Question 1

Psychrometrics: According to Dalton’s law for moist air, the partial pressure of water vapour (p_v) at a given dry-bulb temperature is best expressed as:

Accepted Answer

Introduction / Context:In air-conditioning and refrigeration engineering, we frequently need the partial pressure of water vapour in moist air. This quantity underpins calculations of humidity ratio, dew-point temperature, and wet-bulb temperature and is central to psychrometric analysis.Given Data / Assumptions:Atmospheric air is a mixture of dry air and water vapour.Dalton’s law of partial pressures applies approximately for typical HVAC conditions.φ denotes relative humidity on a 0 to 1 scale. p_ws(T) denotes saturation vapour pressure of water at temperature T.Concept / Approach:Dalton’s law states that the total pressure equals the sum of the partial pressures of the constituents. For water vapour in moist air, the driving concept is that relative humidity is the ratio of actual water vapour partial pressure to the saturation vapour pressure at the same temperature, i.e., φ = p_v / p_ws(T). Rearranging immediately gives p_v.Step-by-Step Solution:Start from the definition: φ = p_v / p_ws(T).Rearrange to isolate p_v: p_v = φ * p_ws(T).Use tables or equations (e.g., Antoine) to find p_ws(T) at the air temperature, then multiply by the measured or specified relative humidity φ.With p_v known, other properties follow: w = 0.622 * p_v / (p_atm − p_v), and dew point via inversion of p_ws(T_dp) = p_v.Verification / Alternative check:Cross-check by computing humidity ratio from the obtained p_v and then reconstructing p_v = p_atm * w / (0.622 + w); both routes should match within rounding error.Why Other Options Are Wrong:p_v = p_atm * (w / (0.622 + w)) is an identity relating p_v and humidity ratio, not the primary expression in terms of φ and p_ws.p_v = ρ * R * T uses mixture properties; it gives total pressure if mixture data are used, not water vapour partial pressure directly without additional splitting.“None of these” is incorrect because a direct and standard expression exists.Common Pitfalls:Confusing saturation vapour pressure at dry-bulb vs. wet-bulb; p_ws must be evaluated at the relevant temperature indicated in φ's definition (usually the same temperature at which φ is stated).Final Answer:p_v = φ * p_ws(T)

Question 2

Heat exchangers: For the same terminal temperatures, the logarithmic mean temperature difference (LMTD) of a counter-current flow heat exchanger is __________ than that of a parallel-flow heat exchanger.

Accepted Answer

Introduction / Context:The LMTD method is a cornerstone of heat-exchanger sizing. Understanding how flow arrangement influences the temperature driving force helps engineers select compact and efficient configurations.Given Data / Assumptions:Terminal temperatures (hot in/out, cold in/out) are fixed and identical for both arrangements.Steady state, negligible heat losses to surroundings.Fluids have sensible heat exchange (no phase change) for this comparison.Concept / Approach:For given terminal temperatures, counter-current flow maintains a more uniform temperature difference along the length. This results in a larger logarithmic mean of the end temperature differences than parallel flow, where the temperature driving force decays more rapidly.Step-by-Step Solution:Define LMTD: ΔT_lm = (ΔT_1 − ΔT_2) / ln(ΔT_1 / ΔT_2).For the same terminal temperatures, compute ΔT values for both arrangements; counter-current typically yields a higher average.Because the driving force is higher, for the same duty and overall U, counter-current requires less area.Hence, LMTD_counter > LMTD_parallel.Verification / Alternative check:Test a numeric example: hot 100→60°C, cold 30→70°C. Counter: ΔT_1 = 100−70 = 30, ΔT_2 = 60−30 = 30 ⇒ LMTD = 30. Parallel: ΔT_1 = 100−30 = 70, ΔT_2 = 60−70 = −10 (sign handled via magnitudes) ⇒ effective LMTD smaller. This illustrates the advantage.Why Other Options Are Wrong:“Same” is incorrect; textbook results show counter-current advantage.“Less” contradicts the fundamental LMTD comparison.“Indeterminate” is unnecessary when terminal temperatures are specified.Common Pitfalls:Confusing LMTD with effectiveness (ε) or assuming phase change without noting terminal conditions.Final Answer:greater

Question 3

Psychrometrics: For a fixed dry-bulb temperature, as relative humidity __________, the wet-bulb temperature will increase.

