Question

A refrigerator has a second-law efficiency of 28 percent, and heat is removed from the refrigerated space at a rate of \(800 \mathrm{Btu} / \mathrm{min} .\) If the space is maintained at \(25^{\circ} \mathrm{F}\) while the surrounding air temperature is \(90^{\circ} \mathrm{F}\), determine the power input to the refrigerator.

Short answer

Answer: The power input to the refrigerator is approximately 382.8 Btu/min.

Step-by-step solution

Convert temperatures to an absolute scale

The given temperatures are in Fahrenheit, and we need to convert them to an absolute scale (Rankine) by adding 459.67 to each temperature. $$ T_{low} = 25^{\circ} \mathrm{F} + 459.67 = 484.67\,\mathrm{R} $$ $$ T_{high} = 90^{\circ} \mathrm{F} + 459.67 = 549.67\,\mathrm{R} $$

Calculate the ideal COP

The ideal coefficient of performance (COP) for a refrigerator is given by: $$ \mathrm{COP}_{ideal} = \frac{T_{low}}{T_{high} - T_{low}} $$ $$ \mathrm{COP}_{ideal} = \frac{484.67\,\mathrm{R}}{549.67\,\mathrm{R} - 484.67\,\mathrm{R}} = \frac{484.67}{65} = 7.46 $$

Calculate the actual COP

The second-law efficiency of the refrigerator is given as 28 percent. We can use this value to calculate the actual COP: $$ \mathrm{COP}_{actual} = \mathrm{COP}_{ideal} * \frac{\mathrm{second\text{-}law\, efficiency}}{100\%}$$ $$ \mathrm{COP}_{actual} = 7.46 * \frac{28\%}{100\%} = 2.09 $$

Determine the power input

Now that we know the actual COP, we can calculate the power input to the refrigerator. The rate of heat removed from the refrigerated space is given as \(800\,\mathrm{Btu/min}\). The power input can be calculated using the following formula: $$ \mathrm{Power\, input} = \frac{\mathrm{Heat\, removed}}{\mathrm{COP}_{actual}} $$ $$ \mathrm{Power\, input} = \frac{800\,\mathrm{Btu/min}}{2.09} = 382.8\,\frac{\mathrm{Btu}}{\mathrm{min}} $$ Therefore, the power input to the refrigerator is approximately 382.8 Btu/min.

Key concepts

COP (Coefficient of Performance)

In the realm of thermal systems, particularly refrigeration and air conditioning, the Coefficient of Performance (COP) serves as a critical indicator of efficiency. Mathematically, it's expressed as the ratio of useful heating or cooling provided to the work or energy input required.

For a refrigerator, COP can be interpreted as the amount of heat removed from the inner space (cooling effect) per unit of work (or energy) supplied to achieve this effect. A higher COP signifies a more efficient refrigerator as it extracts more heat for the same amount of power input.

To apply COP in practical situations, one must understand that it is a dimensionless quantity and can greatly affect the operating costs of a refrigeration system. When reviewing the exercise, it's also essential to point out that the second-law efficiency must be taken into account to find the actual COP, which reflects real-world deviations from the ideal behavior in thermodynamic systems.

Rankine temperature scale

The Rankine temperature scale, analogous to the Kelvin scale for metric measurements, is a thermodynamic (absolute) temperature scale used primarily in the engineering domains within the United States. It's named after the Scottish engineer William John Macquorn Rankine.

Similar to Kelvin, zero on the Rankine scale, denoted as 0 R, is absolute zero—a theoretical state where no thermal energy exists. The main difference is the unit increments, which are equivalent to Fahrenheit, not Celsius as with Kelvin. Converting from Fahrenheit to Rankine is straightforward—add 459.67 to the Fahrenheit value. This step ensures that all temperature measurements are interpreted correctly in calculating thermodynamic properties, as seen in the solutions provided for the exercise question.

Heat transfer rate

The heat transfer rate is a measure of the thermal energy movement per unit time. It's an essential concept in thermodynamics and, by extension, in the functioning of refrigerators and air conditioners. In the context of the exercise, the term refers to the amount of heat energy removed from the refrigerated space every minute.

The rate at which heat is removed, usually expressed in British Thermal Units per minute (Btu/min) or watts in the metric system, is a vital parameter in calculating COP and subsequently, the energy efficiency of a refrigerator. An effective refrigeration system should maximize the heat transfer rate while minimizing energy input, which directly ties to lower operating costs and environmental impact.

Thermodynamic temperature conversion

Temperature conversions between different scales are foundational for thermodynamic calculations because most formulas require absolute temperatures. The provided exercise demonstrates the essential conversion from the Fahrenheit to the Rankine scale—an absolute measurement necessary for more complex thermodynamic equations.

Understanding these conversions is crucial, not just for academic exercises but also for real-world applications where precise thermal calculations are necessary. For instance, the efficiency of a refrigerator and the determination of power input hinge upon accurate temperature measurements and conversions, ensuring the safety, effectiveness, and sustainability of thermal systems.