrogen.
87. A major difficulty in using hydrogen as a vehicle fuel is that:
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Per kg it is excellent, per m³ not.
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Answer: B. its energy density per unit volume is low, making storage difficult
Hydrogen has a very high energy per kg (~120 MJ/kg) but very low volumetric density, requiring high-pressure or cryogenic storage.
88. The complete combustion of 1 kg of carbon (C + O₂ → CO₂) produces CO₂ of mass:
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Use molecular masses 12 and 44.
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Answer: C. 3.67 kg
12 kg C gives 44 kg CO₂, so 1 kg C gives 44/12 = 3.67 kg CO₂.
89. Which energy source is associated with acid rain mainly through emission of sulphur dioxide?
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Which one involves burning sulphur-bearing fuel?
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Answer: C. burning high-sulphur coal
Sulphur in coal burns to SO₂, which forms sulphurous/sulphuric acid in the atmosphere.
90. A major environmental concern of nuclear power generation, apart from accident risk, is:
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Waste lasts for millennia.
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Answer: B. long-lived radioactive waste disposal
Spent fuel remains radioactive for thousands of years and needs secure storage; CO₂ emissions in operation are small.
9.4 Combustion and combustion products
30 questions · AMeE0904
91. The three elements of the fire triangle are:
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Think of what is needed to start any fire.
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Answer: B. fuel, oxygen (oxidiser) and heat
Combustion needs a combustible material, an oxidiser and sufficient heat (ignition energy); removing any one extinguishes the fire.
92. The minimum temperature at which a fuel gives off enough vapour to form a flammable mixture that flashes momentarily on application of a flame is the:
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The flame flashes and goes out.
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Answer: A. flash point
Flash point: a flash appears but does not sustain; fire point (slightly higher) is when burning continues; auto-ignition needs no external flame.
93. For a given liquid fuel the correct ascending order of temperatures is:
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Auto-ignition needs the most heat.
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Answer: C. flash point < fire point < auto-ignition temperature
A fuel flashes first, sustains burning (fire point) a few degrees higher, and self-ignites without a spark at a much higher temperature.
94. Taking air to be 23% oxygen by mass, the stoichiometric air required to burn 1 kg of carbon completely to CO₂ is about:
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Convert O₂ requirement to air using the mass fraction.
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Answer: D. 11.6 kg
C + O₂ → CO₂: 12 kg C needs 32 kg O₂, so 1 kg needs 2.667 kg O₂; air = 2.667/0.23 = 11.6 kg.
95. The stoichiometric air–fuel ratio by mass for methane (CH₄ + 2O₂ → CO₂ + 2H₂O), with air 23% O₂ by mass, is about:
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Oxygen required per kg fuel divided by 0.23.
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Answer: C. 17.4
O₂ required = 2×32/16 = 4 kg/kg CH₄; air = 4/0.23 = 17.4 kg/kg.
96. A petrol engine runs with an actual air–fuel ratio of 18 where the stoichiometric ratio is 14.7. The percentage excess air is about:
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Excess air is relative to the stoichiometric requirement.
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Answer: A. 22.4%
Excess air = (18 − 14.7)/14.7 × 100 = 22.4% (λ = 1.22, a lean mixture).
97. An engine running with equivalence ratio φ > 1 is operating with a:
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Compare fuel with what the air can burn.
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Answer: D. rich mixture
φ = (A/F)stoich/(A/F)actual; φ > 1 means more fuel than required, i.e. rich.
98. The stage of CI engine combustion in which the injected fuel accumulates before burning begins is called the:
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The 'waiting' stage.
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Answer: B. ignition delay period
Between start of injection and start of combustion is the delay period; the fuel accumulated then burns rapidly (uncontrolled phase).
99. Diesel knock is caused by:
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Think of the delay period.
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Answer: B. a long ignition delay which lets too much fuel accumulate before burning
A long delay causes sudden burning of a large accumulated charge, giving a sharp pressure rise. End-gas auto-ignition is SI knock.