Accepted Answer

Introduction / Context:Wet-bulb temperature (WBT) is a key indicator of the moisture content and evaporative cooling potential of air. It reflects the adiabatic saturation process and is central to cooling tower and comfort calculations.Given Data / Assumptions:Dry-bulb temperature (DBT) is held constant.Relative humidity (RH) varies between 0 and 100%.Pressure is near atmospheric.Concept / Approach:At fixed DBT, higher RH means the air is closer to saturation, so there is less capacity to absorb additional moisture via evaporation. Because WBT is the equilibrium temperature reached by evaporative cooling, as RH increases, the gap DBT − WBT narrows, implying WBT rises toward DBT.Step-by-Step Solution:At RH ≈ 0%: maximal evaporative potential ⇒ large DBT − WBT difference ⇒ low WBT.At intermediate RH: reduced evaporative potential ⇒ WBT increases.At RH = 100%: air is saturated; no evaporation possible ⇒ WBT = DBT, the maximum possible for given DBT.Therefore, WBT increases as RH increases for constant DBT.Verification / Alternative check:Read any psychrometric chart at fixed DBT: moving along a vertical DBT line toward higher RH moves upward along constant WBT lines, confirming higher WBT.Why Other Options Are Wrong:Decreases: opposite of chart behavior.Remains constant: only true if DBT and humidity ratio both fixed.Non-monotonic change: no such trend exists under these conditions.Common Pitfalls:Confusing WBT with dew point temperature; dew point increases strongly with humidity ratio, whereas WBT approaches DBT as RH approaches 100%.Final Answer:increases

Question 4

Refrigeration components: In a vapour-compression system, the line drier's primary purpose is to remove moisture from the refrigerant circuit.

Accepted Answer

Introduction / Context:Moisture inside a refrigeration circuit causes ice formation at expansion devices, chemical reactions, and acid formation that damages compressors. A drier (often combined with a filter) is essential to system reliability.Given Data / Assumptions:System is a sealed vapour-compression loop.Drier contains desiccant (e.g., molecular sieve, activated alumina).Installed in the liquid line after the condenser/receiver.Concept / Approach:The drier removes dissolved and free moisture from the liquid refrigerant and traps particulates. Moisture removal prevents freeze-ups at the throttle device and limits hydrolysis that can create acids in the presence of POE oils and high temperatures.Step-by-Step Solution:Liquid refrigerant exits the condenser and passes through the filter-drier.Desiccant adsorbs water molecules, lowering moisture content to very low ppm levels.Clean, dry refrigerant then enters the expansion device, minimizing risk of ice formation and corrosion.Regular replacement after contamination events or burnouts is recommended.Verification / Alternative check:Sight-glass indicators often include a moisture dot. If it signals wet conditions, service practice is to replace the drier and evacuate the system properly.Why Other Options Are Wrong:“Incorrect” ignores standard refrigeration practice and manufacturer guidelines.“Only in ammonia systems” is false; all refrigerants benefit from moisture control.“Only for low temperature” is false; even comfort cooling needs dry circuits.Common Pitfalls:Assuming evacuation alone removes all moisture; oil and surfaces retain water which the drier scavenges during operation.Final Answer:Correct

Question 5

Moist air properties: The wet-bulb temperature is a practical indicator of the moisture content (humidity) of air.

Accepted Answer

Introduction / Context:Wet-bulb temperature (WBT) reflects the adiabatic saturation process and is widely used in HVAC, meteorology, and cooling tower design to infer humidity levels and evaporative cooling potential.Given Data / Assumptions:Air at near-atmospheric pressure.Psychrometric relationships hold (Dalton’s law approximation).WBT measured with adequate air velocity over a wetted wick.Concept / Approach:WBT depends on both dry-bulb temperature and moisture content. For a fixed DBT, higher humidity yields a higher WBT (closer to DBT). Conversely, drier air produces more evaporative cooling at the wick, lowering the WBT. Thus, WBT is a practical indicator of air moisture content.Step-by-Step Solution:At low humidity: strong evaporation from the wetted wick removes more latent heat ⇒ WBT much lower than DBT.As humidity increases: evaporation decreases ⇒ WBT rises toward DBT.At saturation (100% RH): no evaporation ⇒ WBT = DBT.Hence, observing WBT trends reveals moisture content qualitatively and, with charts, quantitatively.Verification / Alternative check:Using a psychrometric chart or equations, plot points with constant DBT and increasing humidity ratio; WBT values increase monotonically.Why Other Options Are Wrong:“Disagree” is incorrect because WBT is standard for indicating humidity.“Only true at 0% RH” and “Only true at 100% RH” misunderstand the continuous relationship; WBT indicates humidity across the full range.Common Pitfalls:Equating WBT directly with humidity ratio; WBT is an indicator that must be interpreted with DBT or charts for exact values.Final Answer:Agree

