Question 1

Refrigeration fundamentals: During subcooling (liquid cooling below its saturation temperature at condenser pressure), is sensible heat removed from the liquid refrigerant?

Accepted Answer

Introduction / Context: Subcooling is a common technique in vapour-compression systems to improve capacity and avoid flashing in the liquid line. Knowing what type of heat is removed clarifies enthalpy changes and energy balances across system components. Given Data / Assumptions: Liquid refrigerant leaves the condenser at saturation temperature corresponding to condenser pressure. It is further cooled below this saturation temperature at essentially constant pressure. No phase change occurs during subcooling. Concept / Approach: Sensible heat refers to temperature change without phase change. During subcooling, the refrigerant temperature decreases below saturation at the same pressure, so the removed energy is sensible heat. This lowers the enthalpy at the expansion device inlet and increases the refrigeration effect per kilogram in the evaporator. Step-by-Step Solution: Start state: saturated liquid at condenser pressure.Process: cool the liquid at (approximately) constant pressure to a lower temperature (no boiling).Energy removed: sensible heat → enthalpy decreases by Δh = c_p,liq * ΔT.Result: higher net refrigeration effect in the evaporator after throttling. Verification / Alternative check: On a pressure–enthalpy (p–h) diagram, subcooling is a leftward movement along the liquid line below the saturated liquid point. The enthalpy change directly shows sensible heat removal. Why Other Options Are Wrong: Phase change is not involved in subcooling; therefore it is not latent heat removal.Throttling and evaporation are separate processes; subcooling occurs before the expansion device.Two-phase region is avoided during subcooling; the fluid remains liquid. Common Pitfalls: Confusing subcooling with superheating; superheating applies to vapour above saturation temperature, while subcooling applies to liquid below it. Final Answer: Yes

Question 2

Comparative emissivity: The emissivity of a polished silver surface, when compared to an ideal black body at the same temperature, is generally considered to be…

Accepted Answer

Introduction / Context: Emissivity strongly influences radiative heat transfer. Highly reflective metals like polished silver are widely used in thermal insulation and radiation shields precisely because of their low emissivity. Given Data / Assumptions: Comparison at the same temperature and broad thermal radiation wavelengths. Black body emissivity is 1 by definition. Polished silver has high reflectivity in the infrared spectrum. Concept / Approach: Emissivity is the ratio of a real surface’s emissive power to that of a black body at the same temperature. A polished silver surface reflects most incident radiation and emits poorly. Thus, its emissivity is much less than 1 and is commonly referred to as very low (on the order of 0.02–0.05 depending on finish and temperature). Step-by-Step Solution: Define ε_black = 1 (maximum possible).Metals like silver have high ρ (reflectivity) → low ε (Kirchhoff’s law connections under equilibrium).Therefore ε_silver ≪ 1 → commonly termed very low. Verification / Alternative check: Radiation heat-transfer handbooks list emissivity values for polished metals significantly below 0.1, confirming the “very low” classification compared with a black body. Why Other Options Are Wrong: Same/high/greater than 1 contradict both measurement and theory.“Low” is qualitatively true, but “very low” better reflects the known magnitude for polished silver. Common Pitfalls: Assuming shiny appearance alone defines emissivity across all wavelengths; surface oxidation or roughness can raise emissivity, but polished silver remains very low relative to a black body. Final Answer: very low

Question 3

Definition check: A grey body is defined as one whose absorptivity is independent of wavelength (spectrally constant). Is it also independent of temperature?

