Question 1

Radiation exchange between two infinite parallel plates For two infinite parallel planes with emissivities ε1 and ε2, what is the correct interchange factor (radiation network factor) multiplying σA(T1^4 − T2^4)?

Accepted Answer

Introduction / Context:Radiative heat transfer between large, parallel, diffuse–gray plates is a textbook case for using the radiation network method. The net exchange depends on a combined “interchange factor” that accounts for both emissivities and view factors (unity here).Given Data / Assumptions:Two very large (infinite) parallel plates facing each other.Diffuse–gray surfaces with emissivities ε1 and ε2.View factors F12 = F21 = 1; surroundings do not participate.Concept / Approach:For two infinite parallel plates, the net radiative heat flow is Q = σ A (T1^4 − T2^4) / R_rad, where R_rad is the radiation resistance. The equivalent interchange factor, often denoted as 1/R_rad, is 1 / ( (1/ε1) + (1/ε2) − 1 ). This comes from series addition of surface resistances 1/ε − 1 and accounts for multiple reflections between the plates.Step-by-Step Solution:Write network: R_total = (1 − ε1)/(ε1 A) + 1/(A) + (1 − ε2)/(ε2 A).For infinite parallel plates, geometric resistance (1/A) simplifies to 1/A since F12 = 1, leading to the classic denominator form.Invert R_total/A to get the interchange factor: 1 / ( (1/ε1) + (1/ε2) − 1 ).Verification / Alternative check:Check limiting cases: if ε1 = ε2 = 1 (black bodies), factor becomes 1, so Q = σ A (T1^4 − T2^4), as expected.Why Other Options Are Wrong:(b) is an algebraic convenience used in some reflectivity relations but not the correct network factor. (c) and (d) ignore multiple reflections. (e) lacks the proper reciprocal structure and does not reduce to unity for black bodies.Common Pitfalls:Forgetting that surface resistances add in series, which introduces reciprocals and the “−1” term.Final Answer:1 / ( (1/ε1) + (1/ε2) − 1 )

Question 2

Electrolux (ammonia–water–hydrogen) refrigerator Why is hydrogen circulated in the Electrolux absorption refrigerator?

Accepted Answer

Introduction / Context:The Electrolux absorption refrigerator uses ammonia as the refrigerant, water as the absorbent, and hydrogen as an inert gas. Understanding hydrogen’s role clarifies how cooling is achieved without a mechanical compressor.Given Data / Assumptions:Evaporator contains liquid ammonia.Hydrogen is essentially insoluble in ammonia and water and acts as an inert gas.Total pressure in the low-pressure side is nearly constant.Concept / Approach:Hydrogen reduces the partial pressure of ammonia in the evaporator while keeping the total pressure nearly constant. Lower ammonia partial pressure reduces its saturation temperature, allowing it to evaporate rapidly at a very low temperature, thus absorbing heat from the refrigerated space.Step-by-Step Solution:Total pressure = p_total = p_NH3 + p_H2 (approximately).Introduce H2 → p_NH3 drops for the same p_total.Lower p_NH3 → lower saturation temperature → faster evaporation → more cooling.Verification / Alternative check:Dalton’s law in gas mixtures explains the partition of total pressure; reducing ammonia’s partial pressure directly lowers its boiling point at the evaporator.Why Other Options Are Wrong:(a) Equalising pressures is not the purpose; the system already operates near uniform low-side pressure. (b) Opposite to the actual effect. (d) No moving parts in the low-side that need hydrogen lubrication. (e) Hydrogen does not raise condenser pressure.Common Pitfalls:Confusing total system pressure with ammonia’s partial pressure; it is the latter that sets the evaporation temperature.Final Answer:to increase the rate of ammonia evaporation in the evaporator

Question 3

Refrigeration fundamentals — identify the correct statements. Which of the following statements about absorption/compression systems and sub-cooling is/are correct?

