Q: Why aircraft use air-cycle refrigeration Aircraft commonly employ air (reversed Brayton) refrigeration cycles primarily because:

Introduction / Context: Air-cycle (reversed Brayton) systems are favored in aviation. Understanding the engineering trade-offs explains why COP is not the sole selection criterion. Given Data / Assumptions: Aircraft already carry air compressors/turbomachinery (bleed air/ram air). Weight, reliability, and maintenance are critical constraints. Concept / Approach: Although air-cycle COP can be lower than that of vapour-compression, the system offers major advantages: light weight, no refrigerant handling, tolerance to leaks, and effective use of available ram air and high-altitude low ambient temperatures. Step-by-Step Solution: Assess competing drivers: mass and simplicity dominate in aircraft.Air cycle uses air as the working fluid; no evaporator with phase change refrigerant is required, reducing complexity and weight.Altitude provides cold sink air to enhance cooling performance. Verification / Alternative check: Aircraft environmental control systems (ECS) widely use bootstrap/air-cycle packs that prioritize weight and reliability over peak COP. Why Other Options Are Wrong: Highest heat-transfer coefficients: not the primary justification. “Only” low ambient temperature: it helps but is not the sole reason. Higher COP than vapour cycles: generally untrue. No compressor: air cycle still uses a compressor or utilizes bleed/ram compression. Common Pitfalls: Equating selection solely with COP while ignoring weight and system integration. Final Answer: They provide low weight per tonne of refrigeration and simple, robust hardware.

Q: Naming pressures in a refrigeration compressor What is the standard term for the pressure measured at the inlet of a refrigerant compressor (i.e., the evaporator pressure at the suction line)?

Introduction / Context: Correct terminology helps in reading P–h charts, service gauges, and troubleshooting logs. The compressor has a low-pressure side (suction) and a high-pressure side (discharge). Given Data / Assumptions: Standard vapour-compression layout with evaporator, compressor, condenser, and expansion device. Inlet line to compressor originates from evaporator outlet/suction accumulator. Concept / Approach: The pressure at the compressor inlet is called suction pressure. It corresponds to the evaporating pressure and is read on the low-side gauge in service practice. Some texts also refer to evaporator pressure as back pressure; however, “suction pressure” is the unambiguous term at the compressor inlet. Step-by-Step Solution: Identify component: compressor inlet.Associate pressure name: suction pressure (low side).Note: discharge pressure is the high-side pressure at compressor outlet; critical pressure refers to fluid property, not line pressure. Verification / Alternative check: Service manifolds label LP gauge for suction (evaporator) pressure and HP gauge for discharge (condensing) pressure. Why Other Options Are Wrong: Discharge pressure: outlet, not inlet. Critical pressure: property at critical point, unrelated to component location. Back pressure: sometimes synonymous with evaporator pressure, but the standard inlet term is suction pressure. Common Pitfalls: Using “back pressure” interchangeably can cause confusion in multi-stage or cascade systems. Final Answer: Suction pressure

Q: Power requirement per tonne of refrigeration (PTR) For a vapour-compression plant, express the power per tonne of refrigeration (in kW per TR) in terms of the coefficient of performance (COP). Take 1 TR = 3.5 kW of cooling.

Introduction / Context: Linking COP to input power per unit cooling capacity is essential for comparing chiller and refrigeration plant efficiencies across sizes. Given Data / Assumptions: Definition: COP = Q_L / W_in. One tonne of refrigeration (1 TR) = 3.5 kW of cooling capacity. Concept / Approach: For 1 TR, Q_L = 3.5 kW. Rearranging the COP definition gives W_in = Q_L / COP. Therefore, power per TR is 3.5 / COP (kW per TR). Step-by-Step Solution: Start with COP = Q_L / W_in.Rearrange: W_in = Q_L / COP.For 1 TR, W_in = 3.5 / COP kW. Verification / Alternative check: Numerical example: COP = 3 → PTR = 3.5/3 ≈ 1.167 kW/TR. This matches typical plant data. Why Other Options Are Wrong: COP/3.5 or 3.5COP invert or misapply the relationship. 1/(3.5COP) has wrong dimensions and magnitude. Common Pitfalls: Confusing COP with EER or SEER; forgetting that PTR decreases as COP increases. Final Answer: 3.5 / COP

Question 1

Primary functions of a refrigerant compressor  In a vapour-compression refrigeration system, what does the compressor accomplish?