Accepted Answer

Introduction / Context: Grey-body assumptions simplify radiation problems by treating absorptivity and emissivity as constant over wavelength. However, temperature dependence is a separate matter and is often misunderstood in exam questions. Given Data / Assumptions: Grey body: spectral dependence removed (ε and α do not vary with wavelength). No statement is made about variation with temperature. Concept / Approach: The formal definition of a grey body addresses spectral uniformity: ε(λ) = constant with respect to λ. It does not require that ε be constant with temperature. In practice, surface properties can and do change with temperature (oxidation state, phase changes, or electronic effects). Step-by-Step Solution: State definition: grey → wavelength independence.Temperature is a different axis of variability.Therefore, the statement “does not vary with temperature and wavelength” is incorrect. Verification / Alternative check: Material data sheets show emissivity tables versus temperature for many materials assumed “grey” over a band; values still change with temperature even though treated as spectrally uniform in models. Why Other Options Are Wrong: Qualifiers like “only for metals/vacuum/room temperature” misunderstand the definition; spectral constancy is the key, not the environment. Common Pitfalls: Equating “grey” with “constant in every way.” Grey is a modeling convenience for wavelength, not a guarantee against temperature dependence. Final Answer: False

Question 4

Air-conditioning process insight: During purely sensible cooling of air (no moisture removal), how does the wet-bulb temperature behave?

Accepted Answer

Introduction / Context: On the psychrometric chart, different HVAC processes trace different paths. For sensible cooling, the air's dry-bulb temperature drops without changing its humidity ratio, a key concept for diagnosing coil performance and reheating needs. Given Data / Assumptions: No condensation occurs (constant humidity ratio). Process line is horizontal to the left on the chart. Air properties are measured in steady state. Concept / Approach: The wet-bulb temperature is a measure related to the adiabatic saturation state. For a constant humidity ratio, the adiabatic saturation temperature stays fixed, and so does the wet-bulb temperature. Therefore, WB remains the same while DB reduces in a sensible-only process. Step-by-Step Solution: Identify process: sensible-only → constant humidity ratio (horizontal line).Lines of constant wet-bulb are nearly parallel to constant enthalpy in comfort ranges.With no moisture change, WB remains effectively constant while DB falls. Verification / Alternative check: Compute humidity ratio from measured DB/WB pairs before and after. If the ratio is unchanged and no condensate forms, WB should remain the same within instrument accuracy. Why Other Options Are Wrong: Increase/decrease implies moisture exchange, which contradicts the definition of sensible cooling.“Varies unpredictably” ignores psychrometric relationships.Equalling dew point happens only at saturation; not the case here. Common Pitfalls: Assuming WB always follows DB; in reality, WB is tied to moisture content and will not change in truly sensible-only cooling. Final Answer: remains constant

Question 5

Service practice: Using a halide torch to detect a leak of Freon-type refrigerant (old CFC/HCFC), the flame colour typically changes to which hue?

Accepted Answer

Introduction / Context: Before modern electronic leak detectors, technicians often used halide torches to identify CFC/HCFC leaks. Recognizing the characteristic flame colour change is part of historical service knowledge and safety lessons. Given Data / Assumptions: Halogenated refrigerant (e.g., R-12, R-22) is present in the air near the torch intake. Combustion products interact with halogens, producing a coloured flame. The method is discussed for education; it is not recommended due to safety/environmental concerns. Concept / Approach: When halogen vapours are drawn into the halide torch flame, copper halides form and emit a greenish colour. This bright green or blue-green tint indicates the presence of a halogenated refrigerant in the surrounding air. Step-by-Step Solution: Operate halide torch per instructions (historical practice).Move intake near suspected leak path.Observe flame colour; a greenish shift signifies halogen vapours. Verification / Alternative check: Modern electronic detectors (heated diode, infrared) are safer and more sensitive. They confirm leaks without combustion by-products. Why Other Options Are Wrong: Yellow/red/orange/violet are not the characteristic halide indication for CFC/HCFC leaks. Common Pitfalls: Using open flames near refrigerant or oil can be hazardous; always follow contemporary codes and use approved electronic detectors. Final Answer: bright green

Question 6

Absorption refrigeration: In common absorption vapour systems, is R-12 the widely used refrigerant, or are other working pairs standard?

