Q: Thermal diffusivity comparison: “The thermal diffusivity of solids is generally less than that of liquids and gases.” Is this statement correct?

Introduction / Context: Thermal diffusivity (often written as alpha_th) indicates how quickly a material responds to temperature changes. It combines thermal conductivity, density, and specific heat and is central to transient heat transfer problems. Given Data / Assumptions: Thermal diffusivity definition: alpha_th = k / (rho * c). Typical orders of magnitude: metals have high k and often high alpha_th; liquids like water have lower k and thus lower alpha_th; gases can have comparable or higher alpha_th than some liquids, depending on conditions. We compare broad classes: solids, liquids, gases. Concept / Approach: Since alpha_th rises with conductivity and falls with heat capacity and density, metals (solids) frequently exhibit large thermal diffusivities (fast temperature equalization). Water and many liquids have much lower alpha_th. Gases can have alpha_th on the order of 10^-5 m^2/s, while water is ~10^-7 m^2/s; many metals exceed 10^-5 m^2/s. Step-by-Step Solution: Recall alpha_th = k / (rho * c).Metals: high k → high alpha_th.Liquids (e.g., water): modest k and significant rho * c → low alpha_th.Therefore, the blanket statement “solids generally less than liquids and gases” is not true. Verification / Alternative check: Compare typical values: aluminum ~ 8.4e-5 m^2/s, copper ~ 1.1e-4 m^2/s; water ~ 1.4e-7 m^2/s; air ~ 2e-5 m^2/s. Solids (metals) are not “generally less.” Why Other Options Are Wrong: “True” and conditional versions contradict well-known property tables; temperature dependences do not reverse the general trend for common ranges. Common Pitfalls: Confusing conductivity with diffusivity or assuming all solids behave like insulators. Materials vary widely; always use alpha_th = k/(rho*c). Final Answer: False

Q: Cooling and heating coils — by-pass factor for sensible heating: Given td1 = dry-bulb temperature of air entering the heating coil, td2 = dry-bulb temperature leaving the coil, and td3 = coil (surface) dry-bulb temperature, select the correct expres…

Introduction / Context: The by-pass factor (BPF) quantifies the fraction of air that effectively does not contact (or equilibrate with) the coil surface. It is crucial for coil performance prediction in HVAC design. Given Data / Assumptions: Sensible heating only (no moisture addition or removal). td3 represents the effective apparatus surface temperature for heating. Steady operation with uniform mixing downstream. Concept / Approach: For sensible heating, the leaving air temperature approaches the coil temperature. The by-pass factor is defined as the ratio of the temperature “gap” that remains after the coil to the initial gap at coil entry. For heating, the correct form is BPF = (td3 − td2) / (td3 − td1). Step-by-Step Solution: Define approach gap at outlet: td3 − td2.Define approach gap at inlet: td3 − td1.Take ratio: BPF = (td3 − td2) / (td3 − td1).A lower BPF indicates better heat exchange (outlet closer to coil temperature). Verification / Alternative check: If td2 equals td3 (perfect contact/equilibration), numerator is zero and BPF = 0, which matches the ideal of no by-pass. If no heating occurred (td2 = td1), BPF = 1, indicating total by-pass. Why Other Options Are Wrong: Forms that swap numerator/denominator or reverse temperatures correspond either to cooling cases or incorrect gaps and do not satisfy the limiting checks above. Common Pitfalls: Using the cooling-coil formula for a heating-coil case. Remember: heating uses (td3 − td2)/(td3 − td1); cooling uses (td2 − td3)/(td1 − td3). Final Answer: BPF = (td3 − td2) / (td3 − td1)

Question 1

Absorption refrigeration (aqua–ammonia): If rectification is incomplete, where does the carried-over water ultimately accumulate in the system?

Accepted Answer

Introduction / Context: In aqua–ammonia absorption systems, the generator liberates ammonia vapour from the strong solution. A rectifier (dephlegmator) removes water vapour to ensure nearly pure ammonia reaches the condenser and evaporator. Incomplete rectification causes operational issues. Given Data / Assumptions: Working pair: ammonia (refrigerant) and water (absorbent). Rectifier is not fully effective; some water vapour passes with ammonia. Standard component sequence: generator → rectifier → condenser → expansion → evaporator → absorber. Concept / Approach: If water passes beyond the rectifier, it condenses with ammonia in the condenser and is then throttled with it to the evaporator. At evaporator conditions, water has an extremely low vapour pressure and tends to remain liquid, progressively accumulating and impairing heat transfer and capacity. Step-by-Step Solution: Trace the path: generator vapour mixture → rectifier (partially effective) → condenser (both NH3 and H2O condense).After expansion, the mixture enters the evaporator at low pressure.Ammonia evaporates readily; water does not and accumulates as liquid in the evaporator.Accumulation reduces effective refrigerant circulation and evaporator performance. Verification / Alternative check: Check vapour pressure data: at evaporator temperatures, water's saturation pressure is too low to evaporate significantly, confirming accumulation. Why Other Options Are Wrong: Condenser drains to the expansion device; absorber receives vapour from evaporator, not liquid water from poor rectification; “none” ignores the described mechanism. Common Pitfalls: Neglecting rectifier performance; assuming any condensate is harmless—water contamination severely degrades refrigeration effect. Final Answer: Evaporator

