Question 1

Psychrometrics: sensible heat factor for cooling and dehumidification  For a process over a cooling coil, with h1 = enthalpy of entering air, h2 = enthalpy of leaving air, and hA = enthalpy at the end of dehumidification (apparatus dew point), choos…

Accepted Answer

Given data h1: enthalpy of entering air h2: enthalpy of leaving air hA: enthalpy at apparatus dew-point (end of dehumidification line) Concept/Approach For cooling and dehumidification, the sensible heat factor is the ratio of sensible heat to total heat for the 1→2 process and, in terms of psychrometric geometry, is given by the segment ratio along the process line. Derivation (sketch-based reasoning) Total heat removed = (h1 − h2); sensible component corresponds to (h1 − hA). SHF = (h1 − hA) / (h1 − h2). Sanity check If there is no dehumidification, hA → h2 and SHF → 1, as expected for purely sensible cooling. Final Answer SHF = (h1 − hA) / (h1 − h2)

Question 2

Human thermal comfort: seasonal change in optimum effective temperature  Compare the optimum effective temperature for human comfort in winter versus summer. Choose the correct relation.

Accepted Answer

Concept/Approach The effective temperature concept accounts for dry-bulb temperature, humidity, and air motion to reflect perceived comfort. Seasonal clothing and acclimatization shift the comfort zone. Reasoning In winter, lower air movement and heavier clothing lead to a slightly higher optimum effective temperature compared to summer, where higher air movement and lighter clothing make a lower effective temperature feel comfortable. Final Answer Higher in winter than in summer

Question 3

Refrigeration maintenance: defrosting by compressor shutdown  Is defrosting of a domestic refrigerator achievable by stopping the compressor for a short period (off-cycle defrost)? Choose the correct truth value.

Accepted Answer

Concept/Approach When the compressor is stopped, coil temperature rises above the frost point, allowing accumulated frost to melt (off-cycle defrost), especially in medium/high evaporating temperature systems. Notes Other methods include electric heaters and hot-gas defrost for low-temperature (freezer) systems, but stopping the compressor is a standard defrost option in suitable applications. Final Answer True

Question 4

Condensing temperature comparison: water-cooled versus air-cooled condenser  For similar load and ambient conditions, a water-cooled condenser typically operates at what relative condensing temperature compared to an air-cooled condenser? Choose the…

Accepted Answer

Concept/Approach Water has higher heat-transfer coefficients and can be supplied cooler than ambient air (via cooling towers or mains), so the condenser saturation temperature (and pressure) can be reduced. Implication Lower condensing temperature reduces compressor work and improves system COP. Final Answer Lower condensing temperature

Question 5

Material selection: condenser tubes for ammonia refrigeration  For ammonia (NH3) refrigerating systems using shell-and-tube condensers, which tube material is appropriate considering compatibility and corrosion? Choose the correct material.

Accepted Answer

Concept/Approach Anhydrous ammonia is incompatible with copper and copper alloys (e.g., brass), which can suffer corrosion/stress corrosion cracking in ammonia service. Therefore Steel tubes are used for condensers/evaporators in ammonia systems.

Question 6

Psychrometrics numeric: compute sensible heat factor from sensible and latent rates  A process adds sensible heat = 30 kJ/s and latent heat = 20 kJ/s to moist air. Calculate the sensible heat factor (SHF) for the process. Choose the correct value.

Accepted Answer

Given data Sensible heat rate = 30 kJ/s Latent heat rate = 20 kJ/s Concept/Approach Sensible heat factor: SHF = (Sensible heat) / (Total heat) = S / (S + L). Step-by-step calculation S + L = 30 + 20 = 50 kJ/sSHF = 30 / 50 = 0.60 Final Answer 0.6

Question 7

Conduction comparison across geometries: thick sphere versus thick cylinder  Assess the statement: Heat transfer by conduction through a thick spherical wall is the same as through a thick cylindrical wall. Choose the correct truth value.