Accepted Answer

Introduction / Context: Absorption refrigeration systems use heat to drive a refrigerant–absorbent pair, unlike vapour-compression systems that use mechanical work. It is essential to distinguish the typical working fluids of each technology. Given Data / Assumptions: Absorption pairs commonly include ammonia–water and water–lithium bromide. R-12 is a CFC refrigerant historically used in vapour-compression (not absorption). Environmental regulations have phased out R-12 in most applications. Concept / Approach: In absorption cycles, refrigerant is absorbed in a liquid and later boiled out by an external heat input. The most common pairs are ammonia (refrigerant) with water (absorbent) for low-temperature applications, and water (refrigerant) with lithium bromide (absorbent) for air conditioning. R-12 lacks an absorbent pair standard in commercial absorption systems and is not the “widely used” refrigerant there. Step-by-Step Solution: Identify technology: absorption vs vapour-compression.List typical pairs: NH3–H2O and H2O–LiBr.Conclude: the statement “R-12 widely used in absorption” is incorrect. Verification / Alternative check: Manufacturer literature for absorption chillers consistently specifies LiBr–water for HVAC and NH3–water for refrigeration/ice plants, confirming the standard practice. Why Other Options Are Wrong: Conditional options about pressure, size, or machine type do not turn R-12 into a standard absorption refrigerant. Common Pitfalls: Confusing the fluid choices between absorption and compression cycles; the presence of a compressor vs an absorber–generator set dictates suitable working pairs. Final Answer: Incorrect

Question 7

Spray washer psychrometrics: If the spray water temperature is higher than the entering air dry-bulb temperature, how will the air state change across the spray washing chamber?

Accepted Answer

Introduction / Context: Air–water contact devices (spray washers, air washers) can heat/cool and humidify/dehumidify air depending on the relative temperatures and vapour pressures. Knowing the direction of heat and mass transfer is crucial for process control. Given Data / Assumptions: Spray water temperature > entering air dry-bulb temperature. Sufficient contact to exchange sensible heat and moisture. Negligible heat losses to surroundings for conceptual analysis. Concept / Approach: When water is warmer than air, heat flows from water to air, increasing the air's sensible temperature. Simultaneously, provided the water has a higher vapour pressure at its temperature than the partial pressure of vapour in air, evaporation occurs, adding moisture to air (humidification). Thus, the air is heated and humidified. Step-by-Step Solution: Compare temperatures: T_water > T_air → heat transfer to air.Vapour pressure at water surface > vapour partial pressure in air → mass transfer of moisture to air.Air outlet condition: higher dry-bulb (heated) and higher humidity ratio (humidified). Verification / Alternative check: Plot the process on a psychrometric chart from the inlet state toward the saturation curve at a higher enthalpy. The trajectory shows both DBT and humidity ratio increasing when water is hotter. Why Other Options Are Wrong: Heated and dehumidified: contradicts mass transfer direction; warm water adds, not removes, moisture.Cooled options: violate the given temperature relation (water hotter than air).Unchanged: ignores clear temperature and vapour pressure gradients. Common Pitfalls: Assuming every air–water contact is adiabatic saturation; the result depends on relative temperatures and flow rates, not a universal rule. Final Answer: heated and humidified

Question 8

Cooling and dehumidification over a coil: Compared with the entering air, the relative humidity of the leaving air is generally…

Accepted Answer

Introduction / Context: During coil cooling with condensation, the air state moves toward the apparatus dew-point. Although moisture is removed, the relative humidity of the remaining air usually increases because temperature drops faster than the vapour content on a percentage basis. Given Data / Assumptions: Cooling coil below the entering air dew-point, causing condensation. Steady flow and standard psychrometric behaviour. No reheat applied immediately after the coil. Concept / Approach: Relative humidity compares actual vapour partial pressure to saturation pressure at the same temperature. When air is cooled, the saturation pressure decreases steeply. Even if some moisture is removed, the ratio often rises, pushing RH closer to 100% at the coil exit (unless followed by reheat). Step-by-Step Solution: Inlet state: T_in, RH_in.Across coil: T decreases; moisture ratio decreases due to condensation.Saturation pressure drops faster than vapour partial pressure → RH increases.Therefore, RH_out is generally higher than RH_in. Verification / Alternative check: Plot inlet and outlet on a psychrometric chart. The path moves down-left toward the saturation curve; RH lines indicate an increase unless a reheat stage follows. Why Other Options Are Wrong: Lower/unchanged: not typical during active dehumidification without reheat.“Indeterminate” disregards the strong, standard trend on typical comfort-cooling coils.Always 100%: only true if leaving state is exactly at saturation; in practice it is high but not necessarily 100%. Common Pitfalls: Confusing absolute moisture removal with relative humidity behaviour; removing water does not automatically mean lower RH if temperature drops significantly. Final Answer: higher

