Analyzing Work And Heat Transfer In A Tank System A Thermodynamic Approach

In the realm of thermodynamics, analyzing energy interactions within a system is crucial for understanding its behavior. This article delves into a specific scenario: a tank containing air stirred by a paddle wheel, where work is input, and heat is transferred to the surroundings. We will explore how to determine the external work done by the system using the fundamental principles of thermodynamics, specifically the first law of thermodynamics. This analysis is essential for various engineering applications, from designing efficient thermal systems to optimizing energy consumption in industrial processes. Understanding the interplay between work, heat, and internal energy changes within a system allows engineers to predict and control system behavior, ensuring optimal performance and energy efficiency.

The problem at hand involves a closed system – a tank containing air – where energy interactions occur via paddle wheel work and heat transfer. The paddle wheel stirs the air, implying work input into the system. Simultaneously, heat is transferred from the tank to the surroundings. Our objective is to determine the net external work done by the system. To tackle this, we need to grasp the first law of thermodynamics, which states that energy is conserved. In simpler terms, the change in internal energy of a system equals the net heat added to the system minus the net work done by the system. Mathematically, this is expressed as: ΔU = Q - W, where ΔU represents the change in internal energy, Q is the heat added to the system, and W is the work done by the system. It's crucial to distinguish between work done on the system and work done by the system, as they have opposite signs in the equation. Heat transfer also follows a sign convention: heat added to the system is positive, while heat leaving the system is negative. Applying these concepts, we can analyze the given scenario and arrive at the correct solution. Further complicating the scenario, we need to consider the implications of a closed system. A closed system, by definition, exchanges energy with its surroundings but not mass. This characteristic is paramount in simplifying our analysis, as we can focus solely on the energy interactions without the complexities of mass flow. The air inside the tank acts as our working fluid, and the paddle wheel serves as the mechanism for work interaction. The heat transfer to the surroundings represents an energy loss from the system, which needs to be accounted for in our calculations. By meticulously tracking these energy exchanges, we can accurately determine the external work done by the system.

To solve this problem, we need to carefully apply the first law of thermodynamics and pay close attention to the sign conventions. First, let's identify the given quantities: The work input to the paddle wheel is 9000 kJ. Since this is work done on the system, we will consider it as negative work. Therefore, W_in = -9000 kJ. The heat transferred to the surroundings is 3000 kJ. Because heat is leaving the system, we'll treat this as negative heat transfer: Q = -3000 kJ. Now, the critical point is to determine the external work done by the system. The paddle wheel does work on the air inside the tank, increasing its internal energy. The heat transfer to the surroundings reduces the internal energy. However, there's no volume change in this scenario. The tank has a fixed volume, meaning the system doesn't expand or contract against external pressure. Therefore, the work done by the system due to volume change (also known as boundary work or PΔV work) is zero. The external work done by the system, in this case, refers specifically to the work done against external forces due to a change in volume. Since the volume remains constant, there is no external work done by the system in that sense. Now, let's consider the first law of thermodynamics: ΔU = Q - W. Here, W represents the net work done by the system. We can rearrange this equation to find the change in internal energy: ΔU = -3000 kJ - (-9000 kJ) = 6000 kJ. This means the internal energy of the air inside the tank increased by 6000 kJ. However, this doesn't directly answer our question about the external work done by the system. As established earlier, since the tank volume is constant, there's no PΔV work. The work done by the paddle wheel increases the internal energy, and the heat transfer reduces it, but neither contributes to external work done by the system in the conventional sense. Therefore, the external work done by the system is zero. In essence, the 9000 kJ of work input is used to increase the internal energy of the air, and the 3000 kJ of heat is dissipated to the surroundings. The system itself does not perform any work against external forces because its volume is fixed. This distinction is crucial for a clear understanding of thermodynamic processes.