Accepted Answer

Concept/Approach In steady radial conduction, the governing area for heat flow varies with radius: Acyl = 2πrL (cylinder) versus Asph = 4πr² (sphere). This leads to different analytical forms for thermal resistance and heat rate. Consequences For a cylinder: Q = 2πkL (T1 − T2) / ln(r2/r1). For a sphere: Q = 4πk (T1 − T2) / (1/r1 − 1/r2). Final Answer False — the conduction relations are different for thick spheres and thick cylinders.

Question 8

Psychrometrics relation: dew-point vs wet-bulb as relative humidity decreases  As the relative humidity of moist air decreases, how does the dew-point temperature compare to the wet-bulb temperature? Choose the correct relation.

Accepted Answer

Given concept For unsaturated air: Tdew ≤ Twb ≤ Tdb. As relative humidity (RH) drops, the mixture is further from saturation, so Tdew falls away from Twb. Reasoning Lower RH → lower vapor partial pressure → lower Tdew. The wet-bulb temperature decreases too, but not as much; thus Tdew stays below Twb. Common pitfall Confusing Tdew with Twb; they are equal only at saturation (RH = 100%). Final Answer Less than the wet-bulb temperature.

Question 9

Heat transfer truths and one false statement: identify the incorrect claim  Four statements about heat transfer and temperature differences are given. Select the statement that is wrong.

Accepted Answer

Concept check (A) True in general: convection dominates bulk heat transfer in fluids (though conduction at the boundary initiates it).(B) True: heat flow depends on thermal conductivity k (a material property) and geometry.(D) True: LMTD ≠ arithmetic mean; LMTD accounts for the varying temperature difference along the exchanger. Why (C) is wrong For most solid metals, lattice/electron scattering increases with temperature, so thermal conductivity decreases as temperature rises (over wide ranges of practical interest). Final Answer “For solid metals, thermal conductivity increases with temperature” is the incorrect statement.

Question 10

Adiabatic saturation of unsaturated air: identify the quantity that remains constant  During an adiabatic saturation process applied to unsaturated air (idealized psychrometric analysis), which standard air property stays essentially constant? Choos…

Accepted Answer

Approach In the ideal adiabatic saturation process (no heat/work interactions with surroundings; negligible KE/PE changes), the air-vapor mixture follows an approximately constant wet-bulb temperature (and constant enthalpy) path on the psychrometric chart. Eliminations Relative humidity varies as the state approaches saturation.Dew-point and dry-bulb temperatures change during adiabatic humidification. Final Answer Wet-bulb temperature remains essentially constant.

Question 11

Centrifugal (turbo) refrigerant compressors are most suitable for refrigerants and duty conditions that require which combination of displacement and condensing pressure?

Accepted Answer

Introduction / Context: Centrifugal compressors are dynamic machines that deliver high flow rates at modest pressure ratios per stage. This question asks which refrigerant duty combination best suits centrifugal compressors. Given Data / Assumptions: Focus on single-stage or few-stage centrifugal machines. Typical refrigeration pressure ratios (per stage) are limited compared with positive-displacement machines. Concept / Approach: Dynamic compressors excel at moving large volumes at relatively low to moderate pressure rises per stage. In refrigeration, they pair well with refrigerants and systems where volumetric flow is large and the required condensing pressure (and thus pressure ratio) per stage is not extreme. Positive-displacement compressors are preferred when very high pressure ratios or small displacements are required. Step-by-Step Solution: 1) Identify centrifugal strengths: high flow capacity, moderate pressure ratio per stage.2) Map to refrigeration duties: large displacement (high volumetric flow) with relatively low condensing pressure demands per stage.3) Thus, the suitable combination is “large displacements and low condensing pressures.” Verification / Alternative check: Industrial centrifugal chillers commonly serve large tonnage with moderate pressure ratios, confirming the selection. Why Other Options Are Wrong: Small displacements: Better matched to reciprocating or screw compressors. High condensing pressures: Drive up pressure ratio per stage, favoring PD machines or multistaging. Negligible displacement: Not meaningful for practical refrigeration duty. Common Pitfalls: Assuming centrifugal compressors suit any pressure ratio; ignoring surge and efficiency penalties at off-design conditions. Final Answer: Large displacements and low condensing pressures

Question 12

In heat-exchanger analysis using the NTU–effectiveness method, the number of transfer units (NTU) is defined in terms of overall heat transfer coefficient U, heat transfer area A, and the minimum heat capacity rate Cmin. Which expression is correct?