Question 9

Psychrometrics – Heating coil bypass factor (BPF)  Atmospheric air at a dry-bulb temperature of 15 °C enters a heating coil whose surface is maintained at 40 °C. The leaving-air dry-bulb temperature is 25 °C. Compute the by-pass factor (BPF) of the …

Accepted Answer

Introduction / Context: Heating and cooling coils in HVAC analysis are often evaluated using the by-pass factor (BPF). For a sensible heating process, BPF quantifies the fraction of air that effectively “bypasses” the coil surface temperature and leaves warmer than the coil surface would otherwise impose. Given Data / Assumptions: Entering air dry-bulb temperature, t_in = 15 °C. Coil surface (apparatus) temperature, t_coil = 40 °C (assumed uniform). Leaving air dry-bulb temperature, t_out = 25 °C. Process is sensible heating (moisture change neglected for BPF relation). Concept / Approach: For heating coils, the standard relation is BPF = (t_coil − t_out) / (t_coil − t_in). This mirrors the cooling-coil form with temperatures appropriately referenced to the coil surface. Step-by-Step Solution: t_coil − t_out = 40 − 25 = 15.t_coil − t_in = 40 − 15 = 25.BPF = 15 / 25 = 0.60.Therefore, the by-pass factor is 0.60. Verification / Alternative check: A higher BPF indicates poorer coil contact (more “bypassing”). Here, leaving air (25 °C) is much cooler than the coil (40 °C), consistent with a relatively high BPF of 0.60. Why Other Options Are Wrong: 0.376, 0.40, 0.24, 0.67: These do not satisfy the simple ratio with the given numbers; only 0.60 matches the formula. Common Pitfalls: Using the cooling-coil form without flipping the temperature order; forgetting temperatures must be absolute only when using thermodynamic ratios (here a simple difference suffices in °C). Final Answer: 0.60

Question 10

Dimensionless groups in heat transfer  Confirm whether the Prandtl number is defined as: Pr = μ * cp / k.

Accepted Answer

Introduction / Context: The Prandtl number connects momentum and thermal diffusivities and is fundamental in convection correlations such as for boundary layers, internal duct flow, and external flow over plates. Given Data / Assumptions: Dynamic viscosity μ (Pa·s), specific heat at constant pressure cp (J/kg·K), thermal conductivity k (W/m·K). Continuum behavior; properties evaluated at film temperature. Concept / Approach: Prandtl number: Pr = ν / α, where ν = μ/ρ is kinematic viscosity and α = k/(ρcp) is thermal diffusivity. Eliminating ρ gives Pr = (μ/ρ) / (k/(ρcp)) = μ * cp / k. Step-by-Step Solution: Start from Pr = ν / α.Substitute ν = μ/ρ and α = k/(ρcp).Simplify: Pr = (μ/ρ) * (ρcp/k) = μcp/k.Thus the given definition is correct. Verification / Alternative check: Check units: μcp/k is dimensionless, satisfying the requirement for a similarity parameter. Why Other Options Are Wrong: Limiting validity to gases or liquids is incorrect; the definition is general. Temperature restrictions are not part of the definition, though property variation with temperature affects numerical values. Common Pitfalls: Confusing Pr with Reynolds or Nusselt numbers; mixing up cp and cv; using inconsistent units causing numerical errors. Final Answer: Yes

Question 11

Modes of heat transfer by molecular collisions  In which phase of matter is conductive heat transfer by molecular collision the smallest, all else being equal?

