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Unformatted text preview: Chapter 7  Section B  NonNumerical Solutions 7.2 ( a ) Apply the general equation given in the footnote on page 260 to the particular derivative of interest here: µ ∂ T ∂ P ¶ S = − µ ∂ T ∂ S ¶ P µ ∂ S ∂ P ¶ T The two partial derivatives on the right are found from Eqs. (6.17) and (6.16); thus, µ ∂ T ∂ P ¶ S = T C P µ ∂ V ∂ T ¶ P For gases, this derivative is positive. It applies to reversible adiabatic expansions and compressions in turbines and compressors. ( b ) Application of the same general relation (page 260) yields: µ ∂ T ∂ V ¶ U = − µ ∂ T ∂ U ¶ V µ ∂ U ∂ V ¶ T The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus, µ ∂ T ∂ V ¶ U = 1 C V · P − T µ ∂ P ∂ T ¶ V ¸ For gases, this may be positive or negative, depending on conditions. Note that it is zero for an ideal gas. It applies directly to the Joule expansion , an adiabatic expansion of gas confined in a portion of a container to fill the entire container. 7.3 The equation giving the thermodynamic sound speed appears in the middle of page 250. At written, it implicitly requires that V represent specific volume. This is easily confirmed by a dimensional analysis. If V is to be molar volume, then the right side must be divided by molar mass: c 2 = − V 2 M µ ∂ P ∂ V ¶ S ( A ) Applying the equation given in the footnote on page 260 to the derivative yields: µ ∂ P ∂ V ¶ S = − µ ∂ P ∂ S ¶ V µ ∂ S ∂ V ¶ P This can also be written: µ ∂ P ∂ V ¶ S = − ·µ ∂ P ∂ T ¶ V µ ∂ T ∂ S ¶ V ¸·µ ∂ S ∂ T ¶ P µ ∂ T ∂ V ¶ P ¸ = − ·µ ∂ T ∂ S ¶ V µ ∂ S ∂ T ¶ P ¸·µ ∂ P ∂ T ¶ V µ ∂ T ∂ V ¶ P ¸ Division of Eq. (6.17) by Eq. (6.30) shows that the first product in square brackets on the far right is the ratio C P / C V . Reference again to the equation of the footnote on page 260 shows that the second product in square brackets on the far right is − (∂ P /∂ V ) T , which is given by Eq. (3.3). Therefore, µ ∂ P ∂ V ¶ S = C P C V µ ∂ P ∂ V ¶ T = C P C V µ − 1 κ V ¶ 573 Substitute into Eq. ( A ): c 2 = VC P M C V κ or c = VC P M C V κ ( a ) For an ideal gas, V = RT / P and κ = 1 / P . Therefore, c ig = RT M C P C V ( b ) For an incompressible liquid, V is constant, and κ = 0, leading to the result: c = ∞ . This of course leads to the conclusion that the sound speed in liquids is much greater than in gases....
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 Spring '10
 brown
 Thermodynamics, Mole, Trigraph, Adiabatic process, ∂z ∂T

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