For a refrigerator or air conditioner, the coefficient of...

For a refrigerator or air conditioner, the coefficient of performance $K$ (often denoted as $\mathrm{COP}$ ) is, as in Eq. (20.9) , the ratio of cooling output $\left|Q_{\mathrm{C}}\right|$ to the required electrical energy input $|W|,$ both in joules. The cocfficicnt of performance is also expressed as a ratio of powers, $$ K=\frac{\left|Q_{\mathrm{c}}\right| / t}{|W| / t} $$ where $\left|Q_{\mathrm{C}}\right| / t$ is the cooling power and $|W| / t$ is the electrical power input to the device, both in watts. The energy efficiency ratio (FER) is the same quantity expressed in units of Btu for $\left|Q_{\mathrm{C}}\right|$ and $\mathrm{W} \cdot \mathrm{h}$ for $|W|$ (a) Derive a general relationship that expresses EER in terms of $K$. (b) For a home air conditioner, FER is generally determined for a $95^{\circ} \mathrm{F}$ outside temperature and an $80^{\circ} \mathrm{F}$ return air temperature. Calculate EER for a Carnot device that operates between $95^{\circ} \mathrm{F}$ and $80 \mathrm{~F}$. (c) You have an air conditioner with an EER of 10.9 . Your home on average requires a total cooling output of $\left|Q_{\mathrm{C}}\right|=1.9 \times 10^{10} \mathrm{~J}$ per year. If electricity costs you 15.3 cents per $\mathrm{kW} \cdot \mathrm{h}$, how much do you spend per year, on average. to operate your air conditioner? (Assume that the unit's EER accurately represcnts the operation of your air conditioner. A seasonal energy efficiency ratio (SFER) is often used. The SFFR 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?

For a refrigerator or air conditioner, the coefficient of performance $K$ (often denoted as $\mathrm{COP}$ ) is, as in Eq. (20.9) , the ratio of cooling output $\left|Q_{\mathrm{C}}\right|$ to the required electrical energy input $|W|,$ both in joules. The cocfficicnt of performance is also expressed as a ratio of powers,
$$
K=\frac{\left|Q_{\mathrm{c}}\right| / t}{|W| / t}
$$
where $\left|Q_{\mathrm{C}}\right| / t$ is the cooling power and $|W| / t$ is the electrical power input to the device, both in watts. The energy efficiency ratio (FER) is the same quantity expressed in units of Btu for $\left|Q_{\mathrm{C}}\right|$ and $\mathrm{W} \cdot \mathrm{h}$ for $|W|$
(a) Derive a general relationship that expresses EER in terms of $K$.
(b) For a home air conditioner, FER is generally determined for a $95^{\circ} \mathrm{F}$ outside temperature and an $80^{\circ} \mathrm{F}$ return air temperature. Calculate EER for a Carnot device that operates between $95^{\circ} \mathrm{F}$ and $80 \mathrm{~F}$. (c) You have an air conditioner with an EER of 10.9 . Your home on average requires a total cooling output of $\left|Q_{\mathrm{C}}\right|=1.9 \times 10^{10} \mathrm{~J}$ per year. If electricity costs you 15.3 cents per $\mathrm{kW} \cdot \mathrm{h}$, how much do you spend per year, on average. to operate your air conditioner? (Assume that the unit's EER accurately represcnts the operation of your air conditioner. A seasonal energy efficiency ratio (SFER) is often used. The SFFR 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?

Key Concepts

Coefficient of Performance (COP)

The coefficient of performance (COP) is a measure of efficiency for refrigeration and air conditioning devices. It is defined as the ratio of the useful cooling (or heating) provided to the work input required to produce that cooling or heating. This concept is fundamental in thermodynamics when analyzing the performance of devices that transfer heat from one location to another rather than directly converting work into heat.

Energy Efficiency Ratio (EER)

The Energy Efficiency Ratio (EER) is a metric commonly used to rate the efficiency of cooling devices, such as air conditioners. It represents the ratio of the cooling output, typically expressed in British Thermal Units (BTUs), to the electrical energy input, often measured in watt-hours. EER provides a practical, unit-specific evaluation of device efficiency that can be easily compared and is directly related to the COP when appropriate conversion factors are used.

Carnot Efficiency

Carnot efficiency refers to the theoretical maximum efficiency that any heat engine, refrigerator, or air conditioner can achieve based on the second law of thermodynamics. For a refrigeration cycle, the Carnot COP sets an ideal performance benchmark, calculated using the temperatures between which the device operates (expressed in absolute units like Kelvin). This concept is crucial for understanding the upper limits of performance and for comparing real devices against the ideal case.

Energy Unit Conversions

Energy unit conversions are essential when relating different efficiency metrics, such as converting between joules, British Thermal Units (BTUs), and kilowatt-hours (kW·h). These conversions enable the expression of cooling outputs and work inputs in compatible units. This is particularly important when deriving relationships between COP and EER or when computing operational costs from energy usage data.

Cost Estimation and Electricity Pricing

Cost estimation involves applying the efficiency metrics, expressed in terms of energy units, to calculate the operational cost of a device over a given period. By converting between energy measurements and applying the price per unit of energy (such as per kW·h), one can evaluate and compare the economic implications of using different air conditioning systems. This process is crucial for practical decision-making regarding energy consumption and equipment replacement.

Transcript

00:01 Hi everyone, this is the problem based on carnot cycle.

00:04 Here it is given that the cooling output and required electrical energy input both are in june.

00:44 So coefficient of performance will be.

01:04 The energy efficiency ratio is given by qc upon w.

01:31 Qc will be taken and vatu.

01:35 What should be taken in? water one watt hour is 3 .41 vehicle so energy efficiency ratio will be 3 .41 k deep part temperature of hot to be 95 finite that is 308 .8.

02:28 108 .15 kelvin temperature of cold junction to be 80 % sorry 85 % that is 299 .82 kelvin.

02:43 So coefficient of performance kish cult to 299 .8 to 35 .93.

03:28 So energy efficiency ratio for card not device working on the given temperature err will be 3 .41 into 30 that is 122 .74 c part given ac with energy efficiency ratio 10 .9 cooling output is 1 .9 into 10 to the power 10 june per year eer is 10 .9 is 10 .9 that will be 3 .41 into into coefficient of performance.

04:54 So coefficient of performance you will get 10 .9 upon 3 .41 which is equal to qc upon qh minus qc substituting the value hence heat transfer to the heart junction you will get 2 .4944 jule per year of 6 .9289 of 6 .9 kilowatt hour per year.

06:09 The cost of electricity is given 15 .3 cents per kilowatt hour.

06:39 So expenses for ac in one year, 15 .3 into 6 .928910 to the power 3.

07:04 So it is 1060 dollars per year.

07:18 Now depart...

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