Question 6

Vapour absorption refrigeration (VAR): Which statement is the most universally true when comparing VAR systems to vapour-compression systems?

Accepted Answer

Introduction / Context:Vapour absorption refrigeration (VAR) systems replace the mechanical compressor with a thermochemical loop using a generator, absorber, pump, and solution heat exchanger, typically driven by heat (steam, gas, or waste heat). They are common where electricity is scarce or waste heat is abundant.Given Data / Assumptions:Typical working pairs: ammonia–water or lithium bromide–water.Few rotating components: mainly solution pump(s).Applications include large HVAC plants and remote or off-grid refrigeration.Concept / Approach:With no high-speed compressor, VAR systems operate quietly. Electrical power consumption is low relative to the cooling duty because most input is thermal. However, achievable temperatures and performance depend on the working pair and design; sub-zero capability is not universal (e.g., LiBr–water cannot cool below 0°C due to water freezing).Step-by-Step Solution:Identify the primary mechanism: thermally driven absorption–desorption cycle replaces compression.Assess acoustic profile: absence of a large compressor means low noise; only pumps and fluid flow contribute modestly.Evaluate temperature capability: depends on refrigerant; not “always below 0°C”.Compare power use: electrical draw is small but not “the same” as compression systems.Verification / Alternative check:Manufacturer data consistently cite low noise and low electrical demand, especially for LiBr chiller plants.Why Other Options Are Wrong:Noisy operation: contrary to the defining advantage of VAR.Always below 0°C: false for water-based VAR (freezing limit).Same electrical power: false; VAR uses mainly thermal energy with small pump power.Common Pitfalls:Assuming all VAR systems can deep-freeze; only ammonia–water systems can reach well below 0°C, and design complexity rises.Final Answer:Gives quiet operation (few moving parts)

Question 7

Psychrometric chart reading: On a standard psychrometric chart, the constant relative humidity lines are:

Accepted Answer

Introduction / Context:Accurate reading of a psychrometric chart is fundamental for HVAC design, cooling tower analysis, and dehumidification calculations. Each family of lines corresponds to a different air property.Given Data / Assumptions:Standard psychrometric chart at 1 atm.Dry-bulb temperature plotted on the abscissa; humidity ratio on the ordinate.Other property lines (wet-bulb, RH, enthalpy, specific volume) are overlaid.Concept / Approach:Constant relative humidity (RH) lines are curved, intersecting dry-bulb verticals and trending toward the saturation curve (100% RH). Their curvature reflects Clausius–Clapeyron behaviour of water vapour pressure versus temperature.Step-by-Step Solution:Identify saturation curve (100% RH) as the left boundary of the chart’s state region.Trace RH = 10%, 30%, 50%, 70%, 90% lines: they bow and converge toward the saturation curve.Contrast with other families: dry-bulb lines are vertical; humidity ratio lines are horizontal; wet-bulb lines are quasi-straight slanted.Thus, RH lines are definitively curved.Verification / Alternative check:Consult any standard chart; the geometry is consistent across references though exact spacing is non-uniform due to property equations.Why Other Options Are Wrong:Vertical/horizontal descriptors apply to DBT and humidity ratio, not RH.“Horizontal non-uniform” describes enthalpy or wet-bulb trends poorly; RH is not horizontal.Common Pitfalls:Mistaking wet-bulb lines (near straight slanted) for RH; always locate the saturation curve first to orient the chart.Final Answer:Curved lines (approximately concave to the humidity ratio axis)

Question 8

Counter-flow heat exchanger with equal heat capacities: A hot stream enters at 100°C and leaves at 60°C. A cold stream enters at 40°C. If C_hot = C_cold, what is the mean temperature difference (LMTD) between the two fluids?

