Vapor-Compression Refrigeration - Thermodynamics

1. In a vapor-compression refrigeration cycle, the refrigerant properties at key state points are given in the table. Based on this data, what is the coefficient of performance of the cycle?
   Answer: 3.6 The correct answer is C. From the table: State 1 (compressor inlet): h₁ = 245 kJ/kg, State 2 (compressor exit): h₂ = 285 kJ/kg, State 3 (condenser exit): h₃ = 100 kJ/kg, State 4 (throttling valve exit): h₄ = 100 kJ/kg. The refrigeration effect is q_evap = h₁ - h₄ = 245 - 100 = 145 kJ/kg. The compression work is w_comp = h₂ - h₁ = 285 - 245 = 40 kJ/kg. The COP = q_evap/w_comp = 145/40 = 3.625 ≈ 3.6.
2. A vapor-compression refrigeration system has a compressor that consumes $8 \text{ kW}$ and provides cooling at a rate of $24 \text{ kW}$. If the system is modified to operate between the same temperature limits but with a Carnot cycle instead, what would be the work input required for the same cooling capacity?
   Answer: The work input cannot be determined without knowing the reservoir temperatures When analyzing refrigeration cycles, you need to distinguish between what information is given versus what's required to make theoretical comparisons. This question tests whether you understand the fundamental relationship between coefficient of performance (COP) and temperature limits. The actual system has a COP of $\text{COP}_{\text{actual}} = \frac{Q_L}{W} = \frac{24}{8} = 3$. For a Carnot refrigerator operating between the same temperatures, the COP would be $\text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L}$, where $T_L$ and $T_H$ are the absolute temperatures of the cold and hot reservoirs. To find the Carnot work input for the same 24 kW cooling, you'd need $W_{\text{Carnot}} = \frac{Q_L}{\text{COP}_{\text{Carnot}}}$. However, without knowing the actual reservoir temperatures, you cannot calculate the Carnot COP or the required work input. Answer A is correct because the reservoir temperatures are essential but missing information. Answer B incorrectly assumes a fixed 25% improvement, which isn't a thermodynamic principle. Answer C suggests you can determine the work using some "efficiency improvement ratio," but this ratio depends on the unknown temperature values. Answer D wrongly assumes that operating between identical temperature limits means identical work requirements, ignoring that Carnot cycles are more efficient than real cycles. Study tip: In thermodynamics problems involving cycle comparisons, always check whether you have sufficient information to calculate the theoretical performance. Temperature limits are crucial for Carnot cycle calculations—if they're missing, you likely cannot solve the problem numerically.
3. In a refrigeration cycle, the condenser rejects $150 \text{ kW}$ to the environment while the evaporator absorbs $120 \text{ kW}$ from the refrigerated space. What is the coefficient of performance (COP) of this refrigeration system?
   Answer: 4.0 When you encounter refrigeration cycle problems, focus on the energy balance and understand what COP actually measures. The coefficient of performance for a refrigerator tells you how much cooling effect you get per unit of work input - essentially, how efficient your refrigerator is. Start with the energy balance: the condenser rejects 150 kW while the evaporator absorbs 120 kW. The difference between these values represents the work input required by the compressor: $W = Q_H - Q_C = 150 - 120 = 30 \text{ kW}$. This makes physical sense because energy must be conserved - the heat rejected equals the heat absorbed plus the work added. For a refrigerator, COP is defined as the desired effect (cooling) divided by the required input (work): $\text{COP} = \frac{Q_C}{W} = \frac{120}{30} = 4.0$. This confirms answer B is correct. Answer A (1.25) represents $\frac{Q_H}{Q_C} = \frac{150}{120}$, which is the ratio of rejected to absorbed heat but has no physical meaning as a COP. Answer C (5.0) might result from incorrectly using $\frac{Q_H}{W} = \frac{150}{30}$, which would be the COP for a heat pump, not a refrigerator. Answer D (0.8) comes from the inverse calculation $\frac{W}{Q_C} = \frac{30}{120}$, which is efficiency-like but not COP. Remember: refrigerator COP always equals cooling effect divided by work input, and it's typically greater than 1 for any practical system.
