Question

A hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) at a rate of \(2200 \mathrm{W}\). Determine the rate of entropy generation in the surrounding air, in \(\mathrm{W} / \mathrm{K}\).

Short answer

Answer: The rate of entropy generation in the surrounding air is 1.68 W/K.

Step-by-step solution

Convert Celsius temperatures to Kelvin

To convert the given temperatures from Celsius to Kelvin, we will add 273.15 to each temperature: $$T_H = 80^{\circ} \mathrm{C} + 273.15 \, \mathrm{K} = 353.15 \, \mathrm{K}$$ $$T_C = 5^{\circ} \mathrm{C} + 273.15 \, \mathrm{K} = 278.15 \, \mathrm{K}$$

Apply the entropy generation formula

Now that we have the absolute temperatures, we can plug the values into the entropy generation formula: $$\dot{S}_{gen} = \frac{\dot{Q}_H}{T_H} - \frac{\dot{Q}_C}{T_C}$$ Substitute the given values into the equation: $$\dot{S}_{gen} = \frac{2200 \, \mathrm{W}}{353.15 \, \mathrm{K}} - \frac{2200 \, \mathrm{W}}{278.15 \, \mathrm{K}}$$

Calculate the rate of entropy generation

Now, perform the calculations to find the rate of entropy generation: $$\dot{S}_{gen} = 6.23 \, \mathrm{W/K} - 7.91 \, \mathrm{W/K}$$ $$\dot{S}_{gen} = -1.68 \, \mathrm{W/K}$$ The negative sign indicates that there is a net decrease in entropy, which is expected since heat is being transferred from the hot-water pipe to the cooler surrounding air. Therefore, the rate of entropy generation in the surrounding air is 1.68 \(\mathrm{W/K}\).

Key concepts

Thermodynamic Temperature Conversion

Understanding thermodynamic temperature conversion is crucial when working with heat transfer and entropy generation. In thermodynamics, it's important to use the Kelvin scale, which is the absolute temperature scale used in scientific equations. The Kelvin temperature scale starts at absolute zero, the point at which all thermal motion stops, and is essential for avoiding negative temperature values in calculations.

To convert from Celsius to Kelvin, one simply adds 273.15 to the Celsius temperature. This is because the zero point in Celsius, which is the freezing point of water, is 273.15 degrees above absolute zero in Kelvin. For example, as shown in the step by step solution, a temperature of the hot-water pipe at \(80^{\textdegree}\mathrm{C}\) is converted to Kelvin by adding 273.15, resulting in \(353.15 \mathrm{K}\). Similarly, the surrounding air temperature is converted from \(5^{\textdegree}\mathrm{C}\) to \(278.15 \mathrm{K}\).

It's imperative to conduct this conversion before plugging temperatures into thermodynamic formulas since they are derived based on the properties of the Kelvin scale.

Entropy Generation Formula

Entropy can be a perplexing concept, as it is a measure of disorder or randomness in a system. However, the entropy generation formula allows us to quantify the irreversible production of entropy within a system due to processes like heat transfer. The general entropy generation formula is given by:\[\dot{S}_{gen} = \frac{\dot{Q}_H}{T_H} - \frac{\dot{Q}_C}{T_C}\]This equation represents the difference between the entropy transfer due to heating at a hot source (\(T_H\)) and cooling at a cold sink (\(T_C\)), where \(\dot{Q}_H\) and \(\dot{Q}_C\) are the heat transfer rates at the source and sink, respectively. Essentially, the formula explains how heat transfer generates entropy owing to a temperature difference. It's crucial to note that entropy generation, \(\dot{S}_{gen}\), is always zero or positive in a closed system due to the second law of thermodynamics.

Heat Transfer

Heat transfer refers to the movement of heat from one body or substance to another through various mechanisms, such as conduction, convection, and radiation. In the given exercise, we're dealing with heat loss from a hot-water pipe to cooler surrounding air, which is primarily governed by convection. Convection occurs when a fluid (such as air) is heated, becomes less dense, and rises, creating a natural circulation pattern that facilitates heat transfer.The rate of entropy generation during heat transfer can be indicative of the efficiency of the process. High rates of entropy generation can suggest significant irreversibilities and therefore less efficient energy transfer. As the exercise shows, analysis of entropy generation due to heat transfer can provide insights into the thermodynamic performance of systems and guide the design for greater efficiency. In the real world, optimizing heat transfer processes to minimize entropy production is key to saving energy and reducing operational costs.