Accepted Answer

Introduction / Context:Refrigeration systems can be broadly classified into vapour compression and vapour absorption types. Sub-cooling in the condenser is another practical aspect that affects system capacity and performance. This question checks basic conceptual understanding.Given Data / Assumptions:Idealized descriptions are acceptable (e.g., adiabatic/isentropic compression as a model).Condenser sub-cooling refers to cooling the liquid below its saturation temperature at the condensing pressure.Concept / Approach:Absorption systems replace mechanical compression by thermal “compression” using an absorber, pump, generator, and rectifier. Sub-cooling increases the net refrigerating effect by raising the enthalpy difference across the expansion device. In vapour compression, the compressor indeed draws vapour and compresses it (ideally adiabatically/isentropically) to the condensing pressure.Step-by-Step Solution:(a) True: absorption cycle avoids mechanical compression of the refrigerant itself.(b) True: better condenser heat transfer (e.g., more water flow or lower coolant temperature) raises sub-cooling.(c) True: fundamental description of the compression process.Hence, all statements are correct.Verification / Alternative check:Cycle diagrams show no compressor in absorption systems; instead, a solution pump is used. p-h charts demonstrate the benefit of sub-cooling.Why Other Options Are Wrong:Choosing anything other than “all of the above” wrongly denies one or more correct fundamentals.Common Pitfalls:Assuming sub-cooling is fixed; it depends on condenser performance and ambient conditions.Final Answer:all of the above

Question 4

Aircraft and rocket refrigeration Which air-cycle cooling system is used for very high-speed applications such as supersonic aircraft and rockets?

Accepted Answer

Introduction / Context:Air-cycle (reverse Brayton) systems in aviation come in multiple configurations. As flight speeds and altitudes increase, higher pressure ratios and more effective heat rejection are needed to achieve the required cabin or equipment cooling.Given Data / Assumptions:Supersonic aircraft and rocket environments demand high effectiveness and lightweight systems.Bleed-air temperatures and ram effects can be very high.Concept / Approach:The regenerative air cooling system incorporates heat exchangers and reheat/recool steps to extract more work in the turbine and thereby increase the temperature drop of the air supplied to the cabin or payload bay. This configuration is suited to very high-speed regimes where simple or even boot-strap systems may not provide adequate temperature depression.Step-by-Step Solution:Compare variants: simple Supersonic/rocket → very high stagnation temperatures → need additional stages/heat exchange → regenerative best choice.Verification / Alternative check:Aerospace texts map system selection to flight envelope; regenerative schemes are cited for the highest speed ranges.Why Other Options Are Wrong:(a) and (b) may suffice for subsonic/transonic transports, not extreme regimes. (c) is a descriptive phrase, not a standard system. (e) evaporative cooling is not used in these aircraft ECS due to icing and water management issues.Common Pitfalls:Assuming the boot-strap system is universal; it dominates transport jets but not necessarily supersonic/rocket applications.Final Answer:regenerative air cooling system

Question 5

Psychrometrics — ratio based on the same volume at the same state What do we call the ratio of the actual mass of water vapour in a given volume of moist air to the mass of water vapour in the same volume when saturated at the same temperature and p…

Accepted Answer

Introduction / Context:Several moisture measures exist for moist air, including humidity ratio, relative humidity, absolute humidity, and degree of saturation. Selecting the correct one requires careful attention to whether the definition uses mass-per-mass, mass-per-volume, or a ratio to saturation at the same state.Given Data / Assumptions:Comparison is for the same volume of moist air.Reference state is saturation at the same temperature and pressure.Ideal gas behavior is assumed for the common definitions.Concept / Approach:Relative humidity is the ratio of actual water vapour amount to the maximum possible at the same temperature and pressure. It can be expressed via partial pressures (p_v/p_vs) or via mass in the same volume at the same state, which is equivalent since mass is proportional to partial pressure for ideal gases.Step-by-Step Solution:Identify definition: ratio to saturation for the same T and p.Map to standard terms: that is relative humidity.Confirm against alternatives: humidity ratio is per kg of dry air; absolute humidity is mass per unit volume; degree of saturation differs from RH.Verification / Alternative check:Psychrometric charts show RH lines as fractions/percentages of saturation at the same T and p.Why Other Options Are Wrong:(a) humidity ratio is m_vapour per kg dry air. (c) absolute humidity is m_vapour per m^3 of moist air. (d) degree of saturation is related but not identical to RH.Common Pitfalls:Confusing relative humidity with humidity ratio because both use “humidity” in the name.Final Answer:relative humidity

Question 6

Why is a thermos flask evacuated? Choose the correct statement about the role of vacuum between the walls of a thermos (vacuum) flask.