Question 2

Radiation basics: Evaluate the statement “In a black body, all radiations are rejected.” Choose the correct assessment and reasoning.

Accepted Answer

Introduction / Context: Blackbody radiation is a cornerstone concept in heat transfer and thermodynamics. It defines ideal absorption and emission characteristics and underpins Planck’s law and Stefan–Boltzmann law. Given Data / Assumptions: “Black body” refers to an ideal surface with absorptivity equal to 1 at all wavelengths and directions. Thermal equilibrium concepts apply. Concept / Approach: A black body absorbs all incident radiation (it does not reflect or transmit). It also is the most efficient emitter at a given temperature. Therefore, the statement that it “rejects” all radiation is the opposite of the definition. Step-by-Step Solution: Define black body: absorptivity = 1, reflectivity = 0, transmissivity = 0.Kirchhoff’s law connects high absorptivity to high emissive power at the same temperature.Conclude that a black body does not reject radiation; it absorbs all and emits maximally.Hence, the statement is false. Verification / Alternative check: Compare to “white body,” which reflects; black body exhibits the opposite behaviour. Why Other Options Are Wrong: Conditional “true” options conflict with the universal definition; wavelength or medium does not reverse the black body’s absorbing nature. Common Pitfalls: Confusing “black” with non-emitting; in thermal radiation, a perfect absorber is also a perfect emitter at that temperature. Final Answer: False

Question 3

Aircraft environmental control: An ordinary passenger aircraft typically requires an air-conditioning (cooling) capacity closest to which tonnage?

Accepted Answer

Introduction / Context: Aircraft environmental control systems (ECS) provide ventilation, pressurization, and cooling. For sizing and academic problems, a representative cooling tonnage is often cited for “ordinary” passenger aircraft. Given Data / Assumptions: “Ordinary passenger aircraft” refers to small-to-medium capacity passenger airplanes in textbook contexts. Cooling tonnage is the refrigeration rate; 1 TR = 3.517 kW of cooling. Loads include people, equipment, solar and skin heating, and ventilation air cooling. Concept / Approach: Rule-of-thumb classroom values place ECS cooling capacity for such aircraft on the order of several tons of refrigeration. Among the offered choices, 10 TR aligns with widely used illustrative estimates and permits handling peak cabin and avionics loads with margin. Step-by-Step Solution: Convert 10 TR to kW: 10 * 3.517 ≈ 35.17 kW.Compare to smaller options (2–8 TR), which may be insufficient for typical occupant and ventilation loads.Larger values (e.g., 20 TR) exceed the simple “ordinary” case described.Therefore, 10 TR is the best match among given options. Verification / Alternative check: Typical ECS pack capabilities for small commuter jets or turboprops fall in the tens of kW, supporting the 10 TR order-of-magnitude figure for instructional problems. Why Other Options Are Wrong: 2–8 TR can undersize peak latent/sensible loads; 20 TR may be more suitable for larger platforms or special operating conditions. Common Pitfalls: Confusing TR with electrical input power; or comparing aircraft ECS (air-cycle) directly to building vapour-compression systems without considering different COPs and operating envelopes. Final Answer: 10 TR

Question 4

Humidification process insight: During humidification of air, how does the specific humidity (humidity ratio) change?

