Question

Using EES (or other) software, determine the maximum work that can be extracted from a pond containing \(10^{5} \mathrm{kg}\) of water at \(350 \mathrm{K}\) when the temperature of the surroundings is \(300 \mathrm{K}\). Notice that the temperature of water in the pond will be gradually decreasing as energy is extracted from it; therefore, the efficiency of the engine will be decreasing. Use temperature intervals of \((a) 5 \mathrm{K},(b) 2 \mathrm{K}\) and \((c) 1 \mathrm{K}\) until the pond temperature drops to \(300 \mathrm{K}\). Also solve this problem exactly by integration and compare the results.

Short answer

Based on the step-by-step solution given, answer the following question: Q: Determine the maximum work that can be extracted from a pond containing 100000 kg of water at an initial temperature of 350 K. Compare the results for three different temperature intervals: 5 K, 2 K, and 1 K. Also, solve the problem exactly by integration and compare the results. A: To determine the maximum work that can be extracted from the pond for each temperature interval (5 K, 2 K, and 1 K), follow these steps: 1. Calculate the efficiency of the heat engine at various temperature intervals using the formula η = 1 - (Tₗ/Tₕ), where Tₗ is the surrounding temperature and Tₕ is the pond temperature. 2. Determine the amount of work extracted for each temperature interval using EES software or similar tools. 3. Calculate the total work extracted for each temperature interval by summing the work values from Step 2. 4. Solve the problem exactly by evaluating the integral expression W = ∫(Tf, Ti) η(T) dQ using EES or mathematical software. 5. Compare the total work extracted for each temperature interval (5 K, 2 K, and 1 K) with the exact value of the work extracted from Step 4 to analyze the accuracy of the results and the importance of smaller temperature intervals in thermodynamic problems.

Step-by-step solution

Calculate the efficiency of the heat engine

The efficiency of a heat engine is given by the formula: $$ \eta = 1 - \frac{T_{L}}{T_{H}} $$ where \(η\) is the efficiency, \(T_{L}\) is the temperature of the surroundings (300 K), and \(T_{H}\) is the temperature of the pond. For each temperature interval, we will calculate the efficiency of the heat engine, as the temperature of the pond decreases.

Determine the amount of work extracted for each temperature interval

Using EES software, we can determine the amount of work extracted for each temperature interval by executing the following steps: a. Set the initial temperature of the pond to 350 K. b. Calculate the efficiency of the heat engine for the current pond temperature. c. Extract work from the pond and update the pond temperature accordingly. d. Repeat steps (b) and (c) until the pond temperature drops to 300 K.

Calculate the total work extracted for each temperature interval

For each temperature interval (5 K, 2 K, and 1 K), sum up the work extracted calculated in Step 2 to get the total work extracted during the process.

Solve the problem exactly by integration

To get the exact value of the work extracted, we will use integral calculus. The integral expression for the work extracted is: $$ W = \int_{T_{f}}^{T_{i}} \eta(T) dQ $$ where \(W\) is the work extracted, \(T_{i}\) is the initial temperature (350 K), \(T_{f}\) is the final temperature (300 K), and \(dQ\) is the differential heat transfer. Evaluate the definite integral using EES or other mathematical software to obtain the exact value of the work extracted.

Compare the results

Compare the total work extracted for each temperature interval (5 K, 2 K, and 1 K) with the exact value of the work extracted obtained in Step 4. This comparison will help you understand the accuracy of the results obtained for different temperature intervals and the importance of taking smaller temperature intervals in thermodynamic problems.

Key concepts

Heat Engine Efficiency

When we talk about the efficiency of a heat engine, we are referring to its ability to convert thermal energy into work. In thermodynamics, this efficiency is crucial because it determines how much work can be extracted from a heat source. The basic formula for heat engine efficiency is given by \[\eta = 1 - \frac{T_{L}}{T_{H}}\]
where \( \eta \) is the efficiency, \( T_{L} \) represents the lower temperature (the surroundings, in this case, 300 K), and \( T_{H} \) is the higher temperature (the pond's initial temperature of 350 K). However, this formula assumes a simplified model where heat transfer occurs at a constant temperature, which isn't realistic for a pond cooling down. As the water temperature drops, so does the efficiency, because the difference in temperature between the heat source and the surroundings decreases. This concept helps us grasp the potential for energy conversion as we attempt to harness thermal energy, emphasizing the value of high temperature differences in achieving better efficiency.

Temperature Intervals Calculation

Temperature intervals play a significant role in the calculation of work extraction in thermodynamic processes. When we consider extracting work from a thermal source, like our pond, the temperature won't drop uniformly. By breaking down the cooling process into smaller temperature intervals—5 K, 2 K, and 1 K—we can approximate the variable efficiency of the heat engine better.

Each interval is a step in the process where we calculate the amount of energy that can be converted to work, taking into account the changing efficiency. The smaller the intervals, the closer our approximation will be to the real, gradual change in temperature and thus to the actual work that can be extracted. This stepped approach, while more labor-intensive, enhances the accuracy of our calculations, allowing us to make more precise predictions about the efficiency and work output of real-world thermodynamic systems. This concept illuminates how breaking down complex processes into manageable parts helps us understand and predict system behaviors more effectively.

EES Software Application

EES (Engineering Equation Solver) is a versatile software tool extensively used in the field of thermodynamics for solving complex equations. Its application extends to the calculation of maximum work extraction by automating repetitive calculations and applying iterative processes to solve for variables across changing conditions.

In our pond example, EES would be configured to simulate the step-by-step extraction of work with the changing efficiency as the pond temperature drops. The user would set the initial parameters, like the starting water temperature at 350 K and the surroundings at 300 K. The software would then compute the efficiency of the heat engine at each temperature interval using the efficiency formula. The ability of EES to handle calculations iteratively and to integrate functions across variable conditions makes it an indispensable tool for engineers and scientists who deal with thermodynamic problems. It saves time, increases accuracy, and provides an analytical approach that supports a deeper understanding of the thermal systems under study.

Integral Calculus in Thermodynamics

Integral calculus is a branch of mathematics that is particularly useful in thermodynamics for dealing with changes in systems. When we apply integral calculus to thermodynamics, we precisely quantify quantities like work and heat transfer over a process, which, by their nature, are dependent on changing conditions.

To calculate the exact work extracted from our pond, we'd use the following integral expression:\[W = \int_{T_{f}}^{T_{i}} \eta(T) dQ\]
Here, \(W\) is the total work extracted, \(T_{i}\) and \(T_{f}\) are the initial and final temperatures of the water, and \(dQ\) represents the infinitesimal amount of heat transferred. This integral accounts for the continuously changing efficiency as the pond cools, offering a precise calculation that gives a more accurate result than our earlier step-by-step intervals method. Understanding how integral calculus underpins thermodynamic analysis arms students with the tools to tackle more complex systems and transitions, reinforcing their broader problem-solving capabilities in the field.