Accepted Answer

Introduction / Context: The NTU–effectiveness method is a standard framework for analyzing heat exchangers without detailed temperature profiles. The definition of NTU is central to selecting charts/correlations for effectiveness based on flow arrangement and capacity-rate ratio. Given Data / Assumptions: U is the overall heat transfer coefficient (W/m^2·K). A is the effective heat transfer area (m^2). Cmin is the smaller of the two capacity rates (W/K). Concept / Approach: NTU is a nondimensional parameter quantifying the size/conductance of a heat exchanger relative to the minimum capacity rate. By definition, NTU = (U * A) / Cmin. This links exchanger size and thermal conductance to the stream that most limits the temperature change. Step-by-Step Solution: 1) Write the definition: NTU = (overall conductance) / (minimum capacity rate).2) Overall conductance = U * A.3) Therefore, NTU = U * A / Cmin. Verification / Alternative check: Dimensional check: (W/m^2·K * m^2) / (W/K) → (W/K) / (W/K) = dimensionless, as required for NTU. Why Other Options Are Wrong: Using Cmax in the denominator gives a smaller NTU than correct and is not the standard definition. Reciprocal forms invert the dimensionless group and are inconsistent with standard charts. U/(ACmin) or A/(UCmin) are dimensionally incorrect for NTU. Common Pitfalls: Mixing up Cmin and Cmax; forgetting that NTU must be dimensionless; confusing NTU with effectiveness (ε), which depends on NTU and capacity-rate ratio (Cmin/Cmax). Final Answer: NTU = U * A / Cmin

Question 13

Ammonia (R-717) as a refrigerant has several distinct characteristics relevant to safety and materials. Which combination of statements is correct?

Accepted Answer

Introduction / Context: Ammonia (R-717) is a widely used industrial refrigerant prized for efficiency and low environmental impact, but it has important safety and materials-compatibility considerations. This question consolidates commonly taught properties relevant to plant design and operation. Given Data / Assumptions: Comparative statement versus typical fluoro-carbon refrigerants (CFC/HCFC/HFC families). Standard industrial evaporating/condensing conditions. General properties rather than a specific plant case. Concept / Approach: Ammonia is toxic and has a sharp odor; materials of construction must avoid copper and many copper alloys due to chemical attack. In many exam and handbook treatments, ammonia is also noted to require larger compressor displacement per ton than certain halocarbons because of its suction specific volume characteristics, despite high mass-based latent heat; this affects volumetric sizing and selection of machines. Step-by-Step Solution: 1) Safety: Ammonia is toxic and irritates mucous membranes; leak management is critical.2) Materials: Avoid copper and many copper alloys; use appropriate steels and compatible materials.3) Volumetric effect: For comparable duties, ammonia may demand larger volumetric displacement in many comparisons with fluoro-carbons; machine selection must account for this.4) Therefore, all three statements are accepted teaching points in refrigeration engineering. Verification / Alternative check: Design guides and safety resources emphasize ammonia’s toxicity, copper incompatibility, and the importance of volumetric capacity in compressor selection, supporting the combined statement. Why Other Options Are Wrong: Individual statements alone understate the full set of concerns/traits. None of these: Contradicts well-documented ammonia properties. Common Pitfalls: Confusing mass-based efficiency (high latent heat per kg) with volumetric sizing (specific volume and suction density drive displacement). Also, overlooking materials compatibility when retrofitting systems. Final Answer: All of these

Question 14

Psychrometrics — during a humidification process, which property of moist air must increase? In heating, ventilation, and air-conditioning (HVAC) practice, humidification means adding moisture to the air stream. Considering standard adiabatic or ste…