Accepted Answer

Introduction / Context: Conduction arises from molecular interactions (collisions and energy exchange). The efficiency of this mechanism depends strongly on intermolecular spacing and bonding, which vary across solids, liquids, and gases. Given Data / Assumptions: No bulk motion (pure conduction comparison). Similar temperature gradients and characteristic length. Typical engineering materials at standard conditions. Concept / Approach: Thermal conductivity k reflects how readily energy is transferred microscopically. In solids, lattice vibrations and free electrons (in metals) enable high k. In liquids, closer molecular proximity yields moderate k. In gases, large mean free paths and low density give the smallest k under comparable conditions. Step-by-Step Solution: Compare typical k values: metals >> nonmetal solids ≈ liquids > gases.Gases have fewer collisions per unit time due to low density.Therefore, the smallest conductive transfer by molecular collision occurs in gases. Verification / Alternative check: Practical evidence: still air is a good insulator; adding convection (air motion) dramatically increases heat transfer, confirming conduction alone is weak. Why Other Options Are Wrong: Solids/liquids exhibit higher conductivities than gases in most cases. “None of these” contradicts empirical data. Common Pitfalls: Confusing conduction with convection; assuming water (a liquid) always conducts poorly—while not high, it is superior to air. Final Answer: Gases

Question 12

Heat conduction through a two-layer composite slab (equal thicknesses)  Two layers of different materials with conductivities k1 and k2 are placed in series; each layer has the same thickness. What is the equivalent thermal conductivity of the compo…

Accepted Answer

Introduction / Context: Composite walls are common in insulation and heat-exchanger design. Determining an equivalent thermal conductivity allows engineers to treat multiple layers as a single homogeneous slab for steady, one-dimensional conduction. Given Data / Assumptions: Two layers in series, equal thickness L for each. Thermal conductivities: k1 and k2. Same cross-sectional area A; steady 1-D conduction; contact resistance neglected. Concept / Approach: The total thermal resistance in series adds: R_total = L/(k1A) + L/(k2A). Define an equivalent single slab of thickness 2L and conductivity k_eq such that R_total = 2L/(k_eqA). Step-by-Step Solution: R_total = L/(k1A) + L/(k2A) = (L/A) * (1/k1 + 1/k2).Set 2L/(k_eqA) = (L/A) * (1/k1 + 1/k2).Divide by L/A: 2/k_eq = 1/k1 + 1/k2.Solve for k_eq: k_eq = 2k1k2 / (k1 + k2). Verification / Alternative check: Check symmetry: if k1 = k2 = k, then k_eq = 2kk/(2k) = k, as expected. Why Other Options Are Wrong: (k1 + k2)/2 and sqrt(k1k2) are for parallel or other averaging concepts, not series conduction with equal thickness. k1*k2 and (k1^2 + k2^2)/(k1 + k2) do not satisfy resistance addition. Common Pitfalls: Confusing series and parallel arrangements; forgetting that thickness doubles in the equivalent single-layer model. Final Answer: k_eq = 2 * k1 * k2 / (k1 + k2)

Question 13

Psychrometrics terminology  The wet-bulb depression (DBT − WBT) is primarily an indicator of which humidity measure of air?

Accepted Answer

Introduction / Context: Wet-bulb depression—the difference between dry-bulb temperature (DBT) and wet-bulb temperature (WBT)—is widely used in field measurements to estimate moisture content and comfort conditions. Given Data / Assumptions: DBT measured with a standard dry thermometer. WBT measured with a wetted wick and adequate air motion. Near-atmospheric pressure; ideal gas approximation for moist air. Concept / Approach: As relative humidity (RH) decreases (air becomes drier), evaporation from the wet bulb increases, lowering WBT and enlarging the depression. Conversely, as RH approaches 100%, WBT → DBT and the depression tends to zero. Step-by-Step Solution: Compute depression: ΔT = DBT − WBT.Interpretation: large ΔT → strong evaporative potential → low RH.Therefore, wet-bulb depression is primarily an indicator of relative humidity. Verification / Alternative check: Psychrometric tables convert DBT and WBT directly to RH; a one-to-one mapping exists for fixed barometric pressure. Why Other Options Are Wrong: Absolute/specific humidity depend on total moisture content but are not read directly from the depression alone without charts/tables. Adiabatic saturation temperature is a related concept but not the quantity indicated by the depression per se. Common Pitfalls: Interpreting depression without accounting for barometric pressure in precise calculations; however, qualitatively it reflects RH. Final Answer: Relative humidity

Question 14

Thermal radiation terminology  If the energy radiated per second per square centimetre for wavelengths between λ and λ + dλ is eλ dλ, what is eλ called?