100. Knocking in a spark-ignition engine is due to:
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Where does the unburnt charge sit?
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Answer: D. auto-ignition of the unburnt end gas ahead of the flame front
The end gas, compressed by the expanding flame, ignites spontaneously and produces pressure waves (pinging).
101. A fire involving flammable liquids such as petrol or oil is classified (in the usual A–D scheme) as:
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B stands for burning liquids.
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Answer: A. Class B
Class A: ordinary solids; B: flammable liquids; C: gases (or electrical in some systems); D: combustible metals.
102. Water should not be used on an oil fire because:
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Oil and water do not mix.
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Answer: C. oil floats and burning oil spreads with the water
Oil is lighter and immiscible; water sinks beneath it or splashes the burning oil, spreading the fire. Foam or dry chemical should be used.
103. Fire is extinguished by smothering when:
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Which side of the triangle is removed?
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Answer: B. the supply of oxygen to the fuel is cut off
Smothering (blankets, foam, CO₂) separates fuel from oxygen; cooling removes heat and starving removes fuel.
104. For fires in live electrical equipment, the most suitable extinguisher is:
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The agent must not conduct current.
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Answer: D. CO₂ or dry chemical powder
CO₂ and dry powder are non-conductive and leave no residue; water and foam conduct electricity.
105. Carbon monoxide in combustion products is dangerous because it:
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Think of blood.
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Answer: C. combines with haemoglobin and reduces oxygen transport
CO has about 200+ times the affinity of O₂ for haemoglobin, forming carboxyhaemoglobin.
106. Sulphur dioxide formed by burning sulphur-containing fuels mainly causes:
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Sulphur → acid.
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Answer: A. acid rain and respiratory problems
SO₂ forms sulphuric acid in the atmosphere and irritates the respiratory tract.
107. Thermal NOₓ in engine exhaust is formed mainly when:
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Heat and air together.
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Answer: C. combustion temperatures are high and oxygen is available
Nitrogen and oxygen of air react at high temperature (Zeldovich mechanism); NOₓ is highest at slightly lean mixtures where temperature and oxygen are both high.
108. Exhaust gas recirculation (EGR) reduces NOₓ emission because it:
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What does NOₓ formation depend on?
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Answer: B. lowers peak combustion temperature
Recirculated inert exhaust gas dilutes the charge and absorbs heat, cutting the peak temperature and thermal NOₓ formation.
109. A three-way catalytic converter on a petrol engine simultaneously reduces:
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Three pollutants, three-way.
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Answer: D. CO, HC and NOₓ
It oxidises CO and HC and reduces NOₓ, working best near the stoichiometric air–fuel ratio (with an oxygen sensor feedback).
110. Leaded petrol must not be used in vehicles with catalytic converters because lead:
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The catalyst is damaged.
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Answer: A. poisons the catalyst
Lead compounds coat the catalyst surface, deactivating it; leaded petrol has also been phased out for health reasons.
111. The black smoke from a diesel engine is mainly:
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Rich zones form it.
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Answer: A. unburnt carbon particles (soot) formed in fuel-rich zones
Locally rich regions with insufficient oxygen crack the fuel to soot; a particulate filter is used to trap it.
112. Unburnt hydrocarbon (HC) emission from a spark-ignition engine is mainly due to:
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Cold surfaces extinguish the flame.
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Answer: B. flame quenching at cold walls and crevices, and misfire
Flame cannot enter narrow crevices or cold boundary layers, leaving unburnt fuel; very rich mixtures and misfires also raise HC.
113. Burning 1 kg of octane (C₈H₁₈, molar mass 114) completely gives CO₂ of mass:
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Eight CO₂ molecules per octane molecule.
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Answer: C. 3.09 kg
C₈H₁₈ + 12.5O₂ → 8CO₂ + 9H₂O; 114 kg fuel gives 8 × 44 = 352 kg CO₂, so 1 kg gives 3.09 kg.