Accepted Answer

Introduction / Context: Humidification is a core psychrometric process used in air-conditioning systems to ensure occupant comfort and protect sensitive materials. It involves introducing additional water vapor into an air stream. Understanding which air properties must increase (and which may increase or decrease) is essential for correct system design, control, and troubleshooting. Given Data / Assumptions: Moist air is treated as a mixture of dry air and water vapor. Humidification implies adding moisture (mass of water vapor) to the air stream. Process may be adiabatic (e.g., evaporative pads) or involve steam injection; fan work and duct heat gains are small. Concept / Approach: The defining variable for moisture content is the humidity ratio (also termed specific humidity), usually denoted w = mass of water vapor per unit mass of dry air. By definition, any humidification process increases the mass of water vapor while the dry air mass remains practically unchanged, so w must rise. Other properties such as dry-bulb temperature, wet-bulb temperature, and relative humidity can move in different directions depending on whether latent heat comes from air (evaporation) or from injected steam, and on heat exchange with surroundings. Step-by-Step Solution: Recognize that humidification adds water vapor → m_v increases while m_da is constant.Compute humidity ratio: w = m_v / m_da. Since m_da is constant and m_v increases, w increases.For evaporative (adiabatic) humidification, dry-bulb temperature typically decreases while wet-bulb remains approximately constant; for steam injection, dry-bulb often increases.Relative humidity may increase or even decrease transiently depending on heat effects; it is not the defining must-increase variable. Verification / Alternative check: On a psychrometric chart, humidification traces move to the right (higher w). Evaporative cooling follows a line of nearly constant wet-bulb temperature, while steam injection trends upward and to the right. In all cases, the x-axis (humidity ratio) increases. Why Other Options Are Wrong: (b) Relative humidity does not necessarily increase; it depends on temperature changes. (c) Dry-bulb may either rise (steam) or fall (evaporation). (d) Wet-bulb does not always decrease; in adiabatic humidification it is approximately constant. (e) Static pressure change is a duct/ fan matter, not a defining psychrometric effect of humidification. Common Pitfalls: Equating relative humidity with moisture content; RH is temperature dependent. Always check humidity ratio when diagnosing humidification behavior. Final Answer: specific humidity (humidity ratio)

Question 15

Modes of heat transfer — precise definition of conduction Which statement best defines heat transfer by conduction within a body or between bodies in contact, in classical thermal engineering terms?

Accepted Answer

Introduction / Context: Engineering heat transfer recognizes three fundamental modes: conduction, convection, and radiation. Correctly distinguishing them is essential for selecting correlations, sizing insulation, and designing heat exchangers. Given Data / Assumptions: No phase change in the described mode. Continuum assumption applies; microscopic details are abstracted. Steady or unsteady processes both permitted. Concept / Approach: Conduction is microscopic energy transfer due to temperature gradients, arising from molecular agitation and electron transport. It does not require net bulk motion of the material. In contrast, convection is energy transfer augmented by macroscopic fluid motion. Radiation is electromagnetic energy exchange that requires no medium and travels in straight lines in free space. Step-by-Step Solution: Identify key phrase “without bulk motion” → characteristic of conduction.Option (a) matches: particle-to-particle energy migration via collisions/electrons across a temperature gradient.Option (b) describes convection (forced or natural).Option (c) describes thermal radiation traveling through space or transparent media. Verification / Alternative check: Fourier’s law q_x = -k * dT/dx formalizes conduction: heat flux proportional to temperature gradient and thermal conductivity, independent of bulk flow. Why Other Options Are Wrong: (b) Bulk fluid motion is convection, not conduction. (c) Ray-like transfer without medium interaction is radiation. (d) Incorrect as (a) is correct. (e) Conduction occurs in solids, liquids, and gases; not “only in fluids at rest.” Common Pitfalls: Confusing “conduction in fluids” with “convection”; both occur simultaneously in fluids, but convection requires bulk motion. Final Answer: Transfer from one particle to another without bulk motion of the particles

Question 16

Industrial safety — properties of sulphur dioxide (SO₂) Evaluate the statement: “Sulphur dioxide is extremely toxic and non-flammable.” Choose the correct option.