Accepted Answer

Introduction / Context: The compressor is the work-input device of a vapour-compression cycle. Recognizing all of its effects clarifies why condenser and expansion device conditions are what they are. Given Data / Assumptions: Compression of superheated or saturated vapour from evaporator pressure to condenser pressure. Real compressors have isentropic efficiency < 1, so discharge temperature rises. Concept / Approach: By compressing the vapour, the compressor increases its pressure to the condensing level, simultaneously increasing temperature (due to compression work) and providing the driving force for circulation around the closed loop. Step-by-Step Solution: Pressure: suction (low) → discharge (high), enabling condensation at ambient conditions.Temperature: rises due to work input; crucial for superheat management and oil return.Circulation: the compressor draws vapour from the evaporator and pushes it through condenser and expansion device. Verification / Alternative check: Energy balance: W_in to the compressor equals the enthalpy rise across it (neglecting heat loss), explaining temperature increase. Why Other Options Are Wrong: Any single-function choice is incomplete; the compressor performs all three. “Only maintains mass balance” ignores the required pressure rise and work. Common Pitfalls: Overlooking that discharge temperature depends on isentropic efficiency and suction superheat. Final Answer: All of the above

Question 2

Why aircraft use air-cycle refrigeration  Aircraft commonly employ air (reversed Brayton) refrigeration cycles primarily because:

Accepted Answer

Introduction / Context: Air-cycle (reversed Brayton) systems are favored in aviation. Understanding the engineering trade-offs explains why COP is not the sole selection criterion. Given Data / Assumptions: Aircraft already carry air compressors/turbomachinery (bleed air/ram air). Weight, reliability, and maintenance are critical constraints. Concept / Approach: Although air-cycle COP can be lower than that of vapour-compression, the system offers major advantages: light weight, no refrigerant handling, tolerance to leaks, and effective use of available ram air and high-altitude low ambient temperatures. Step-by-Step Solution: Assess competing drivers: mass and simplicity dominate in aircraft.Air cycle uses air as the working fluid; no evaporator with phase change refrigerant is required, reducing complexity and weight.Altitude provides cold sink air to enhance cooling performance. Verification / Alternative check: Aircraft environmental control systems (ECS) widely use bootstrap/air-cycle packs that prioritize weight and reliability over peak COP. Why Other Options Are Wrong: Highest heat-transfer coefficients: not the primary justification. “Only” low ambient temperature: it helps but is not the sole reason. Higher COP than vapour cycles: generally untrue. No compressor: air cycle still uses a compressor or utilizes bleed/ram compression. Common Pitfalls: Equating selection solely with COP while ignoring weight and system integration. Final Answer: They provide low weight per tonne of refrigeration and simple, robust hardware.

Question 3

Naming pressures in a refrigeration compressor  What is the standard term for the pressure measured at the inlet of a refrigerant compressor (i.e., the evaporator pressure at the suction line)?

Accepted Answer

Introduction / Context: Correct terminology helps in reading P–h charts, service gauges, and troubleshooting logs. The compressor has a low-pressure side (suction) and a high-pressure side (discharge). Given Data / Assumptions: Standard vapour-compression layout with evaporator, compressor, condenser, and expansion device. Inlet line to compressor originates from evaporator outlet/suction accumulator. Concept / Approach: The pressure at the compressor inlet is called suction pressure. It corresponds to the evaporating pressure and is read on the low-side gauge in service practice. Some texts also refer to evaporator pressure as back pressure; however, “suction pressure” is the unambiguous term at the compressor inlet. Step-by-Step Solution: Identify component: compressor inlet.Associate pressure name: suction pressure (low side).Note: discharge pressure is the high-side pressure at compressor outlet; critical pressure refers to fluid property, not line pressure. Verification / Alternative check: Service manifolds label LP gauge for suction (evaporator) pressure and HP gauge for discharge (condensing) pressure. Why Other Options Are Wrong: Discharge pressure: outlet, not inlet. Critical pressure: property at critical point, unrelated to component location. Back pressure: sometimes synonymous with evaporator pressure, but the standard inlet term is suction pressure. Common Pitfalls: Using “back pressure” interchangeably can cause confusion in multi-stage or cascade systems. Final Answer: Suction pressure

Question 4

Power requirement per tonne of refrigeration (PTR)  For a vapour-compression plant, express the power per tonne of refrigeration (in kW per TR) in terms of the coefficient of performance (COP). Take 1 TR = 3.5 kW of cooling.

Accepted Answer

Introduction / Context: Linking COP to input power per unit cooling capacity is essential for comparing chiller and refrigeration plant efficiencies across sizes. Given Data / Assumptions: Definition: COP = Q_L / W_in. One tonne of refrigeration (1 TR) = 3.5 kW of cooling capacity. Concept / Approach: For 1 TR, Q_L = 3.5 kW. Rearranging the COP definition gives W_in = Q_L / COP. Therefore, power per TR is 3.5 / COP (kW per TR). Step-by-Step Solution: Start with COP = Q_L / W_in.Rearrange: W_in = Q_L / COP.For 1 TR, W_in = 3.5 / COP kW. Verification / Alternative check: Numerical example: COP = 3 → PTR = 3.5/3 ≈ 1.167 kW/TR. This matches typical plant data. Why Other Options Are Wrong: COP/3.5 or 3.5COP invert or misapply the relationship. 1/(3.5COP) has wrong dimensions and magnitude. Common Pitfalls: Confusing COP with EER or SEER; forgetting that PTR decreases as COP increases. Final Answer: 3.5 / COP

Heat Transfer, Refrigeration and Air Conditioning Questions