114. An Orsat analysis of the flue gas from a boiler shows 4.2% O₂ (by volume, dry) and no CO. The approximate percentage excess air is:
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Use the 21% O₂ in air.
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Answer: D. 25%
Excess air ≈ O₂/(21 − O₂) × 100 = 4.2/16.8 × 100 = 25%.
115. Complete combustion of carbon to CO₂ releases about 32.8 MJ/kg, whereas combustion only to CO releases about 9.2 MJ/kg. This shows that:
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Compare the two numbers.
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Answer: D. incomplete combustion wastes much of the fuel's energy
Most of the energy remains in CO as chemical energy (it can still burn to CO₂), so incomplete combustion reduces efficiency and emits toxic CO.
116. Burning 1 kg of methane completely produces CO₂ of mass:
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One CO₂ per CH₄.
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Answer: A. 2.75 kg
CH₄ + 2O₂ → CO₂ + 2H₂O: 16 kg CH₄ gives 44 kg CO₂, so 1 kg gives 2.75 kg.
117. A fuel–air mixture will burn only if its composition lies between:
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Composition limits, not temperature.
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Answer: B. the lower and upper flammability limits
Mixtures too lean (below LEL) or too rich (above UEL) cannot propagate a flame; e.g. methane in air about 5–15% by volume.
118. A petrol engine runs at an air–fuel ratio of 12.5 (stoichiometric 14.7). The equivalence ratio is about:
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Stoichiometric over actual.
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Answer: C. 1.18
φ = (A/F)stoich/(A/F)actual = 14.7/12.5 = 1.18, a rich mixture.
119. A liquid fuel contains 85% carbon and 15% hydrogen by mass. With air 23% O₂ by mass, the minimum (stoichiometric) air required per kg of fuel is about:
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C needs 32/12 kg O₂/kg, H needs 8 kg O₂/kg.
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Answer: D. 15.1 kg
O₂ = 0.85×(32/12) + 0.15×8 = 3.467 kg; air = 3.467/0.23 = 15.1 kg per kg fuel.
120. A fuel needs 15.1 kg of air per kg theoretically. It is burnt with 20% excess air. The actual air supplied per kg of fuel is about:
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Multiply the theoretical air by 1.2.
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Answer: A. 18.1 kg
Actual air = 1.2 × 15.07 = 18.1 kg per kg fuel.
9.5 Engine fuels
30 questions · AMeE0905
121. Petroleum (crude oil) consists mainly of:
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Compounds of carbon and hydrogen.
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Answer: B. a mixture of hydrocarbons (paraffins, naphthenes and aromatics)
Crude oil is a complex mixture of hydrocarbons with small amounts of sulphur, nitrogen and oxygen compounds.
122. The general formula of straight-chain saturated hydrocarbons (paraffins/alkanes) is:
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Methane is CH₄.
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Answer: C. CₙH₂ₙ₊₂
Alkanes such as methane CH₄ and octane C₈H₁₈ follow CₙH₂ₙ₊₂; olefins and naphthenes are CₙH₂ₙ.
123. Benzene (C₆H₆) and toluene belong to the hydrocarbon family called:
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Ring structure with alternating double bonds.
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Answer: D. aromatics
Aromatics contain a benzene ring; they have high octane ratings but poor cetane numbers.
124. The reference fuels for the octane number scale are:
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Think of the SI engine knock scale.
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Answer: A. iso-octane (100) and n-heptane (0)
Octane number is the percentage by volume of iso-octane in a blend with n-heptane that matches the knock behaviour of the test fuel.
125. A reference blend contains 80% iso-octane and 20% n-heptane by volume. Its octane number is:
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Percentage of iso-octane.
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Answer: B. 80
Octane number equals the volume percentage of iso-octane in the blend = 80.
126. The reference fuels for the cetane number scale are:
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Think of the CI engine ignition quality scale.
