Nepal Engineering Council · Mechanical Engineering · Chapter 2
Engineering Thermodynamics
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180 questions in 6 syllabus topics.
2.1 Thermodynamics basics
30 questions · AMeE0201
1. A system in which energy can cross the boundary but mass cannot is called a
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Think about whether mass is allowed across the boundary.
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Answer: D. closed system
A closed system (control mass) exchanges only energy (heat and work) with its surroundings, while its mass stays fixed.
2. A system that exchanges neither mass nor energy with its surroundings is called
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No interaction of any kind with the surroundings.
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Answer: C. an isolated system
In an isolated system neither mass nor energy (heat or work) crosses the boundary.
3. A steam turbine running in a power plant is best analysed as
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Does steam keep flowing through it?
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Answer: A. an open system (control volume)
Steam continuously enters and leaves the turbine, so mass crosses the boundary and a control volume is used.
4. Which of the following is an intensive property?
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Ask whether the value halves when the system is cut in half.
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Answer: A. Density
Intensive properties do not depend on the mass of the system; density, pressure and temperature are intensive, while volume, enthalpy and internal energy are extensive.
5. Which of the following is a path function?
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Heat and work are energy in transit, not stored properties.
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Answer: B. Work
Work (and heat) depends on the path followed between two states, whereas enthalpy, entropy and internal energy are properties (state functions).
6. For any thermodynamic property P, the cyclic integral ∮dP is
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The system ends at the same state it started from.
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Answer: C. zero
A property depends only on the state, so after a complete cycle it returns to its initial value and the net change is zero.
7. The measurement of temperature by a thermometer is based on which law of thermodynamics?
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The law that defines thermal equilibrium.
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Answer: C. Zeroth law
The zeroth law (thermal equilibrium transitivity) allows a thermometer to be used as the third body that compares temperatures.
8. The zeroth law of thermodynamics states that
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Think of A, B and a third body C.
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Answer: A. if two bodies are each in thermal equilibrium with a third body, they are in thermal equilibrium with each other
This transitive property of thermal equilibrium is the zeroth law and forms the basis of temperature measurement.
9. A system is said to be in complete thermodynamic equilibrium when it is in
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More than two kinds of equilibrium are required.
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Answer: D. mechanical, thermal and chemical equilibrium simultaneously
Thermodynamic equilibrium requires the absence of unbalanced forces, temperature differences and chemical potential differences at the same time.
10. The equation of state of an ideal gas of mass m is
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Pressure times volume equals mass times R times T.
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Answer: C. pV = mRT
For an ideal gas pV = mRT, or pv = RT on a unit-mass basis, where R is the characteristic gas constant.
11. The value of the universal gas constant is approximately
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It is expressed per kilomole, not per kilogram.
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Answer: B. 8.314 kJ/kmol·K
Ru = 8.314 kJ/(kmol·K), the same for every gas. The characteristic constant is R = Ru/M, in kJ/(kg·K).
12. Real gases behave most nearly like an ideal gas at
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Molecules should be far apart and fast moving.
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Answer: D. low pressure and high temperature
At low pressure molecules are far apart and at high temperature kinetic energy dominates, so intermolecular forces and molecular volume become negligible.
13. The quality (dryness fraction) x of a wet mixture is defined as
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It is a mass ratio based on the vapour part.
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Answer: A. mass of vapour / total mass of the mixture
x = m_vapour/(m_liquid + m_vapour); x = 0 for saturated liquid and x = 1 for saturated vapour.
14. The quality of a substance has physical meaning only when the substance is
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Which region of the T–v diagram has both phases present?
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Answer: D. a saturated liquid–vapour mixture
Quality describes the proportion of vapour in a two-phase mixture, so it applies only in the wet region between x = 0 and x = 1.
15. At the critical point of a pure substance
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The top of the saturation dome.
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Answer: B. the saturated liquid and saturated vapour states are identical
At the critical point the liquid and vapour lines meet, so hfg = 0 and there is no distinction between the two phases.
16. The temperature of the triple point of water is
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It is slightly above the ice point.
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Answer: A. 273.16 K
Water's triple point is 0.01 °C = 273.16 K at 0.6117 kPa; the ice point at 1 atm is 273.15 K.
17. A vapour whose temperature is higher than the saturation temperature corresponding to its pressure is called
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It has been heated beyond the point of being just-saturated.
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Answer: D. a superheated vapour
Superheated vapour has T > Tsat at the given pressure (equivalently, p < psat at the given temperature).
18. During constant-pressure heating of a wet mixture of water (between x = 0 and x = 1), the temperature
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Think of boiling water at fixed pressure.
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Answer: C. remains constant while the quality rises
Within the two-phase region, T equals Tsat(p) and stays fixed; added heat is used as latent heat to increase the quality.
19. A temperature of 27 °C is equal to how many kelvin?
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Add the offset between the Celsius and Kelvin scales.
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Answer: B. 300 K
T(K) = 27 + 273 = 300 K.
20. A rigid tank of 2 m³ holds air at 200 kPa and 300 K (R = 0.287 kJ/kg·K). The mass of air is approximately
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Use m = pV/(RT) with consistent units (kPa and kJ/kg·K).
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Answer: B. 4.65 kg
m = pV/RT = (200 × 2)/(0.287 × 300) = 4.65 kg.
21. At 100 °C, saturated water has vf = 0.001043 m³/kg and vg = 1.6729 m³/kg. The specific volume of a mixture of quality 0.8 is about
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Use v = vf + x·vfg.
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Answer: A. 1.339 m³/kg
v = vf + x(vg − vf) = 0.001043 + 0.8 × 1.671857 = 1.339 m³/kg.
22. A pressure gauge on a tank reads 150 kPa. If the atmospheric pressure is 101.3 kPa, the absolute pressure is
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Add the atmospheric pressure.
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Answer: D. 251.3 kPa
p_abs = p_gauge + p_atm = 150 + 101.3 = 251.3 kPa.
23. The characteristic gas constant of oxygen (molar mass 32 kg/kmol) is approximately
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R = Ru/M.
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Answer: C. 0.260 kJ/kg·K
R = Ru/M = 8.314/32 = 0.260 kJ/kg·K.
