Q: Refrigerant property recall — normal boiling point of R-12 (dichlorodifluoromethane) Is the approximate normal boiling point of R-12 about −30 °C under 1 bar atmospheric pressure?

Introduction / Context: Older vapor-compression systems frequently referenced R-12 (CCl2F2) in teaching materials. Knowing its normal boiling point helps understand why it was suitable for subzero evaporator temperatures at moderate suction pressures. Given Data / Assumptions: Normal boiling point refers to saturation at approximately 1 bar (standard atmospheric pressure). R-12 property value commonly cited near −29.8 °C. Concept / Approach: A refrigerant with a low normal boiling point can evaporate at low temperatures while maintaining manageable pressures, enabling efficient refrigeration for comfort and medium-temperature applications. Step-by-Step Solution: Recall property: T_b,normal for R-12 ≈ −29.8 °C.Compare to the statement “about −30 °C.”Conclusion: the statement is correct within engineering rounding. Verification / Alternative check: Reference pressure–temperature charts for R-12 show saturation around −30 °C at approximately 1 bar absolute. Why Other Options Are Wrong: Values like −10 °C or above 0 °C would imply much higher saturation temperatures inconsistent with known R-12 data; restricting to “reduced pressure only” misunderstands the definition of normal boiling point. Common Pitfalls: Confusing R-12 with R-22 or R-134a, which have different boiling points; always confirm the specific refrigerant. Final Answer: True

Q: Air-conditioning process selection on the psychrometric chart Outside air at 45 °C dry-bulb and 70% RH must be conditioned to a room state of 25 °C dry-bulb and 50% RH. Which practical treatment best achieves this target?

Introduction / Context: Determining the correct HVAC air-treatment sequence is a core psychrometrics skill. We compare initial hot-humid outdoor air with the desired cooler, moderately humid room condition. Given Data / Assumptions: Outdoor state: T_db = 45 °C, RH = 70% (high moisture and temperature). Room state: T_db = 25 °C, RH = 50%. Standard atmospheric pressure and well-mixed air streams. Concept / Approach: Moving from a hot, very humid point to a cooler, less humid point generally requires cooling air below its dew point, condensing moisture, then reheating if needed to the final dry-bulb. That combined action is called cooling and dehumidification (often via a cooling coil with surface temperature below dew point). Step-by-Step Solution: Plot 45 °C, 70% RH → high humidity ratio and enthalpy.Target is 25 °C, 50% RH → both lower temperature and lower humidity ratio.To reduce humidity ratio, pass across saturation by cooling below dew point → moisture condenses.Then, if the dry-bulb leaves too cold, apply sensible reheating to reach 25 °C at ≈50% RH. Verification / Alternative check: On the chart, a path leftward and downward (cooling and dehumidification) reaches a state near the target; simple dehumidification without cooling is not physically achievable for air without a desiccant process. Why Other Options Are Wrong: Dehumidification only: air temperature and humidity ratio cannot both be reduced without cooling below dew point or chemical drying.Cooling and humidification: humidity ratio would rise, not desired.Dehumidification then sensible heating: final 25 °C requires prior cooling below dew point; however, the net description is still “cooling and dehumidification.”Adiabatic humidification: adds moisture, opposite of need. Common Pitfalls: Stopping at sensible cooling only; this raises RH and can approach saturation, but does not lower humidity ratio to the required value. Final Answer: cooling and dehumidification

Q: Identify the law — heat transfer proportional to area and temperature difference The statement “heat flow rate between two bodies is directly proportional to the exposed surface area and the temperature difference between them” refers to which law?

Introduction / Context: Engineers often linearize convective heat transfer using an overall heat transfer coefficient. This linear form is rooted in a classical empirical law relating heat flow to temperature difference. Given Data / Assumptions: Surface temperature T_s and fluid bulk temperature T_∞. Convective heat transfer area A and coefficient h. Concept / Approach: Newton’s law of cooling states q = h * A * (T_s − T_∞). It treats the convective boundary layer as a lumped resistance, proportional to area and linearly dependent on temperature difference, with proportionality constant h capturing fluid properties and flow regime. Step-by-Step Solution: Match form: proportional to A and ΔT → q ∝ A * ΔT.Identify convection-specific constant → h.Therefore, the correct law is Newton’s law of cooling. Verification / Alternative check: Contrast: Fourier’s law is q = −k * A * dT/dx (conduction); Stefan–Boltzmann gives q = ε * σ * A * (T_s^4 − T_sur^4) (radiation), not linear in ΔT for large differences. Why Other Options Are Wrong: First law asserts energy conservation, not a transfer rate relation. “Newton’s law of heating” is not a standard term; heating or cooling both follow the same convection relation. Stefan’s law has T^4 dependence, not linear. Common Pitfalls: Applying Newton’s law outside its valid range (very large temperature differences with strong radiation or rapidly varying h). Final Answer: Newton's law of cooling

Q: Define absolute humidity precisely Absolute humidity is defined as the mass of water vapour contained in a unit volume of the air–vapour mixture. Is the statement “mass of water vapour present in 1 m³ of dry air” correct?

