Refrigerating machine and heat pump

Explain in your own words how a refrigerator can 'remove cold' if what it actually does is move heat: where it takes it from, where it expels it, and which machine provides the necessary work.

The figure shows the refrigeration cycle with its four components, but one is covered. State which component is missing, what happens to the refrigerant inside it (change of state and pressure), and where it is physically located in your refrigerator at home.

Put the four steps of the refrigeration cycle in order as the refrigerant goes through them: (a) it evaporates at low pressure, absorbing heat; (b) it expands and its temperature drops sharply; (c) it is compressed, increasing pressure and temperature; (d) it condenses, releasing heat to the outside. Explain which component performs each step.

Define the coefficient of performance (COP) of a refrigerator and explain what a COP of $3$ means in terms of heat extracted per unit of electricity consumed. Why does a higher COP indicate a more efficient appliance?

A refrigerator extracts $600\ \text{kJ}$ of heat from its interior every hour, consuming $200\ \text{kJ}$ of electrical energy. a) Calculate its efficiency or coefficient of performance, $\mathrm{COP} = \frac{Q_1}{W}$. b) Calculate the total heat released to the kitchen in that hour. c) Explain why the COP can be greater than $1$ without violating the conservation of energy.

Four cans of soft drink at $25\ ^\circ\text{C}$ are placed in a refrigerator that can extract $100\ \text{W}$ of heat from its interior. Estimate the minimum time to cool them to $5\ ^\circ\text{C}$ (total mass of liquid $1{,}4\ \text{kg}$, approximate specific heat capacity of water $4\,180\ \text{J}/(\text{kg}\cdot\text{K})$) and comment on two reasons why the actual time will be longer.

A refrigerator that extracts $600\ \text{kJ}$ of heat from its interior every hour consuming $200\ \text{kJ}$ of electrical energy is used in winter as a heat pump to heat a dwelling. Calculate its COP as a heat pump, $\mathrm{COP}_{bc} = \frac{Q_2}{W}$, and explain why a heat pump heats with more thermal energy than an electric heater of the same power output.

An air conditioner consumes $1{,}2\ \text{kWh}$ of electricity in an afternoon and extracts $3\,600\ \text{kcal}$ from the air in the house. a) Convert the heat to $\text{kWh}$ ($1\ \text{kWh} = 860\ \text{kcal}$) and calculate the COP. b) Calculate the heat released to the outside. c) Compare with a fan: where does the electrical energy go in each appliance?

A single-family home has a heating demand of $8\ \text{kW}$. Two alternatives are evaluated: a natural gas boiler (seasonal efficiency of $90\,\%$, fuel cost $0,07\ \text{€/kWh}$) and an air-source heat pump (electricity cost $0,18\ \text{€/kWh}$). a) Determine the minimum $\mathrm{COP}$ that the heat pump must achieve so that the hourly economic cost is lower than that of the boiler. b) If the electricity emission factor is $0,25\ \text{kg CO}_2/\text{kWh}$ and that of natural gas is $0,20\ \text{kg CO}_2/\text{kWh}$, calculate the $\mathrm{COP}$ above which the heat pump reduces $\text{CO}_2$ emissions. c) Knowing that at $-7\ ^\circ\text{C}$ outdoors the actual $\mathrm{COP}$ drops to $2,1$, evaluate which of the two systems is more advantageous under these extreme conditions and what technical solution is usually added to air-source heat pumps in very cold climates.

The commercial datasheet for a domestic freezer claims that «thanks to its inverter control, the unit achieves a cooling $\mathrm{COP}$ of $6,5$ while maintaining the chamber at $-18\ ^\circ\text{C}$ with an ambient kitchen temperature of $25\ ^\circ\text{C}$». a) Prove that this claim is thermodynamically impossible by calculating the theoretical maximum Carnot $\mathrm{COP}$ between both thermal reservoirs. b) In a real vapor-compression cycle, to transfer heat with a reasonable heat exchanger size, the refrigerant must evaporate at $5\ ^\circ\text{C}$ below the chamber and condense at $10\ ^\circ\text{C}$ above ambient temperature. Calculate the new Carnot $\mathrm{COP}$ with the actual refrigerant temperatures. c) If the actual measured $\mathrm{COP}$ on a test bench is $1,9$, calculate the exergetic efficiency relative to the Carnot cycle in part b and justify two internal irreversibilities that cause this discrepancy.

A commercial facility must replace an $\text{R-404A}$ central refrigeration unit ($\mathrm{GWP} = 3922$) that has a charge of $50\ \text{kg}$, annual leaks of $8\,\%$, and an electricity consumption of $40\,000\ \text{kWh/year}$. Two options are being studied: 1) Retrofit to $\text{R-449A}$ ($\mathrm{GWP} = 1397$), which maintains the leakage rate but increases electricity consumption by $3\,\%$ due to lower thermodynamic performance. 2) New installation of $\text{R-290}$ (propane, $\mathrm{GWP} = 3$), which reduces leaks to $1\,\%$ and electricity consumption by $6\,\%$. With an electricity emission factor of $0,25\ \text{kg CO}_2/\text{kWh}$: a) Calculate the total annual environmental impact in equivalent $\text{CO}_2$ emissions (direct from leaks and indirect from consumption) of each option and compare them with the original situation. b) Justify why the concept of Total Equivalent Warming Impact (TEWI) shows that a gas with a lower $\mathrm{GWP}$ is not always the best solution if it impairs efficiency. c) Explain the safety requirements imposed by the F-Gas regulation and the safety regulations for refrigeration systems when choosing propane.