Accepted Answer

Introduction / Context: Humidification is common in comfort conditioning, textile mills, and process industries. The key variable is specific humidity (also called humidity ratio), defined as kilograms of water vapour per kilogram of dry air. Given Data / Assumptions: Moisture is added to air (e.g., steam or evaporation). Pressure is near atmospheric and constant for chart use. No dehumidifying surfaces present. Concept / Approach: Adding moisture increases the mass of water vapour per unit mass of dry air. Hence, the humidity ratio must rise. Depending on the method (evaporative vs steam injection), dry-bulb temperature may fall or rise, but the specific humidity will increase. Step-by-Step Solution: Define specific humidity: w = m_vapour / m_dry_air.Add water vapour: m_vapour increases while m_dry_air is unchanged.Therefore, w increases monotonically during humidification.Conclude that the correct option is “Increases.” Verification / Alternative check: On a psychrometric chart, humidification moves the state point to the right (higher w), either up-left (evaporative cooling) or up-right (steam injection) depending on sensible heat exchange. Why Other Options Are Wrong: “Remains constant” and “decreases” contradict the definition of humidification; zero or negative values are nonphysical under the stated process. Common Pitfalls: Confusing relative humidity (a percentage) with humidity ratio; RH can decrease during steam injection if temperature rises rapidly, even though w increases. Final Answer: Increases

Question 5

Planck’s law applicability: Does Planck’s radiation law hold for all “coloured” (real) bodies exactly as stated for a black body?

Accepted Answer

Introduction / Context: Planck’s law gives the spectral distribution of emissive power of a perfect black body. Real (coloured) bodies deviate from blackbody behaviour and are described through emissivity. Given Data / Assumptions: Black body: ideal reference with emissivity = 1 at all wavelengths. Real body: emissivity ε(λ, T) < 1 and wavelength dependent. Concept / Approach: For real bodies, spectral emissive power E(λ, T) = ε(λ, T) * E_b(λ, T), where E_b is given by Planck’s law. Thus, Planck’s law applies to the ideal blackbody spectrum, while real bodies follow it only after scaling by their emissivity function. Step-by-Step Solution: State Planck’s law as the blackbody spectral distribution.Introduce emissivity ε(λ, T) for real surfaces.Relate real-body emission to blackbody by multiplication with ε(λ, T).Conclude that the statement “holds good for all coloured bodies” is false. Verification / Alternative check: Kirchhoff’s law links emissivity and absorptivity; experimental spectra for paints, metals, and ceramics confirm departures from the blackbody curve. Why Other Options Are Wrong: Restrictions to vacuum, temperature range, or material do not convert a real body into a black body; emissivity remains less than unity and wavelength dependent. Common Pitfalls: Using blackbody tables directly for engineering surfaces without applying emissivity corrections leads to significant errors. Final Answer: False

Question 6

Aircraft air-cycle systems: The “reduced ambient” air-cooling system configuration contains which combination of turbines and heat exchangers?

Accepted Answer

Introduction / Context: Aircraft environmental control systems use variants of the air-cycle (reversed Brayton) process: simple, bootstrap, regenerative, and reduced ambient. Each configuration uses different numbers of turbines and heat exchangers to meet cabin cooling at varying flight conditions. Given Data / Assumptions: Reduced ambient system is intended to achieve lower turbine inlet temperatures by further cooling the compressed air. Ram air availability and pressure ratios are typical of transport aircraft. Components are arranged to maximize precooling before expansion. Concept / Approach: Reduced ambient systems employ an additional cooling/expansion stage to reduce the effective “ambient” temperature seen by the main expansion turbine. This requires two heat exchangers (primary and secondary) and two turbines arranged to stepwise drop the temperature and extract work. Step-by-Step Solution: Compressed bleed air is first cooled in a primary heat exchanger.It may be expanded in an auxiliary turbine to reduce its temperature further.Air is then cooled again in a secondary heat exchanger using the lower “ambient,” enhancing effectiveness.Finally, it expands in the main cooling turbine to produce the cabin supply at low temperature. Verification / Alternative check: Cycle schematics used in textbooks show two HXs and two turbines for the reduced ambient variant, distinct from simple (1 HX, 1 turbine) and bootstrap (2 HXs, 1 turbine + secondary compressor) systems. Why Other Options Are Wrong: Options with only one turbine or one heat exchanger describe simpler cycles that cannot attain the same low supply temperatures under comparable conditions. Common Pitfalls: Mixing up “bootstrap” and “reduced ambient”; the former uses a secondary compressor driven by a turbine, while the latter uses additional cooling and expansion stages. Final Answer: Two cooling turbines and two heat exchangers

Question 7

Psychrometrics — definition check: Relative humidity (often denoted by φ or RH) is defined using partial pressures. Given pv = partial pressure of water vapour, pd = saturation pressure at the dry-bulb temperature, pb = barometric pressure, ps = pre…