24. A 5 kg wet mixture of water and steam contains 1.5 kg of vapour. The quality is
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Vapour mass over total mass.
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Answer: D. 0.30
x = m_vapour/m_total = 1.5/5 = 0.30.
25. The density of air (R = 0.287 kJ/kg·K) at 100 kPa and 300 K is approximately
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ρ = p/(RT).
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Answer: B. 1.161 kg/m³
ρ = p/RT = 100/(0.287 × 300) = 1.161 kg/m³.
26. Air at 100 kPa in a 1 m³ cylinder is compressed isothermally to 0.25 m³. The final pressure is
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Boyle's law.
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Answer: A. 400 kPa
Isothermal ideal gas: p1V1 = p2V2, so p2 = 100 × 1/0.25 = 400 kPa.
27. A gas at 300 K and 1 m³ is heated at constant pressure until its volume is 1.5 m³. The final temperature is
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At constant pressure V is proportional to T.
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Answer: A. 450 K
Charles's law: V1/T1 = V2/T2, so T2 = 300 × 1.5 = 450 K.
28. The mass of 8 kg of oxygen (M = 32 kg/kmol) expressed in kilomoles is
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Number of kmol = mass ÷ molar mass.
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Answer: C. 0.25 kmol
n = m/M = 8/32 = 0.25 kmol.
29. Water at 150 °C has a saturation pressure of 476 kPa. If the water is at 150 °C and 1000 kPa, its state is
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Compare the actual pressure with the saturation pressure at that temperature.
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Answer: B. compressed (subcooled) liquid
The actual pressure (1000 kPa) exceeds psat at 150 °C (476 kPa), so the water is a compressed liquid.
30. A 0.5 m³ rigid vessel contains 2 kg of an ideal gas at 400 kPa. The gas constant R is 0.2 kJ/kg·K. The gas temperature is
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T = pV/(mR).
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Answer: B. 500 K
T = pV/(mR) = (400 × 0.5)/(2 × 0.2) = 500 K.
2.2 1st Law of thermodynamics
30 questions · AMeE0202
31. For a closed system undergoing a process, the first law of thermodynamics is written as
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Energy balance: heat in = stored energy change + work out.
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Answer: D. Q = ΔU + W
Heat added to the system equals the increase in internal energy plus the work done by the system (sign convention: Q in, W out positive).
32. The first law of thermodynamics is essentially a statement of
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Which quantity is conserved?
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Answer: B. conservation of energy
The first law states that energy can be neither created nor destroyed; it can only change form.
33. A perpetual motion machine of the first kind (PMM1) is impossible because it would
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Work output with zero input.
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Answer: B. produce work without any energy input, violating the first law
PMM1 creates energy from nothing, which contradicts the first law. (The second-kind machine violates the second law.)
34. The internal energy of an ideal gas depends only on its
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Joule's free-expansion experiment.
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Answer: B. temperature
Joule's law: for an ideal gas, u = u(T) alone, so du = cv dT in any process.
35. Enthalpy of a system is defined as
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Internal energy plus the product of pressure and volume.
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Answer: A. H = U + pV
Enthalpy combines internal energy with flow (pV) work: H = U + pV.
36. For an ideal gas the specific heats are related by
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Mayer's relation.
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Answer: A. cp − cv = R
Mayer's relation: cp − cv = R, and the ratio γ = cp/cv is always greater than 1.
37. The ratio of specific heats γ = cp/cv for a monatomic ideal gas such as helium is
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Only three translational degrees of freedom.
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Answer: A. 1.67
For a monatomic gas cv = (3/2)R and cp = (5/2)R, giving γ = 5/3 ≈ 1.67 (air is about 1.4).
38. The work done by a closed system in a constant-volume (isochoric) process is
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No boundary movement.
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Answer: B. zero
W = ∫p dV and dV = 0, so W = 0 and all the heat goes to internal energy: Q = ΔU.
39. In an isothermal expansion of an ideal gas
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Internal energy depends only on temperature.
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Answer: C. the heat added equals the work done by the gas
For an ideal gas T constant means ΔU = 0, so Q = W.
40. In a reversible adiabatic process on an ideal gas, the pressure and volume are related by
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The polytropic index equals γ.
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Answer: C. pV^γ = constant
For an isentropic ideal-gas process pV^γ = constant, with γ = cp/cv.
41. A polytropic process pV^n = constant with n = 0 represents a
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Put n = 0 in the equation.
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Answer: C. constant-pressure process
With n = 0, pV^0 = p = constant (isobaric). n = 1 is isothermal, n = γ is adiabatic and n = ∞ is isochoric.
42. The boundary work in a polytropic process pV^n = C (n ≠ 1) between states 1 and 2 is
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The denominator is n − 1.
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Answer: A. (p1V1 − p2V2)/(n − 1)
W = ∫p dV = (p1V1 − p2V2)/(n − 1), valid for n ≠ 1; for n = 1 the result is p1V1 ln(V2/V1).
43. For steady flow through a nozzle (negligible heat transfer, work and potential energy change), the energy equation reduces to
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Enthalpy converts into kinetic energy.
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Answer: C. h1 + V1²/2 = h2 + V2²/2
With q = w = 0 and Δz = 0, the SFEE gives h1 + V1²/2 = h2 + V2²/2, so enthalpy drops as kinetic energy rises.
44. A throttling process (valve) of a real fluid with no heat loss is considered
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No work, no heat, velocities small.
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Answer: B. isenthalpic
With q = 0, w = 0 and negligible kinetic energy change, h1 = h2 across the throttle, although entropy increases.
45. For an adiabatic steam turbine with negligible changes in kinetic and potential energy, the specific work output is
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Work equals the enthalpy drop.
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Answer: D. h1 − h2
SFEE with q = 0 gives w = h1 − h2, where state 1 is the inlet and state 2 is the exit.
46. For steady flow of mass through a control volume with one inlet and one outlet, the continuity equation gives
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Mass flow rate is constant.
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Answer: B. ρ1A1V1 = ρ2A2V2
Conservation of mass for steady flow: ṁ = ρAV is the same at both sections.
47. Which of the following is an unsteady-flow (transient) process?
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Look for a case where the mass inside changes.