Accepted Answer

Introduction / Context:The logarithmic mean temperature difference (LMTD) is a key parameter for sizing heat exchangers when the overall heat transfer coefficient and duty are known. Special simplifications occur when the capacity rates of the two fluids are equal.Given Data / Assumptions:Counter-flow heat exchanger.Hot fluid: T_h,in = 100°C, T_h,out = 60°C.Cold fluid: T_c,in = 40°C.Equal capacity rates: C_h = C_c ⇒ ΔT changes linearly and the cold outlet temperature rise equals the hot temperature drop.Concept / Approach:With equal capacity rates, energy balance dictates that the cold stream temperature rise equals the hot stream temperature drop (40°C here). Therefore T_c,out = 40 + 40 = 80°C. Once the terminal temperatures are known, compute LMTD for counter-flow using the two end temperature differences.Step-by-Step Solution:Energy balance with C_h = C_c: C*(100 − 60) = C*(T_c,out − 40) ⇒ T_c,out = 80°C.Counter-flow end differences: ΔT_1 = T_h,in − T_c,out = 100 − 80 = 20°C.ΔT_2 = T_h,out − T_c,in = 60 − 40 = 20°C.LMTD = (ΔT_1 − ΔT_2) / ln(ΔT_1 / ΔT_2) = (20 − 20) / ln(20/20) which evaluates to 20°C in the equal-endcase limit.Verification / Alternative check:If ΔT_1 = ΔT_2, the logarithmic mean reduces to that common difference directly; thus LMTD = 20°C without further calculation.Why Other Options Are Wrong:40°C or higher values do not reflect the computed terminal differences.66.7°C is unrelated to these endpoints and overestimates the driving force.Common Pitfalls:Using parallel-flow formula or assuming the cold outlet is unknown; with equal capacities it is immediately 80°C, making the end differences equal.Final Answer:20°C

Question 9

Thermal radiation law: The wavelength corresponding to maximum monochromatic emissive power varies inversely with absolute temperature according to which law?

Accepted Answer

Introduction / Context:Blackbody radiation describes how ideal emitters radiate energy as a function of wavelength and temperature. Several fundamental laws describe different aspects of this emission.Given Data / Assumptions:We consider the wavelength λ_max at which the spectral emissive power is maximum for a given absolute temperature T.Emphasis is on the inverse proportionality λ_max * T ≈ constant.Concept / Approach:Wien’s displacement law states λ_max * T = b, where b is Wien’s constant. This gives the simple inverse relationship and explains why hotter bodies radiate at shorter peak wavelengths (e.g., the Sun peaks in the visible, a hot stove in the infrared).Step-by-Step Solution:Identify the target: relation between peak wavelength and temperature.Recall the formula: λ_max * T = b (b ≈ 2.898 × 10^−3 m·K).Conclude the governing law: Wien’s displacement law.Contrast with other laws: Stefan–Boltzmann relates total emissive power to T^4; Planck gives full spectral distribution; Kirchhoff relates emissivity and absorptivity at thermal equilibrium.Verification / Alternative check:Apply to two temperatures to see the shift: doubling T halves λ_max, confirming the inverse proportionality.Why Other Options Are Wrong:Kirchhoff’s law: not about peak wavelength.Stefan–Boltzmann: total power, proportional to T^4, not spectral peak.Planck’s law: full spectrum formula; the inverse result emerges from differentiating Planck’s law, but the named simple relation is Wien’s.Common Pitfalls:Confusing total emissive power (Stefan–Boltzmann) with the wavelength of maximum emission (Wien’s).Final Answer:Wien’s displacement law

Question 10

Refrigerant property of ammonia: The __________ point of ammonia (NH₃) at standard pressure is −77.7°C.