Accepted Answer

Introduction / Context: Relative humidity is a core concept in HVAC and meteorology. It compares the actual moisture present in air to the maximum possible moisture the air could hold at the same temperature. Using the correct pressure terms avoids large errors in coil sizing, comfort prediction, and psychrometric chart work. Given Data / Assumptions: pv = partial pressure of water vapour actually present in the air. pd = saturation pressure of water vapour at the current dry-bulb temperature. pb = barometric (total) pressure of the air–vapour mixture. ps = pressure of saturated air (contextual, but not needed for RH). pw = saturation pressure at the wet-bulb temperature. μ = degree of saturation (a different humidity measure). Concept / Approach: Relative humidity φ is the ratio of the current vapour partial pressure to the saturation vapour pressure at the same dry-bulb temperature. Therefore, φ = pv / pws, where pws here is denoted as pd. If φ is required in percent, multiply by 100. Values range from 0 to 1 (or 0% to 100%). Step-by-Step Solution: Identify the target: relative humidity (φ).Use definition: φ = actual vapour pressure / saturation vapour pressure at DBT.Map symbols: actual vapour pressure = pv; saturation at DBT = pd.Hence φ = pv / pd (dimensionless or ×100 for %). Verification / Alternative check: On a psychrometric chart, for a given dry-bulb temperature, lines of constant φ correspond to pv proportional to pws at that temperature. Measured φ must always be ≤ 1 since pv ≤ pd. Why Other Options Are Wrong: φ = (pb − pv) / pb: mixes total pressure with vapour pressure; not RH.φ = pv / pw: uses saturation at wet-bulb temperature, not at dry-bulb; incorrect.φ = μ * ps / pb: degree of saturation μ is not equal to relative humidity and requires additional relations.none of these: incorrect since φ = pv / pd is valid. Common Pitfalls: Confusing relative humidity with humidity ratio (mass basis) or using wet-bulb saturation pressure instead of dry-bulb saturation pressure. Always pair φ with the saturation pressure at the same temperature. Final Answer: φ = pv / pd

Question 8

Thermal diffusivity comparison: “The thermal diffusivity of solids is generally less than that of liquids and gases.” Is this statement correct?

Accepted Answer

Introduction / Context: Thermal diffusivity (often written as alpha_th) indicates how quickly a material responds to temperature changes. It combines thermal conductivity, density, and specific heat and is central to transient heat transfer problems. Given Data / Assumptions: Thermal diffusivity definition: alpha_th = k / (rho * c). Typical orders of magnitude: metals have high k and often high alpha_th; liquids like water have lower k and thus lower alpha_th; gases can have comparable or higher alpha_th than some liquids, depending on conditions. We compare broad classes: solids, liquids, gases. Concept / Approach: Since alpha_th rises with conductivity and falls with heat capacity and density, metals (solids) frequently exhibit large thermal diffusivities (fast temperature equalization). Water and many liquids have much lower alpha_th. Gases can have alpha_th on the order of 10^-5 m^2/s, while water is ~10^-7 m^2/s; many metals exceed 10^-5 m^2/s. Step-by-Step Solution: Recall alpha_th = k / (rho * c).Metals: high k → high alpha_th.Liquids (e.g., water): modest k and significant rho * c → low alpha_th.Therefore, the blanket statement “solids generally less than liquids and gases” is not true. Verification / Alternative check: Compare typical values: aluminum ~ 8.4e-5 m^2/s, copper ~ 1.1e-4 m^2/s; water ~ 1.4e-7 m^2/s; air ~ 2e-5 m^2/s. Solids (metals) are not “generally less.” Why Other Options Are Wrong: “True” and conditional versions contradict well-known property tables; temperature dependences do not reverse the general trend for common ranges. Common Pitfalls: Confusing conductivity with diffusivity or assuming all solids behave like insulators. Materials vary widely; always use alpha_th = k/(rho*c). Final Answer: False

Question 9

Cooling and heating coils — by-pass factor for sensible heating: Given td1 = dry-bulb temperature of air entering the heating coil, td2 = dry-bulb temperature leaving the coil, and td3 = coil (surface) dry-bulb temperature, select the correct expres…

Accepted Answer

Introduction / Context: The by-pass factor (BPF) quantifies the fraction of air that effectively does not contact (or equilibrate with) the coil surface. It is crucial for coil performance prediction in HVAC design. Given Data / Assumptions: Sensible heating only (no moisture addition or removal). td3 represents the effective apparatus surface temperature for heating. Steady operation with uniform mixing downstream. Concept / Approach: For sensible heating, the leaving air temperature approaches the coil temperature. The by-pass factor is defined as the ratio of the temperature “gap” that remains after the coil to the initial gap at coil entry. For heating, the correct form is BPF = (td3 − td2) / (td3 − td1). Step-by-Step Solution: Define approach gap at outlet: td3 − td2.Define approach gap at inlet: td3 − td1.Take ratio: BPF = (td3 − td2) / (td3 − td1).A lower BPF indicates better heat exchange (outlet closer to coil temperature). Verification / Alternative check: If td2 equals td3 (perfect contact/equilibration), numerator is zero and BPF = 0, which matches the ideal of no by-pass. If no heating occurred (td2 = td1), BPF = 1, indicating total by-pass. Why Other Options Are Wrong: Forms that swap numerator/denominator or reverse temperatures correspond either to cooling cases or incorrect gaps and do not satisfy the limiting checks above. Common Pitfalls: Using the cooling-coil formula for a heating-coil case. Remember: heating uses (td3 − td2)/(td3 − td1); cooling uses (td2 − td3)/(td1 − td3). Final Answer: BPF = (td3 − td2) / (td3 − td1)

