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Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Question
For a rigid tank with a volume of contains air at , . The tank is heated to , . The specific heat are , . What is the boundary work ?_________
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
Transcribed Image Text:**Problem Statement:**
For a rigid tank with a volume of \( V = 3 \, \text{m}^3 \), containing air at \( T_1 = 27^\circ \text{C} \) and \( P_1 = 200 \, \text{kPa} \), the tank is heated to \( T_2 = 102^\circ \text{C} \) and \( P_2 = 250 \, \text{kPa} \). The specific heat values are \( c_v = 0.718 \, \text{kJ/kg} \cdot \text{K} \) and \( c_p = 1.005 \, \text{kJ/kg} \cdot \text{K} \). What is the boundary work \( W_b \)?
**Options:**
- **A:** \( \frac{(P_1 + P_2)V}{2} \)
- **B:** \( c_v(T_2 - T_1) \)
- **C:** \( c_p(T_2 - T_1) \)
- **D:** 0
**Explanation:**
The problem is asking for the boundary work \( W_b \) done by the system when the air inside the rigid tank is heated. The given options offer different formulas to calculate this work based on the provided parameters.
- **Option A** suggests using the average pressure multiplied by the volume for the work done.
- **Option B** considers the change in internal energy using specific heat at constant volume.
- **Option C** involves the change in enthalpy using specific heat at constant pressure.
- **Option D** suggests that no boundary work is done.
**Analysis of the Rigid Tank System:**
In a rigid tank, the volume does not change during the process. Therefore, the boundary work done in such a system is typically zero, since boundary work \( W_b = P \Delta V \), and \( \Delta V = 0 \).
**Conclusion:**
The correct answer is **D: 0**, as no boundary work is performed in a rigid tank process.
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