4. A refrigeration system has an evaporator that operates at $-10°C$ and a condenser at $40°C$. If the system uses an ideal vapor-compression cycle with isentropic compression, what factor most significantly affects the cycle's coefficient of performance?
   Answer: The difference between the saturation pressures at the evaporator and condenser temperatures When analyzing vapor-compression refrigeration cycles, you need to understand that the coefficient of performance (COP) fundamentally depends on the thermodynamic properties at the operating conditions, not the physical system parameters. The COP for an ideal vapor-compression cycle is determined by the enthalpy differences across the cycle components, which are directly tied to the saturation pressures at the evaporator and condenser temperatures. A larger pressure difference between the high-pressure condenser (40°C) and low-pressure evaporator (-10°C) means the compressor must do more work to pump refrigerant between these states, reducing the COP. This pressure difference drives the entire thermodynamic cycle and establishes the work input required. Looking at the incorrect options: (A) The specific heat ratio affects the isentropic compression process details, but it's the pressure levels themselves that determine the overall cycle performance, not the path between them. (C) Mass flow rate is a design parameter that affects the cooling capacity but doesn't change the cycle efficiency—COP remains constant regardless of how much refrigerant circulates. (D) Heat exchanger surface areas influence heat transfer rates and approach temperatures but don't directly affect the ideal cycle COP, which assumes perfect heat exchange. Study tip: For thermodynamic cycle problems, always distinguish between intensive properties (like pressures and temperatures that determine cycle efficiency) and extensive properties (like mass flow rates and physical dimensions that affect capacity but not efficiency). The COP depends on the thermodynamic state points, which are set by the operating pressures.
5. A refrigeration system operates between a cold reservoir at $5°C$ and a hot reservoir at $35°C$. If the actual coefficient of performance (COP) is $3.2$, what percentage of the Carnot COP is achieved by this system?
   Answer: 34.6% When you encounter refrigeration efficiency problems, you're comparing real-world performance to the theoretical maximum given by the Carnot cycle. This requires calculating both the actual COP and the ideal Carnot COP. First, convert temperatures to Kelvin: Cold reservoir = 5°C + 273 = 278 K, Hot reservoir = 35°C + 273 = 308 K. For a Carnot refrigerator, the coefficient of performance is: $COP_{Carnot} = \frac{T_C}{T_H - T_C} = \frac{278}{308 - 278} = \frac{278}{30} = 9.27$ The efficiency percentage is: $\text{Efficiency} = \frac{COP_{actual}}{COP_{Carnot}} \times 100\% = \frac{3.2}{9.27} \times 100\% = 34.6\%$ This confirms answer B is correct. Looking at the wrong answers: A (32.8%) likely results from calculation errors in the temperature conversion or Carnot COP formula. C (38.2%) might come from incorrectly using Celsius temperatures instead of Kelvin, which dramatically changes the denominator in the Carnot formula. D (41.7%) could result from reversing the temperature difference or mixing up refrigerator and heat pump COP formulas. Study tip: Always convert to Kelvin immediately when working with thermodynamic cycles – this is the most common source of errors. Also, memorize that refrigerator COP uses $T_C$ in the numerator, while heat pump COP uses $T_H$. Set up your Carnot calculation first, then compare to actual performance.
6. A vapor-compression refrigeration system experiences a gradual increase in condensing pressure over several months of operation while evaporating pressure remains constant. The superheat at the compressor inlet decreases from 8°C to 3°C, and subcooling at the condenser exit drops from 5°C to 2°C. What is the most likely cause of these observations, and how would this affect the system COP?
   Answer: Condenser fouling reduces heat transfer effectiveness, increasing condensing pressure and reducing COP by 10-20% due to higher compression ratios Condenser fouling (dirt, scale, or reduced airflow) is the most likely cause of gradually increasing condensing pressure with constant evaporating conditions. Fouling reduces heat transfer, requiring higher condensing temperatures to reject heat. This increases compression work significantly. Reduced superheat and subcooling indicate the system is working harder but less effectively. COP typically drops 10-20% as compression ratio increases. Choice B is wrong because overcharge would increase subcooling. Choice C incorrectly links compressor degradation to these specific symptoms. Choice D misidentifies the cause and incorrectly suggests COP maintenance.