Accepted Answer

Introduction / Context:Thermos flasks reduce heat transfer so that contents remain hot or cold. The evacuated space between the double walls is the key design feature limiting two of the three heat-transfer modes.Given Data / Assumptions:Two closely spaced walls with the interspace evacuated.Reflective coatings reduce radiation (though not mentioned explicitly, typical in flasks).Concept / Approach:Conduction and convection both require matter. By evacuating the interspace, the flask essentially eliminates gas conduction and internal natural convection. Radiation remains the principal mode, which is mitigated with low-emissivity surfaces.Step-by-Step Solution:Conduction: k_medium ~ 0 in vacuum → negligible conduction.Convection: no fluid in the gap → no convective currents.Radiation: minimized via reflective surfaces.Net effect: vacuum is a poor (bad) conductor, reducing heat leak.Verification / Alternative check:Heat-leak calculations for vacuum insulation show orders-of-magnitude reduction versus air-filled gaps.Why Other Options Are Wrong:(a) Opposite of true; vacuum conducts neither well nor at all. (c) With no fluid, convection cannot occur. (d) Radiation is not increased by vacuum; it is unchanged by the medium, but surfaces are designed to reduce it. (e) Vacuum does not increase wall conductivity.Common Pitfalls:Assuming radiation is eliminated by vacuum; only conduction and convection are.Final Answer:Vacuum is a bad conductor of heat

Question 7

Critical temperature comparison Evaluate the statement: “Among common legacy refrigerants, R-11 has one of the highest critical temperatures.”

Accepted Answer

Introduction / Context:The critical temperature of a refrigerant strongly influences condenser design and achievable cycle performance in warm climates. Higher critical temperatures generally make condensation easier at high ambient conditions.Given Data / Assumptions:Comparison among common legacy single-component refrigerants like R-11, R-12, R-22, etc.No detailed numerical values are required; qualitative ranking is sufficient.Concept / Approach:R-11 (trichlorofluoromethane) has a notably high critical temperature compared with many other legacy refrigerants (e.g., R-12, R-22). This historically favored its use in large centrifugal chillers for comfort cooling, where high condensing temperatures can occur.Step-by-Step Solution:Recognize R-11’s property set: low pressures, high molecular weight, and high critical temperature.Relate to application: easier condensation in warm climates makes chiller design more forgiving.Therefore, the qualitative statement is correct.Verification / Alternative check:Property tables show R-11 critical temperature significantly higher than R-12 and R-22, aligning with historical use cases.Why Other Options Are Wrong:Options adding conditions (c), (d), (e) are irrelevant to the intrinsic property. Saying “Incorrect” ignores well-documented data.Common Pitfalls:Conflating critical temperature with normal boiling point; both influence design, but the critical value is the upper limit for condensation.Final Answer:Correct

Question 8

Where reciprocating refrigerant compressors excel Select the operating range for which reciprocating refrigerant compressors are most suitable.

Accepted Answer

Introduction / Context:Compressor selection depends on flow rate and pressure ratio. Reciprocating (piston) compressors are positive displacement machines particularly suited to certain combinations of capacity and discharge pressure in refrigeration service.Given Data / Assumptions:Comparing reciprocating machines with centrifugal/axial or screw compressors.Refrigeration applications requiring moderate to high condensing pressures.Concept / Approach:Reciprocating compressors handle small to moderate volumetric flows with high pressure ratios efficiently. They develop high discharge pressures at acceptable isentropic efficiencies in compact packages, making them ideal for small systems, low-temperature plants, and high head applications.Step-by-Step Solution:Identify capacity range: small displacement volumes per unit time.Identify pressure range: capable of high condensing pressures and large pressure ratios.Thus, the best fit is “small displacements and high condensing pressures.”Verification / Alternative check:Industry practice uses reciprocating units for small cold rooms, freezers, and laboratory refrigeration, while centrifugal/screw units dominate high-flow, lower pressure-ratio HVAC chillers.Why Other Options Are Wrong:(a) Low condensing pressures do not require the strengths of reciprocating compressors. (b) Large displacements are better handled by rotary or dynamic machines. (d) Same issue as (a). (e) Very large flows with low ratios are the domain of centrifugal/axial machines.Common Pitfalls:Assuming “bigger is always better”; machine type must match both flow and head requirements.Final Answer:small displacements and high condensing pressures

Question 9

Refrigeration fundamentals: The relative coefficient of performance (relative C.O.P.) is defined as the ratio of the actual coefficient of performance to the theoretical (Carnot) coefficient of performance for the same temperature limits.