Accepted Answer

Introduction / Context:Ammonia (R-717) is a widely used refrigerant in industrial systems. Knowing its key thermophysical properties—boiling, freezing, and critical points—is essential for safe and reliable design.Given Data / Assumptions:Standard atmospheric pressure conditions.Ammonia’s normal boiling point is approximately −33.3°C.The stated temperature is −77.7°C.Concept / Approach:Compare the given temperature with known characteristic temperatures. Because −77.7°C is far below the normal boiling point, it aligns with the freezing (melting) point of ammonia, the temperature at which liquid ammonia solidifies at 1 atm.Step-by-Step Solution:Recall key points: T_boil,1atm ≈ −33.3°C; T_freeze ≈ −77.7°C; T_critical ≈ 132.4°C (critical temperature).Match the given −77.7°C to these known values.Conclude that −77.7°C corresponds to the freezing (melting) point.Therefore the blank should be “freezing (melting)”.Verification / Alternative check:Consult standard property tables for R-717 to confirm the numerical values; industrial design texts list these constants consistently.Why Other Options Are Wrong:Boiling point is −33.3°C at 1 atm, not −77.7°C.Critical point temperature is above 100°C, not negative.Dew point refers to saturation temperature at a given pressure for vapour mixtures, not a fixed constant for a pure substance at 1 atm.Common Pitfalls:Confusing boiling and freezing points for substances with negative values; always cross-reference known constants.Final Answer:freezing (melting)

Question 11

Environmental protection: Why have international agencies advised against the use of chlorofluorocarbon (CFC) refrigerants in HVAC and refrigeration systems?

Accepted Answer

Introduction / Context:CFCs (chlorofluorocarbons) were once widely used as refrigerants because they are non-flammable, non-corrosive, and stable. However, environmental protection agencies worldwide, supported by international agreements such as the Montreal Protocol, moved to phase them out due to their severe impact on the ozone layer, which shields life from harmful ultraviolet-B radiation.Given Data / Assumptions:CFC molecules are chemically stable in the lower atmosphere.Over years, they diffuse upward to the stratosphere.Under high-energy UV, they release chlorine radicals.Ozone depletion increases UV-B reaching Earth’s surface.Concept / Approach:The key mechanism is catalytic: a single chlorine radical can participate in repeated cycles converting ozone (O3) into oxygen (O2) without being consumed. This dramatically reduces ozone concentration, particularly in polar springtime, creating “ozone hole” conditions. The decision to eliminate CFCs stems from this long-lived, global-scale risk.Step-by-Step Solution:Stability in troposphere → CFCs persist and migrate upward.UV photolysis in stratosphere → Cl• radicals are released.Catalytic cycles → Cl• + O3 → ClO• + O2; then ClO• + O → Cl• + O2.Net effect → O3 + O → 2 O2 (ozone destroyed repeatedly).Verification / Alternative check:Observed global declines in stratospheric ozone and recovery trends after the CFC phase-out corroborate the mechanism. Laboratory photochemistry and satellite measurements align with the catalytic depletion pathway.Why Other Options Are Wrong:Acid rain: primarily linked to SO2 and NOx, not CFCs.Greenhouse effect from plant reaction: greenhouse forcing is due to gas radiative properties, not plant chemistry; CFCs are potent greenhouse gases but the question focuses on ozone depletion.Reaction with oxygen: not a primary atmospheric pathway for CFCs.Surface smog: associated with NOx and volatile organics under sunlight, not stable CFCs condensing.Common Pitfalls:Confusing ozone depletion (stratospheric) with acid rain or urban smog (tropospheric). Also, assuming rapid breakdown of CFCs near the ground—CFCs are too stable there; the issue arises aloft.Final Answer:They reach the stratosphere and catalytically destroy ozone, thinning the ozone layer

Question 12

Identify the situation where heat transfer occurs simultaneously by conduction, convection, and radiation in thermal systems.

Accepted Answer

Introduction / Context:Engineers must recognize dominant heat-transfer modes to design efficient equipment. In many real systems, more than one mode operates simultaneously. Boiler furnaces are a classic example where conduction, convection, and radiation all play significant roles.Given Data / Assumptions:Combustion gases at very high temperature are present.Metal enclosures, refractory linings, and boiler tubes exist.Flowing gas sweeps over heated surfaces.Concept / Approach:Radiation dominates heat transfer from the luminous flame and hot gas to furnace walls and tube banks. Convection accompanies the gas motion over the tubes. Conduction transmits absorbed heat through tube walls to the working fluid (water/steam). Recognizing all three modes is essential for accurate thermal calculations and material selection.Step-by-Step Solution:Inside furnace → hot flame and gas emit radiation → radiant heat to tubes/walls.Gas motion → forced convection from gas to external tube surfaces.Within solids → heat conducts through tube metal into water/steam side.Result → simultaneous conduction, convection, and radiation.Verification / Alternative check:Heat-flux measurements show radiation fractions are high near burners, while convective contributions increase downstream; metal wall temperature gradients confirm conduction through solids.Why Other Options Are Wrong:Melting ice: mainly convection in liquid and conduction within the solid; radiation is negligible without hot radiative surroundings.Surface condenser: dominant condensation convection with conduction through tubes; radiation is negligible at low temperatures.“No process involves all three”: contradicted by furnace practice and textbooks.Common Pitfalls:Underestimating radiation at high temperatures or ignoring conduction through metal walls when focusing on fluid-side coefficients only.Final Answer:Boiler furnaces in steam power plants