Accepted Answer

Introduction / Context: Radiative heat transfer depends on how emission varies with wavelength. Distinguishing between total and spectral quantities is fundamental for applying laws such as Planck’s, Wien’s, and Stefan–Boltzmann. Given Data / Assumptions: Definition: energy in the wavelength interval [λ, λ + dλ] is eλ dλ per unit time per unit area. Surface is isothermal and diffuse for simplicity. Concept / Approach: Total emissive power E is the integral over all wavelengths. The integrand eλ is the spectral (monochromatic) emissive power, giving the distribution of emission with wavelength. Step-by-Step Solution: Write E = ∫ eλ dλ over the spectrum.Identify eλ as the rate per unit wavelength interval.Hence, eλ is termed spectral emissive power. Verification / Alternative check: Dimensional check: eλ has units of W·m⁻²·μm⁻¹ (or per metre), consistent with a spectral density; integrating over wavelength yields W·m⁻². Why Other Options Are Wrong: Emissive power without “spectral” denotes the total over all wavelengths. Emissivity is a dimensionless property (ratio to a blackbody). Absorptive power and irradiation refer to absorption and incident radiation, respectively. Common Pitfalls: Confusing spectral emissive power with emissivity; mixing per-wavelength and total quantities. Final Answer: Spectral (monochromatic) emissive power

Question 15

Coil performance in sensible cooling  For a sensible cooling process across a cooling coil, the coil efficiency equals which of the following expressions in terms of by-pass factor (BPF)?

Accepted Answer

Introduction / Context: Cooling coil performance is often expressed via by-pass factor (BPF) and coil efficiency. Understanding their relationship streamlines quick estimates of outlet air conditions relative to the apparatus dew point (ADP). Given Data / Assumptions: Sensible cooling (moisture change negligible for the relation). Standard definition of BPF and ADP; uniform face velocity and distribution. Concept / Approach: By definition, for cooling: BPF = (t_out − t_ADP) / (t_in − t_ADP). Coil efficiency measures approach to the ADP: efficiency = (t_in − t_out) / (t_in − t_ADP) = 1 − BPF. Step-by-Step Solution: Write BPF in terms of temperatures.Rearrange for (t_in − t_out) / (t_in − t_ADP).Obtain coil efficiency = 1 − BPF. Verification / Alternative check: If BPF = 0 (perfect coil with no bypass), efficiency = 1 (100%); if BPF = 1 (complete bypass), efficiency = 0—consistent with the formula. Why Other Options Are Wrong: BPF − 1 and 1 + BPF can yield negative or >1 values, not physically meaningful. Equaling BPF or 1/BPF does not match the standard definition. Common Pitfalls: Mixing heating and cooling forms; forgetting that efficiency measures approach to ADP, not to entering air. Final Answer: Coil efficiency = 1 − BPF

Question 16

Refrigeration safety and controls  When discharge (head) pressure in a refrigeration system rises excessively, a high-pressure control should stop the compressor. State whether this is correct.

Accepted Answer

Introduction / Context: High-pressure (HP) controls are critical protective devices in vapour-compression systems, preventing dangerous operating conditions and equipment damage when condensing pressure becomes excessive. Given Data / Assumptions: Compressor with appropriate HP cut-out switch. Potential causes: condenser fouling, cooling water failure, fan failure, overcharge, non-condensables. Concept / Approach: Compressor power and discharge temperature rise with head pressure. HP controls monitor discharge pressure and open the control circuit when pressure exceeds a safe limit, shutting the compressor to avert damage. Step-by-Step Solution: Detect abnormal rise in condensing pressure.HP switch trips at setpoint (e.g., corresponding to safe saturation temperature).Compressor stops, allowing investigation and correction (restore cooling water/airflow, clean condenser, purge non-condensables). Verification / Alternative check: Manufacturers specify HP cut-out settings in service manuals; safety codes require these protections for most systems. Why Other Options Are Wrong: Limiting correctness to a specific condenser type or expansion device is incorrect; HP protection is universal best practice. Common Pitfalls: Bypassing a tripping HP control without diagnosing the root cause; mis-setting the differential leading to short-cycling. Final Answer: Correct