7. A vapor-compression refrigeration cycle operates with refrigerant entering the compressor at state 1 and exiting at state 2. If the specific entropy increases from $s_1 = 1.7 \text{ kJ/kg}\cdot\text{K}$ to $s_2 = 1.8 \text{ kJ/kg}\cdot\text{K}$, what can be concluded about the compression process?
   Answer: The compression is irreversible and the isentropic efficiency is less than 100% When analyzing compression processes in refrigeration cycles, entropy changes reveal crucial information about process irreversibility. An ideal compressor would operate isentropically (constant entropy), but real compressors always have inefficiencies that increase entropy. The entropy increase from $s_1 = 1.7$ to $s_2 = 1.8 \text{ kJ/kg}\cdot\text{K}$ indicates irreversibilities within the compressor, such as friction, heat transfer, and fluid turbulence. This entropy generation means the actual compression process deviates from the ideal isentropic path, resulting in an isentropic efficiency less than 100%. The isentropic efficiency compares actual work required to the theoretical minimum work for isentropic compression between the same pressure limits. Answer A incorrectly suggests the entropy change is "minimal" – any entropy increase during compression indicates irreversibility, regardless of magnitude. Answer C confuses compression with heat addition processes; while isothermal heat addition does increase entropy, compression involves pressure rise, not constant temperature. The process isn't isothermal because compressor work typically raises both pressure and temperature. Answer D contradicts itself by claiming both adiabatic operation and entropy change – truly adiabatic reversible processes are isentropic (constant entropy), but entropy increased here. Therefore, B correctly identifies that compression is irreversible with efficiency below 100%. Study tip: Remember that entropy always increases in irreversible processes. For compressors, any entropy rise from inlet to outlet immediately tells you the process has inefficiencies and isn't operating at ideal isentropic conditions.
8. A vapor-compression heat pump system operates in heating mode with R-410A. The outdoor coil (evaporator) operates at 0°C while the indoor coil (condenser) operates at 45°C. As outdoor temperature drops to -10°C, the system automatically switches to defrost mode every 90 minutes. What is the primary thermodynamic reason for this requirement, and how does it impact overall heating performance?
   Answer: Frost formation reduces evaporator effectiveness and airflow, decreasing refrigeration effect and requiring periodic removal to maintain capacity and COP At low outdoor temperatures with high humidity, frost forms on the evaporator coil surfaces when surface temperature is below 0°C. This frost acts as an insulating layer, reducing heat transfer effectiveness and blocking airflow, which decreases both the refrigeration effect and heat transfer coefficient. The evaporator becomes less effective at absorbing heat from outdoor air, reducing system capacity and COP. Defrost temporarily reverses the cycle to melt accumulated frost. Choice B incorrectly describes refrigerant behavior. Choice C misidentifies the cause as compressor efficiency rather than heat transfer. Choice D incorrectly suggests initial heat transfer improvement and wrong failure mode.
9. A refrigeration cycle operating with R-134a has an evaporator temperature of $0°C$ and condenser temperature of $50°C$. If the refrigerant enters the compressor as saturated vapor and exits the condenser as saturated liquid, which process in the cycle contributes most significantly to the system's irreversibility?
   Answer: Throttling process due to the irreversible pressure drop across the expansion valve When analyzing irreversibility in refrigeration cycles, you need to identify which process generates the most entropy due to its inherently irreversible nature. Irreversibility is directly related to entropy generation, which occurs when processes deviate from ideal, reversible conditions. The throttling process across the expansion valve is fundamentally irreversible because it's an isenthalpic (constant enthalpy) process where pressure drops dramatically while enthalpy remains constant. This creates a significant entropy increase as the refrigerant transitions from high-pressure liquid to low-pressure liquid-vapor mixture. Unlike other processes that could theoretically approach reversibility under ideal conditions, throttling is inherently irreversible by design. Option A is incorrect because while compression does involve some irreversibility from friction and heat transfer, modern compressors can approach isentropic (reversible adiabatic) conditions relatively closely, making this less significant than throttling. Option B is wrong because condensation, though involving finite temperature differences, can approach reversibility when temperature differences are minimized. The entropy decrease of the refrigerant is largely offset by entropy increase in the cooling medium. Option D is incorrect for similar reasons as B - evaporation can approach reversible heat transfer with small temperature differences between the refrigerant and refrigerated space. Study tip: Remember that throttling processes are always the largest source of irreversibility in vapor compression cycles because they're designed to be irreversible. When comparing cycle modifications, replacing throttling with expansion work (like in a turbine) is often the most effective way to improve efficiency.
