ENG_sgwhk05 - -5th lecture- 2.7 Joule-Thomson effect (-)...

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1 The application of the 1 st law in throttling flow process of a real gas 2.7 Joule-Thomson effect ( 焦耳 - 汤姆逊效应 ) From previous class, we have The key point is to obtain -5 th lecture- V [——] p T C p -C V = {[——] T + P} V U P [——] V T C p V = { V -[——] T } P H and T T UH Vp ⎛⎞ ∂∂ ⎜⎟ ⎝⎠
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2 From U U ( V , T ) Joule coefficient μ J ( —— ) U T V V J V U T C T U V T V U μ = = 1 = V U T U T T V V U
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3 From H H ( p , T ) Joule-Thomson coefficient μ J-T ( —— ) H T p p T J p H T C T H p T p H μ = = 1 = p H T H T T p p H Isothermal Joule- Thomson coefficient
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4 Joule experiment (Air expands to vacuum Not sensitive not precise 1 The heat capacity of water in the calorimeter vessel is much larger than that of the gas 2 It does exist onflow Δ T =0
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5 Joule-Thomson effect (1852), the T, V have been changed after the throttling flow (1)Joule-Thomson experiment ( 焦耳 汤姆生实验 ) Throttling flow process ( 节流过程 ) ( p 1 > p 2 ) 多孔塞 P 1 V 1 p 2 V 2 p 1 p 2 开始 结束 绝热筒
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6 Because all changes to the gas occur adiabatically: Q = 0 . The gas on the left is compressed isothermally and the work done: W L = - p i (0 - V i ) = p i V i The work done in right: W R = - p f ( V f - 0) = - p f V f The total work done: W = W L + W R = p i V i - p f V f The thermodynamic basis of Joule-Thomson expansion. (2) Characteristics of the throttling flow
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7 Since Q = 0 , then Δ U= W + Q = W . the change of U of the gas as it moves from one side of the throttle to the other is: U f - U i = W = p i V i - p f V f U f + p f V f = U i + p i V i H f = H i The expansion occurs without change of enthalpy: an isenthalpic process ( 焓过程 ) . The thermodynamic basis of Joule-Thomson expansion.
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