Question 10

Refrigeration performance from energy flows: A condenser rejects 120 kW of heat while the compressor consumes 30 kW of shaft power. What is the coefficient of performance (COP) of the refrigeration system?

Accepted Answer

Introduction / Context: The coefficient of performance (COP) of a refrigerator measures how much cooling effect is obtained per unit of input work. Interpreting heat rejection and compressor power correctly is fundamental for energy audits and sizing. Given Data / Assumptions: Heat rejected at condenser, Q_H = 120 kW. Compressor power input, W = 30 kW. Steady-state operation with Q_H = Q_L + W. Concept / Approach: For a refrigerator, COP_R = Q_L / W, where Q_L is the cooling (evaporator) load. Since the condenser rejects the sum of the evaporator load and compressor work, we find Q_L by subtraction. Step-by-Step Solution: Compute evaporator load: Q_L = Q_H − W = 120 − 30 = 90 kW.Compute COP: COP_R = Q_L / W = 90 / 30 = 3.Therefore, COP = 3. Verification / Alternative check: The energy balance must hold: Q_H = 120 kW = 90 kW + 30 kW, which is satisfied. Typical COP values of 2–5 are realistic for many systems, so 3 is plausible. Why Other Options Are Wrong: 1/4 and 1/3 invert the ratio; they would correspond to W > Q_L, which contradicts the data.4 would require Q_L = 120 kW with W = 30 kW, but Q_L is only 90 kW here.2 would imply Q_L = 60 kW, not matching the balance. Common Pitfalls: Mistaking the condenser heat flow Q_H for the cooling load Q_L. Always subtract compressor work to find Q_L from Q_H. Final Answer: 3

Question 11

Domestic refrigerator sizing: Select the typical capacity range (in tonnes of refrigeration, TR) for a household refrigerator.

Accepted Answer

Introduction / Context: Household refrigerators are small refrigeration units compared to comfort-cooling or industrial systems. Knowing the typical capacity range helps students relate theory to real appliances. Given Data / Assumptions: Capacity is quoted in tonnes of refrigeration (TR), where 1 TR = 3.517 kW of cooling. Domestic fridges serve small enclosures with intermittent door openings. Concept / Approach: Compact, insulated household units require modest cooling capacities, far below split ACs or chillers. Typical single-door and double-door home refrigerators fall in the fraction-of-a-ton range. Step-by-Step Solution: Convert mental benchmark: 0.1 TR ≈ 0.35 kW; 0.3 TR ≈ 1.06 kW.These levels align with compressor sizes used in domestic refrigerators.Thus, 0.1–0.3 TR is the correct range. Verification / Alternative check: Compare to room air conditioners (commonly 1–2 TR): far larger than a fridge. Industrial or central plant capacities (3–7 TR and above) are orders of magnitude higher. Why Other Options Are Wrong: 1–3 TR and above represent space conditioning or small commercial refrigeration, not typical household units.0.5–1.0 TR still overestimates most domestic fridges. Common Pitfalls: Confusing electrical power (kW input) with cooling capacity (kW or TR). Always compare like quantities. Final Answer: 0.1 to 0.3 TR

Question 12

Carbon dioxide (CO₂) as a gas — select the correct set of properties relevant to fire and safety considerations.

Accepted Answer

Introduction / Context: Basic gas properties are essential in refrigeration, fire suppression, and safety engineering. CO₂ is widely used as a fire extinguishing agent and as a refrigerant in transcritical systems. Given Data / Assumptions: CO₂ at ordinary conditions (non-cryogenic). Properties considered: colour, odour, and flammability. Concept / Approach: CO₂ is colourless and odourless. It does not support combustion and is used to smother fires by diluting oxygen concentration. While CO₂ is asphyxiant at high concentrations, it is not classified as a flammable gas. Step-by-Step Solution: Check colour: CO₂ is colourless.Check odour: CO₂ is odourless.Check flammability: CO₂ is non-flammable and used in extinguishers.Therefore, “all of these” is correct. Verification / Alternative check: Fire codes and data sheets list CO₂ as non-flammable and suitable for certain classes of fires, confirming these properties. Why Other Options Are Wrong: Picking any single property ignores the others; “toxic and highly flammable” is incorrect. Common Pitfalls: Confusing CO₂ with CO (carbon monoxide). CO is toxic; CO₂ is an asphyxiant at high concentrations but not flammable. Final Answer: all of these

Question 13

Humidity terminology: The difference between dry-bulb temperature and dew-point temperature is called what?