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Answer: A. Filling a rigid evacuated tank from a supply line
In charging or discharging a tank the mass and energy inside the control volume change with time, so the SFEE alone does not apply.
48. The first law for an open system (control volume), written on a rate basis for unsteady flow, equates the rate of energy accumulation to
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Include flow energy of the streams.
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Answer: D. net energy carried in by mass flow plus heat transfer in minus work out
dE_cv/dt = Q̇ − Ẇ + Σṁin(h + V²/2 + gz) − Σṁout(h + V²/2 + gz).
49. A diffuser is a steady-flow device that
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It is the opposite of a nozzle.
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Answer: C. decelerates the flow and increases its pressure
A diffuser converts kinetic energy into enthalpy (and pressure), the reverse of a nozzle.
50. A gas expands at a constant pressure of 200 kPa from 0.5 m³ to 1.2 m³. The work done by the gas is
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W = p·ΔV.
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Answer: D. 140 kJ
W = pΔV = 200 × (1.2 − 0.5) = 140 kJ.
51. A closed system receives 50 kJ of heat and does 30 kJ of work on the surroundings. The change in internal energy is
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ΔU = Q − W.
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Answer: A. +20 kJ
ΔU = Q − W = 50 − 30 = +20 kJ.
52. 1 kg of air (R = 0.287 kJ/kg·K) expands isothermally at 300 K to twice its volume. The work done is approximately
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W = mRT ln(V2/V1).
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Answer: D. 59.7 kJ
W = mRT ln(V2/V1) = 0.287 × 300 × ln 2 = 59.7 kJ.
53. Air (γ = 1.4) at 500 kPa and 0.1 m³ expands reversibly and adiabatically to 100 kPa. The work done is approximately
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Find V2 first, then use (p1V1 − p2V2)/(γ − 1).
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Answer: C. 46.1 kJ
V2 = 0.1(500/100)^(1/1.4) = 0.3157 m³; W = (p1V1 − p2V2)/(γ − 1) = (50 − 31.57)/0.4 = 46.1 kJ.
54. The heat needed to raise 3 kg of air (cv = 0.718 kJ/kg·K) from 300 K to 400 K in a rigid vessel is
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No work at constant volume, so use cv.
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Answer: B. 215.4 kJ
Constant volume: Q = mcvΔT = 3 × 0.718 × 100 = 215.4 kJ.
55. For an ideal gas with R = 0.287 kJ/kg·K and γ = 1.4, cp is approximately
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Combine cp − cv = R and cp/cv = γ.
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Answer: C. 1.004 kJ/kg·K
cp = γR/(γ − 1) = 1.4 × 0.287/0.4 = 1.004 kJ/kg·K.
56. Steam is expanded in a nozzle with an enthalpy drop of 100 kJ/kg. If the inlet velocity is negligible and the flow is adiabatic, the exit velocity is approximately
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V = √(2Δh) with Δh in J/kg.
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Answer: D. 447 m/s
V2 = √(2Δh) = √(2 × 100 000) = 447 m/s (Δh in J/kg).
57. An adiabatic steam turbine receives 5 kg/s with h1 = 3200 kJ/kg and exhausts at h2 = 2400 kJ/kg. Neglecting kinetic energy changes, the power output is
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Mass flow times enthalpy drop.
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Answer: A. 4000 kW
Ẇ = ṁ(h1 − h2) = 5 × 800 = 4000 kW.
58. A gas initially at 100 kPa and 1 m³ is compressed according to pV^1.3 = constant to 0.25 m³. The work transfer is approximately
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Find p2 first and watch the sign.
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Answer: D. 172 kJ done on the gas
p2 = 100 × 4^1.3 = 606 kPa; W = (p1V1 − p2V2)/(n − 1) = (100 − 151.6)/0.3 = -172 kJ; the negative sign means work is done on the gas.
59. Water (1000 kg/m³) flows steadily through a 100 mm diameter pipe at 5 m/s. The mass flow rate is approximately
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ṁ = ρAV with A in m².
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Answer: A. 39.3 kg/s
ṁ = ρAV = 1000 × (π/4)(0.1)² × 5 = 39.3 kg/s.
60. In an adiabatic rigid container, 2 kg of air receives 100 kJ of paddle-wheel work. With cv = 0.718 kJ/kg·K, the temperature rise is approximately
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All work input becomes internal energy.
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Answer: B. 69.6 K
Q = 0 and the work input raises U: ΔT = W/(m cv) = 100/(2 × 0.718) = 69.6 K.
2.3 2nd Laws of thermodynamics
30 questions · AMeE0203
61. The Kelvin–Planck statement of the second law says that it is impossible to
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Think of a heat engine with no sink.
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Answer: B. construct a device operating in a cycle that produces work while exchanging heat with a single reservoir only
Kelvin–Planck: no cyclic heat engine can convert all the heat received from a single reservoir into work; some heat must be rejected to a sink.
62. The Clausius statement of the second law says that
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It concerns refrigerators and heat pumps.
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Answer: C. it is impossible for a cyclic device to transfer heat from a colder body to a hotter body without work input
Clausius: heat will not flow from a low-temperature body to a high-temperature body on its own, without external work (as in a refrigerator).
63. The Kelvin–Planck and Clausius statements of the second law are
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They are two forms of one law.
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Answer: B. equivalent: violation of one implies violation of the other
It can be shown by combining a hypothetical violating device with a real one that breaking either statement breaks the other.
64. The thermal efficiency of a heat engine is defined as
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Desired output over cost input.
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Answer: D. net work output / heat supplied
η = W_net/Q_H = 1 − Q_L/Q_H.
65. The coefficient of performance of a refrigerator is defined as
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The 'useful' effect is cooling.
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Answer: D. heat extracted from the cold space / work input
COP_R = Q_L/W. The desired output is the heat removed from the refrigerated space.
66. For the same reservoirs, the COPs of a heat pump and a refrigerator operating on the same cycle are related by
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Q_H = Q_L + W.
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Answer: A. COP_HP = COP_R + 1
Since Q_H = Q_L + W, COP_HP = Q_H/W = (Q_L + W)/W = COP_R + 1. So a heat pump COP always exceeds 1.