Introduction / Context: Psychrometric terminology distinguishes between absolute humidity, specific humidity (humidity ratio), and relative humidity. Precision matters in calculations of loads and coil selection. Given Data / Assumptions: Absolute humidity (sometimes called absolute moisture content) is mass of water vapour per unit volume of the moist air mixture. Specific humidity (humidity ratio) is mass of water vapour per unit mass of dry air. Concept / Approach: The phrase “per 1 m³ of dry air” belongs to humidity ratio, not absolute humidity. Absolute humidity uses the actual volume occupied by the moist mixture (air plus vapour) at its state point. Step-by-Step Solution: Identify the property: absolute humidity → kg_vapour per m³ of moist air.Identify the incorrect phrase: “per m³ of dry air” corresponds to humidity ratio when expressed as kg_vapour per kg_dry_air.Hence, the given statement is incorrect. Verification / Alternative check: Psychrometric chart axes and standard formulas separate volumetric measures (absolute humidity) from mass ratios (specific humidity); textbooks consistently define them this way. Why Other Options Are Wrong: Adding conditions like saturation or standard pressure does not change the fundamental definition; the denominator remains the moist mixture volume for absolute humidity. Common Pitfalls: Using terms “absolute” and “specific” interchangeably; this can lead to significant sizing errors because volume and mass bases differ. Final Answer: Disagree

Q: Absorption refrigeration pair identification — LiBr system In a lithium–bromide absorption refrigeration system used for air-conditioning, which working-pair role assignment is correct?

Introduction / Context: Absorption refrigeration systems use a refrigerant–absorbent pair instead of a mechanical compressor to circulate refrigerant. Lithium–bromide systems are common in large HVAC applications. Given Data / Assumptions: Application: air-conditioning (chilled water), not low-temperature freezing. LiBr–H2O pair operates under vacuum with water as the refrigerant. Concept / Approach: In LiBr systems, water evaporates in the evaporator (producing cooling) and is absorbed by a strong lithium–bromide solution in the absorber. The solution is then pumped to the generator where heat drives off the absorbed water vapour, which condenses and returns through the expansion device to the evaporator. Step-by-Step Solution: Identify refrigerant: the species that evaporates to produce cooling → water.Identify absorbent: the species that absorbs refrigerant vapour → lithium bromide solution.Therefore, the correct assignment is water as refrigerant and LiBr as absorbent. Verification / Alternative check: Contrast with ammonia–water systems, where ammonia is the refrigerant and water is the absorbent; LiBr cannot practically absorb ammonia for HVAC use. Why Other Options Are Wrong: Reversing roles contradicts the operating principle; ammonia/CO2 pairs are not the LiBr air-conditioning system. Common Pitfalls: Assuming the higher-mass fraction component is the refrigerant; in absorption, the low-boiling constituent is the refrigerant regardless of mass fraction in solution. Final Answer: water is the refrigerant and lithium bromide is the absorbent

Q: Control characteristic — another name for a thermostatic expansion valve (TEV) In vapor-compression systems, the thermostatic expansion valve is commonly referred to by which functional description?

Introduction / Context: Expansion devices regulate refrigerant flow to the evaporator. The thermostatic expansion valve (TEV or TXV) uses a sensing bulb to modulate flow according to evaporator outlet superheat. Given Data / Assumptions: TEV senses temperature at the evaporator outlet and compares it to pressure via an internal diaphragm. Goal: maintain a nearly constant superheat to avoid liquid floodback and ensure coil utilization. Concept / Approach: “Constant superheat valve” aptly describes the TEV’s control objective. As load increases, outlet superheat rises; the valve opens to feed more liquid. As load drops, superheat falls; the valve throttles to prevent liquid carryover. Step-by-Step Solution: Define superheat: T_actual − T_saturation at evaporator pressure.TEV modulates flow so superheat ≈ setpoint (e.g., 5–8 K).Hence, TEV ≡ “constant superheat valve.” Verification / Alternative check: System commissioning procedures specify setting or checking TEV superheat, confirming its functional name. Why Other Options Are Wrong: Constant pressure or temperature valves describe other regulators; fixed-orifice and float valves are different device types not using a thermostatic bulb. Common Pitfalls: Assuming TEVs hold evaporator pressure constant; that is influenced by compressor capacity control, not the TEV. Final Answer: constant superheat valve

Q: Psychrometrics — what decreases during sensible heating? During sensible heating of air at essentially constant moisture content, which property decreases as temperature rises?

Introduction / Context: Sensible heating raises the dry-bulb temperature without changing the amount of water vapour in the air. Understanding which psychrometric properties increase or decrease is crucial for HVAC control. Given Data / Assumptions: No moisture is added or removed (humidity ratio w constant). Standard atmospheric pressure; air behaves as an ideal mixture. Concept / Approach: Relative humidity (RH) is the ratio of the partial pressure of water vapour to the saturation pressure at the same temperature. As temperature increases at constant vapour content, saturation pressure increases strongly, so RH decreases. Step-by-Step Solution: Sensible heating: T_db increases, w constant.Saturation pressure p_ws(T) rises with T → denominator in RH increases.Therefore RH = p_v / p_ws decreases. Verification / Alternative check: On the psychrometric chart, sensible heating moves horizontally to the right; constant w lines indicate RH contours dropping as temperature rises. Why Other Options Are Wrong: Dry-bulb temperature increases, not decreases. Specific humidity (humidity ratio) remains constant in sensible processes. Wet-bulb temperature typically increases somewhat with sensible heating at constant w, not decreases. Common Pitfalls: Equating RH with moisture content; RH can fall even though the actual moisture in the air is unchanged. Final Answer: relative humidity

Question 1

Heating and humidification of air: The final relative humidity of the leaving air (after adding heat and moisture) is generally:

Accepted Answer

Introduction / Context: Heating and humidification is common in winter air-conditioning, where dry cold air is warmed and moisture is added. Predicting the final relative humidity (RH) requires understanding of how both dry-bulb temperature and absolute moisture content change simultaneously. Given Data / Assumptions: Humidifier adds water vapour (steam or evaporative), while a heater raises dry-bulb temperature. Process may be simultaneous (e.g., spray washer with heating) or sequential (heater then humidifier). Pressure remains near constant. Concept / Approach: RH is defined as p_w / p_ws(T). Adding moisture increases p_w; heating increases p_ws(T). Depending on the relative magnitudes of these changes, RH can rise or fall. On a psychrometric chart, heating shifts right; humidification shifts up; combined, the path can approach or recede from the saturation curve depending on rates of heat and moisture addition. Step-by-Step Solution: Consider strong heating with little humidification: T increases substantially, p_ws(T) rises, p_w rises only slightly → RH may decrease.Consider modest heating with strong humidification: p_w increases markedly while T rises a little → RH may increase.General outcome: RH can be higher or lower than at inlet; it is not fixed.Exact result requires mass and energy balances across the equipment. Verification / Alternative check: Plotting initial and final states on the chart confirms that different heater/humidifier capacities produce different RH results for the same inlet state. Why Other Options Are Wrong: Always higher / always lower: Over-generalizations; outcome depends on process intensity. Always 100%: Only occurs if the process reaches saturation, which is not guaranteed. Always unchanged: RH is sensitive to both T and moisture changes. Common Pitfalls: Assuming RH tracks moisture addition alone; ignoring the strong influence of temperature on saturation pressure. Final Answer: can be lower or higher than that of the entering air

Question 2

Overall heat transfer coefficient (U): In which type of problems do we use the overall heat transfer coefficient concept?

Accepted Answer

Introduction / Context: Real heat exchangers involve multiple layers and mechanisms: conduction through walls and fouling layers, plus convection on fluid sides. To simplify analysis, we combine these effects into a single parameter called the overall heat transfer coefficient U. Given Data / Assumptions: Steady-state heat transfer. Planar or cylindrical walls separating fluids. Thermal resistances can be added in series. Concept / Approach: The overall coefficient U encapsulates all resistances between two bulk fluids. For a flat wall, 1/U = 1/h_hot + R_cond + 1/h_cold (+ fouling terms). For a tube, appropriate logarithmic mean area is used. U enables the compact expression Q = U * A * ΔT_mean for exchanger design or rating. Step-by-Step Solution: Write individual resistances: R_conv,hot = 1/(h_hot * A), R_cond = L/(k * A), R_conv,cold = 1/(h_cold * A).Total resistance R_total = R_conv,hot + R_cond + R_conv,cold (+ fouling).Define U from Q = ΔT_overall / R_total = U * A * ΔT_mean, hence 1/U = (R_total / A).Apply appropriate area (inside/outside/log-mean) for cylindrical walls. Verification / Alternative check: Compare predictions with detailed CFD or layer-by-layer models; the series-resistance model provides good engineering estimates for design. Why Other Options Are Wrong: Using U for a single pure mechanism (conduction only or convection only) is unnecessary; radiation alone also does not require U unless combined with other modes in a composite resistance network. Common Pitfalls: Mixing areas when summing resistances in cylindrical systems; neglecting fouling, which significantly lowers U in service. Final Answer: Combined conduction and convection (resistances in series)

Question 3

Electrolux (Platen–Munters) absorption refrigerator: Which combination of working fluids is used?

Accepted Answer

Introduction / Context: The classic Electrolux (Platen–Munters) absorption refrigerator is a diffusion-absorption system widely used in off-grid or RV applications because it requires no mechanical compressor. Understanding the fluid trio clarifies how refrigeration is achieved without moving parts. Given Data / Assumptions: Absorber and generator handle an ammonia–water solution. Hydrogen is used as an inert gas to reduce ammonia partial pressure in the evaporator. Heat input at the generator replaces the work input of a compressor. Concept / Approach: In the Electrolux system, ammonia (refrigerant) is absorbed in water (absorbent). Heating the strong solution in the generator liberates ammonia vapour, which is condensed and then expanded into the evaporator. Hydrogen circulates on the low-pressure side to lower the partial pressure of ammonia, enabling ammonia to evaporate at a low temperature and absorb heat (refrigeration effect). Step-by-Step Solution: Generator: apply heat → ammonia separates from water (strong to weak solution).Condenser: ammonia vapour rejects heat and becomes liquid.Evaporator: liquid ammonia meets hydrogen; low ammonia partial pressure allows low-temperature evaporation, absorbing heat from the refrigerated space.Absorber: weak solution absorbs ammonia again, closing the loop. Verification / Alternative check: Mass and partial pressure balances show that the inert hydrogen permits evaporation at reduced ammonia partial pressure without a mechanical compressor. Why Other Options Are Wrong: Pairs like “water and hydrogen” or “ammonia and hydrogen” omit the necessary absorbent (water) or refrigerant; “ammonia and CO2” is unrelated to the Electrolux configuration. Common Pitfalls: Confusing the role of hydrogen (it is not condensed; it provides partial pressure control) and thinking water is the refrigerant (it is the absorbent). Final Answer: Ammonia, water, and hydrogen

Question 4

Liquid refrigerant cooled below its saturation temperature after condensation but before throttling is said to be undergoing:

Accepted Answer

Introduction / Context: Sub-cooling is a common enhancement in vapour-compression systems. Cooling the condensed liquid refrigerant below the saturation temperature at condenser pressure improves system capacity and reduces throttling losses. Given Data / Assumptions: Refrigerant exits condenser as saturated liquid at condenser pressure. A sub-cooler or liquid line heat exchanger removes additional sensible heat. Expansion device follows (valve/capillary). Concept / Approach: Sub-cooling means T_liquid < T_sat at the same pressure. This ensures the refrigerant entering the expansion device is fully liquid, minimizing flash vapour before the evaporator and increasing the refrigeration effect per unit mass. Step-by-Step Solution: Identify condenser outlet: saturated liquid at T_sat, P_cond.Remove additional sensible heat to reach T_out < T_sat (sub-cooled liquid).Throttle to evaporator pressure: reduced flash gas fraction; more liquid available to evaporate in the evaporator → higher capacity.Overall COP can improve slightly due to increased refrigeration effect for similar compressor work. Verification / Alternative check: h_subcooled < h_sat, so refrigeration effect (h_evap.out − h_exp.out) increases. P-h diagrams clearly show the benefit. Why Other Options Are Wrong: “Super-cooling” is non-standard wording; “normal cooling” is vague; “flash gas formation” occurs after throttling, which sub-cooling helps to reduce; “wet compression” refers to liquid carryover into the compressor and is undesirable. Common Pitfalls: Confusing sub-cooling (liquid below T_sat) with superheating (vapour above T_sat) and misplacing the process relative to the expansion valve. Final Answer: sub-cooling or under-cooling

Question 5

Parallel-flow heat exchanger concept check In a parallel-flow (co-current) heat exchanger, both fluids enter from the same end. Considering the temperature driving force along the length, is it correct to say that the instantaneous heat-transfer rat…

Accepted Answer

Introduction / Context: Heat exchangers transfer energy between two streams. In a parallel-flow (co-current) arrangement, both hot and cold fluids enter at the same end and move in the same direction. A frequent conceptual question is where the temperature driving force and cooling/heating rates are largest along the exchanger. Given Data / Assumptions: Two-stream, steady-state, parallel-flow heat exchanger. Negligible heat loss to surroundings. Conventional definition: local heat-transfer rate q' = U * ΔT_local per unit area, where U is an overall coefficient. Concept / Approach: In parallel flow, the temperature difference ΔT between hot and cold streams is largest at the inlet because the hot fluid is at its hottest while the cold fluid is at its coldest. Since local q' is proportional to ΔT_local, the instantaneous rate of heat transfer per area and the rate at which each stream's temperature changes with distance are maximum at the inlet and decrease downstream as the two temperatures approach each other. Step-by-Step Solution: Recognize co-current entry → hot at maximum T_h,in meets cold at minimum T_c,in.Compute local driving force: ΔT_inlet = T_h,in − T_c,in (largest in the exchanger).Relate local heat flux: q' = U * ΔT_local → maximum at inlet.Downstream, ΔT_local decreases → smaller q' and smaller temperature gradients. Verification / Alternative check: For the same terminal temperatures, the LMTD for parallel flow is lower than for counter flow; however, this does not contradict the fact that within a parallel exchanger the maximum local ΔT occurs right at the inlet. Why Other Options Are Wrong: No: ignores the inlet driving-force peak inherent to co-current entry.Only for gases / equal heat capacity rates / low Reynolds number: the inlet ΔT peak is geometric, not limited to specific fluids or flow regimes. Common Pitfalls: Confusing local inlet behavior with overall effectiveness; counter flow is generally more effective overall, yet parallel flow still has its maximum local driving force at the inlet. Final Answer: Yes

Question 6

Refrigerant property recall — normal boiling point of R-12 (dichlorodifluoromethane) Is the approximate normal boiling point of R-12 about −30 °C under 1 bar atmospheric pressure?

Accepted Answer

Introduction / Context: Older vapor-compression systems frequently referenced R-12 (CCl2F2) in teaching materials. Knowing its normal boiling point helps understand why it was suitable for subzero evaporator temperatures at moderate suction pressures. Given Data / Assumptions: Normal boiling point refers to saturation at approximately 1 bar (standard atmospheric pressure). R-12 property value commonly cited near −29.8 °C. Concept / Approach: A refrigerant with a low normal boiling point can evaporate at low temperatures while maintaining manageable pressures, enabling efficient refrigeration for comfort and medium-temperature applications. Step-by-Step Solution: Recall property: T_b,normal for R-12 ≈ −29.8 °C.Compare to the statement “about −30 °C.”Conclusion: the statement is correct within engineering rounding. Verification / Alternative check: Reference pressure–temperature charts for R-12 show saturation around −30 °C at approximately 1 bar absolute. Why Other Options Are Wrong: Values like −10 °C or above 0 °C would imply much higher saturation temperatures inconsistent with known R-12 data; restricting to “reduced pressure only” misunderstands the definition of normal boiling point. Common Pitfalls: Confusing R-12 with R-22 or R-134a, which have different boiling points; always confirm the specific refrigerant. Final Answer: True

Question 7

Air-conditioning process selection on the psychrometric chart Outside air at 45 °C dry-bulb and 70% RH must be conditioned to a room state of 25 °C dry-bulb and 50% RH. Which practical treatment best achieves this target?