Accepted Answer

Introduction / Context: Sulphur dioxide (SO₂) is encountered in flue gases, acid manufacturing, and some refrigeration histories. Knowing its hazard profile is essential for ventilation, leak response, and personal protective equipment selection. Given Data / Assumptions: SO₂ at typical industrial temperatures and pressures. Focus on toxicity and flammability only. Concept / Approach: SO₂ is a corrosive, highly irritating gas with a pungent odor. It is toxic by inhalation and contact with moist tissues (forming sulphurous acid). However, it does not support combustion and is generally classified as non-flammable. Thus the combined statement is correct. Step-by-Step Solution: Assess toxicity: SO₂ causes respiratory distress even at low ppm; at higher concentrations it is dangerous.Assess flammability: SO₂ lacks a flammable range in air under normal conditions and does not sustain flames.Conclude the statement “extremely toxic and non-flammable” holds. Verification / Alternative check: Safety data sheets (SDS) list SO₂ hazard statements for acute toxicity and corrosion but do not assign flammability ratings typical of hydrocarbons. Why Other Options Are Wrong: (b) contradicts the known properties. (c), (d), (e) add conditions that do not change the toxicity/flammability assessment in normal practice. Common Pitfalls: Confusing SO₂ with SO₃ or with combustible sulphur vapors; only elemental sulphur dust can pose combustion risks, not SO₂ gas. Final Answer: True

Question 17

Reversed Carnot devices — relating COP of heat pump and refrigerator A heat pump operating on a reversed Carnot cycle has a COP (heat pump) of 5. If the same device now operates as a refrigerator with a work input of 1 kW, what is the refrigerating …

Accepted Answer

Introduction / Context: For any reversed cycle, the coefficient of performance (COP) takes different forms depending on whether the device is viewed as a heat pump (delivering heat to a warm space) or a refrigerator (removing heat from a cold space). The two COPs are simply related for the same temperature limits. Given Data / Assumptions: Reversed Carnot cycle with the same temperature limits. COP_heat pump = 5. As a refrigerator, work input W = 1 kW. Concept / Approach: The fundamental relation is COP_HP = COP_R + 1. This derives from energy balance: Q_H = Q_L + W. Dividing by W gives Q_H/W = Q_L/W + 1, or COP_HP = COP_R + 1. Once COP_R is known, the cooling capacity follows as Q_L = COP_R * W. Step-by-Step Solution: Given COP_HP = 5 → COP_R = COP_HP - 1 = 4.Work input as refrigerator W = 1 kW.Refrigerating effect Q_L = COP_R * W = 4 * 1 kW = 4 kW. Verification / Alternative check: Using Carnot expressions: COP_R = T_L / (T_H - T_L) and COP_HP = T_H / (T_H - T_L). Their difference is 1, confirming the rule used. Why Other Options Are Wrong: (a), (b), (c) understate capacity; (e) confuses COP_HP with cooling capacity; COP of 5 is not power directly. Common Pitfalls: Forgetting the relation COP_HP = COP_R + 1 and mixing up which heat flow (Q_H or Q_L) a COP refers to. Final Answer: 4 kW

Question 18

Radiation fundamentals — emissivity of an ideal black body What is the emissivity value assigned to a perfect black body in thermal radiation analysis?

Accepted Answer

Introduction / Context: Emissivity quantifies how effectively a surface emits thermal radiation compared to an ideal emitter at the same temperature. Black bodies define the theoretical upper limit for emission and absorption, forming the basis for many radiative heat transfer models. Given Data / Assumptions: Idealized black body concept per classical radiation theory. Diffuse-gray simplifications are standard unless wavelength dependence is discussed separately. Concept / Approach: A black body is defined as one that absorbs all incident radiation (absorptivity = 1) and, by Kirchhoff’s law, has emissivity equal to absorptivity for surfaces in thermal equilibrium. Therefore, a perfect black body has emissivity epsilon = 1, the maximum possible value. Step-by-Step Solution: Definition: black body → absorptivity = 1.Kirchhoff’s law: emissivity = absorptivity at the same wavelength and temperature.Hence, epsilon_black = 1. Verification / Alternative check: Stefan–Boltzmann law for a real surface is q = epsilon * sigma * T^4; for a black body, epsilon = 1, giving the maximum radiant exitance. Why Other Options Are Wrong: (a), (b), (c) are sub-unity values for gray bodies, not black bodies. (e) While real surfaces have spectral emissivity, the ideal black body has emissivity 1 at all wavelengths. Common Pitfalls: Confusing “black paint” with “black body.” Color and visual blackness are not reliable indicators of emissivity across thermal infrared wavelengths. Final Answer: 1