Accepted Answer

Introduction / Context:The “relative coefficient of performance” is a handy performance index in refrigeration that benchmarks a real plant against the best possible ideal cycle working between the same temperature limits. It tells how close the real system comes to the Carnot refrigerator's performance.Given Data / Assumptions:Actual C.O.P. refers to Q_L / W for the real system.Theoretical (Carnot) C.O.P. is the C.O.P. of an ideal reversible refrigerator operating between the same evaporator and condenser temperatures.Temperatures are steady and heat leaks are negligible.Concept / Approach:Performance comparison must be dimensionless and normalized to the best theoretical limit. Dividing the actual C.O.P. by the Carnot C.O.P. yields a number between 0 and 1 (typically), directly indicating effectiveness relative to the ideal.Step-by-Step Solution:Write the definition: Relative C.O.P. = (Actual C.O.P.) / (Carnot C.O.P.).Recall Carnot refrigerator: C.O.P._Carnot = T_L / (T_H − T_L) when temperatures are in kelvin.Compute Actual C.O.P. from test data: C.O.P._actual = Q_L / W_input.Form the ratio to obtain the relative C.O.P.Verification / Alternative check:If a plant operates ideally, Actual C.O.P. equals Carnot C.O.P. and the relative C.O.P. becomes 1. Any irreversibilities will make it less than 1, which matches intuition.Why Other Options Are Wrong:The reciprocal (Theoretical / Actual) can exceed 1 and does not measure closeness to the ideal correctly.The product Actual × Theoretical has incorrect units and no physical meaning.“None of these” is wrong because a standard definition exists.The difference divided by theoretical gives “percentage shortfall,” not relative C.O.P.Common Pitfalls:Using Celsius instead of kelvin in Carnot formulas; mixing up ideal heat pump C.O.P. with refrigerator C.O.P.; and comparing to the wrong theoretical temperature limits.Final Answer:Relative C.O.P. = Actual C.O.P. / Theoretical (Carnot) C.O.P.

Question 10

Pressure–enthalpy (p–h) chart interpretation: The region between the saturated liquid line and the saturated vapour line represents the two-phase (wet) mixture, not the subcooled liquid region.

Accepted Answer

Introduction / Context:Technicians and students regularly read pressure–enthalpy (p–h) charts to analyze vapour-compression cycles. Correct identification of chart regions is essential for locating evaporator, compressor, condenser, and expansion valve states.Given Data / Assumptions:A p–h chart shows saturated liquid and saturated vapour curves enclosing the two-phase dome.Left of the saturated liquid line is the subcooled (compressed) liquid region.Right of the saturated vapour line is the superheated vapour region.Concept / Approach:Inside the saturated dome (between the two saturation lines) the refrigerant exists as a mixture of liquid and vapour at equilibrium. Quality (dryness fraction) defines the phase split. Subcooled liquid states occur at temperatures below saturation for the same pressure and therefore lie to the left of the saturated liquid line.Step-by-Step Solution:Identify the saturation dome bounded by the saturated liquid and saturated vapour curves.Any state point between these curves is a wet mixture with 0 Subcooled liquid points must lie left of the saturated liquid curve (lower h at the same p).Therefore, the statement “the space between the saturated lines is subcooled liquid” is incorrect.Verification / Alternative check:Trace a condenser path: from superheated vapour, through desuperheating to saturated vapour, then across the dome to saturated liquid, and finally into subcooling outside the dome on the left—this path visually confirms the layout.Why Other Options Are Wrong:The claim is false universally (for both water–steam and refrigerants), independent of absolute pressure and not tied to throttling; throttling typically moves the state into the dome, not “subcooled” within it.Common Pitfalls:Confusing subcooling with “more liquid present.” Subcooling is a single-phase liquid state at T below saturation, not a two-phase region.Final Answer:False

Question 11

Air refrigeration cycles: An air refrigerator (as used in aircraft) operates on which idealized cycle?