Question 17

Critical temperature (thermodynamics)  Select the correct statement that defines the critical temperature of a substance.

Accepted Answer

Introduction / Context: Critical properties are pivotal for refrigeration, power cycles, and supercritical fluids. The critical temperature establishes the boundary between subcritical condensation and supercritical behavior. Given Data / Assumptions: Pure substance; equilibrium thermodynamics. Critical point corresponds to Tc, pc, and critical specific volume. Concept / Approach: At temperatures above the critical temperature Tc, distinct liquid and vapor phases cease to exist; compression alone cannot induce liquefaction. Below Tc, sufficient pressure can condense the vapor. Step-by-Step Solution: Identify Tc as the upper limit for phase coexistence.Interpretation: for T > Tc, only a supercritical fluid exists; applying pressure does not cross a phase boundary.Therefore, the correct definition is that above Tc, a gas cannot be liquefied by pressure alone. Verification / Alternative check: Examine CO₂: Tc ≈ 31 °C. At 40 °C, even very high pressures do not yield a liquid phase—consistent with the definition. Why Other Options Are Wrong: Obedience to gas laws is unrelated; “always liquefied below Tc” is false without sufficient pressure; explosion is not part of the definition; maximum density statement is unrelated. Common Pitfalls: Confusing “cannot liquefy above Tc” with “cannot compress”—compression still increases density but without phase change. Final Answer: Above it, a gas will never be liquefied by pressure alone

Question 18

Why vapour-compression systems use a throttle valve instead of an expander  In a vapour compression refrigeration system, a throttle (expansion) valve is used in place of an ideal expander primarily because:

Accepted Answer

Introduction / Context: Real vapour-compression systems expand refrigerant through a throttling device (capillary/TEV/EEV) rather than a turbine-type expander. Understanding why helps clarify practical system design and economics. Given Data / Assumptions: Typical refrigerants at HVAC/R temperature lifts. Expansion occurs from condenser pressure to evaporator pressure. Two-phase region often encountered during expansion. Concept / Approach: While an isentropic expander could, in principle, recover some work and reduce throttling losses, the recoverable shaft work is small for common refrigerants and conditions. The mechanical complexity, cost, reliability issues, and two-phase nature of flow make expanders unattractive in standard systems. Step-by-Step Solution: Consider ideal expansion work: W_rec ≈ h_in − h_out (isentropic).For typical refrigerants, this enthalpy drop is modest across the valve.Balance recovered work against added capital, efficiency penalties, and maintenance.Conclusion: throttling is preferred; expander use is limited to specialized cycles. Verification / Alternative check: Exergy analysis shows throttling destroys availability, but the absolute value of recoverable work in standard HVAC ranges is too small to justify turbines; contrast with cryogenic plants where expanders are indeed used. Why Other Options Are Wrong: Mass reduction and condenser size claims are not the primary drivers. Cost argument alone is incomplete without acknowledging the negligible work potential in typical applications. Expander does not increase evaporator pressure. Common Pitfalls: Assuming any reduction in throttling loss is automatically economical; ignoring two-phase expansion and erosion risks in micro-turbomachinery. Final Answer: The positive work obtainable from isentropic expansion across the valve is very small

Question 19

In a standard vapor-compression refrigeration system, what is the thermodynamic state of the refrigerant immediately before it enters the condenser (i.e., at the compressor discharge)?