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Answer: D. n-cetane (100) and α-methylnaphthalene (0)
Cetane number is the percentage of n-cetane (hexadecane) in a blend with α-methylnaphthalene that has the same ignition delay as the fuel.
127. A good diesel fuel should have:
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The two ratings are roughly opposite.
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Answer: A. a high cetane number and low octane number
Diesel fuels should auto-ignite readily (high cetane), which corresponds to poor knock resistance in SI engines (low octane).
128. In petroleum refining, cracking is the process of:
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It increases the gasoline yield.
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Answer: C. breaking large hydrocarbon molecules into smaller, lighter ones
Thermal or catalytic cracking converts heavy fractions into gasoline-range hydrocarbons; reforming raises octane by restructuring molecules.
129. Catalytic reforming of naphtha mainly:
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Octane improvement.
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Answer: D. raises the octane number by converting paraffins to aromatics and iso-paraffins
Reforming restructures low-octane straight-chain molecules into high-octane aromatics and branched isomers.
130. For a fuel to be used safely at higher compression ratios in an SI engine, it must have a higher:
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Resistance to knock.
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Answer: B. octane number
A higher compression ratio increases end-gas temperature and pressure, so greater knock resistance (octane) is needed.
131. The main function of a carburettor is to:
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It makes the mixture.
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Answer: A. mix fuel and air in the proper ratio for the engine load and speed
The carburettor atomises fuel into the air stream through a venturi and supplies the mixture strength needed at start, idle, cruise and full load.
132. The pressure depression at the venturi of a carburettor is 5 kPa and air density is 1.2 kg/m³. Neglecting losses, the air velocity at the throat is about:
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ΔP = ½ρv².
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Answer: C. 91 m/s
From Bernoulli, ΔP = ½ρv²; v = √(2×5000/1.2) = 91 m/s.
133. The choke valve in a carburettor is used to:
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Cold engines need more fuel.
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Answer: A. enrich the mixture for cold starting
Closing the choke reduces airflow, increasing the venturi depression and fuel flow, giving a rich mixture for cold starting.
134. In a multi-point fuel injection (MPFI) petrol engine, fuel is injected:
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'Multi-point' means one injector per cylinder.
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Answer: B. into the intake port of each cylinder
MPFI uses one injector per cylinder at the inlet port; GDI injects directly into the cylinder.
135. In a diesel engine, the component that raises the fuel pressure to the value needed for atomisation is the:
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High pressure is needed for atomisation.
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Answer: C. fuel injection pump
The lift pump feeds fuel at low pressure; the injection pump (or common-rail pump) delivers it at high pressure to the injectors.
136. The function of the governor in a diesel injection pump is to:
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There is no throttle in a diesel.
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Answer: D. control the quantity of fuel injected according to load and speed
Because diesel engines have no throttle, the governor adjusts fuel delivery to maintain the required speed under varying load.
137. A 4-cylinder, 4-stroke diesel engine running at 2000 rpm consumes 8 kg/h of fuel. Fuel injected per cylinder per cycle is about:
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Each cylinder fires once every two revolutions.
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Answer: A. 33.3 mg
Injections per minute = 4 × 2000/2 = 4000; fuel per minute = 8000/60 = 133.3 g; per injection = 33.3 mg.
138. Which of the following is NOT a function of the lubricating oil in an engine?
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Look for the function not related to oil.
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Answer: B. increasing the compression ratio
Lubricants reduce friction, carry heat away, seal and clean; they do not alter the compression ratio.
139. In a wet-sump lubrication system:
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'Wet' means oil is inside the pan.
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Answer: C. the oil is stored in the crankcase sump at the bottom of the engine
A wet sump keeps oil in the crankcase pan; a dry sump scavenges oil to a separate tank, used on racing and aircraft engines.
140. The Viscosity Index (VI) of an oil indicates:
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Temperature sensitivity.