Accepted Answer

Introduction / Context: Accurate HVAC diagnostics depend on clear temperature terminology. Dry-bulb, wet-bulb, and dew-point temperatures each describe different aspects of an air–water vapour mixture. Given Data / Assumptions: T_DB = dry-bulb temperature (ordinary thermometer reading). T_WB = wet-bulb temperature (adiabatic saturation proxy). T_DP = dew-point temperature (onset of condensation). Concept / Approach: The “depression” relative to dew point is simply T_DB − T_DP, commonly called dew-point depression. It indicates how close the air is to saturation; smaller values signal higher moisture content and higher relative humidity. Step-by-Step Solution: Identify difference: DB − DP.Definition: this is dew-point depression.Therefore select “dew point depression.” Verification / Alternative check: On a psychrometric chart, for a fixed DB, moving toward the saturation curve decreases DP depression; at saturation, DB = DP and the depression is zero. Why Other Options Are Wrong: Dry bulb depression usually describes thermometer shielding effects and is not a standard psychrometric term here.Wet bulb depression is DB − WB, not DB − DP.Degree of saturation and relative humidity are different measures, not simple temperature differences. Common Pitfalls: Confusing WB and DP differences; remember WB reflects adiabatic saturation while DP reflects actual vapour content relative to saturation at lower temperature. Final Answer: dew point depression

Question 14

Modes of heat transfer — identify the definition: “Conduction is heat transfer from one particle to another by actual motion of the heated particles.” Is this statement correct?

Accepted Answer

Introduction / Context: There are three basic heat transfer modes: conduction, convection, and radiation. Clear definitions are vital to selecting the correct correlations and boundary conditions in engineering problems. Given Data / Assumptions: The quoted statement attributes heat transfer “by actual motion of the heated particles” to conduction. No special micro-scale effects are invoked. Concept / Approach: Conduction is heat transfer through a medium without bulk motion of the medium, primarily via molecular interactions, lattice vibrations, or electron transport (in metals). Convection is heat transfer accompanied by bulk fluid motion (natural or forced). The statement given actually describes convection, not conduction. Step-by-Step Solution: Recall definitions: conduction → no bulk motion; convection → bulk motion carries energy.The phrase “actual motion of heated particles” is the hallmark of convection.Therefore the statement about conduction is false. Verification / Alternative check: Fourier’s law (q = −k * dT/dx) is the governing relation for conduction and contains no velocity term. Convection uses Newton’s law of cooling (q = h * A * ΔT) and relies on flow. Why Other Options Are Wrong: Any conditional “true” option misclassifies the transport mechanism; conduction does not become convection at high temperature by definition. Common Pitfalls: Equating microscopic molecular motion in solids with macroscopic fluid motion. Conduction considers energy transfer due to gradients without bulk displacement. Final Answer: False

Question 15

Effect of evaporator temperature on performance: If condenser temperature is held constant, how does the refrigeration cycle COP change as the evaporator temperature increases?

Accepted Answer

Introduction / Context: The coefficient of performance (COP) of a vapour-compression system depends strongly on the evaporating and condensing temperatures. Understanding these trends informs set-point selection and defrost strategies. Given Data / Assumptions: Condenser temperature (or pressure) is fixed. Evaporator temperature is increased (warmer evaporator). Same refrigerant and general cycle configuration. Concept / Approach: For idealized cycles, COP_R increases as the temperature lift (T_cond − T_evap) decreases. Raising the evaporator temperature reduces this lift and increases the ratio of cooling effect to compressor work. Step-by-Step Solution: Let T_H be condenser temperature, T_L be evaporator temperature.Temperature lift = T_H − T_L; increasing T_L reduces lift.Lower lift → lower compressor work per unit cooling → higher COP. Verification / Alternative check: For a simple ideal cycle, COP_R ≈ T_L / (T_H − T_L) on absolute scale. Increasing T_L while T_H constant increases COP_R. Why Other Options Are Wrong: “No/unchanged” contradicts the standard dependence on temperature lift.Non-monotonic or zero COP outcomes do not apply here. Common Pitfalls: Comparing COP across different refrigerants or mixing Celsius and absolute temperatures; always use absolute temperatures in theoretical relations. Final Answer: Yes

Question 16

Radiative properties — diathermanous body: Given absorptivity α, reflectivity ρ, and transmissivity τ, select the correct set of values for a perfectly diathermanous body (perfectly transparent to thermal radiation).