Question 13

Cooling coils and air-side design: How does the by-pass factor (BPF) of a cooling coil change with fin spacing and the number of tube rows?

Accepted Answer

Introduction / Context:The by-pass factor (BPF) of a coil is the fraction of air that effectively “misses” perfect contact with cold surface and leaves closer to its entering condition. Designers aim for a low BPF so that supply air approaches the apparatus dew-point for better dehumidification and cooling.Given Data / Assumptions:BPF depends on surface area, face velocity, fin geometry, and number of rows.Other variables (airflow rate, coil cleanliness) are assumed constant for comparison.Concept / Approach:More surface area and longer air path improve heat and mass transfer. Decreasing fin spacing (more fins per metre) increases area and surface contact. Increasing the number of rows extends contact time and path length. Both measures reduce BPF by allowing more of the airstream to approach coil surface conditions.Step-by-Step Solution:Define BPF qualitatively: BPF = (T_out − T_surface) / (T_in − T_surface) for sensible cooling.Increase area by tighter fin spacing → T_out moves closer to T_surface → BPF falls.Increase rows → more residence time and area → further reduction in BPF.Therefore, smallest BPF occurs with decreased fin spacing and increased rows.Verification / Alternative check:Manufacturer coil performance data show BPF trending downward with higher fins per metre and additional rows at the same face velocity.Why Other Options Are Wrong:Increasing fin spacing reduces area, raising BPF.Reducing rows shortens air contact, also raising BPF.Independence from geometry is incorrect; geometry is a primary driver.Common Pitfalls:Forgetting that extremely tight fin spacing can increase air-side pressure drop and risk frosting; practical design balances BPF against fan power and maintenance.Final Answer:It decreases when fin spacing is decreased and the number of rows is increased

Question 14

In a refrigerator, what is the primary control function of the expansion (metering) valve?

Accepted Answer

Introduction / Context:The expansion valve (capillary, thermostatic, or electronic) meters liquid refrigerant from the high-pressure condenser/liquid line into the low-pressure evaporator. It is crucial for maintaining proper evaporator feeding and pressure levels.Given Data / Assumptions:Single-stage vapour-compression system.Condenser provides subcooled liquid to the metering device.Evaporator requires controlled flow to match heat load.Concept / Approach:The metering device creates a pressure drop from condenser pressure to evaporator pressure. It controls mass flow so that the evaporator receives the right amount of refrigerant, preventing starvation (low capacity) and flooding (risk of liquid slugging). TEVs further modulate flow to maintain a target superheat.Step-by-Step Solution:Upstream: high pressure, moderate temperature liquid.Across valve: throttling → pressure drop; enthalpy remains essentially constant.Downstream: low-pressure mixture enters evaporator to absorb heat.Valve modulates/set orifice defines flow to match load.Verification / Alternative check:System pressure and superheat measurements confirm correct metering: stable evaporator pressure and target outlet superheat indicate proper control.Why Other Options Are Wrong:Compressor start/stop, condenser pressure regulation, and non-condensable purging are separate functions handled by controls, head-pressure devices, or service procedures.Common Pitfalls:Misadjusting superheat on a TEV or selecting too small a capillary causing high superheat and low capacity.Final Answer:Correct

Question 15

Kirchhoff’s radiation law: Is a body’s emissivity always equal to its absorptivity at all temperatures and conditions?