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Answer: D. how little its viscosity changes with temperature
A high VI means viscosity changes slowly with temperature, which is desirable for engine oils.
141. In the oil grade SAE 20W-50, the 'W' indicates:
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W is a season.
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Answer: A. suitability for winter (low-temperature) use
The number before W is the cold-cranking viscosity grade; 50 is the grade at 100 °C. Multigrade oils meet both.
142. The API classification letters 'S' (e.g. SN) and 'C' (e.g. CK) in engine oil specification indicate that the oil is suited respectively for:
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S for spark, C for compression.
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Answer: B. spark-ignition (petrol) and compression-ignition (diesel) engines
API 'S' = service category for gasoline engines; 'C' = commercial/diesel engines.
143. Detergent-dispersant additives are added to lubricating oil to:
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Cleaning action.
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Answer: D. keep engine parts clean by suspending contaminants
Detergents and dispersants prevent sludge and deposits by holding combustion products in suspension until the oil is changed.
144. An engine develops 30 kW brake power while consuming 8 kg/h of fuel. The brake specific fuel consumption is:
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bsfc = ṁ_f / BP.
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Answer: C. 0.267 kg/kWh
bsfc = fuel rate / brake power = 8/30 = 0.267 kg/kWh.
145. An engine has a brake specific fuel consumption of 0.25 kg/kWh and the fuel has a calorific value of 44,000 kJ/kg. The brake thermal efficiency is:
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1 kWh = 3600 kJ.
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Answer: D. 32.7%
η = 3600/(bsfc × CV) = 3600/(0.25 × 44,000) = 32.7%.
146. An engine producing 50 kW with bsfc 0.3 kg/kWh uses petrol of density 740 kg/m³. The fuel consumption in litres per hour is about:
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Mass divided by density gives volume (kg ÷ kg/L).
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Answer: B. 20.3 L/h
Mass flow = 50 × 0.3 = 15 kg/h; volume = 15/0.74 = 20.3 L/h.
147. A bsfc of 0.30 kg/kWh expressed in kg per brake horsepower-hour (1 hp = 0.7457 kW) is:
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One horsepower is smaller than a kilowatt.
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Answer: C. 0.224 kg/bhp·h
1 hp·h = 0.7457 kWh, so 0.30 × 0.7457 = 0.224 kg per hp·h.
148. Using °API = 141.5/SG − 131.5, the API gravity of a diesel fuel of specific gravity 0.85 is about:
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Insert SG into the given formula.
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Answer: A. 35°API
°API = 141.5/0.85 − 131.5 = 35.0.
149. Fuels with higher volatility (high Reid vapour pressure) increase the risk of:
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Fuel boils in the line.
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Answer: B. vapour lock in the fuel supply system
Highly volatile fuel can vaporise in the fuel line or pump, blocking flow (vapour lock), especially in hot weather.
150. An engine takes in 0.1 kg/s of air and 0.007 kg/s of fuel. The air–fuel ratio is about:
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Ratio of mass flows.
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Answer: C. 14.3
A/F = 0.1/0.007 = 14.3.
9.6 Environment and pollution control
30 questions · AMeE0906
151. Which of the following is a secondary air pollutant?
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Not emitted directly.
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Answer: D. ground-level ozone
Secondary pollutants form in the atmosphere by reactions of primary pollutants; ozone is formed from NOₓ and VOCs in sunlight.
152. PM2.5 refers to particulate matter with an aerodynamic diameter:
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µm, not mm.
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Answer: C. of 2.5 µm or less
PM2.5 are fine particles ≤ 2.5 µm that penetrate deep into the lungs.
153. Photochemical smog is formed by the action of sunlight on:
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Typical of sunny, traffic-heavy cities.
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Answer: B. nitrogen oxides and volatile organic compounds
NOₓ and VOCs react in sunlight to form ozone, PAN and aldehydes, the components of photochemical smog.
154. An electrostatic precipitator removes particulates from flue gas by:
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Electric field.