10. In a vapor-compression refrigeration cycle analysis, the expansion valve (throttling process) is often considered the least efficient component. If this throttling process were replaced with an ideal expander (turbine) that produces work, what would be the theoretical impact on cycle performance, and why is this replacement not practical in real systems?
    Answer: COP would improve by 8-15% due to work recovery, but expanders cannot handle two-phase flow effectively and are economically unfeasible Replacing throttling with ideal expansion would recover work equal to the enthalpy difference across the expansion, typically improving COP by 8-15% in conventional cycles. However, practical expanders struggle with the wet vapor conditions typical after expansion, suffer from erosion and efficiency losses with two-phase flow, and add significant cost and complexity for relatively small energy recovery. Choice B overstates the improvement and incorrectly suggests technology limitations. Choice C is wrong because expanders generate rather than consume work. Choice D incorrectly dismisses the potential benefits of expansion work recovery.
11. A vapor-compression refrigeration cycle uses $2 \text{ kg/s}$ of refrigerant. The specific enthalpy values are: compressor inlet $h_1 = 250 \text{ kJ/kg}$, compressor exit $h_2 = 290 \text{ kJ/kg}$, condenser exit $h_3 = 95 \text{ kJ/kg}$, and evaporator inlet $h_4 = 95 \text{ kJ/kg}$. What is the rate of heat rejection in the condenser?
    Answer: 390 kW When analyzing vapor-compression refrigeration cycles, focus on applying energy conservation principles to each component. The condenser's job is to reject heat from the hot, high-pressure refrigerant vapor entering from the compressor. To find the condenser heat rejection rate, apply the steady-flow energy equation to the condenser. Heat flows out of the refrigerant as it condenses from vapor to liquid. The rate of heat rejection equals the mass flow rate times the change in specific enthalpy across the condenser: $\dot{Q}_{condenser} = \dot{m}(h_2 - h_3) = 2 \text{ kg/s} \times (290 - 95) \text{ kJ/kg} = 2 \times 195 = 390 \text{ kW}$ This confirms answer C) 390 kW is correct. A) 190 kW incorrectly uses only the enthalpy difference (195 kJ/kg) without multiplying by the mass flow rate, forgetting that power requires considering the rate of mass processing. B) 310 kW might result from using the wrong enthalpy difference, perhaps confusing condenser inlet/outlet with other cycle points or making an arithmetic error in the enthalpy calculation. D) 580 kW could come from incorrectly adding enthalpies instead of finding their difference, or from using the total enthalpy at the condenser inlet without accounting for the outlet conditions. Remember: For any heat exchanger analysis, always identify the correct inlet and outlet states, then multiply the mass flow rate by the enthalpy change. The condenser removes energy, so heat rejection equals mass flow rate times the enthalpy decrease across the component.
12. A vapor-compression refrigeration system operates with refrigerant R-134a between evaporator and condenser pressures of 200 kPa and 1200 kPa, respectively. The refrigerant exits the evaporator as saturated vapor and exits the condenser as saturated liquid. If the compressor has an isentropic efficiency of 85% and the refrigeration capacity is 10 kW, what is the coefficient of performance (COP) of this system?
    Answer: COP = 3.2, indicating moderate efficiency with significant compressor irreversibilities reducing performance For this vapor-compression cycle: At 200 kPa (evaporator), saturated vapor h₁ = 244.5 kJ/kg. At 1200 kPa (condenser), saturated liquid h₃ = 117.8 kJ/kg, saturated vapor h₂s = 278.9 kJ/kg. For non-ideal compression: h₂ = h₁ + (h₂s - h₁)/η = 244.5 + (278.9 - 244.5)/0.85 = 284.9 kJ/kg. Refrigeration effect: qₑ = h₁ - h₃ = 244.5 - 117.8 = 126.7 kJ/kg. Compressor work: wc = h₂ - h₁ = 284.9 - 244.5 = 40.4 kJ/kg. COP = qₑ/wc = 126.7/40.4 = 3.14 ≈ 3.2. The other values result from errors in property lookup or efficiency calculations.