Accepted Answer

Introduction / Context:Kirchhoff’s law of thermal radiation is often summarized as “emissivity equals absorptivity.” This is precise only under specific conditions. Misapplying it leads to errors in radiative heat-transfer calculations and material selection.Given Data / Assumptions:Thermal equilibrium between a body and its surroundings.Same temperature for exchanging surfaces.Diffuse–gray surface idealization.Concept / Approach:Kirchhoff’s law states that, at thermal equilibrium and for a given wavelength and direction, emissivity equals absorptivity. Outside equilibrium, or when spectral/Directional effects matter, equality need not hold. Thus, the blanket statement “equal at all temperatures” is incorrect unless the system is in equilibrium and comparisons are like-for-like (same wavelength and direction).Step-by-Step Solution:Define ε(λ,θ,φ) and α(λ,θ,φ): spectral–directional properties.At equilibrium: ε = α for each wavelength and direction.Out of equilibrium or across bands: ε̄ may differ from ᾱ.Therefore, “always equal at all temperatures” → false.Verification / Alternative check:Experimental data show materials with different apparent band-averaged emissivity and absorptivity when measured against sources at different temperatures (thus different spectra).Why Other Options Are Wrong:Restrictions to metals, vacuum, or high temperature are unnecessary; the governing condition is thermal equilibrium and spectral matching, not material class or environment alone.Common Pitfalls:Using room-temperature absorptivity to predict emissive performance at furnace temperatures without spectral correction.Final Answer:No

Question 16

By-pass factor (sensible cooling): Select the correct expression in terms of dry-bulb temperatures td1 (entering air), td2 (leaving air), and td3 (coil surface/apparatus dew-point).

Accepted Answer

Introduction / Context:The by-pass factor (BPF) quantifies how closely the leaving air approaches the coil surface temperature during sensible cooling. A low BPF indicates effective contact and good approach to the apparatus dew-point, important for comfort cooling and dehumidification.Given Data / Assumptions:Steady air flow across a cooling coil.td1 = entering dry-bulb, td2 = leaving dry-bulb, td3 = coil surface (apparatus dew-point).Negligible reheat between coil and measurement point.Concept / Approach:For sensible cooling, the fraction of air that effectively “bypasses” the coil behaves as if it were mixed with air cooled to the coil surface temperature. From the standard mixing relation, BPF is the ratio of the leaving-to-entering temperature differences measured relative to the coil surface temperature.Step-by-Step Solution:Start with mixing concept: td2 is between td1 and td3.Define BPF = (td2 − td3) / (td1 − td3).Check bounds: If td2 = td3 → BPF = 0 (perfect contact). If td2 = td1 → BPF = 1 (complete bypass).Therefore, the correct expression is BPF = (td2 − td3) / (td1 − td3).Verification / Alternative check:The equation satisfies dimensional consistency and limiting cases; it is the standard textbook form used in coil selection and performance rating.Why Other Options Are Wrong:Other forms invert or misplace temperature differences, leading to values outside the 0–1 range or violating limits at perfect contact.Common Pitfalls:Using mixed air temperature after a reheat coil as td2, which inflates BPF; always use the air temperature immediately leaving the cooling coil.Final Answer:BPF = (td2 − td3) / (td1 − td3)

Question 17

Psychrometrics: What do we call a mixture of dry air and water vapour that has diffused the maximum possible moisture at its temperature?

Accepted Answer

Introduction / Context:Air-conditioning analysis requires precise terminology. “Saturated air” is a fundamental state used to define dew point, wet-bulb behavior, and the saturation curve on the psychrometric chart.Given Data / Assumptions:Mixture at a specified barometric pressure.Water vapour present up to its maximum capacity at the given temperature.No liquid droplets initially (just at the verge of condensation).Concept / Approach:At saturation, the partial pressure of water vapour equals the saturation vapour pressure at that temperature. Any further addition of moisture or drop in temperature causes condensation. Relative humidity is then 100%.Step-by-Step Solution:Define saturated condition: p_vapour = p_sat(T).Resulting psychrometric properties: DBT = WBT = DPT at perfect saturation.Therefore, the mixture is called saturated air.Verification / Alternative check:On the psychrometric chart, saturated states lie on the 100% RH curve bordering the two-phase region of air–water systems.Why Other Options Are Wrong:Dry air: contains negligible water vapour.Moist air: generic term for any air containing water vapour, not necessarily at saturation.Specific humidity: a property (humidity ratio), not a state name.Common Pitfalls:Confusing “saturated air” with fog; fog implies presence of suspended droplets after supersaturation or cooling below dew point.Final Answer:Saturated air

Question 18

Refrigeration compressors: What is the pressure at the outlet of a refrigerant compressor commonly called?

