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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 ?_________     A.     B.     C.

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 ?_________     A.     B.     C.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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 ?_________

 

  A.

 
  B.

 
  C.

 
  D.

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**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.
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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Givenrigid tankV=3m3T1=27oCP1=200kPaT2=102oCP2=250 kPaCv=0.718 kJ/kgKCp=1.005 kJ/kgK

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