Question 19

Gas properties — temperature effect on thermal conductivity of air Evaluate the statement: “Thermal conductivity of air increases with rise in temperature.”

Accepted Answer

Introduction / Context: Thermal conductivity, k, indicates a substance’s ability to conduct heat by molecular transport. In gases, k is strongly tied to molecular motion and collision dynamics, which change with temperature. Given Data / Assumptions: Ideal gas behavior approximation for air over common engineering ranges. No phase change; composition approximately constant. Concept / Approach: For gases, as temperature rises, molecular speeds increase and mean free path changes such that energy is transferred more rapidly between molecules. Kinetic theory therefore predicts an increase in thermal conductivity with temperature for most gases at low to moderate pressures, including air. Step-by-Step Solution: Consider kinetic theory: k ∝ c_p * density * molecular speed * mean free path factors.With temperature up, average molecular speed rises, increasing k.Empirical property tables for air confirm k increases from near 0.024 W/m·K at ~300 K to higher values at elevated temperatures. Verification / Alternative check: Heat transfer handbooks show monotonic increase of air’s thermal conductivity with T at atmospheric pressure across typical engineering temperatures. Why Other Options Are Wrong: (b) contradicts data. (c) and (d) impose arbitrary thresholds that are not fundamental. (e) While property charts often tabulate at 1 atm, the trend with temperature is not restricted to “constant pressure only.” Common Pitfalls: Confusing gas behavior with liquids; many liquids show decreasing k with temperature, opposite to gases like air. Final Answer: Correct

Question 20

Heat exchangers — minimum surface area with phase change In a heat exchanger where one of the fluids is undergoing evaporation or condensation (approximately constant temperature on that side), which flow arrangement generally requires the least hea…

Accepted Answer

Introduction / Context: Exchanger area depends on the required heat duty, overall heat transfer coefficient, and the logarithmic mean temperature difference (LMTD). When one stream changes phase, its temperature is nearly constant, which changes LMTD behavior across flow arrangements. Given Data / Assumptions: One side evaporating or condensing, so its bulk temperature stays essentially constant. Uniform overall heat transfer coefficient for comparison. Same inlet/outlet temperatures targeted. Concept / Approach: Counter flow maximizes the temperature driving force at each location, producing the highest LMTD among the common arrangements (for the same terminal temperatures). A higher LMTD means, for a fixed duty and U, the necessary area A = Q/(U*LMTD) is minimized. Step-by-Step Solution: For parallel flow, hot and cold streams enter from the same end; temperature driving force decays quickly, lowering LMTD.For counter flow, streams move oppositely; the local temperature difference stays higher, increasing LMTD.With a phase-changing side, counter flow further avoids “pinch” at the cold end, giving the largest LMTD → smallest area. Verification / Alternative check: Worked examples with typical terminal temperatures show counter flow LMTD exceeds parallel/cross, hence smaller area. Why Other Options Are Wrong: (a) Parallel has lower LMTD → larger area. (c) Cross flow lies between parallel and counter depending on mixing; seldom beats pure counter flow. (d) Not true because LMTD differs. (e) Not a conventional recuperative exchanger arrangement. Common Pitfalls: Assuming “phase change means infinite U so area is minimal anyway.” U may be high, but LMTD still governs area for a specified Q. Final Answer: Counter flow

Heat Transfer, Refrigeration and Air Conditioning Questions