13. In a vapor-compression refrigeration cycle, the refrigerant exits the condenser at $30°C$ as saturated liquid and enters the evaporator through a throttling valve. If the evaporator operates at $-5°C$, what is the most likely state of the refrigerant immediately after the throttling process?
    Answer: Two-phase mixture at -5°C, with quality determined by the constant enthalpy process When analyzing refrigeration cycles, the throttling valve (expansion valve) is crucial because it's an isenthalpic process—enthalpy remains constant while pressure drops dramatically. This pressure reduction fundamentally changes the refrigerant's state. The refrigerant enters the throttling valve as saturated liquid at 30°C with relatively high enthalpy. As it expands through the valve, pressure drops to match the evaporator conditions at -5°C, but the enthalpy stays the same. Here's the key insight: at -5°C and lower pressure, that same enthalpy value now falls within the two-phase region on the refrigerant's property diagram. Since the enthalpy is higher than that of saturated liquid at -5°C but lower than saturated vapor at -5°C, the refrigerant exists as a mixture of liquid and vapor. Choice A incorrectly assumes throttling only affects temperature while maintaining the liquid state—this ignores the pressure reduction's effect on phase change. Choice B suggests complete vaporization to superheated vapor, which would require much higher enthalpy than available from saturated liquid at 30°C. Choice D similarly assumes complete vaporization but to saturated vapor, still requiring more enthalpy than the throttling process provides. The quality (vapor fraction) of this two-phase mixture depends on the specific refrigerant and operating conditions, calculated using $h = h_f + x \cdot h_{fg}$ where $x$ is quality. Remember: throttling valves create two-phase mixtures in refrigeration cycles. If you see "throttling" or "expansion valve" with temperature/pressure changes, immediately think isenthalpic process and check which phase region that enthalpy corresponds to at the new conditions.
14. A vapor-compression refrigeration system is modified by adding a heat exchanger that subcools the liquid refrigerant leaving the condenser using the vapor refrigerant leaving the evaporator. If the original COP was 3.5, what is the most likely effect of this modification?
    Answer: The COP will increase because subcooling increases the refrigeration effect in the evaporator When analyzing modifications to vapor-compression refrigeration cycles, focus on how changes affect the two key COP components: refrigeration effect (cooling capacity) and compressor work input. Adding a heat exchanger that subcools the liquid leaving the condenser while superheating the vapor leaving the evaporator creates what's called a liquid-suction heat exchanger. This modification increases the COP because subcooling the liquid refrigerant increases the enthalpy difference across the evaporator. When colder liquid enters the throttling valve and then the evaporator, more heat can be absorbed per unit mass of refrigerant, directly increasing the refrigeration effect. The slight increase in compressor work due to superheating is typically much smaller than the gain in cooling capacity. Option A incorrectly suggests complexity alone reduces efficiency - the thermodynamic benefit far outweighs any minor losses from an additional heat exchanger. Option C misunderstands that while total cycle heat transfer may be similar, the distribution changes favorably: more useful cooling occurs in the evaporator relative to the work input. Option D overemphasizes the superheating effect - while superheating does slightly increase compression work, this small penalty is overwhelmed by the refrigeration effect improvement from subcooling. Remember this pattern: in refrigeration cycle modifications, subcooling the liquid before expansion almost always improves performance because it maximizes the enthalpy drop available for useful cooling. Look for the dominant effect when multiple changes occur simultaneously.
15. In analyzing a vapor-compression refrigeration cycle, an engineer finds that the compressor work input is $30 \text{ kW}$ and the condenser heat rejection is $120 \text{ kW}$. If the system is designed to maintain a cold space at $-15°C$ while rejecting heat to the environment at $35°C$, what can be concluded about this system's performance?