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Answer: A. charging the particles and collecting them on oppositely charged plates
Corona discharge charges the dust particles and the electric field drives them to collecting electrodes, giving >99% efficiency.
155. Using a molar volume of 24.45 L/mol (25 °C, 1 atm), a CO concentration of 9 ppm (volume) corresponds to about:
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Multiply by molecular mass, divide by molar volume.
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Answer: A. 10.3 mg/m³
mg/m³ = ppm × M / 24.45 = 9 × 28.01 / 24.45 = 10.3 mg/m³.
156. On the US EPA Air Quality Index (AQI), values above 300 are described as:
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The worst category.
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Answer: C. hazardous
AQI categories: 0–50 good, 51–100 moderate, 101–150 unhealthy for sensitive groups, 151–200 unhealthy, 201–300 very unhealthy, 301+ hazardous.
157. When the environmental lapse rate is greater than the dry adiabatic lapse rate (superadiabatic), the atmosphere is:
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Parcel buoyancy.
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Answer: D. unstable, favouring vertical mixing and dispersion
A rising parcel stays warmer than surroundings and keeps rising, which promotes dispersion (looping plume).
158. The ground temperature is 30 °C. Taking the dry adiabatic lapse rate as 9.8 °C/km, the temperature of a dry rising air parcel at 500 m is:
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Convert 500 m to km.
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Answer: B. 25.1 °C
T = 30 − 9.8 × 0.5 = 25.1 °C.
159. A temperature inversion in the atmosphere is hazardous for air quality because it:
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Stable layer acts like a lid.
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Answer: B. suppresses vertical mixing and traps pollutants near the ground
Warm air above cool air is stable; pollutants cannot rise, causing episodes such as winter smog in valleys (e.g. Kathmandu Valley).
160. The ground-level centreline concentration from an elevated source is C = Q/(π u σy σz)·exp(−H²/2σz²). For Q = 100 g/s, u = 5 m/s, σy = 50 m, σz = 25 m, H = 50 m, C is about:
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Evaluate the exponent as −H²/(2σz²) = −2.
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Answer: A. 689 µg/m³
Q/(π u σyσz) = 100/(π×5×50×25) = 5.09×10⁻³ g/m³; exp(−2500/1250) = e⁻² = 0.1353; C = 689 µg/m³.
161. Increasing the effective stack height H (with other factors constant) will:
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Higher stack disperses over a wider region.
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Answer: D. reduce the ground-level concentration
The exponent exp(−H²/2σz²) gets smaller as H increases, so ground-level concentration near the source decreases.
162. The Biochemical Oxygen Demand (BOD) test is normally conducted at:
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Standard incubation.
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Answer: C. 20 °C for 5 days
The standard BOD₅ test incubates the sample in the dark at 20 °C for five days.
163. For the same wastewater sample, which relation is generally true?
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Chemical oxidation measures more than bacteria do.
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Answer: C. COD ≥ BOD
COD includes both biodegradable and non-biodegradable matter that can be chemically oxidised, so it is at least as large as BOD.
164. A 10 mL wastewater sample is diluted to 300 mL in a BOD bottle. The initial DO is 8.0 mg/L and the DO after 5 days is 4.0 mg/L. The BOD₅ of the wastewater is:
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Multiply by the dilution factor.
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Answer: B. 120 mg/L
BOD₅ = (DO_i − DO_f) × (bottle volume/sample volume) = 4.0 × 300/10 = 120 mg/L.
165. A wastewater has an ultimate BOD of 200 mg/L and a first-order BOD rate constant k = 0.23 d⁻¹ (base e). The BOD at 5 days is about:
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Use BOD_t = L₀(1 − e^(−kt)).
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Answer: A. 137 mg/L
BOD₅ = L₀(1 − e^(−kt)) = 200(1 − e^(−1.15)) = 137 mg/L.
