ThermoSolutions-CHAPTER12

# ThermoSolutions-CHAPTER12 - 12-1 Chapter 12 THERMODYNAMIC...

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12-1 Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations 12-1C +, +, y x z z dz dy y dx x w . w { w { w 12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as the other variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables. 12-3C ( a ) ( w x ) y = d x ; ( b ) ( w z ) d d z ; and ( c ) d z = ( w z ) x + ( w z ) 12-4C Only when ( w z / w x ) = 0. That is, when z does not depend on y and thus z = z ( x ). 12-5C It indicates that z does not depend on y. That is, z = z ( x ). 12-6C Yes. 12-7C Yes. y ( w z ) x ( w z ) y d x d z x +d x y + d y z d y x y x

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12-2 12-8 Air at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Air is an ideal gas Properties The gas constant of air is R = 0.287 kPa·m 3 /kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/ v . Noting that R is a constant and P = P ( T , ), 2 d T R dT R dv T P dT T P dP 0 ¸ ¹ · ¨ © § w w . ¸ ¹ · ¨ © § w w ( a ) The change in T can be expressed as d T #' T = 400 u 0.01 = 4.0 K. At = constant, +, kPa 276 . 1 ¢ ¢ /kg m 0.90 K) K)(4.0 /kg m kPa (0.287 3 3 dT R dP ( b ) The change in v can be expressed as d = 0.90 u 0.01 = 0.009 m 3 /kg. At T = constant, kPa 276 . 1 0 ¢ ¢ 0 0 2 3 3 3 2 /kg) m (0.90 /kg) m K)(0.009 K)(400 /kg m kPa (0.287 d T R dP T ( c ) When both v and T increases by 1%, the change in P becomes 0 0 . . ) 276 . 1 ( 276 . 1 ) ( ) ( T dP dP dP Thus the changes in T and balance each other. 12-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Helium is an ideal gas Properties The gas constant of helium is R = 2.0769 kPa·m 3 /kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/ . Noting that R is a constant and P = P ( T , ), 2 d T R dT R d T P dT T P dP 0 ¸ ¹ · ¨ © § w w . ¸ ¹ · ¨ © § w w ( a ) The change in T can be expressed as dT T = 400 u 0.01 = 4.0 K. At = constant, kPa 231 9 . /kg m 0.90 K) K)(4.0 /kg m kPa (2.0769 3 3 ¢ ¢ dT R dP ( b ) The change in can be expressed as d = 0.90 u 0.01 = 0.009 m 3 /kg. At T = constant, kPa 231 9 . /kg) m (0.90 ) m K)(0.009 K)(400 /kg m kPa (2.0769 2 3 3 3 2 0 ¢ ¢ 0 d T R dP T ( c ) When both v and T increases by 1%, the change in P becomes 0 ) 231 . 9 ( 231 . 9 ) ( ) ( 0 . . T dP dP dP Thus the changes in T and balance each other.
12-3 12-10 It is to be proven for an ideal gas that the P = constant lines on a T - v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis ( a ) For an ideal gas P = RT or T = P / R . Taking the partial derivative of T with respect to holding P constant yields R P T P ¸ ¹ · ¨ © § w w T P = const which remains constant at P = constant. Thus the derivative ( w T / w ) P , which represents the slope of the P = const. lines on a T - diagram, remains constant. That is, the P = const. lines are straight lines on a T - diagram.

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## This note was uploaded on 09/21/2009 for the course ME 311 taught by Professor Ferrenberg during the Fall '09 term at Nevada.

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ThermoSolutions-CHAPTER12 - 12-1 Chapter 12 THERMODYNAMIC...

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