67. According to Carnot's theorem, the efficiency of a reversible heat engine operating between two given reservoirs
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Only the reservoir temperatures matter.
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Answer: B. is the maximum possible and is independent of the working substance
All reversible engines between the same two reservoirs have the same efficiency, which no irreversible engine can exceed.
68. The Carnot cycle consists of
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Constant T for heat exchange, no heat in the other two.
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Answer: D. two reversible isothermal and two reversible adiabatic processes
The ideal Carnot cycle: isothermal heat addition, adiabatic expansion, isothermal heat rejection, adiabatic compression.
69. On a T–s diagram, the Carnot cycle appears as a
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Horizontal and vertical lines.
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Answer: B. rectangle
Isothermal processes are horizontal and reversible adiabatic (isentropic) processes are vertical lines, giving a rectangle.
70. The Clausius inequality states that for any cycle
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Equality holds for reversible cycles.
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Answer: B. ∮ δQ/T ≤ 0
The cyclic integral of δQ/T is zero for reversible cycles and negative for irreversible ones.
71. The increase of entropy principle states that, for an isolated system, the entropy
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Natural processes have a preferred direction.
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Answer: C. never decreases
ΔS_isolated ≥ 0: equality for reversible processes, strict inequality for irreversible ones.
72. A reversible adiabatic process is also called
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The entropy change is zero.
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Answer: C. an isentropic process
With δQ = 0 and no irreversibility, dS = δQ_rev/T = 0, so entropy is constant.
73. Which of the following is an irreversible process?
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A process without any quasi-static control.
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Answer: D. Free expansion of a gas into a vacuum
Unrestrained (free) expansion has no controlled work and cannot be reversed without external effect; the others are idealised reversible processes.
74. Area under a reversible process curve on a T–s diagram represents
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Q = ∫T ds.
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Answer: A. heat transfer
δQ_rev = T ds, so ∫T ds is the heat transfer in a reversible process.
75. The isentropic efficiency of a turbine is defined as
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Real turbines produce less than the ideal.
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Answer: D. actual work output / isentropic work output
η_T = (h1 − h2)/(h1 − h2s) < 1 because actual expansion is irreversible.
76. The isentropic efficiency of a compressor is defined as
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The ideal needs less work than the real.
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Answer: A. isentropic work input / actual work input
η_C = (h2s − h1)/(h2 − h1) < 1 because the real compressor needs more work than the ideal one.
77. Entropy generation in an actual process is always
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Related to irreversibility.
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Answer: C. greater than or equal to zero
S_gen ≥ 0: zero for reversible processes and positive for irreversible ones. A negative value would violate the second law.
78. For a given sink temperature, the efficiency of a Carnot engine can be increased by
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Look at the formula 1 − TL/TH.
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Answer: A. raising the source temperature
η = 1 − TL/TH increases with TH (and falls with TL); working fluid does not matter.
79. The Carnot COP of a refrigerator operating between TL and TH is
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Cold reservoir temperature over the temperature lift.
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Answer: D. TL/(TH − TL)
COP_R,Carnot = TL/(TH − TL); the heat pump value is TH/(TH − TL).
80. A Carnot engine operates between 600 K and 300 K. Its thermal efficiency is
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η = 1 − TL/TH.
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Answer: A. 50%
η = 1 − TL/TH = 1 − 300/600 = 0.5.
81. A Carnot engine works between a source at 527 °C and a sink at 27 °C. Its efficiency is
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Convert temperatures to kelvin first.
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Answer: A. 62.5%
Convert to kelvin: TH = 800 K, TL = 300 K; η = 1 − 300/800 = 0.625.
82. A heat engine receives 500 kJ of heat and rejects 300 kJ to the sink. The net work output is
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W = QH − QL.
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Answer: B. 200 kJ
W = QH − QL = 500 − 300 = 200 kJ (η = 40%).
83. A refrigerator removes 12 kW from the cold space while consuming 4 kW of power. Its COP is
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COP = cooling effect/work.
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Answer: D. 3.0
COP = QL/W = 12/4 = 3.
84. A reversible refrigerator operates between −13 °C and 27 °C. Its COP is
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Use absolute temperatures.
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Answer: C. 6.5
TL = 260 K, TH = 300 K; COP = TL/(TH − TL) = 260/40 = 6.5.
85. A heat pump with COP of 4 draws 2 kW of electrical power. The rate of heat delivered to the room is
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COP_HP = QH/W.
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Answer: A. 8 kW
Q̇H = COP × W = 4 × 2 = 8 kW.
86. 600 kJ of heat is added reversibly and isothermally to a system at 300 K. The entropy change of the system is
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ΔS = Q_rev/T.
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Answer: C. 2 kJ/K
ΔS = Q/T = 600/300 = 2 kJ/K.
87. 1 kg of air (cp = 1.005 kJ/kg·K) is heated at constant pressure from 300 K to 600 K. The entropy change is approximately
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ΔS = m cp ln(T2/T1).
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Answer: C. 0.697 kJ/K
ΔS = cp ln(T2/T1) = 1.005 × ln 2 = 0.697 kJ/K.
88. An actual turbine produces 800 kJ/kg of work while the isentropic work between the same states is 1000 kJ/kg. Its isentropic efficiency is
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Actual over ideal for a turbine.
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Answer: B. 80%
η = actual/isentropic = 800/1000 = 0.80.
89. The maximum work obtainable from 1000 kJ of heat taken from a source at 800 K when the sink is at 300 K is
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Multiply by the Carnot efficiency.
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Answer: A. 625 kJ
W_max = Q(1 − TL/TH) = 1000 × (1 − 300/800) = 625 kJ.
90. The source temperature of a Carnot engine is raised from 600 K to 750 K while the sink stays at 300 K. The efficiency changes from
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Compute 1 − TL/TH before and after.
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Answer: B. 50% to 60%
Before: 1 − 300/600 = 0.50; after: 1 − 300/750 = 0.60.
2.4 Thermodynamic cycles
30 questions · AMeE0204
91. The ideal Rankine cycle consists of the following processes in order:
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Four devices: pump, boiler, turbine, condenser.