Accepted Answer

Introduction / Context:Aircraft air-cycle machines and simple air refrigerators use the working fluid (air) directly. Their thermodynamic model is the Bell–Coleman or reversed Brayton cycle, not the vapour-compression cycle used with refrigerants like ammonia or R-134a.Given Data / Assumptions:Working fluid is air (treated as an ideal gas).Processes are ideally isentropic compression/expansion with heat exchange at constant pressure.Engine bleed air or ram air often serves as the heat source/sink.Concept / Approach:The Bell–Coleman cycle consists of two isentropic processes (compression and expansion in a turbine) and two constant-pressure heat transfer processes (cooling after compression and heating after expansion), producing refrigeration by expansion work.Step-by-Step Solution:List the ideal steps: 1–2 isentropic compression; 2–3 constant-pressure heat rejection; 3–4 isentropic expansion; 4–1 constant-pressure heat absorption (refrigeration).Cooling effect occurs during 4–1 as cold air absorbs heat from the space.This directly matches the reversed Brayton layout.Hence, the correct cycle is Bell–Coleman.Verification / Alternative check:Temperature–entropy diagrams and p–h surrogates for air show the characteristic Brayton rectangles/curves, not the dome typical of vapour systems.Why Other Options Are Wrong:Reversed Carnot is the theoretical upper limit, but not the practical air-cycle model.“Both” is incorrect because the standard air refrigerator is specifically the reversed Brayton.Reversed Rankine applies to vapour-compression, not air-cycle machines.Common Pitfalls:Confusing “reversed” with “Carnot” generically; always map processes to the actual hardware (compressor, cooler, turbine, heat exchanger).Final Answer:Bell–Coleman (reversed Brayton) cycle

Question 12

Thermostatic expansion valves (TEVs): Most factory settings target a superheat of approximately how many degrees Celsius at the evaporator outlet?

Accepted Answer

Introduction / Context:The thermostatic expansion valve (TEV or TXV) regulates refrigerant flow to maintain a set superheat at the evaporator outlet, preventing liquid slugging of the compressor while ensuring good evaporator utilization.Given Data / Assumptions:Superheat is defined as T_vapour_out − T_saturation at evaporating pressure.Factory or commissioning targets are sought for comfort cooling systems.Line losses and bulb placement are correct.Concept / Approach:A modest superheat ensures that all liquid has boiled off before the suction line, while not starving the coil. Industry practice typically targets about 5 K (≈ 5°C) at the evaporator outlet, with acceptable ranges often 4–8 K depending on application.Step-by-Step Solution:Identify the control objective: maintain a small, stable superheat.Select a commonly recommended setpoint: around 5°C for many A/C systems.Adjust as needed for long lines, low-temperature applications, or special evaporators (ranges may move toward 7–10°C).Confirm correct bulb strapping and insulation to avoid hunting.Verification / Alternative check:Commissioning guidelines from OEMs and service manuals typically cite superheat in the 4–8°C band, with 5°C as a frequent nominal value for initial setting.Why Other Options Are Wrong:10°C or 15–20°C superheat is sometimes used for special cases, but “most” valves are not set that high by default.2°C is too low and risks liquid carryover.Common Pitfalls:Measuring superheat at the wrong location; not correcting for pressure drops; or chasing charge issues with TXV adjustments.Final Answer:5°C

Question 13

Carbon dioxide property check: The commonly cited temperature −78.3°C for CO₂ at 1 atm corresponds to which “boiling” or phase-change point?

Accepted Answer

Introduction / Context:Unlike many refrigerants, carbon dioxide (R-744) has a triple point above atmospheric pressure. Therefore, at 1 atm CO₂ does not “boil”; it sublimates, transitioning directly between solid and gas.Given Data / Assumptions:Pressure is 1 atm.Triple point of CO₂ is at about 5.18 bar; critical point is around 31.1°C and 73.8 bar.Given temperature is −78.3°C (often quoted as −78.5°C).Concept / Approach:Below the triple point pressure, liquid CO₂ cannot exist. Thus, at 1 atm the phase change occurs directly between solid and vapour at the sublimation temperature, approximately −78.3°C.Step-by-Step Solution:Recognize that “boiling point” at 1 atm is not applicable to CO₂.Identify −78.3°C as the sublimation temperature at 1 atm.Relate to dry ice behavior: it “smokes” due to sublimation, not melting then boiling at 1 atm.Conclude that the correct interpretation is the sublimation point.Verification / Alternative check:Phase diagrams of CO₂ show 1 atm lying left of the triple point, confirming no liquid phase is possible and the change is solid–gas.Why Other Options Are Wrong:Other temperatures listed do not match standard constants for CO₂ at 1 atm; −63.3°C is closer to oxygen’s boiling point, not CO₂.Common Pitfalls:Using the term “boiling point” for substances that sublimate at 1 atm; always check the triple point.Final Answer:−78.3°C (sublimation point at 1 atm)

