HW9s - Spring 2006 Thermodynamics Homework #9 Solutions 1....

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Spring 2006 1 ME 201 Thermodynamics Homework #9 Solutions 1. A rigid wall, insulated container is divided into two regions by a removable wall. One region contains 1 lb m of kerosene at 100 ° F, while the other region contains 2 lb m of kerosene at 150 ° F. A stirrer is inserted into the container and when the wall between the two regions is removed the stirrer provides a shaft work input of 70 Btu. Determine the final temperature of the kerosene. Solution: System Type: Closed System Working Fluid: kerosene (incompressible substance) Process: Constant Volume Initial State Final State State 1 State 2 State 3 T 1 = 100 ° F = 560 R T 2 = 150 ° F = 610 R T 3 = 640 R = 182 ° F m 1 = 1 lb m m 2 = 2 lb m m 3 = 3 lb m Bold values are calculated. Q = 0 W sh = -70 Btu (into system) W bnd = 0 Initial State: Fixed Final State: UNKNOWN Conservation of Mass: m 1 + m 2 = m 3 1st Law: U final - U initial = -W sh Approach: To fix the final state we will use our process description along with the conservation of mass to determine the final mass. Then using the conservation of
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW9s - Spring 2006 Thermodynamics Homework #9 Solutions 1....

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