    Answer: The COP is 3.0 and the system operates at 58.1% of Carnot efficiency When analyzing refrigeration cycles, you need to apply both the first law of thermodynamics and understand the relationship between actual and theoretical performance limits. Start by applying energy conservation to find the cooling load. For any refrigeration system, the first law requires: $Q_{\text{evaporator}} + W_{\text{compressor}} = Q_{\text{condenser}}$. Therefore: $Q_{\text{evaporator}} = 120 - 30 = 90 \text{ kW}$. The coefficient of performance is $\text{COP} = \frac{Q_{\text{evaporator}}}{W_{\text{compressor}}} = \frac{90}{30} = 3.0$. For the Carnot refrigerator operating between the same temperatures: $\text{COP}_{\text{Carnot}} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} = \frac{258.15}{308.15 - 258.15} = 5.163$. The efficiency relative to Carnot is: $\frac{3.0}{5.163} = 58.1\%$. Answer A is wrong because the energy balance actually confirms the first law is satisfied (90 + 30 = 120). Answer B correctly calculates COP = 3.0 but incorrectly states 48.0% Carnot efficiency instead of 58.1%. Answer C has the wrong COP (4.0 instead of 3.0) and wrong Carnot efficiency (77.4%). Answer D is incorrect because you can fully evaluate performance using the given energy flows and temperatures. The correct answer must be E, as none of the provided options correctly combine both the COP calculation and Carnot efficiency. Study tip: Always verify refrigeration problems using the first law energy balance first, then calculate both actual COP and Carnot COP to assess performance. Watch for calculation errors in efficiency comparisons.
16. A two-stage vapor-compression refrigeration system with intercooling operates between 100 kPa and 1600 kPa. The intermediate pressure is optimized to minimize total compression work. If the low-stage compressor has an isentropic efficiency of 82% and the high-stage compressor has an isentropic efficiency of 78%, what is the primary advantage of this configuration compared to single-stage compression, and what is the approximate work savings?
    Answer: Reduced irreversibilities and lower discharge temperatures result in 15-20% work savings compared to single-stage compression Two-stage compression with intercooling reduces the work per stage and keeps discharge temperatures manageable, reducing irreversibilities from excessive superheat. The optimal intermediate pressure (geometric mean ≈ 400 kPa) minimizes total work. With the given efficiencies and pressure ratio of 16, two-stage compression typically saves 15-20% work compared to single-stage. Choice B incorrectly focuses on capacity rather than work. Choice C overstates savings and incorrectly mentions throttling elimination. Choice D understates the benefits and complexity - two-stage systems require additional equipment and controls.
17. A commercial refrigeration system operates between $-20°C$ and $40°C$ with an actual COP of $2.1$. The system is proposed to be replaced with a heat pump that would heat a building by extracting heat from the outside air at $-20°C$ and delivering it to the building at $40°C$. What would be the COP of this heat pump if it has the same thermodynamic efficiency as the original refrigeration system?
    Answer: 3.1, since the heat pump COP equals the refrigeration COP plus one When analyzing refrigeration and heat pump systems, you need to understand that both are reverse heat engines operating between the same temperature reservoirs, but with different performance metrics. The key insight is the relationship between their coefficients of performance (COP). For any system with the same thermodynamic efficiency, the heat pump COP always equals the refrigeration COP plus one. This stems from their definitions: refrigeration COP = $Q_L/W$ (cooling effect per work input), while heat pump COP = $Q_H/W$ (heating effect per work input). Since $Q_H = Q_L + W$ by energy conservation, the heat pump COP = $(Q_L + W)/W = Q_L/W + 1$. With the refrigeration system's COP of 2.1, the heat pump COP becomes $2.1 + 1 = 3.1$. Answer A incorrectly assumes both systems have identical COPs simply because they operate between the same temperatures, missing that COP definitions differ between refrigeration and heat pump modes. Answer C suggests doubling the refrigeration COP, which has no thermodynamic basis and confuses the additive relationship. Answer D incorrectly claims heat pumps are inherently less efficient, when in fact they deliver more energy (heating) per unit of work input than refrigerators remove (cooling). Study tip: Remember the "plus one" rule - for identical efficiency between the same temperature limits, heat pump COP always equals refrigeration COP plus one. This relationship appears frequently on thermodynamics exams and reflects the fundamental difference in what each system delivers as its useful output.