Accepted Answer

Introduction / Context: Identifying the correct state of the refrigerant at key points in the vapor-compression refrigeration cycle is fundamental for design, control, and diagnostics. The point immediately before the condenser is the compressor discharge, and understanding this state helps explain why condensers are sized for both desuperheating and condensation duties. Given Data / Assumptions: Cycle sequence: evaporator → compressor → condenser → expansion device. Compressor handles low-pressure vapour and delivers high-pressure vapour. Real compressors add sensible superheat to the discharge stream. Concept / Approach: During compression, work input raises both pressure and temperature of the refrigerant. Because the gas is compressed from an evaporator outlet state (usually slightly superheated to avoid liquid slugging) to a higher pressure, the discharge temperature exceeds the saturation temperature corresponding to the condenser pressure. Therefore, the refrigerant at the condenser inlet is superheated vapour, not saturated vapour or liquid. Step-by-Step Solution: 1) Evaporator outlet: low-pressure vapour, typically slightly superheated to protect the compressor.2) Compressor action: pressure increases sharply; so does temperature due to work input.3) At condenser inlet (compressor discharge): temperature is above saturation temperature at the new high pressure → superheated vapour.4) In the condenser: first, desuperheating reduces temperature to saturation; then condensation removes latent heat; finally, possible subcooling of liquid. Verification / Alternative check: A temperature–entropy (T–s) or pressure–enthalpy (p–h) diagram shows the maximum temperature at compressor discharge, located to the right of the saturation dome (superheated region) at condenser pressure. Why Other Options Are Wrong: Saturated liquid (a) and subcooled liquid (e) occur after condensation, not before the condenser. Wet vapour (b) is two-phase; discharge lines do not contain two-phase under normal operation. Dry saturated vapour (c) would mean no superheat; real compressors deliver superheat due to inefficiencies and work input. Common Pitfalls: Confusing the point “before the condenser” with “after the condenser.” Remember, the condenser handles hot, high-pressure superheated vapour first, not liquid. Final Answer: Superheated vapour

Question 20

In psychrometry, what do we call the temperature indicated by a thermometer whose bulb is covered with a wetted wick and exposed to moving air?

Accepted Answer

Introduction / Context: Wet-bulb temperature is a cornerstone concept in HVAC, meteorology, and drying processes. It reflects the combined effects of sensible heat and latent cooling due to evaporation and is essential for calculating humidity ratio, enthalpy, and equipment loads using the psychrometric chart. Given Data / Assumptions: A thermometer bulb is wrapped with a wet wick (wick kept moist). Air stream moves across the wetted bulb to enable evaporation. Steady ambient pressure is assumed. Concept / Approach: As air flows over the wetted wick, water evaporates from the wick surface. Evaporation requires latent heat, which is drawn from the water-film/thermometer bulb, reducing its temperature. The temperature indicated after reaching equilibrium is the wet-bulb temperature. At 100% relative humidity, no net evaporation occurs and wet-bulb equals dry-bulb temperature; at lower humidity, wet-bulb is lower than dry-bulb by an amount related to moisture deficit. Step-by-Step Solution: 1) Start with a wetted wick on the thermometer bulb.2) Pass air across the bulb: evaporation removes heat from the bulb film.3) The bulb temperature drops until heat removed by evaporation equals heat gained from sensible transfer from the air.4) The stable reading is the wet-bulb temperature. Verification / Alternative check: A sling psychrometer uses two thermometers: one dry-bulb and one wet-bulb. Their readings, along with psychrometric relations, yield humidity ratio, relative humidity, and enthalpy of moist air. Why Other Options Are Wrong: Dry-bulb temperature (b) is measured with a normal thermometer unaffected by evaporation. Dew-point (c) is the saturation temperature at the existing vapour partial pressure, not measured by a wetted wick. Adiabatic saturation temperature (d) is a related theoretical construct but not the direct thermometer reading with a wick. Globe temperature (e) measures radiant heat effects with a black globe, not evaporation. Common Pitfalls: Assuming wet-bulb equals dew-point universally; they are equal only at saturation (100% RH) when both also equal dry-bulb. Final Answer: Wet-bulb temperature

Heat Transfer, Refrigeration and Air Conditioning Questions