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For a refrigerator or air conditioner, the coefficient of performance \(K\) (often denoted as COP) is, as in Eq. (20.9), the ratio of cooling output \(Q_C\) 0 to the required electrical energy input \(W\) , both in joules. The coefficient of performance is also expressed as a ratio of powers, $$K = {(Q_C ) /t \over (W) /t}$$ where \(Q_C /t\) is the cooling power and \(W /t\) is the electrical power input to the device, both in watts. The energy efficiency ratio (\(EER\)) is the same quantity expressed in units of Btu for \(Q_C\) and \(W \cdot h\) for \(W\) . (a) Derive a general relationship that expresses \(EER\) in terms of \(K\). (b) For a home air conditioner, \(EER\) is generally determined for a 95\(^\circ\)F outside temperature and an 80\(^\circ\)F return air temperature. Calculate \(EER\) for a Carnot device that operates between 95\(^\circ\)F and 80\(^\circ\)F. (c) You have an air conditioner with an \(EER\) of 10.9. Your home on average requires a total cooling output of \(Q_C = 1.9 \times 10^{10} J\) per year. If electricity costs you 15.3 cents per \(kW \cdot h\), how much do you spend per year, on average, to operate your air conditioner? (Assume that the unit's \(EER\) accurately represents the operation of your air conditioner. A \(seasonal\) \(energy\) \(efficiency\) \(ratio\) (\(SEER\)) is often used. The \(SEER\) is calculated over a range of outside temperatures to get a more accurate seasonal average.) (d) You are considering replacing your air conditioner with a more efficient one with an \(EER\) of 14.6. Based on the \(EER\), how much would that save you on electricity costs in an average year?

Step by step solution

01

Derive the Relationship between EER and COP (K)

The \(EER\) is expressed in Btu/W·h, while \(K\) is dimensionless. To convert \(K\) to \(EER\), we use the conversion factor: 1 watt = 3.412 Btu/h. Therefore, \(EER = \frac{Q_C}{W} = K \times 3.412\).

02

Calculate EER for a Carnot Cycle

The Carnot efficiency between the two temperatures is given by \( K = \frac{T_c}{T_h - T_c}\). Convert temperatures from Fahrenheit to Kelvin: \( T_h = 308.15 \text{ K} \) and \( T_c = 299.82 \text{ K} \). Then calculate \( K \) and plug it into the equation: \( EER = K \times 3.412\).

03

Calculate Yearly Cost of Current AC

First, find \( W \) using \( EER \): \( EER = \frac{Q_C}{W} \Rightarrow W = \frac{Q_C}{EER}\). Calculate the energy in kWh, then find the cost by multiplying by the cost per kWh: \( Cost = \frac{Q_C}{EER} \times \frac{1}{3600} \times 0.153 \). Simplify to get the total yearly cost.

04

Calculate Savings with Efficient AC

Repeat Step 3 using the new \( EER = 14.6 \). Find the power usage \( W' \), and calculate the new cost. Determine savings by subtracting the new total cost from the previous total cost.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Efficiency Ratio (EER)

The Energy Efficiency Ratio, or EER, is a key metric when evaluating air conditioning and refrigeration systems. It measures how efficiently these systems convert energy into cooling output. Specifically, EER is defined as the ratio of the cooling output measured in British thermal units (Btu) to the electrical energy input measured in watt-hours.The formula can be expressed as:\[EER = \frac{Q_C}{W}\]where \(Q_C\) is the cooling energy output in Btu, and \(W\) is the electrical energy input in watt-hours. To derive EER from the Coefficient of Performance (COP), known as \(K\), we apply a conversion factor because EER involves different units. Since 1 watt equals 3.412 Btu/hour, we have:\[EER = K \times 3.412\]Understanding EER helps homeowners save energy and money. Higher values signify more efficient performance, making units less costly over time. It's vital to consider EER when comparing appliances, especially in hot climates where air conditioners run extensively.

Carnot Cycle

The Carnot cycle is a theoretical model that sets the efficiency benchmark for heat engines, including refrigeration systems. Named after the French physicist Sadi Carnot, it describes an idealized cycle where a system operates between two thermal reservoirs, taking advantage of temperature differentials to deliver work.For refrigeration systems, the Carnot cycle helps understand the maximum achievable efficiency. The Coefficient of Performance (COP) for a Carnot refrigerator is expressed by:\[K = \frac{T_c}{T_h - T_c}\]Where \(T_c\) is the temperature of the cold reservoir, and \(T_h\) is the temperature of the hot reservoir, both measured in Kelvin.By converting temperatures from Fahrenheit to Kelvin, as shown in our problem, you can calculate the COP and subsequently the EER of a Carnot-based device. This evaluation reveals why real-world devices can't reach Carnot's efficiency, owing to practical losses such as friction and non-reversible processes. Nonetheless, the Carnot cycle remains a crucial reference point for designing efficient cooling systems.

Refrigeration and Air Conditioning

Refrigeration and air conditioning systems are essential for maintaining comfortable living environments and preserving perishable goods. They rely on cyclic processes to absorb heat from a space and expel it outside, thus cooling the interior environment. These systems predominantly use mechanical vapor-compression cycles, although other technologies like absorption refrigeration exist. The efficiency of these systems is pivotal, not only for minimizing electricity usage but also for reducing environmental impact. Some key features and processes include:

• **Compressor:** which pressurizes the refrigerant, increasing its temperature.
• **Condenser:** where the refrigerant releases heat to the external environment.
• **Expansion Valve:** lowers the pressure of the refrigerant, cooling it further.
• **Evaporator:** where the refrigerant absorbs heat from the indoor space, cooling it down.

To assess and improve system efficiency, metrics such as the EER and COP are commonly used. These metrics provide insight into how well a system performs relative to its energy consumption, guiding consumers in making sustainable and cost-effective choices. The design and choice of components significantly influence both the immediate experience and operational costs, leading to considerable long-term savings.