Accepted Answer

Introduction / Context: In radiation heat transfer, extreme idealizations clarify real-world behavior. A diathermanous body is an ideal transparent medium for thermal radiation, analogous to clear glass for certain wavelength bands. Given Data / Assumptions: α = absorptivity; ρ = reflectivity; τ = transmissivity. Energy conservation for radiation at a surface: α + ρ + τ = 1. “Perfectly diathermanous” means fully transmitting at the considered wavelengths. Concept / Approach: A perfectly diathermanous surface transmits all incident radiation. Therefore τ = 1 and neither absorbs nor reflects radiation: α = 0 and ρ = 0. This satisfies the conservation relation trivially. Step-by-Step Solution: Start from α + ρ + τ = 1.For perfect transparency, set τ = 1.Then α = 0 and ρ = 0 to satisfy the sum = 1. Verification / Alternative check: Compare with black body (α = 1, ρ = 0, τ = 0) and white (ρ → 1, α → 0, τ → 0). The diathermanous case is the complementary extreme where transmission dominates. Why Other Options Are Wrong: α = 1, ρ = 0, τ = 0: black body.α = 0, ρ = 1, τ = 0 and α + ρ = 1 with τ = 0: reflective/opaque cases, not transparent.α = 0.5, ρ = 0.5, τ = 0: violates transparency. Common Pitfalls: Assuming “transparent” implies some reflection. In the ideal diathermanous definition, reflectivity is zero within the specified band. Final Answer: α = 0, ρ = 0, τ = 1

Question 17

Central HVAC systems vs. individual (unit) systems  In building services engineering, how does the overall energy efficiency of a centralized air-conditioning system compare with that of multiple individual room/packaged units serving the same load?

Accepted Answer

Introduction / Context: Central air-conditioning (HVAC) plants typically include high-efficiency chillers or central heat pumps, cooling towers, large air handlers, and centralized controls. The question tests conceptual understanding of why a centralized plant often achieves better overall efficiency than scattered individual units serving the same building or campus. Given Data / Assumptions: Equal comfort setpoints for both approaches. Comparable envelope, internal loads, and occupancy schedules. Properly designed distribution (ducts/piping) and controls for the central system. Concept / Approach: Central plants benefit from diversity of loads, superior component efficiencies at larger scales, heat-recovery opportunities, and optimized part-load operation. Individual room units lack load diversity and often cycle on/off inefficiently with poorer maintenance and control. Step-by-Step Solution: Load diversity: Not all zones peak simultaneously, so the central plant can run smaller than the sum of nameplates.Equipment efficiency: Large chillers/heat pumps typically have better COP/EER than window or small split units.Integrated controls: Central BMS optimizes staging, setpoints, and chilled-water/air temperatures to minimize energy.Heat recovery: Central systems can reclaim condenser heat for reheat or domestic hot water, improving seasonal performance. Verification / Alternative check: Benchmarking data and energy models consistently show lower kWh per ton-hour for central plants in medium-to-large buildings when distribution is well designed and maintained. Why Other Options Are Wrong: Lower/same: ignores diversity, scale, and control advantages. Depends only on outdoor temperature: efficiency depends on many variables (control, load profile, equipment type). Cannot be compared: they are routinely compared in design studies and audits. Common Pitfalls: Assuming any central plant is automatically efficient—poorly balanced pumps/fans, excessive static pressure, or fouled heat exchangers can erode benefits. Conversely, assuming small units are always inefficient—modern variable-speed splits can be very good for small, highly zoned buildings. Final Answer: higher

Question 18

Heat exchanger analysis – Is LMTD equal to AMTD?  State whether the logarithmic mean temperature difference (LMTD) is the same as the arithmetic mean temperature difference for heat exchangers.