Question 14

Desirable refrigerant properties for vapour-compression systems include which of the following?

Accepted Answer

Introduction / Context:Selecting a refrigerant involves balancing thermodynamic, environmental, safety, and materials considerations. Thermodynamically favorable properties help reduce compressor work, heat-exchanger size, and operating pressures.Given Data / Assumptions:Single-stage vapour-compression cycle context.Ambient condenser cooling near 25–45°C.Safe and compatible materials assumed separately from thermodynamics here.Concept / Approach:Low boiling point permits low evaporator temperatures without excessive vacuum. A high critical temperature allows condensation at practical temperatures and pressures. High latent heat reduces required mass flow for a given capacity, improving efficiency and component sizes.Step-by-Step Solution:Relate low boiling point to attainable coil temperatures and pressure levels.Relate high critical temperature to condenser approach temperature and head pressure.Relate high latent heat to smaller compressors and heat exchangers for the same tonnage.Combine these to see that each property is beneficial; hence “All of these.”Verification / Alternative check:Compare common refrigerants: ammonia has high latent heat; R-134a has favorable pressures and critical temperature; CO₂ has very low critical temperature making transcritical operation necessary in warm climates—illustrating the importance of T_critical.Why Other Options Are Wrong:Each single item is indeed desirable; choosing any one misses the holistic requirement. “None” contradicts standard design guidance.Common Pitfalls:Ignoring non-thermodynamic factors (toxicity, flammability, GWP, ODP) which also affect refrigerant choice.Final Answer:All of these

Question 15

Psychrometric processes: On a standard chart, adiabatic humidification (evaporative cooling of unsaturated air) is represented by which type of line?

Accepted Answer

Introduction / Context:Understanding how processes plot on the psychrometric chart is crucial for sizing evaporative coolers, spray washers, and air-handling equipment. Adiabatic humidification adds moisture while air gives up sensible heat, moving toward saturation.Given Data / Assumptions:Atmospheric pressure is near 1 atm.Water temperature approximates the entering air wet-bulb temperature.Heat exchange with surroundings is negligible (adiabatic).Concept / Approach:For adiabatic humidification, the air enthalpy remains nearly constant while humidity ratio increases and dry-bulb temperature decreases. On a psychrometric chart, constant enthalpy and constant wet-bulb lines are nearly straight slanted lines; the process thus follows an inclined path up and to the left toward the saturation curve.Step-by-Step Solution:Start at the initial state point (DBT high, moderate humidity ratio).As water evaporates, latent heat is supplied by air's sensible heat, so DBT drops while humidity ratio rises.Because the process is adiabatic, enthalpy stays approximately constant.Plotting on the chart gives an inclined line toward the saturation curve.Verification / Alternative check:Compare with measured psychrometer data: states with equal wet-bulb temperature lie nearly on straight lines; evaporative cooler performance follows those lines.Why Other Options Are Wrong:Horizontal implies constant humidity ratio (no moisture addition), which contradicts humidification.Vertical implies constant dry-bulb (pure steam injection at matched sensible heat), not adiabatic evaporative cooling.Curved constant-dew-point is not the evaporative path.Common Pitfalls:Confusing steam injection (often increases DBT) with evaporative cooling (decreases DBT); and misreading wet-bulb vs dew point lines.Final Answer:Inclined line (approximately straight along near-constant wet-bulb/enthalpy)

Question 16

Definitions: The ratio of the heat extracted in a refrigerator to the work input to the refrigerant is called the __________.