Accepted Answer

Introduction / Context: Determining the correct HVAC air-treatment sequence is a core psychrometrics skill. We compare initial hot-humid outdoor air with the desired cooler, moderately humid room condition. Given Data / Assumptions: Outdoor state: T_db = 45 °C, RH = 70% (high moisture and temperature). Room state: T_db = 25 °C, RH = 50%. Standard atmospheric pressure and well-mixed air streams. Concept / Approach: Moving from a hot, very humid point to a cooler, less humid point generally requires cooling air below its dew point, condensing moisture, then reheating if needed to the final dry-bulb. That combined action is called cooling and dehumidification (often via a cooling coil with surface temperature below dew point). Step-by-Step Solution: Plot 45 °C, 70% RH → high humidity ratio and enthalpy.Target is 25 °C, 50% RH → both lower temperature and lower humidity ratio.To reduce humidity ratio, pass across saturation by cooling below dew point → moisture condenses.Then, if the dry-bulb leaves too cold, apply sensible reheating to reach 25 °C at ≈50% RH. Verification / Alternative check: On the chart, a path leftward and downward (cooling and dehumidification) reaches a state near the target; simple dehumidification without cooling is not physically achievable for air without a desiccant process. Why Other Options Are Wrong: Dehumidification only: air temperature and humidity ratio cannot both be reduced without cooling below dew point or chemical drying.Cooling and humidification: humidity ratio would rise, not desired.Dehumidification then sensible heating: final 25 °C requires prior cooling below dew point; however, the net description is still “cooling and dehumidification.”Adiabatic humidification: adds moisture, opposite of need. Common Pitfalls: Stopping at sensible cooling only; this raises RH and can approach saturation, but does not lower humidity ratio to the required value. Final Answer: cooling and dehumidification

Question 8

Identify the law — heat transfer proportional to area and temperature difference The statement “heat flow rate between two bodies is directly proportional to the exposed surface area and the temperature difference between them” refers to which law?

Accepted Answer

Introduction / Context: Engineers often linearize convective heat transfer using an overall heat transfer coefficient. This linear form is rooted in a classical empirical law relating heat flow to temperature difference. Given Data / Assumptions: Surface temperature T_s and fluid bulk temperature T_∞. Convective heat transfer area A and coefficient h. Concept / Approach: Newton’s law of cooling states q = h * A * (T_s − T_∞). It treats the convective boundary layer as a lumped resistance, proportional to area and linearly dependent on temperature difference, with proportionality constant h capturing fluid properties and flow regime. Step-by-Step Solution: Match form: proportional to A and ΔT → q ∝ A * ΔT.Identify convection-specific constant → h.Therefore, the correct law is Newton’s law of cooling. Verification / Alternative check: Contrast: Fourier’s law is q = −k * A * dT/dx (conduction); Stefan–Boltzmann gives q = ε * σ * A * (T_s^4 − T_sur^4) (radiation), not linear in ΔT for large differences. Why Other Options Are Wrong: First law asserts energy conservation, not a transfer rate relation. “Newton’s law of heating” is not a standard term; heating or cooling both follow the same convection relation. Stefan’s law has T^4 dependence, not linear. Common Pitfalls: Applying Newton’s law outside its valid range (very large temperature differences with strong radiation or rapidly varying h). Final Answer: Newton's law of cooling

Question 9

Define absolute humidity precisely Absolute humidity is defined as the mass of water vapour contained in a unit volume of the air–vapour mixture. Is the statement “mass of water vapour present in 1 m³ of dry air” correct?

Accepted Answer

Introduction / Context: Psychrometric terminology distinguishes between absolute humidity, specific humidity (humidity ratio), and relative humidity. Precision matters in calculations of loads and coil selection. Given Data / Assumptions: Absolute humidity (sometimes called absolute moisture content) is mass of water vapour per unit volume of the moist air mixture. Specific humidity (humidity ratio) is mass of water vapour per unit mass of dry air. Concept / Approach: The phrase “per 1 m³ of dry air” belongs to humidity ratio, not absolute humidity. Absolute humidity uses the actual volume occupied by the moist mixture (air plus vapour) at its state point. Step-by-Step Solution: Identify the property: absolute humidity → kg_vapour per m³ of moist air.Identify the incorrect phrase: “per m³ of dry air” corresponds to humidity ratio when expressed as kg_vapour per kg_dry_air.Hence, the given statement is incorrect. Verification / Alternative check: Psychrometric chart axes and standard formulas separate volumetric measures (absolute humidity) from mass ratios (specific humidity); textbooks consistently define them this way. Why Other Options Are Wrong: Adding conditions like saturation or standard pressure does not change the fundamental definition; the denominator remains the moist mixture volume for absolute humidity. Common Pitfalls: Using terms “absolute” and “specific” interchangeably; this can lead to significant sizing errors because volume and mass bases differ. Final Answer: Disagree

Question 10

Absorption refrigeration pair identification — LiBr system In a lithium–bromide absorption refrigeration system used for air-conditioning, which working-pair role assignment is correct?

Accepted Answer

Introduction / Context: Absorption refrigeration systems use a refrigerant–absorbent pair instead of a mechanical compressor to circulate refrigerant. Lithium–bromide systems are common in large HVAC applications. Given Data / Assumptions: Application: air-conditioning (chilled water), not low-temperature freezing. LiBr–H2O pair operates under vacuum with water as the refrigerant. Concept / Approach: In LiBr systems, water evaporates in the evaporator (producing cooling) and is absorbed by a strong lithium–bromide solution in the absorber. The solution is then pumped to the generator where heat drives off the absorbed water vapour, which condenses and returns through the expansion device to the evaporator. Step-by-Step Solution: Identify refrigerant: the species that evaporates to produce cooling → water.Identify absorbent: the species that absorbs refrigerant vapour → lithium bromide solution.Therefore, the correct assignment is water as refrigerant and LiBr as absorbent. Verification / Alternative check: Contrast with ammonia–water systems, where ammonia is the refrigerant and water is the absorbent; LiBr cannot practically absorb ammonia for HVAC use. Why Other Options Are Wrong: Reversing roles contradicts the operating principle; ammonia/CO2 pairs are not the LiBr air-conditioning system. Common Pitfalls: Assuming the higher-mass fraction component is the refrigerant; in absorption, the low-boiling constituent is the refrigerant regardless of mass fraction in solution. Final Answer: water is the refrigerant and lithium bromide is the absorbent

Question 11

Control characteristic — another name for a thermostatic expansion valve (TEV) In vapor-compression systems, the thermostatic expansion valve is commonly referred to by which functional description?

