id at constant volume and and AT is the temperature difference during the process Example 3-1:A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid Determine the final internal energy of the fluid. Neglect the energy stored in the Qout 500KJ %| paddle wheel.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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heat capacity of ifhe flüid at constant vofume and
process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on
e respectively, and AT is the temperature difference during the process.
paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling
Example 3-1: A rigid tank contains a hot fluid that is cooled while being stirred by a
the fluid Determine the final internal energy
of the fluid. Neglect the energy stored in the
Qout = 500 kJ
paddle wheel.
Wahin
= 100 kJ
Solution : A fluid in a rigid tank looses heat
U= 800 kJ
U = ?
while being stirred. The final internal energy
Fluid
of the fluid is to be determined.
Assumptions :
Transcribed Image Text:heat capacity of ifhe flüid at constant vofume and process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on e respectively, and AT is the temperature difference during the process. paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling Example 3-1: A rigid tank contains a hot fluid that is cooled while being stirred by a the fluid Determine the final internal energy of the fluid. Neglect the energy stored in the Qout = 500 kJ paddle wheel. Wahin = 100 kJ Solution : A fluid in a rigid tank looses heat U= 800 kJ U = ? while being stirred. The final internal energy Fluid of the fluid is to be determined. Assumptions :
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