Accepted Answer

Introduction / Context:Performance metrics distinguish refrigerators, freezers, and heat pumps. The most common is the coefficient of performance (C.O.P.), which appropriately measures benefit per unit input for devices that move heat rather than convert it to work.Given Data / Assumptions:Refrigerator objective: remove heat Q_L from a low-temperature space.Work input to the cycle is W.Steady-state operation, negligible kinetic/potential energy changes.Concept / Approach:Define C.O.P._ref = Q_L / W. This differs from a heat pump's C.O.P._hp = Q_H / W, which counts heat delivered to the warm space. The relative C.O.P. normalizes actual C.O.P. by the Carnot limit, but the basic definition asked here is simply Q_L divided by W.Step-by-Step Solution:Write the energy balance of a cycle: W = Q_H − Q_L.Define refrigerator performance: C.O.P._ref = Q_L / W.Recognize that larger C.O.P. indicates better performance for a given cooling capacity.Do not confuse with heat pump performance where the numerator is Q_H.Verification / Alternative check:For a Carnot refrigerator, C.O.P._Carnot = T_L / (T_H − T_L), which when multiplied by W yields Q_L, reinforcing the definition.Why Other Options Are Wrong:Heat pump C.O.P. uses Q_H, not Q_L.Relative C.O.P. is a ratio to the Carnot value, not the base definition.“Refrigerating efficiency” is an informal term and ambiguous here.Common Pitfalls:Swapping numerators between refrigerator and heat pump; using Celsius instead of kelvin in Carnot expressions.Final Answer:coefficient of performance of refrigeration (C.O.P.)

Question 17

Dehumidification trend: In a cooling and dehumidification process across a coil, how does the relative humidity of the air typically change?

Accepted Answer

Introduction / Context:Air-handling coils often cool air below its dew point so that moisture condenses on the fins. Understanding how relative humidity (RH) changes through this process is essential for comfort control and latent load calculations.Given Data / Assumptions:Coil surface temperature is below the air dew point.Air is cooled sensibly and then along the saturation curve as condensation begins.Process is steady with negligible heat leaks.Concept / Approach:As the air is cooled at nearly constant humidity ratio until it reaches dew point, RH rises. Once dew point is met, further cooling removes moisture; the air state moves along a path close to the saturation curve, where RH approaches 100%. Therefore, dehumidification commonly results in an overall increase in RH, despite a decrease in humidity ratio.Step-by-Step Solution:Initial segment: DBT drops while humidity ratio is nearly constant ⇒ RH increases.At dew point: air becomes saturated (RH = 100%).Condensing segment: both DBT and humidity ratio decrease; the exit air typically has RH close to 100% unless reheated.Thus, across the coil without reheat, RH increases.Verification / Alternative check:Plot on a psychrometric chart: the process line trends toward and often along the saturation curve, confirming higher RH at the outlet.Why Other Options Are Wrong:“Remains constant” applies only to special controlled reheat or desiccant cases.“Decreases” contradicts the basic cooling-to-saturation behavior.“Indeterminate” is incorrect with stated conditions.Common Pitfalls:Confusing relative humidity (a ratio) with humidity ratio; RH can increase even while total moisture per kg dry air decreases.Final Answer:Increases

Question 18

Evaporator function: After the expansion valve, the evaporator converts the low-pressure refrigerant into which state as it absorbs heat from the space?

Accepted Answer

Introduction / Context:The evaporator is the cooling coil of a vapour-compression system. It is where the refrigerant absorbs heat from the conditioned space and changes phase from liquid to vapour at low pressure.Given Data / Assumptions:Refrigerant exits the expansion device as a low-pressure, low-temperature mixture.Heat is absorbed from air or a secondary fluid flowing over/through the coil.Proper superheat is maintained at the outlet to protect the compressor.Concept / Approach:Boiling at nearly constant pressure in the evaporator removes heat from the medium. By the outlet, the design goal is dry saturated or slightly superheated vapour to ensure no liquid enters the compressor.Step-by-Step Solution:After throttling: state is low-pressure two-phase mixture entering the evaporator.During heat absorption: liquid fraction evaporates at nearly constant pressure and temperature.At the outlet: aim for complete evaporation plus a small superheat (e.g., 5°C).Thus, the evaporator delivers low-pressure vapour refrigerant to the compressor suction.Verification / Alternative check:Pressure–enthalpy plotting shows a horizontal (constant-pressure) path across the saturated dome from mixture to saturated vapour, then a slight superheat rise.Why Other Options Are Wrong:High-pressure liquid does not occur in the evaporator; it is a low-pressure component.While the inlet is a mixture, the question asks for the state produced by the evaporator (outlet), which is vapour.High-pressure superheated vapour is a compressor discharge condition, not evaporator outlet.Common Pitfalls:Confusing inlet (mixture) and outlet (vapour) states; overlooking the need for slight superheat as a control target.Final Answer:Low-pressure vapour refrigerant (at outlet, after complete evaporation)