Accepted Answer

Introduction / Context: Expansion devices regulate refrigerant flow to the evaporator. The thermostatic expansion valve (TEV or TXV) uses a sensing bulb to modulate flow according to evaporator outlet superheat. Given Data / Assumptions: TEV senses temperature at the evaporator outlet and compares it to pressure via an internal diaphragm. Goal: maintain a nearly constant superheat to avoid liquid floodback and ensure coil utilization. Concept / Approach: “Constant superheat valve” aptly describes the TEV’s control objective. As load increases, outlet superheat rises; the valve opens to feed more liquid. As load drops, superheat falls; the valve throttles to prevent liquid carryover. Step-by-Step Solution: Define superheat: T_actual − T_saturation at evaporator pressure.TEV modulates flow so superheat ≈ setpoint (e.g., 5–8 K).Hence, TEV ≡ “constant superheat valve.” Verification / Alternative check: System commissioning procedures specify setting or checking TEV superheat, confirming its functional name. Why Other Options Are Wrong: Constant pressure or temperature valves describe other regulators; fixed-orifice and float valves are different device types not using a thermostatic bulb. Common Pitfalls: Assuming TEVs hold evaporator pressure constant; that is influenced by compressor capacity control, not the TEV. Final Answer: constant superheat valve

Question 12

Psychrometrics — what decreases during sensible heating? During sensible heating of air at essentially constant moisture content, which property decreases as temperature rises?

Accepted Answer

Introduction / Context: Sensible heating raises the dry-bulb temperature without changing the amount of water vapour in the air. Understanding which psychrometric properties increase or decrease is crucial for HVAC control. Given Data / Assumptions: No moisture is added or removed (humidity ratio w constant). Standard atmospheric pressure; air behaves as an ideal mixture. Concept / Approach: Relative humidity (RH) is the ratio of the partial pressure of water vapour to the saturation pressure at the same temperature. As temperature increases at constant vapour content, saturation pressure increases strongly, so RH decreases. Step-by-Step Solution: Sensible heating: T_db increases, w constant.Saturation pressure p_ws(T) rises with T → denominator in RH increases.Therefore RH = p_v / p_ws decreases. Verification / Alternative check: On the psychrometric chart, sensible heating moves horizontally to the right; constant w lines indicate RH contours dropping as temperature rises. Why Other Options Are Wrong: Dry-bulb temperature increases, not decreases. Specific humidity (humidity ratio) remains constant in sensible processes. Wet-bulb temperature typically increases somewhat with sensible heating at constant w, not decreases. Common Pitfalls: Equating RH with moisture content; RH can fall even though the actual moisture in the air is unchanged. Final Answer: relative humidity

Question 13

Mode of heat transfer by buoyancy-driven motion Heat transfer caused by the actual movement of fluid particles due to density differences produced by temperature variations is called:

Accepted Answer

Introduction / Context: Heat in fluids moves by both diffusion (molecular conduction) and advection (bulk motion). When bulk motion arises naturally from buoyancy, the mechanism has a specific name with distinct correlations. Given Data / Assumptions: Temperature gradients create density gradients in a gravitational field. No external devices (fans or pumps) drive the flow. Concept / Approach: Free (natural) convection occurs when buoyancy forces due to density differences overcome viscous resistance, causing motion and convective heat transfer. The characteristic dimensionless groups are Grashof and Rayleigh numbers, contrasting with Reynolds for forced convection. Step-by-Step Solution: Identify driver: density difference from temperature → buoyancy.Absence of external forcing → “free” or “natural.”Hence, the correct term is free convection. Verification / Alternative check: Examples include heating of still air by a radiator, cooling of a hot vertical plate, and plume rise from a candle—each occurs without fans or pumps. Why Other Options Are Wrong: Conduction involves energy transfer by molecular interaction without bulk motion. Forced convection requires an external driver (fan, pump). Radiation transfers energy by electromagnetic waves and does not require a medium. Common Pitfalls: At low Grashof numbers, natural convection may be weak and easily swamped by small external drafts—then the regime becomes mixed convection. Final Answer: free convection

Question 14

Refrigeration piping materials for halocarbon (Freon) refrigerants For systems using common halocarbon refrigerants (e.g., R-12, R-22, R-134a), which piping material is generally preferred for refrigerant lines?