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Answer: D. isentropic pump, constant-pressure heat addition, isentropic turbine, constant-pressure heat rejection
Pump (liquid, isentropic), boiler (isobaric), turbine (isentropic), condenser (isobaric) form the ideal Rankine cycle.
92. Which of the following changes does NOT increase the thermal efficiency of a Rankine cycle?
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Think of average temperature of heat rejection.
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Answer: D. Raising the condenser pressure
Higher condenser pressure raises the mean temperature of heat rejection and so reduces net work and efficiency.
93. The main purpose of reheating steam in a Rankine cycle is to
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Wet steam erodes the last-stage blades.
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Answer: A. reduce the moisture content at the turbine exhaust
Reheat allows high-pressure expansion followed by reheating, so exhaust dryness is higher, protecting turbine blades.
94. Regenerative feedwater heating in a steam power cycle improves efficiency because it
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Preheating the water before it reaches the boiler.
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Answer: B. raises the mean temperature at which heat is added
Bleeding steam to preheat feedwater reduces the heat required from the external source at low temperature, raising the average heat-addition temperature.
95. In a Rankine cycle, the pump work compared with the turbine work is
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Liquids are nearly incompressible with small specific volume.
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Answer: D. very small
Pumping a liquid needs only w = v(p2 − p1), which is a tiny fraction of the turbine work; the back-work ratio is typically 1% or less.
96. The ideal Brayton cycle consists of
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Gas turbine cycle.
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Answer: D. two isentropic and two constant-pressure processes
Isentropic compression, constant-pressure heat addition, isentropic expansion and constant-pressure heat rejection.
97. The thermal efficiency of an ideal Brayton cycle depends on
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Look at the efficiency formula.
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Answer: B. the pressure ratio only (for a given gas)
η = 1 − 1/rp^((γ−1)/γ), which depends only on the pressure ratio and γ.
98. The Otto cycle is also known as the
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Spark-ignition engines.
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Answer: D. constant-volume cycle
In the ideal Otto cycle, heat is added and rejected at constant volume.
99. In the ideal Diesel cycle, heat is added at constant ______ and rejected at constant ______.
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Only the Otto cycle adds heat at constant volume.
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Answer: A. pressure; volume
Diesel cycle: combustion approximated by isobaric heat addition; heat rejection at constant volume.
100. For the same compression ratio and same heat input, the efficiency of the ideal
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Compare the formulas at equal r.
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Answer: A. Otto cycle is higher than that of the Diesel cycle
For equal compression ratios the Otto cycle is more efficient; Diesel engines gain advantage in practice because they use higher compression ratios.
101. In the Diesel cycle, with a fixed compression ratio, increasing the cut-off ratio
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Longer constant-pressure heat addition.
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Answer: B. decreases the thermal efficiency
The factor (ρ^γ − 1)/(γ(ρ − 1)) rises with cut-off ratio ρ, lowering η = 1 − (1/r^(γ−1))·[...].
102. The correct sequence of components in a simple vapour-compression refrigeration cycle is
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Follow the refrigerant from the compressor discharge.
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Answer: B. compressor → condenser → expansion valve → evaporator
Vapour is compressed, condensed, throttled and then evaporated, absorbing heat from the cold space.
103. The expansion (throttle) valve in a vapour-compression cycle is ideally an
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Throttling.
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Answer: C. isenthalpic process
The expansion valve involves no work and no heat transfer, so h3 = h4; entropy increases.
104. In the ideal vapour-compression cycle, the refrigerant enters the compressor as
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The compressor must not receive liquid.
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Answer: D. saturated vapour
In the standard ideal cycle the evaporator exit is saturated vapour, which is compressed isentropically to the condenser pressure.
105. In a vapour-absorption refrigeration system, the compressor of the vapour-compression system is replaced by
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Heat rather than mechanical work drives the cycle.
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Answer: B. an absorber, pump and generator
Refrigerant vapour is absorbed in a liquid, pumped to high pressure and driven off by heat in the generator.
106. In an ammonia–water vapour-absorption system, the refrigerant and the absorbent are respectively
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Ammonia dissolves readily in water.
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Answer: A. ammonia and water
NH3 is the refrigerant (evaporates at low temperature) and water absorbs the ammonia vapour. In LiBr–water systems, water is the refrigerant.
107. Compared with a vapour-compression system, a vapour-absorption system
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Think of the energy input.
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Answer: C. can use low-grade heat energy as the main input and has fewer moving parts
Absorption systems use heat (waste heat, solar, gas) in the generator, with only a small pump, but have lower COP.
108. The refrigerating effect of a vapour-compression cycle (evaporator exit state 1, evaporator inlet state 4) per kg of refrigerant is
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Heat absorbed across the evaporator.
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Answer: A. h1 − h4
The heat absorbed in the evaporator is q = h1 − h4, while compressor work is h2 − h1.
109. The reversed Brayton cycle is used for
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Air as refrigerant.
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Answer: B. air refrigeration (for example in aircraft cooling)
The Bell–Coleman (reversed Brayton) cycle uses air as the refrigerant, compresses and expands it in a turbine.
110. The thermal efficiency of an ideal Otto cycle with compression ratio 8 and γ = 1.4 is approximately
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η = 1 − 1/r^(γ−1).
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Answer: A. 56.5%
η = 1 − r^(1−γ) = 1 − 8^(−0.4) = 0.5647.
111. An ideal Brayton cycle with pressure ratio 10 and γ = 1.4 has thermal efficiency of about
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η = 1 − 1/rp^((γ−1)/γ).
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Answer: C. 48.2%
η = 1 − rp^(−(γ−1)/γ) = 1 − 10^(−0.2857) = 0.482.
112. An ideal Diesel cycle has compression ratio 18, cut-off ratio 2 and γ = 1.4. The efficiency is approximately
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The Otto formula needs a cut-off correction factor.
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Answer: D. 63.2%
η = 1 − (1/r^(γ−1))·(ρ^γ − 1)/(γ(ρ − 1)) = 1 − 0.3148 × 1.171 = 0.632.
113. In a simple Rankine cycle, the turbine produces 1000 kJ/kg, the pump consumes 40 kJ/kg and the boiler supplies 2800 kJ/kg. The thermal efficiency is approximately
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Net work over heat input.