Accepted Answer

Introduction / Context:Correct terminology helps technicians interpret gauges and troubleshoot systems. Compressor ports are commonly referred to as suction (inlet) and discharge (outlet) sides.Given Data / Assumptions:Vapour-compression refrigeration cycle.Compressor increases refrigerant pressure and temperature.Outlet feeds the condenser.Concept / Approach:The compressor outlet is the high-pressure side that delivers hot, superheated vapour to the condenser. Its pressure is therefore called discharge pressure. Conversely, the inlet is the low-pressure suction side connected to the evaporator.Step-by-Step Solution:Identify inlet → suction (low pressure).Identify outlet → discharge (high pressure).Thus, outlet pressure is discharge pressure.Verification / Alternative check:Most gauge sets have a high-side (discharge) and low-side (suction) connection, reflecting this terminology.Why Other Options Are Wrong:Suction pressure: inlet side only.Critical pressure: thermodynamic property of a fluid, not a line pressure.Back pressure: often refers to downstream resistance in other contexts; in refrigeration it is not the standard name for compressor outlet.Common Pitfalls:Confusing absolute vs gauge readings; discharge pressure is typically discussed as gauge pressure referenced to ambient.Final Answer:Discharge pressure

Question 19

For unsaturated air (relative humidity less than 100%), how does the dew-point temperature compare with the wet-bulb temperature?

Accepted Answer

Introduction / Context:Interpreting psychrometric measurements requires knowing the order of characteristic temperatures. In unsaturated air, these temperatures follow a consistent inequality that aids quick checks and diagnostics.Given Data / Assumptions:Air–water vapour mixture at RH Proper instrument use and steady conditions.Concept / Approach:For unsaturated air, the standard relationship is: dew-point temperature (DPT) Step-by-Step Solution:Recognize unsaturated condition → potential for evaporation from the wet-bulb.Evaporation cools the wet-bulb → WBT below DBT.Dew point is colder still because cooling to DPT is required for saturation at current moisture content.Hence DPT Verification / Alternative check:Plot a point inside the psychrometric chart’s field (not on the saturation curve): lines of constant DPT lie left of the state point, and constant WBT lines intersect with WBT greater than DPT but less than DBT.Why Other Options Are Wrong:Equality holds only at saturation (RH = 100%).“Greater than” contradicts established ordering in unsaturated air.“Unrelated” is incorrect; the values are physically linked.Common Pitfalls:Instrument lag or poor ventilation around the psychrometer wick can distort WBT readings; still, the theoretical order remains DPT Final Answer:Less than the wet-bulb temperature

Question 20

Function of a thermostatic expansion valve (TEV): Which description best captures its role in a refrigeration system?

Accepted Answer

Introduction / Context:A TEV (TXV) is the most common metering device in comfort and commercial refrigeration. Understanding its control objective is essential for diagnosing issues like hunting, flooding, and starvation.Given Data / Assumptions:Sensor bulb at evaporator outlet measures suction line temperature.Valve diaphragm senses bulb pressure against evaporator pressure and spring force.Target is to maintain a set superheat at the evaporator exit under varying load.Concept / Approach:By modulating the orifice opening in response to superheat, the TEV meters just enough refrigerant so the evaporator is well utilized (effectively filled) without allowing liquid carryover to the compressor. This capability makes TEVs suitable for variable-load systems.Step-by-Step Solution:Measure outlet superheat SH = T_suction − T_sat(evap pressure).If SH rises (starvation) → bulb pressure increases valve opening → more flow.If SH falls (flooding risk) → valve throttles to reduce flow.Result → stable superheat and effective evaporator fill under changing loads.Verification / Alternative check:Field commissioning targets a superheat setpoint (e.g., 5–8 K). Stable SH and evaporator pressure indicate correct operation.Why Other Options Are Wrong:Only constant-load suitability is wrong; TEVs excel at handling variable loads.They do not directly control evaporator temperature; they control superheat (an indirect safeguard).They are not relief valves nor condenser subcooling controllers.Common Pitfalls:Improper bulb placement or poor thermal contact leads to hunting. Excessive pressure drop in distributors can also destabilize control.Final Answer:It meters refrigerant so the evaporator is properly fed and effectively filled across varying loads

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