18. A refrigeration cycle uses $0.5 \text{ kg/s}$ of refrigerant and removes $100 \text{ kW}$ of heat from the cold space. If the compressor work input is $25 \text{ kW}$, what is the difference in specific enthalpy between the evaporator inlet and outlet?
    Answer: 200 kJ/kg When analyzing refrigeration cycles, you need to apply energy conservation principles to each component. The evaporator is where the refrigerant absorbs heat from the cold space, causing a change in the refrigerant's specific enthalpy. For the evaporator, apply the steady-flow energy equation. The heat removed from the cold space equals the mass flow rate times the change in specific enthalpy: $\dot{Q}_{evap} = \dot{m} \times (h_{out} - h_{in})$. Here, $\dot{Q}_{evap} = 100 \text{ kW}$ and $\dot{m} = 0.5 \text{ kg/s}$. Solving: $100 = 0.5 \times (h_{out} - h_{in})$, so $(h_{out} - h_{in}) = 200 \text{ kJ/kg}$. This confirms answer C is correct. Let's examine why the other options are wrong. Choice A (50 kJ/kg) would result if you mistakenly used the compressor work (25 kW) instead of the evaporator heat transfer: $25/0.5 = 50$. Choice B (150 kJ/kg) might come from incorrectly subtracting the compressor work from the evaporator heat: $(100-25)/0.5 = 150$. Choice D (250 kJ/kg) could result from adding the compressor work to the evaporator heat: $(100+25)/0.5 = 250$. Remember that in refrigeration problems, each component has its own energy balance. The compressor work affects the condenser and overall cycle efficiency, but the evaporator enthalpy change depends only on the heat absorbed and mass flow rate. Don't mix up values from different components when applying conservation equations.
19. Which of the following modifications to a vapor-compression refrigeration cycle would most likely result in the greatest improvement in coefficient of performance (COP)?
    Answer: Increasing the evaporator temperature while keeping the condenser temperature constant When analyzing vapor-compression refrigeration cycles, remember that the coefficient of performance (COP) is defined as $COP = \frac{Q_L}{W_{net}}$, where $Q_L$ is the cooling effect and $W_{net}$ is the compressor work input. To maximize COP, you want to either increase the cooling effect or decrease the work required. Option A is correct because increasing the evaporator temperature while keeping the condenser temperature constant reduces the pressure difference the compressor must overcome. This decreases the compressor work significantly while maintaining nearly the same cooling capacity, resulting in a higher COP. The smaller temperature lift makes the cycle more efficient. Option B would actually worsen the COP. While decreasing condenser temperature might slightly increase cooling capacity, it dramatically increases the pressure ratio and compressor work because the system must reject heat at a lower temperature while still cooling at the same evaporator temperature. Option C represents a common misconception. Installing a larger compressor increases the system's cooling capacity (tons of refrigeration) but doesn't improve the COP. The efficiency remains the same – you're just moving more refrigerant through the same thermodynamic cycle. Option D might provide a small benefit by reducing parasitic heat gains, but this improvement would be minimal compared to the thermodynamic gains from reducing the pressure ratio. Study tip: For refrigeration COP questions, focus on how changes affect the pressure ratio between evaporator and condenser. Lower pressure ratios (achieved by raising evaporator temperature or lowering condenser temperature moderately) generally improve efficiency more than equipment modifications.
20. In a vapor-compression refrigeration cycle, the refrigerant enters the compressor at state 1 and exits at state 2. If the specific entropy increases from s₁ = 1.75 kJ/kg·K to s₂ = 1.89 kJ/kg·K, and the temperature rises from T₁ = 5°C to T₂ = 65°C, what can be concluded about the compression process and its impact on cycle performance?
    Answer: The compression is irreversible with entropy generation, reducing COP by approximately 8-12% compared to ideal compression The entropy increase (Δs = 0.14 kJ/kg·K) indicates irreversible compression with internal friction, heat transfer, and other losses. For ideal isentropic compression, entropy would remain constant. The irreversibilities increase the required compression work compared to the ideal case, typically reducing COP by 8-12% for this level of entropy generation. Choice B is wrong because any entropy increase indicates significant irreversibility. Choice C is incorrect because entropy can increase during compression due to irreversibilities. Choice D is wrong because isothermal compression would show much larger entropy changes and is not practical for vapor compression.

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