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Answer: A. 34.3%
η = (wT − wP)/qin = 960/2800 = 0.3429.
114. In a vapour-compression cycle, h1 = 400 kJ/kg (compressor inlet), h2 = 440 kJ/kg (compressor exit) and h3 = h4 = 250 kJ/kg. The COP is
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COP = refrigerating effect/compressor work.
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Answer: B. 3.75
COP = (h1 − h4)/(h2 − h1) = 150/40 = 3.75.
115. A refrigerating capacity of 2 tons of refrigeration (1 TR = 3.517 kW) is equal to
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1 TR ≈ 3.5 kW.
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Answer: C. 7.03 kW
2 × 3.517 = 7.03 kW.
116. Water is pumped isentropically from 100 kPa to 3000 kPa. With v = 0.001 m³/kg, the pump work is about
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wP = vΔp.
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Answer: B. 2.90 kJ/kg
wP = v(p2 − p1) = 0.001 × 2900 = 2.90 kJ/kg.
117. The compression ratio needed for an ideal Otto cycle (γ = 1.4) to reach 60% efficiency is approximately
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Rearrange 1 − r^(1−γ) = η.
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Answer: C. 9.9
1 − r^(−0.4) = 0.6 → r^0.4 = 2.5 → r = 2.5^2.5 = 9.9.
118. A refrigerator has COP 3.5 and requires 2 kW of compressor power. The refrigerating capacity is
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Cooling = COP × work.
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Answer: C. 7 kW
Q_L = COP × W = 3.5 × 2 = 7 kW.
119. An ideal Otto cycle receives 1000 kJ/kg heat and has an efficiency of 55%. The heat rejected per kg is
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Heat rejected = heat in − work.
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Answer: A. 450 kJ/kg
w = 0.55 × 1000 = 550 kJ/kg, so q_rej = 1000 − 550 = 450 kJ/kg.
120. A heat engine on a Brayton cycle has compressor work of 300 kJ/kg and turbine work of 700 kJ/kg. The back-work ratio is
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Compressor work over turbine work.
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Answer: C. 0.43
Back-work ratio = wC/wT = 300/700 = 0.43.
2.5 Internal combustion engines
30 questions · AMeE0205
121. In a spark-ignition (SI) engine, combustion is initiated by
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'Spark' is in the name.
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Answer: B. an electric spark from the spark plug
SI engines compress an air–fuel mixture and ignite it with a timed spark; CI engines rely on compression-ignition of injected fuel.
122. In a compression-ignition (CI) engine, the fuel is ignited by
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No ignition system is needed.
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Answer: D. the high temperature of air compressed in the cylinder
The compression ratio (typically 14–22) raises the air temperature above the fuel's self-ignition temperature.
123. The correct order of strokes in a four-stroke engine is
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Starts with filling the cylinder.
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Answer: D. suction, compression, power, exhaust
Intake draws in charge, compression raises its pressure, combustion drives the power stroke, and exhaust expels the burnt gases.
124. In a four-stroke engine, one complete cycle takes
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Four strokes, 180° each.
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Answer: C. two revolutions of the crankshaft
Four strokes of 180° each give 720° = two crankshaft revolutions per cycle.
125. In a two-stroke engine, one power stroke occurs for every
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Two strokes per cycle.
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Answer: A. one revolution of the crankshaft
Intake/exhaust are combined with compression/power strokes, so the cycle is completed in 360° of crank rotation.
126. In a four-stroke engine, the camshaft rotates at
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One valve event per two revolutions.
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Answer: B. half the speed of the crankshaft
Each valve opens once per cycle (720°), so the camshaft turns once for every two crankshaft revolutions.
127. The main function of the flywheel in an internal combustion engine is to
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Inertia.
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Answer: A. store energy during the power stroke and smooth out speed fluctuations
The flywheel's inertia absorbs energy during power strokes and supplies it during the other strokes, giving uniform rotation.
128. The connecting rod in an engine
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It links the piston and crankshaft.
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Answer: B. transmits the force from the piston to the crankshaft, converting reciprocating motion into rotary motion
It joins the piston (via the gudgeon pin) to the crankpin of the crankshaft.
129. The main function of piston rings is to
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Rings fill the gap between piston and liner.
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Answer: D. seal the combustion gases and control the lubricating oil on the cylinder wall
Compression rings prevent blow-by; oil rings scrape excess oil from the cylinder wall.
130. In an SI engine, the device that supplies a correctly proportioned air–fuel mixture is the
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Mixing air and petrol.
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Answer: B. carburettor (or a fuel-injection system)
The carburettor atomises petrol and mixes it with air in proportion to the load and speed.
131. In a diesel (CI) engine, the intake stroke draws in
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Fuel is introduced later.
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Answer: C. air only
Diesel fuel is injected near the end of compression; the intake stroke admits air alone.
132. The typical range of compression ratio for a diesel engine is
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Much higher than in petrol engines.
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Answer: C. 14 to 22
CI engines need high compression to reach the auto-ignition temperature of the fuel; SI engines are limited to roughly 6–10 by knock.
133. The octane number of a fuel indicates its
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SI-engine fuel quality.
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Answer: C. resistance to knock in SI engines
A higher octane number means better anti-knock performance; cetane number is for diesel fuels.
134. The cetane number of a diesel fuel is a measure of its
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CI engine fuel quality.
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Answer: A. ignition quality (readiness to auto-ignite)
A higher cetane number means shorter ignition delay.
135. Knocking in an SI engine is caused by
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Spontaneous ignition of unburnt mixture.
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Answer: B. auto-ignition of the unburnt end gas ahead of the flame front
End-gas auto-ignition produces pressure waves and the characteristic pinging sound that can damage the engine.
136. Scavenging in a two-stroke engine means
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Cleaning the cylinder.
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Answer: D. driving out the burnt gases from the cylinder with fresh charge
Fresh charge entering through the transfer port pushes the exhaust gases out through the exhaust port.
137. In a conventional crankcase-scavenged two-stroke petrol engine, the air–fuel mixture is first compressed in the
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Below the piston.