Accepted Answer

Introduction / Context: Piping material selection in refrigeration depends on compatibility with refrigerant and oil, corrosion resistance, ease of joining, pressure capability, and cleanliness. Halocarbon (Freon-family) systems dominate comfort cooling and many commercial applications. Given Data / Assumptions: Closed vapor-compression system with halocarbon refrigerant and mineral or POE oils. Standard pressure ranges for comfort cooling and medium-temperature applications. Concept / Approach: Deoxidized, high-phosphorus copper (Type L or K) tubing is widely used because it is readily brazed, cleanable, corrosion-resistant in typical HVAC environments, and compatible with halocarbons and oils. It also withstands common operating pressures and allows good leak-tight joints. Step-by-Step Solution: Assess compatibility: copper + halocarbon refrigerants → compatible.Assess fabrication: copper allows reliable brazed joints and standard fittings.Thus, copper is the preferred line material for most Freon systems. Verification / Alternative check: Manufacturer installation standards specify copper tubing for suction, liquid, and discharge lines; steel is reserved more often for large ammonia systems or very high-pressure CO2 transcritical lines. Why Other Options Are Wrong: Brass is heavier and mainly used for valves/fittings, not full line runs. Steel is common with ammonia or industrial duties but less convenient for halocarbons in smaller systems. Aluminium and PVC lack the pressure/compatibility and joining robustness required. Common Pitfalls: Using soft solder instead of proper brazing alloys for high-pressure or high-temperature segments; always follow code requirements and purge with inert gas while brazing. Final Answer: copper

Question 15

Two-layer insulation on a steam pipe  A steam pipeline is to be insulated using two concentric layers made of different materials with different thermal conductivities. For minimum heat transfer from the hot pipe to the surroundings, where should th…

Accepted Answer

Introduction / Context: Composite cylindrical insulation is common in steam distribution. This question tests understanding of radial conduction and how arranging layers of different thermal conductivity affects overall thermal resistance and thus heat loss. Given Data / Assumptions: Two insulation layers of fixed inner radius r1 (pipe outer surface) and fixed outer radius r3 (overall insulation OD). Layer conductivities: k1 and k2, with one being the better insulator (lower k). Steady one-dimensional radial conduction; contact resistance neglected. Concept / Approach: For a hollow cylinder, radial conduction resistance is R = ln(r_out/r_in) / (2 * pi * k * L). For two layers, total resistance is the sum of individual resistances. For minimum heat transfer (maximum resistance), we choose the layer ordering that maximizes the sum. Step-by-Step Solution: Let r2 be the interface radius. Then R_total = ln(r2/r1)/(2pik1L) + ln(r3/r2)/(2pik2L).Maximize R_total with respect to r2 for fixed r1 and r3. The derivative with respect to r2 is proportional to (1/k1 − 1/k2)/r2.If k1 < k2 (layer 1 is the better insulator), then 1/k1 > 1/k2, so increasing r2 increases R_total. The maximum occurs when layer 1 occupies the larger radial span, i.e., it is placed inside adjacent to the pipe.Therefore, the lower-k material should be nearer the hot surface to maximize total resistance and minimize heat transfer. Verification / Alternative check: Numerical check with r1 = 1, r3 = 2, k_better = 0.05, k_poorer = 0.1 and r2 = 1.5 shows R_total is larger when the lower-k layer is the inner layer, confirming the derivative result. Why Other Options Are Wrong: The better insulation outside: reduces the logarithmic leverage of the low-k layer; gives lower total resistance. Either arrangement same: false; cylindrical geometry breaks this symmetry. Depends primarily on steam temperature: placement effect is geometric and conductivity-driven, not temperature-driven (within property limits). Higher-k inside to avoid stresses: not a heat-loss minimization argument. Common Pitfalls: Assuming series layers behave like plane walls (where order does not matter); forgetting the ln(r) dependence in cylinders. Final Answer: The better insulation must be put inside (closer to the pipe wall).

Question 16

Modes of heat transfer  What do we call the process in which heat is transferred within a fluid (or along a surface to a fluid) due to the actual bulk motion of the heated particles?

Accepted Answer

Introduction / Context: Understanding the three primary modes of heat transfer—conduction, convection, and radiation—is foundational in thermal engineering and HVAC system design. Given Data / Assumptions: Fluids can move, creating bulk transport of energy. No special participating media effects are assumed. Concept / Approach: Convection is heat transfer between a surface and an adjacent moving fluid, or within a fluid, due to combined effects of conduction at the microscopic level and bulk motion (advection) of fluid parcels. The governing empirical relation is Newton’s law of cooling: q = h * A * (T_s − T_∞). Step-by-Step Solution: Identify the mechanism described: “actual motion of heated particles.”This points to convection (forced or natural), not purely molecular conduction or photon-based radiation.Therefore, select convection. Verification / Alternative check: Without motion, heat transfer through a stationary fluid is by conduction across the boundary layer. When buoyancy or external forces drive motion, convection dominates. Why Other Options Are Wrong: Conduction: energy transfer via molecular interactions without bulk motion. Radiation: energy transfer via electromagnetic waves, no medium required. Phase change: a thermal process but not a “mode” by itself—evaporation/condensation accompany convection. Common Pitfalls: Using the term “conduction in fluids” for what is actually convection; forgetting that h depends on flow regime and properties. Final Answer: Convection

Question 17

Vapour-compression cycle temperature levels  In a standard vapour compression refrigeration system, at which process point is the refrigerant at the lowest temperature during the cycle?

