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Consider the springtime melting of ice in the mountains, which provides cold water running in a river at \(34 \mathrm{~F}\) while the air temperature is \(68 \mathrm{~F}\). What is the availability of the water relative to the temperature of the ambient?

Short Answer

Expert verified
The availability of the water relative to the ambient temperature is 6.64%.

Step by step solution

01

Understanding Temperature Conversion

First, we need to convert the temperatures from Fahrenheit to Kelvin. The formula for converting Fahrenheit to Kelvin is \( K = (F - 32) \times \frac{5}{9} + 273.15 \). This will allow us to perform calculations using a thermodynamic temperature scale.
02

Convert Water Temperature to Kelvin

Convert the water temperature from Fahrenheit: \( T_{ ext{water}} = 34 \). Using the conversion formula, we get:\[ K_{ ext{water}} = (34 - 32) \times \frac{5}{9} + 273.15 = 273.71 \, K \]
03

Convert Air Temperature to Kelvin

Convert the air temperature from Fahrenheit: \( T_{ ext{air}} = 68 \). Using the conversion formula, we get:\[ K_{ ext{air}} = (68 - 32) \times \frac{5}{9} + 273.15 = 293.15 \, K \]
04

Calculating Availability

Availability refers to the maximum useful work obtainable (exergy) as the system reaches equilibrium with the environment. Here, since we consider only the thermal part, it's essentially calculating the Carnot efficiency relative to the ambient temperature:\[ \text{Availability} = 1 - \frac{T_{ ext{water}}}{T_{ ext{air}}} \]Substituting the Kelvin temperatures we found:\[ \text{Availability} = 1 - \frac{273.71}{293.15} = 0.0664 \text{ or } 6.64\% \]

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Key Concepts

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

Temperature Conversion
When discussing thermodynamics, we often need to convert temperatures to the Kelvin scale to accurately perform our calculations. This is because Kelvin is an absolute temperature scale widely used in scientific applications. The formula for converting Fahrenheit to Kelvin is simple: subtract 32 from the Fahrenheit temperature, multiply by \(\frac{5}{9}\), and then add 273.15.
For example:
  • To convert 34°F, the calculation is \(34 - 32 = 2\); multiplied by \(\frac{5}{9}\) gives about 1.11. Adding 273.15 brings us to roughly 273.71 K.
  • Similarly, for 68°F, subtract 32 to get 36, multiply by \(\frac{5}{9}\) yielding 20, and add 273.15 to reach 293.15 K.
By doing this conversion, we can better understand and use the temperatures in thermodynamic equations.
Carnot Efficiency
Carnot efficiency is a key concept in thermodynamics and reflects the maximum possible efficiency of a heat engine operating between two temperatures. It's a theoretical value that determines how well we can convert heat into work.
The Carnot efficiency \(\eta\) is computed using the formula: \[ \eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] where \(T_{\text{cold}}\) and \(T_{\text{hot}}\) are the absolute temperatures of the cold and hot reservoirs, respectively.
In our exercise, the water represents the cold reservoir (273.71 K) and the air the hot reservoir (293.15 K). Applying the Carnot efficiency formula gives us: \[ \eta = 1 - \frac{273.71}{293.15} = 0.0664 \] meaning a maximum theoretical efficiency of 6.64%. This value tells us the fraction of the thermal energy that can be transformed into work.
Exergy
Exergy is a term used to describe the maximum useful work that can be extracted from a system as it comes into equilibrium with its environment. It's a measure of a system's potential to cause change. In thermodynamics, exergy is considered when calculating the theoretical limits to the efficiency of processes.
The concept ties closely to Carnot efficiency, as it's about optimizing energy use. Exergy depends on the state of both the system and its environment.
By calculating the availability in the exercise, we're measuring the exergy of the water at 34°F relative to the air temperature at 68°F. The availability calculation aligns with the Carnot efficiency, as it also formulates the ratio of the temperature difference: \[ \text{Availability} = 1 - \frac{T_{\text{water}}}{T_{\text{air}}} = 6.64\% \] This value represents the "useful" part of the energy that can be harnessed, illustrating the limits imposed by thermodynamic laws.

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Most popular questions from this chapter

A refrigerator should remove \(1.5 \mathrm{~kW}\) from the cold space at \(-10^{\circ} \mathrm{C}\) while it rejects heat to the kitchen at \(25^{\circ} \mathrm{C}\). Find the reversible work.

An air conditioner on a hot summer day removes \(8 \mathrm{~kW}\) of energy from a house at \(21^{\circ} \mathrm{C}\) and pushes energy to the outside, which is at \(31^{\circ} \mathrm{C}\). The house has a 15000 -kg mass with an average specific heat of \(0.95 \mathrm{~kJ} / \mathrm{kgK}\). In order to do this, the cold side of the air conditioner is at \(5^{\circ} \mathrm{C}\) and the hot side is at \(40^{\circ} \mathrm{C}\). The air conditioner (refrigerator) has a \(\mathrm{COP}\) that is \(60 \%\) that of a corresponding Carnot refrigerator. Find the actual air conditioner COP, the power required to run the air conditioner, the rate of exergy destruction inside the air conditioner, and the total rate of exergy destruction due to the air conditioner and the house.

An adiabatic and reversible air compressor takes air in at \(100 \mathrm{kPa}, 310 \mathrm{~K}\). The air exits at \(600 \mathrm{kPa}\) at the rate of \(0.4 \mathrm{~kg} / \mathrm{s}\). Determine the minimum \(\mathrm{com}-\) pressor work input and repeat for an inlet at \(295 \mathrm{~K}\) instead. Why is the work less for a lower inlet \(T ?\)

A heat engine receives \(1 \mathrm{~kW}\) heat transfer at \(1000 \mathrm{~K}\) and gives out \(400 \mathrm{~W}\) as work, with the rest as heat transfer to the ambient. Find its first- and second-law efficiencies.

A piston/cylinder has forces on the piston, so it maintains constant pressure. It contains \(2 \mathrm{~kg}\) of ammonia at \(1 \mathrm{MPa}, 40^{\circ} \mathrm{C}\) and is now heated to \(100^{\circ} \mathrm{C}\) by a reversible heat engine that receives heat from a \(200^{\circ} \mathrm{C}\) source. Find the work out of the heat engine using the exergy balance equation.

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