Question 19

Radiative properties check: If a surface has absorptivity α = 1, reflectivity ρ = 0, and transmissivity τ = 0, then in heat transfer terminology this surface is classified as which type of body?

Accepted Answer

Introduction / Context:In thermal radiation, every surface is described by three fundamental radiative properties: absorptivity (α), reflectivity (ρ), and transmissivity (τ). Understanding the extreme cases helps students classify real materials and approximate them in engineering calculations.Given Data / Assumptions:Absorptivity α = 1 (the surface absorbs all incident radiation).Reflectivity ρ = 0 (no portion is reflected).Transmissivity τ = 0 (no portion passes through).Energy conservation for radiation: α + ρ + τ = 1.Concept / Approach:The definition of a black body is a surface that absorbs all incident radiant energy at every wavelength and direction. Therefore its α = 1. Because total must sum to 1, a black body necessarily has ρ = 0 and τ = 0. It is also the ideal emitter: its emissive power at a given temperature is the maximum possible among all bodies.Step-by-Step Solution:Start with conservation: α + ρ + τ = 1.Given α = 1 → ρ + τ = 0 → ρ = 0 and τ = 0.A surface with α = 1 by definition is a black body.Therefore the correct classification is black body.Verification / Alternative check:Kirchhoff’s law at thermal equilibrium states emissivity equals absorptivity. If α = 1, emissivity = 1, matching the notion that a black body is also a perfect emitter (reference curve for radiation exchange).Why Other Options Are Wrong:Grey body: absorptivity/emissivity are constant with wavelength but less than 1.Opaque body: simply means τ = 0; it can still reflect. Here ρ = 0, so this is a special case, not the general term.White body: ideal reflector with ρ = 1, not consistent with α = 1.Translucent body: implies τ > 0, which contradicts τ = 0.Common Pitfalls:Confusing “black” in radiation with visible color; a surface may appear shiny black visually yet not behave as an ideal black body across all wavelengths.Final Answer:black body

Question 20

Psychrometrics definition: The mass of water vapour contained in 1 m³ of moist (wet) air is known as which property?

Accepted Answer

Introduction / Context:HVAC design and atmospheric science rely on precise humidity terminology. Mixing up absolute humidity with specific humidity or relative humidity leads to calculation errors in coil sizing and comfort analysis.Given Data / Assumptions:A fixed volume reference of 1 m³ of moist air is considered.We want the mass of water vapour contained within that volume.Standard psychrometric definitions apply at a stated pressure and temperature.Concept / Approach:Absolute humidity is defined as the mass of water vapour per unit volume of moist air, typically kg/m³. This is distinct from specific humidity (also called humidity ratio), which is the mass of water vapour per unit mass of dry air (kg/kg of dry air), and from relative humidity, which is a percentage comparing actual vapour pressure to saturation vapour pressure at the same temperature.Step-by-Step Solution:Identify target quantity: mass of vapour per unit volume.By definition, this corresponds to absolute humidity.Therefore, for 1 m³ of wet air, the mass of water vapour present is absolute humidity.Verification / Alternative check:Using the ideal gas relation for water vapour, m_v = (p_v * V) / (R_v * T), where p_v is partial pressure of water vapour, gives an absolute humidity value (kg/m³) that changes with temperature and barometric pressure.Why Other Options Are Wrong:Specific humidity (humidity ratio): mass of vapour per kg of dry air, not per m³ of moist air.Relative humidity: a dimensionless ratio (percentage), not a mass measure.Degree of saturation and dew-point temperature are different psychrometric concepts unrelated to mass per unit volume.Common Pitfalls:Using absolute humidity in place of humidity ratio when doing air mixing or coil calculations; most HVAC formulas are written in terms of humidity ratio (kg/kg of dry air), not kg/m³.Final Answer:absolute humidity

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