Accepted Answer

Introduction / Context: The LMTD method is fundamental for sizing and rating heat exchangers. It accounts for the exponential temperature approach of two fluids exchanging heat across a surface. Confusing LMTD with a simple arithmetic mean can lead to serious sizing errors. Given Data / Assumptions: Steady state, no phase change, constant specific heats (typical LMTD assumptions). Any common flow arrangement (parallel, counter, cross with correction). Concept / Approach: The driving force varies along the exchanger length. Because heat transfer is proportional to local temperature difference, the correct average is the logarithmic mean, not the arithmetic mean, unless the temperature differences at each end are equal. Step-by-Step Solution: Define end temperature differences: Δt1 and Δt2.LMTD = (Δt1 − Δt2) / ln(Δt1 / Δt2).AMTD = (Δt1 + Δt2) / 2.Only when Δt1 = Δt2 do both expressions collapse to the same value; otherwise LMTD ≠ AMTD. Verification / Alternative check: Consider a parallel-flow exchanger with Δt1 = 30 K and Δt2 = 10 K. LMTD = (20)/ln(3) ≈ 18.2 K, while AMTD = 20 K. They are not equal, confirming the statement is false. Why Other Options Are Wrong: “Only true for counter-flow / equal mass flow / equal cp” are unnecessary constraints; equality depends solely on Δt1 = Δt2. Common Pitfalls: Using AMTD in spreadsheets for convenience; forgetting to apply correction factors for crossflow or multipass arrangements; mixing inlet-outlet assignments and getting Δt1, Δt2 swapped. Final Answer: Disagree

Question 19

Refrigerants – Which has the highest critical pressure among the following?  Select the refrigerant with the highest critical pressure from this list: R-11, R-12, R-22, and ammonia.

Accepted Answer

Introduction / Context: Critical properties (critical temperature and pressure) are essential when comparing refrigerants for high-side operating limits, compressor discharge pressures, and component ratings. The question asks which listed refrigerant has the highest critical pressure. Given Data / Assumptions: Standard refrigerant identifiers: R-11 (trichlorofluoromethane), R-12 (dichlorodifluoromethane), R-22 (chlorodifluoromethane), and ammonia (R-717). Typical published critical pressures under standard references. Concept / Approach: Ammonia (R-717) has a comparatively high critical pressure and moderate critical temperature, which drives higher allowable pressure ranges and compact heat exchangers but requires robust, rated components. Step-by-Step Solution: Recall approximate critical pressures: ammonia ≈ 11 MPa (110+ bar), R-22 ≈ 5 MPa, R-12 ≈ 4 MPa, R-11 ≈ 4–5 MPa.Compare the values across the options.Identify the maximum: ammonia. Verification / Alternative check: Check manufacturer or handbook tables; all corroborate ammonia’s critical pressure being significantly higher than common halocarbons listed. Why Other Options Are Wrong: R-22, R-12, and R-11 have substantially lower critical pressures than ammonia. R-134a (extra distractor) also has lower critical pressure than ammonia. Common Pitfalls: Confusing critical temperature with critical pressure; assuming modern HFCs necessarily have higher critical pressure than natural refrigerants—this is not generally true. Final Answer: Ammonia

Question 20

Solar-driven absorption refrigeration – Working pair  For absorption refrigerators using heat from solar collectors, water is used with which absorbent as the standard working pair?

Accepted Answer

Introduction / Context: Absorption refrigeration systems use a refrigerant–absorbent working pair instead of a mechanical compressor. Solar thermal heat is well suited to drive generators in absorption chillers. Knowing the correct pair is critical to understanding cycle design and operating envelopes. Given Data / Assumptions: Low to medium temperature solar heat available (e.g., 80–100 °C). Cooling for air-conditioning, not sub-zero freezing. Common commercial absorption pairs considered. Concept / Approach: The water–lithium bromide (H₂O–LiBr) pair is widely used for comfort cooling. In this pair, water is the refrigerant and lithium bromide is the absorbent. It operates under vacuum with evaporator temperatures typically above 0 °C, making it ideal for chilled water air-conditioning. Step-by-Step Solution: Identify target application: solar-driven building cooling (above freezing).Select pair compatible with above-zero evaporator temperature: H₂O–LiBr.Confirm role: water flashes in the evaporator; LiBr solution absorbs water vapor in the absorber; generator heat (from solar) drives water out of the solution. Verification / Alternative check: Compare with NH₃–H₂O (ammonia–water) systems: those use ammonia as refrigerant and can reach below 0 °C but require higher generator temperatures; H₂O–LiBr is the norm for solar-assisted comfort cooling. Why Other Options Are Wrong: Sulphur dioxide and R-12 are outdated mechanical compression refrigerants. Carbon dioxide is not used as an absorbent with water in standard absorption cycles. “Ammonia as absorbent” is incorrect; in NH₃–H₂O, water is the absorbent and ammonia is the refrigerant. Common Pitfalls: Mixing up which component is refrigerant versus absorbent; assuming LiBr systems can provide sub-zero refrigeration—they cannot due to water’s freezing point and vacuum constraints. Final Answer: Lithium bromide

Heat Transfer, Refrigeration and Air Conditioning Questions