Accepted Answer

Introduction / Context: The temperature profile around the vapour-compression loop determines component sizing and frosting risk. Recognizing the coldest point helps diagnose evaporator performance and subcooling needs. Given Data / Assumptions: Idealized cycle with throttling (isenthalpic) expansion device. Negligible pressure drops in lines and heat exchangers. Concept / Approach: In throttling, the refrigerant undergoes a pressure drop with no work and no heat transfer (h_out ≈ h_in). The Joule–Thomson effect causes a drop in temperature when expanding from condenser pressure to evaporator pressure for typical refrigerants, producing the cycle’s lowest temperature at the expansion device outlet. Step-by-Step Solution: Compressor discharge is at high pressure and high temperature → not the minimum.Condenser rejects heat at near ambient condensing temperature → warmer than evaporator.Expansion valve outlet: sharp pressure drop yields the coldest state.Evaporator then absorbs heat, raising the refrigerant quality/temperature toward suction. Verification / Alternative check: On a T–s diagram, the throttling step is a vertical line with temperature decreasing to the evaporating level, the minimum point of the loop. Why Other Options Are Wrong: Compression raises temperature. Condensation occurs near ambient plus approach temperature. Within the evaporator, temperature increases slightly along the flow as heat is absorbed. Common Pitfalls: Confusing the evaporator outlet (warmer) with the expansion device outlet (coldest). Final Answer: Immediately after expansion/throttling (expansion device outlet)

Question 18

Air–water contact in an insulated spray chamber  Air passes through an insulated chamber containing water sprays maintained at a temperature higher than the entering air’s dew-point but lower than its dry-bulb temperature. The resulting air process …

Accepted Answer

Introduction / Context: Spray chambers are classical psychrometric devices where air exchanges sensible and latent heat with water droplets. The relative position of spray water temperature to the air’s dry-bulb (DBT) and dew point (DPT) governs whether moisture is added or removed. Given Data / Assumptions: Insulated chamber → no external heat gain/loss. Spray water temperature T_w satisfies DPT < T_w < DBT_in. Adequate contact for approach toward adiabatic saturation line. Concept / Approach: If T_w is below the air DBT, the air tends to cool sensibly. If T_w is above the air dew point, the droplets remain unsaturated relative to the entering air, so evaporation from droplets to air can occur, adding moisture and increasing humidity ratio. Thus the combined effect is cooling (sensible) plus humidification (latent). Step-by-Step Solution: Check condensation criterion: since T_w > DPT, the air will not condense moisture on the droplets.Check sensible direction: since T_w < DBT_in, heat flows from air to water → air cools.Mass transfer: with air not at saturation and droplets present, evaporation increases humidity ratio. Verification / Alternative check: On the psychrometric chart, the path trends toward the adiabatic saturation line from the initial state, moving left (cooler DBT) and upward (higher humidity ratio). Why Other Options Are Wrong: Cooled and dehumidified: would require T_w below dew point or a cooled coil to condense moisture. Heated and humidified/dehumidified: inconsistent with T_w relative to DBT and DPT and with chamber insulation. Common Pitfalls: Assuming any spray always humidifies without temperature change; ignoring dew-point criterion for condensation versus evaporation. Final Answer: cooled and humidified

Question 19

Small-tonnage hermetic refrigeration units  In small commercial hermetically sealed machines (historical textbook context), which refrigerant has been most widely used?

Accepted Answer

Introduction / Context: Refrigerant choice for compact hermetic systems balances safety, pressure levels, lubricant compatibility, and compressor design. Historically, one refrigerant dominated small sealed units in textbooks and past exams. Given Data / Assumptions: Legacy perspective (pre-phaseout): domestic and light commercial hermetic units. Focus on commonly cited exam answers rather than today’s low-GWP replacements. Concept / Approach: R-12 provided moderate pressures, chemical stability, non-corrosiveness to most metals, and good miscibility with mineral oils—ideal for small hermetic compressors. Toxicity and very high operating pressures excluded ammonia and CO₂ from widespread domestic/small commercial hermetics of that era. Step-by-Step Solution: Compare safety/pressure: R-12 benign and moderate pressure; NH3 toxic; CO₂ very high pressure; SO₂ toxic/corrosive.Match to hermetic design constraints → R-12 fits best. Verification / Alternative check: Historical manufacturer catalogs and standard texts list R-12 for small sealed systems; many modern systems have shifted to R-600a or HFC/HFOs for environmental reasons. Why Other Options Are Wrong: Ammonia and sulphur dioxide: safety/material issues in small hermetics. CO₂: high pressure hardware, not typical for small hermetics historically. Mixtures listed are nonstandard for such units. Common Pitfalls: Applying current refrigerant trends to legacy exam questions without noting the time frame. Final Answer: R-12 (dichlorodifluoromethane)

Question 20

Absorption versus compression refrigeration  In a vapour absorption refrigeration system, what components collectively replace the mechanical compressor of a vapour-compression system?

Accepted Answer

Introduction / Context: Absorption refrigeration uses thermal energy rather than mechanical work for compression. Understanding the functional substitution of components helps in system selection where waste heat is available. Given Data / Assumptions: Working pair examples: NH3–H2O or LiBr–H2O. Steady operation; throttling devices present similar to compression systems. Concept / Approach: Instead of mechanically compressing refrigerant vapour, absorption systems chemically absorb the refrigerant into an absorbent (absorber), pump the liquid solution to high pressure (liquid pump), and then desorb the refrigerant vapour using heat (generator). The rectifier/analyzer may also be present depending on the pair. Step-by-Step Solution: Absorber: low-pressure vapour is absorbed, releasing heat.Pump: raises pressure of the solution (low power due to liquid pumping).Generator: heat input liberates high-pressure refrigerant vapour, functionally equivalent to compressor discharge. Verification / Alternative check: Thermodynamic cycles show the compressor work in vapour-compression is replaced by pump work plus generator heat input in absorption cycles. Why Other Options Are Wrong: Liquid pump only or generator only cannot accomplish both pressurization and vapour production functions. Absorber and generator without pump cannot raise pressure. Common Pitfalls: Ignoring the crucial role of the liquid pump in achieving high pressure with minimal mechanical work. Final Answer: Absorber, generator, and liquid pump

Heat Transfer, Refrigeration and Air Conditioning Questions