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Work Problem \(3.15\) assuming surfaces parallel to the \(x\)-direction are adiabatic.

Short Answer

Expert verified
In order to solve the work problem assuming surfaces parallel to the x-direction are adiabatic, we need to apply the first law of thermodynamics for adiabatic processes, which states that the change in internal energy \(\Delta U\) is equal to the work done \(W\). The work done in an adiabatic process can be calculated using the formula \(W = P_1V_1\left[ \left(\frac{V_2}{V_1}\right)^{-\gamma} - 1\right]\), where \(P_1\) and \(V_1\) are the initial pressure and volume, \(V_2\) is the final volume, and \(\gamma\) is the adiabatic index. By substituting the relevant variables into the work equation and plugging the work value into the first law of thermodynamics, we can find the change in internal energy and analyze the behavior of the thermodynamic system in various engineering and physics applications.

Step by step solution

01

Understand the problem

First, we have to understand the problem. An adiabatic surface means that no heat transfer occurs through the surface. Therefore, any energy or work done using/consuming/producing heat energy must only interact with the system.
02

Identify the relevant variables and equations

Now, let's identify the relevant variables and equations. We are given that surfaces parallel to the x-direction are adiabatic. The relevant variables in an adiabatic process are the internal energy, work done, and the pressure and volume changes of the system. The equation that relates these variables is the first law of thermodynamics for adiabatic processes: \[ \Delta U = W \] where \(\Delta U\) is the change in internal energy, and \(W\) is the work done by or on the system. Work for an adiabatic process is given by the equation: \[ W = P_1V_1\left[ \left(\frac{V_2}{V_1}\right)^{ -\gamma} - 1\right] \] where \(P_1\) and \(V_1\) are the initial pressure and volume, \(V_2\) is the final volume, and \(\gamma\) is the adiabatic index (ratio of specific heat capacities, \(C_p / C_v\)).
03

Use the equations to find a solution

Assuming that we have the values of \(P_1\), \(V_1\), and \(V_2\), and \(\gamma\), we can substitute them into the work equation to find the work done in the adiabatic process: \[ W = P_1V_1\left[ \left(\frac{V_2}{V_1}\right)^{ -\gamma} - 1\right] \] Next, we can plug this work value into the first law of thermodynamics for adiabatic processes to determine the change in internal energy: \[ \Delta U = W \] Once we have the change in internal energy, we can analyze other properties of the system if needed, such as final temperature or final pressure.
04

Interpret the results and draw conclusions

The solution to the exercise gives us an understanding of how the adiabatic process affects the work done and the change in internal energy. The fact that no heat transfer occurs through surfaces parallel to the x-direction greatly simplifies the problem, allowing us to focus on the work done and the change in internal energy. This property can be used to analyze the behavior of thermodynamic systems in various engineering and physics applications.

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Key Concepts

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

First Law of Thermodynamics
The first law of thermodynamics is a version of the law of conservation of energy tailored for thermodynamic processes. It states that energy can neither be created nor destroyed; rather, it can only be transformed from one form to another. This fundamental principle can be expressed in equation form as
\[\begin{equation}\text{}\Delta U = Q - W\end{equation}\]where \(\Delta U\) represents the change in internal energy of the system, \(Q\) stands for the heat added to the system, and \(W\) is the work done by the system. In an adiabatic process, \(Q\) is zero because no heat is transferred in or out of the system. Consequently, any change in the system's internal energy must come from work done on or by the system. Understanding this relationship is crucial in thermodynamics, particularly when predicting how systems will behave when isolated from any heat interchange.

For students grappling with this concept, it is essential to recognize that in an isolated system such as an adiabatic one, any increase in temperature or other forms of energy must come from work being performed. This could be due to compression, expansion, or other mechanical means.
Internal Energy
Internal energy, symbolized as \(U\), is the total energy contained within a system. It encompasses all forms of energy, including kinetic energy of molecules, potential energy from molecular interactions, and other forms of microscopic energy that contribute to a system's total energy state. Internal energy is a state function, meaning it depends only on the current state of the system and not on the path taken to reach that state.
Changes in internal energy, denoted by \(\Delta U\), occur due to the exchange of heat \(Q\) and work \(W\) with the surroundings. However, in an adiabatic process where the heat exchange \(Q\) is zero, the change in internal energy is directly synonymous with the work done on or by the system, as indicated by the equation \(\Delta U = W\).