The correct answer is (a) Zero. This might seem counterintuitive at first, given the work input from the paddle wheel and the heat transfer. However, it's essential to focus on the definition of external work done by the system. In thermodynamics, external work often refers to the work done by the system against external pressure due to a change in volume (PΔV work). In this case, the tank has a fixed volume. This is the crucial piece of information. Since the volume doesn't change, the system cannot expand or contract against its surroundings. Consequently, the PΔV work is zero. The work input from the paddle wheel increases the internal energy of the air, making the air molecules move faster and collide more frequently. This heightened molecular activity translates to a higher temperature within the tank. The heat transfer to the surroundings is a dissipation of this internal energy, reducing the temperature increase. However, neither the paddle wheel work nor the heat transfer directly translates into external work done by the system. The system isn't pushing against anything or moving any external objects. Think of it like stirring a cup of coffee inside a closed, rigid container. You're doing work on the coffee, and it might get warmer, but the coffee isn't doing any work on the container or the surrounding environment (assuming the container doesn't expand). The same principle applies to the air in the tank. The paddle wheel work is internal work, and the heat transfer is energy dissipation. There's no mechanical work being performed on the surroundings. Therefore, the external work done by the system is zero. This understanding highlights the importance of carefully considering the system's boundaries and the types of work interactions involved. A system can have internal energy changes without performing external work, particularly if its volume remains constant. This concept is fundamental to many thermodynamic analyses and applications.

When tackling thermodynamics problems, there are several common pitfalls that students and engineers often encounter. One significant mistake is confusing work done on the system with work done by the system. In our problem, the paddle wheel does work on the air, which is a negative work term in the energy balance equation. The work we are looking for is the work done by the air on its surroundings. Failing to differentiate between these two can lead to incorrect sign conventions and ultimately, the wrong answer. Another common error is overlooking the significance of a fixed volume. Many problems involve PΔV work, which is the work done due to volume changes. However, in this scenario, the tank's rigid walls prevent any volume change. Thus, the PΔV work is zero, and focusing on this detail is key to correctly answering the question. Students might incorrectly assume that because there's work input (paddle wheel) and heat transfer, there must be some external work done. However, the first law of thermodynamics highlights that these energy interactions can change the internal energy of the system without necessarily resulting in external work. A related pitfall is neglecting the sign conventions for heat and work. Heat added to the system is positive, while heat leaving the system is negative. Work done by the system is positive, and work done on the system is negative. Consistent application of these conventions is crucial for accurate calculations. To avoid these pitfalls, a systematic approach is essential. First, carefully read and understand the problem statement, identifying the given quantities and what needs to be determined. Draw a simple diagram of the system to visualize the energy interactions. Clearly define the system boundaries. Apply the first law of thermodynamics, ensuring correct sign conventions. Finally, double-check your answer and make sure it logically aligns with the problem's physical constraints. By adopting these strategies, you can significantly reduce the likelihood of making mistakes and strengthen your understanding of thermodynamics. Remember, a firm grasp of the fundamental principles and attention to detail are paramount in solving thermodynamics problems effectively.

In conclusion, this problem illustrates the importance of a thorough understanding of thermodynamic principles, especially the first law of thermodynamics and the concept of external work. In a system where air is stirred within a fixed-volume tank, the work input from the paddle wheel primarily increases the internal energy of the air, while the heat transfer to the surroundings represents an energy loss. However, since the tank's volume remains constant, there's no PΔV work. Consequently, the external work done by the system is zero. This example highlights the significance of carefully defining the system boundaries, considering the specific types of work interactions involved, and paying close attention to sign conventions. By avoiding common pitfalls, such as confusing work done on and by the system or neglecting the impact of a fixed volume, we can accurately analyze thermodynamic processes. A systematic approach, combined with a solid grasp of fundamental concepts, ensures a correct interpretation of the energy interactions within a system. This understanding is crucial for engineers and scientists in designing and optimizing various thermodynamic systems, ranging from power plants to refrigeration cycles. The ability to correctly assess the work, heat, and internal energy changes enables informed decision-making, leading to improved energy efficiency and system performance. Ultimately, a strong foundation in thermodynamics empowers professionals to tackle complex engineering challenges and contribute to sustainable energy solutions.