166. Eutrophication of a lake is mainly caused by excess:
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Fertiliser runoff.
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Answer: D. nitrogen and phosphorus nutrients
Nutrient enrichment causes algal blooms; on decay they consume dissolved oxygen, killing fish.
167. Activated sludge process in a sewage treatment plant is classified as:
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Uses microbes.
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Answer: D. secondary (biological) treatment
Primary treatment removes settleable solids physically; secondary treatment uses microorganisms to oxidise organic matter.
168. The sound pressure level in decibels is defined as SPL = 20 log₁₀(p/p₀), where the reference pressure p₀ is:
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Very small.
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Answer: A. 20 µPa
p₀ = 20 µPa, the approximate threshold of hearing at 1 kHz (0 dB).
169. Two identical machines each produce a noise level of 80 dB at a point. The combined level at the point is about:
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Decibels add logarithmically.
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Answer: C. 83 dB
Doubling acoustic power adds 10 log 2 ≈ 3 dB, so the level is 80 + 3 = 83 dB.
170. Two noise sources produce 90 dB and 85 dB separately at a point. The combined level is about:
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Convert to intensities, add, convert back.
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Answer: B. 91.2 dB
L = 10 log(10⁹ + 10⁸·⁵) = 91.2 dB.
171. A point source gives 90 dB at 10 m. In free field (inverse-square law), the level at 40 m is about:
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Each doubling of distance reduces by 6 dB.
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Answer: B. 78 dB
ΔL = 20 log(40/10) = 12 dB, so level = 90 − 12 = 78 dB.
172. The best method for noise control is generally:
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Control it first at the origin.
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Answer: A. reducing noise at the source
Source control (quieter machines, damping, mufflers) is the most effective; path and receiver controls are second and third options.
173. In integrated solid waste management, the preferred order of the waste hierarchy (most to least preferred) is:
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Prevention is best.
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Answer: D. reduce, reuse, recycle, recover, dispose
Waste prevention/reduction comes first, then reuse, recycling, energy recovery, and finally landfill disposal.
174. Leachate from a sanitary landfill is:
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Liquid from the waste pile.
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Answer: C. contaminated liquid formed by water percolating through the waste
Leachate, rich in organics and metals, can pollute groundwater and must be collected and treated; landfill gas is the methane/CO₂ mixture.
175. A town of 1,000,000 people generates 0.4 kg of municipal solid waste per person per day. The total daily waste is:
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Convert kg to tonnes.
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Answer: A. 400 tonnes
1,000,000 × 0.4 = 400,000 kg = 400 tonnes per day.
176. Which gas is chiefly responsible for depletion of the stratospheric ozone layer?
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Old refrigerants and aerosol propellants.
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Answer: B. chlorofluorocarbons (CFCs)
Chlorine released from CFCs by UV catalytically destroys ozone; the Montreal Protocol phases them out.
177. Which of the following is a greenhouse gas?
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Gas from marshes, cattle, landfills.
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Answer: C. methane
CH₄, CO₂, N₂O, water vapour and CFCs absorb infrared radiation; noble gases do not.
178. Rain is considered 'acid rain' when its pH is below about:
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Slightly below neutral.
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Answer: D. 5.6
Natural rain is slightly acidic (pH ≈ 5.6) due to dissolved CO₂; below this indicates SO₂/NOₓ contribution.
179. In the Kathmandu Valley, a widely recognised contributor to winter particulate pollution is:
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Local human sources.
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Answer: B. vehicular exhaust, brick kilns and road dust
Traffic emissions, brick kilns and re-suspended dust, trapped by winter inversions in the bowl-shaped valley, are well-known contributors.
180. Indoor air pollution in rural households using firewood on traditional stoves is mainly due to:
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Smoke in the kitchen.
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Answer: D. smoke containing particulates and carbon monoxide
Incomplete combustion in poorly ventilated kitchens releases PM and CO, causing respiratory disease; improved cookstoves and biogas reduce exposure.