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Answer: B. crankcase
The downward-moving piston compresses the mixture in the crankcase before it is transferred to the cylinder through the transfer port.
138. In actual valve timing of a four-stroke engine, the inlet valve typically opens
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Valve timings use lead and lag.
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Answer: D. a few degrees before TDC
Early opening (valve lead) and late closing after BDC improve the volumetric efficiency; exhaust opens before BDC.
139. Engine load in a CI engine is controlled primarily by
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There is no throttle valve in the CI inlet.
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Answer: C. varying the quantity of fuel injected
A CI engine is quality-governed: air flow is essentially constant while the amount of fuel injected is changed.
140. A single-cylinder engine has a bore of 100 mm and a stroke of 120 mm. The swept volume is approximately
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Vs = (π/4) × bore² × stroke.
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Answer: A. 942 cm³
Vs = (π/4)D²L = (π/4)(10)²(12) = 942 cm³.
141. An engine has swept volume 500 cm³ and clearance volume 50 cm³. The compression ratio is
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r = (Vs + Vc)/Vc.
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Answer: B. 11
r = (Vs + Vc)/Vc = 550/50 = 11.
142. A four-cylinder engine has a bore of 80 mm and a stroke of 90 mm. Its total displacement is approximately
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Multiply by the number of cylinders.
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Answer: D. 1810 cm³
Per cylinder (π/4)(8²)(9) = 452.4 cm³; × 4 = 1810 cm³.
143. A single-cylinder four-stroke engine has bore 100 mm, stroke 120 mm, mean effective pressure 600 kPa and speed 2400 rpm. The indicated power is approximately
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Four-stroke: one power stroke per two revolutions.
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Answer: A. 11.3 kW
IP = pm L A N/(2×60) = 600 × 0.12 × 0.00785 × 2400/120 = 11.3 kW.
144. An engine has an indicated power of 24 kW and a brake power of 18 kW. Its mechanical efficiency is
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Brake power over indicated power.
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Answer: B. 75%
ηmech = BP/IP = 18/24 = 0.75.
145. An engine delivers 20 kW brake power and consumes 6 kg/h of fuel of calorific value 42 000 kJ/kg. The brake thermal efficiency is approximately
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Convert fuel rate to kg/s.
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Answer: C. 28.6%
Heat input = (6/3600) × 42 000 = 70 kW; η = 20/70 = 0.286.
146. An engine develops a torque of 150 N·m at 3000 rpm. The brake power is approximately
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P = 2πNT/60.
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Answer: A. 47.1 kW
P = 2πNT/60 = 2π × 3000 × 150/60 = 47124 W = 47.1 kW.
147. A four-stroke engine runs at 3000 rpm. The number of power strokes per minute in one cylinder is
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One power stroke per cycle, two revolutions per cycle.
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Answer: D. 1500
One power stroke per two revolutions: 3000/2 = 1500.
148. An engine with compression ratio 9 has a swept volume of 400 cm³. The clearance volume is
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Vc = Vs/(r − 1).
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Answer: C. 50 cm³
r = 1 + Vs/Vc, so Vc = Vs/(r − 1) = 400/8 = 50 cm³.
149. An engine consumes 7 kg of fuel per hour while developing 20 kW brake power. The brake specific fuel consumption is
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Fuel per hour divided by kW.
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Answer: A. 0.35 kg/kWh
bsfc = fuel rate/BP = 7/20 = 0.35 kg/kWh.
150. A four-stroke engine has one cylinder with swept volume 0.0008 m³. At a mean effective pressure of 800 kPa the work per cycle is
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Work per cycle = mep × swept volume.
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Answer: A. 640 J
W = pm × Vs = 800 000 × 0.0008 = 640 J.
2.6 Applied thermodynamics
30 questions · AMeE0206
151. In a fire-tube boiler, the hot flue gases
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Which fluid is inside the tubes?
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Answer: A. pass through tubes surrounded by water
In fire-tube boilers (e.g. Cochran, Lancashire, locomotive) combustion gases flow inside tubes immersed in the water of the shell; water-tube boilers are the opposite.
152. Compared with fire-tube boilers, water-tube boilers
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Used in power stations.
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Answer: C. can generate steam at higher pressures and capacities
Small-diameter tubes withstand high pressure, and the small water content allows fast steaming and large capacity.
153. Which of the following is a boiler accessory (not a mounting)?
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Accessories improve performance; mountings ensure safety.
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Answer: C. Economiser
Accessories such as economisers, superheaters, air preheaters and feed pumps improve efficiency; mountings such as safety valves and gauges are essential for safe operation.
154. The function of a fusible plug in a boiler is to
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A safety device sensitive to temperature.
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Answer: B. melt and extinguish the fire when the water level falls dangerously low
The plug's low-melting alloy core melts when exposed to flames due to low water level, releasing steam and warning the operator.
155. The purpose of a superheater in a steam boiler is to
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Heating beyond saturation.
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Answer: B. raise the temperature of steam above its saturation temperature
A superheater adds heat to saturated steam at constant pressure, giving dry superheated steam that reduces turbine blade erosion.
156. For a given pressure ratio, the work required by a reciprocating air compressor is minimum when the compression is
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Keep the gas as cool as possible.
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Answer: D. isothermal
Work ∝ ∫v dp and v is smallest when the gas is kept cool, so isothermal compression takes the least work.
157. The main advantage of multi-stage compression with intercooling is
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Cool the air between stages.
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Answer: A. reduced work of compression and lower delivery temperature
Intercooling brings the compression closer to isothermal, saves work, limits temperature of lubricating oil and improves volumetric efficiency.
158. The clearance volume in a reciprocating compressor
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Re-expansion of trapped gas.
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Answer: A. reduces the volumetric efficiency of the compressor
Compressed gas in the clearance re-expands and occupies part of the suction stroke, reducing the fresh charge drawn in.
159. Centrifugal compressors are generally preferred over reciprocating compressors for
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Think of large air volumes in industry.
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Answer: C. large flow rates at moderate pressure ratios
Rotary dynamic compressors give smooth, continuous flow for high volumes; reciprocating machines handle high pressures at low flow.