To truly understand this concept, one must grasp that internal energy is not directly measurable. Instead, we infer changes in internal energy through observable changes, such as temperature or pressure variations in a system undergoing an adiabatic process.
This aspect is particularly important when dealing with adiabatic systems in exercises and real-life scenarios, as the internal energy change provides insight into how the system will respond to work without external heat interactions.
Work Done in Thermodynamics
Work in thermodynamics refers to the energy transfer that occurs when a force is applied over a distance. It is central to thermodynamic processes and is usually represented as \(W\). In a more general context, work done by or on a system can involve various mechanisms, such as mechanical pistons or rotational shafts.
Particularly in adiabatic processes, the work done is directly related to changes in the system's volume and is calculated with the formula:\[\begin{equation}W = P_1V_1\left[ \left(\frac{V_2}{V_1}\right)^{ -\gamma} - 1\right]\end{equation}\]where \(P_1\) is the initial pressure, \(V_1\) the initial volume, \(V_2\) the final volume, and \(\gamma\) the adiabatic index. Understanding this formula is critical for mechanically inclined students, as it illustrates the direct correlation between work done and the physical change in a system's dimensions.

Remembering that in thermodynamics, work is considered positive when done by the system and negative when done on the system can also enhance grasp of the energetic transactions within the adiabatic process.
Pressure and Volume Changes
Pressure and volume are state variables that describe conditions within a thermodynamic system. Their changes are interdependent and follow established physical laws, such as Boyle's Law for an ideal gas at constant temperature stating that pressure is inversely proportional to volume. In adiabatic processes, no heat is exchanged with the environment, and the relationship between pressure and volume undergoes a unique transformation described by the adiabatic equation:\[\begin{equation}PV^\gamma = \text{constant}\end{equation}\]where \(P\) represents pressure, \(V\) represents volume, and \(\gamma\) is again the adiabatic index. This equation shows that pressure and volume change concurrently in a specific way that maintains the above formula as constant throughout the adiabatic process.

Comprehending how pressure and volume interact in an adiabatic process is essential for students analyzing thermodynamic systems, as it can predict outcomes such as temperature changes or variations in internal energy based purely on mechanical manipulations of the system, such as compression or expansion, without considering heat transfer.

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Most popular questions from this chapter

Annular aluminum fins of rectangular profile are attached to a circular tube having an outside diameter of \(50 \mathrm{~mm}\) and an outer surface temperature of \(200^{\circ} \mathrm{C}\). The fins are \(4 \mathrm{~mm}\) thick and \(15 \mathrm{~mm}\) long. The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) If there are 125 such fins per meter of tube length, what is the rate of heat transfer per unit length of tube?

A \(0.20\)-m-diameter, thin-walled steel pipe is used to transport saturated steam at a pressure of 20 bars in a room for which the air temperature is \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient at the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the heat loss per unit length from the bare pipe (no insulation)? Estimate the heat loss per unit length if a 50 -mm-thick layer of insulation (magnesia, \(85 \%\) is added. The steel and magnesia may each be assumed to have an emissivity of \(0.8\), and the steam-side convection resistance may be neglected. (b) The costs associated with generating the steam and installing the insulation are known to be \(\$ 4 / 10^{9} \mathrm{~J}\) and \(\$ 100 / \mathrm{m}\) of pipe length, respectively. If the steam line is to operate \(7500 \mathrm{~h} / \mathrm{yr}\), how many years are needed to pay back the initial investment in insulation?

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.

A thin electrical heater is inserted between a long circular rod and a concentric tube with inner and outer radii of 20 and \(40 \mathrm{~mm}\). The rod (A) has a thermal conductivity of \(k_{\mathrm{A}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the tube (B) has a thermal conductivity of \(k_{\mathrm{B}}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its outer surface is subjected to convection with a fluid of temperature \(T_{\infty}=-15^{\circ} \mathrm{C}\) and heat transfer coefficient \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal contact resistance between the cylinder surfaces and the heater is negligible. (a) Determine the electrical power per unit length of the cylinders \((\mathrm{W} / \mathrm{m})\) that is required to maintain the outer surface of cylinder \(\mathrm{B}\) at \(5^{\circ} \mathrm{C}\). (b) What is the temperature at the center of cylinder A?

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\), respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-\mathrm{mm}\) thick layer of calcium silicate insulation \(\left(k_{\text {ins }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tube?

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