Chapter 7: Problem 123
A \(200-L\) insulated tank contains nitrogen gas at \(200 \mathrm{kPa}, 300 \mathrm{~K}\). A line with nitrogen at \(500 \mathrm{~K}, 500\) \(\mathrm{kPa}\) adds \(40 \%\) more mass to the tank with a flow through a valve. Use constant specific heats to find the final temperature and the entropy generation.
Short Answer
Step by step solution
Identify Given Information
Assumptions and Properties
Calculate Initial Mass
Calculate Final Mass
Apply Energy Balance to Find Final Temperature
Calculate the Change in Entropy
Conclude with Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
ideal gas law
In the exercise, we used the ideal gas law to calculate the initial mass of nitrogen in a tank. Given the initial pressure, volume, and temperature, we determined the mass as \( m_1 = \frac{P_1 V}{RT_1} \). This step is crucial to determine how much nitrogen was originally present, allowing further calculations when additional mass is added.
The Ideal Gas Law assumes that gases behave ideally, meaning that the interactions between molecules are negligible, and the size of the molecules is much smaller than the distance between them. This assumption works well under normal conditions of temperature and pressure for gases like nitrogen.
adiabatic process
In this scenario, the added nitrogen enters the tank without exchanging heat due to the insulation, making the filling process adiabatic. Consequently, we apply energy conservation principles without the need to account for heat transfer.
For adiabatic processes involving ideal gases, the change in internal energy depends solely on the change in temperature. Therefore, knowing the initial and final conditions allows us to find unknown variables like the final temperature of the gas after mass addition. Understanding how adiabatic processes work helps predict system behavior when designing and analyzing thermodynamic cycles.
entropy generation
In our nitrogen tank exercise, entropy generation was calculated using the formula \( \Delta S = m_1 c_v \ln \frac{T_2}{T_1} + m_{in} c_v \ln \frac{T_2}{T_{line}} \). Here, \( m_1 \) and \( m_{in} \) represent the initial mass and the mass added, respectively. The specific heat capacity \( c_v \) acts as a bridge, linking temperature changes to specific entropy changes.
Entropy generation serves as a key indicator of process inefficiency. In engineering, minimizing entropy generation is essential to improving system performance. The second law of thermodynamics tells us that entropy in an isolated system always increases over time, hence understanding and calculating entropy change helps optimize energy systems.
specific heats for nitrogen
In our exercise, the specific heat capacity at constant volume, \( c_v = 0.743 \text{kJ/kg.K} \), was used. This value helps calculate energy changes during the adiabatic process. For instance, knowing \( c_v \) allows us to determine how the internal energy changes when nitrogen is added and how much entropy is generated.
Understanding specific heats is crucial when analyzing thermal systems, as it facilitates accurate modeling of energy flows and temperature changes. While \( c_v \) pertains to energy changes without volume changes, \( c_p \) is used when pressure remains constant. Both form the backbone of energy calculations in thermodynamic analyses.