160. Which of the following is a positive displacement rotary compressor?
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Positive displacement means trapping a volume.
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Answer: A. Screw compressor
Screw, vane and lobe compressors trap volumes of gas and reduce them mechanically, whereas centrifugal and axial machines are dynamic.
161. The refrigerant designated R-717 is
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7 followed by molecular mass.
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Answer: A. ammonia
R-717 is ammonia (R-7 followed by the molecular weight 17); R-744 is CO2 and R-134a is tetrafluoroethane.
162. Which of the following refrigerants contributes most to ozone depletion?
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Contains chlorine.
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Answer: D. R-12 (a CFC)
CFCs such as R-12 release chlorine in the stratosphere, destroying ozone; HFCs, ammonia and CO2 have zero ODP.
163. An important desirable property of a refrigerant is
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More cooling per kg.
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Answer: C. a high latent heat of vaporisation
High latent heat means a lower mass flow of refrigerant for a given cooling load; a critical temperature well above the condensing temperature is also needed.
164. In a basic HVAC system, the component that cools and dehumidifies air with chilled water or refrigerant is the
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Air passes over cold tubes.
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Answer: C. cooling coil in the air-handling unit
The air handling unit uses a cooling coil to lower the air's temperature and moisture content before it is distributed through ducts.
165. For saturated moist air, the dry-bulb, wet-bulb and dew-point temperatures are
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Relative humidity = 100%.
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Answer: D. all equal
At 100% relative humidity no evaporation can occur from the wet bulb, and no cooling is needed to reach saturation.
166. Relative humidity is defined as the ratio of
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A pressure ratio, not a mass ratio.
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Answer: D. the partial pressure of water vapour in air to the saturation pressure at the same dry-bulb temperature
φ = pv/psat at the dry-bulb temperature; the mass ratio is the specific humidity (humidity ratio).
167. During sensible cooling of moist air, the
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No moisture is removed.
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Answer: B. dry-bulb temperature falls while the specific humidity stays constant
Sensible cooling follows a horizontal line on the psychrometric chart; relative humidity rises as temperature falls.
168. Dehumidification of air by cooling requires cooling the air to a temperature
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Condensation starts at the dew point.
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Answer: B. below its dew-point temperature
Moisture condenses only when the air touches surfaces colder than its dew point.
169. Typical comfort conditions for human occupancy in air conditioning are approximately
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Moderate temperature and humidity.
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Answer: A. 22–26 °C and 40–60% relative humidity
These ranges satisfy the usual thermal comfort standards for sedentary occupants.
170. Moist air at total pressure 101.325 kPa has a water-vapour partial pressure of 2 kPa. The specific humidity is approximately
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ω = 0.622 pv/(p − pv).
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Answer: C. 0.0125 kg/kg dry air
ω = 0.622 pv/(p − pv) = 0.622 × 2/99.325 = 0.0125.
171. Moist air at 25 °C (psat = 3.169 kPa) has a vapour partial pressure of 1.5 kPa. The relative humidity is
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φ = pv/psat.
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Answer: D. 47.3%
φ = pv/psat = 1.5/3.169 = 0.473.
172. A boiler produces 2000 kg/h of steam of enthalpy 2800 kJ/kg from feed water of enthalpy 167.5 kJ/kg. The equivalent evaporation from and at 100 °C (latent heat 2257 kJ/kg) is approximately
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Divide the heat gained by the latent heat at 100 °C.
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Answer: A. 2333 kg/h
Eq. evaporation = m(h − hf)/2257 = 2000 × 2632.5/2257 = 2333 kg/h.
173. A boiler generates 5000 kg/h of steam with an enthalpy gain of 2400 kJ/kg per kg of water, burning 700 kg/h of coal of calorific value 25 000 kJ/kg. The boiler efficiency is approximately
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Useful heat to steam divided by heat in fuel.
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Answer: B. 68.6%
η = m_s Δh/(m_f CV) = 12 000 000/17 500 000 = 0.686.
174. A two-stage reciprocating compressor takes air at 1 bar and delivers at 16 bar. For minimum work, the ideal intercooler pressure is
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Geometric mean of suction and delivery pressures.
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Answer: B. 4 bar
p_i = √(p1p2) = √(1 × 16) = 4 bar.
175. A compressor handles 0.1 kg/s of air (R = 0.287 kJ/kg·K) at 300 K and compresses it isothermally with a pressure ratio of 5. The power required is approximately
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W = ṁRT ln(p2/p1).
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Answer: D. 13.9 kW
W = ṁRT ln(p2/p1) = 0.1 × 0.287 × 300 × ln 5 = 13.9 kW.
176. A reciprocating compressor has clearance ratio 0.05 and compresses air with a pressure ratio of 6 following pV^1.3 = constant. The volumetric efficiency is approximately
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ηv = 1 + c − c(p2/p1)^(1/n).
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Answer: B. 85.2%
ηv = 1 + c − c(p2/p1)^(1/n) = 1 + 0.05 − 0.05 × 6^(0.769) = 0.852.
177. Air at 300 K is compressed polytropically (n = 1.3) with a pressure ratio of 8. The delivery temperature is approximately
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T2/T1 = (p2/p1)^((n−1)/n).
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Answer: A. 485 K
T2 = T1 (p2/p1)^((n−1)/n) = 300 × 8^0.2308 = 485 K.
178. A refrigeration system has a cooling capacity of 10.55 kW. Taking 1 TR = 3.517 kW, this is equal to
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Divide by 3.517 kW per TR.
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Answer: B. 3 TR
10.55/3.517 = 3.0 TR.
179. Moist air at a total pressure of 101.3 kPa contains water vapour at a partial pressure of 3 kPa. By Dalton's law, the partial pressure of the dry air is
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Total pressure is the sum of partial pressures.
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Answer: C. 98.3 kPa
p = pa + pv, so pa = 101.3 − 3 = 98.3 kPa.
180. 2 kg/s of dry air is cooled sensibly from 35 °C to 25 °C (cp = 1.005 kJ/kg·K). The cooling load is approximately
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Q = ṁ cp ΔT.
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Answer: D. 20.1 kW
Q = ṁ cp ΔT = 2 × 1.005 × 10 = 20.1 kW.