ME40 HW10 Solutions

ME40 HW10 Solutions - 12-28 A substance is heated in a...

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Unformatted text preview: 12-28 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The boiling temperature of this substance at a different pressure is to be estimated. Analysis From the Clapeyron equation, h fg (5 kPa ⋅ m 3 )/(0.002 kg) dP = = 14.16 kPa/K = dT sat Tv fg (353 K)(1× 10 −3 m 3 )/(0.002 kg) Using the finite difference approximation, P − P1 dP ≈ 2 dT sat T2 − T1 sat Weight Solving for T2, T2 = T1 + P2 − P1 (180 − 200) kPa = 353 K + = 351.6 K dP / dT 14.16 kPa/K 200 kPa 80°C 2 grams Sat. liquid Q 12-65 It is to be demonstrated that the Joule-Thomson coefficient is given by µ = Analysis From Eq. 12-52 of the text, T 2 ∂ (v / T ) . c p ∂T P cp = 1 µ ∂v −v T ∂T P Expanding the partial derivative of v/T produces 1 ∂v v ∂v / T = −2 ∂T P T ∂T P T When this is multiplied by T2, the right-hand side becomes the same as the bracketed quantity above. Then, µ= T2 cp ∂ (v / T ) ∂T P 12-68 The Joule-Thomson coefficient of refrigerant-134a at a given state is to be estimated. Analysis The Joule-Thomson coefficient is defined as µ = ∂T ∂P h We use a finite difference approximation as µ≅ T2 − T1 (at constant enthalpy) P2 − P1 At the given state (we call it state 1), the enthalpy of R-134a is T1 = 20°C P1 = 200 kPa h1 = 270.18 kJ/kg (Table A - 13) T2 = 19.51°C (Table A - 13) The second state will be selected for a pressure of 180 kPa. At this pressure and the same enthalpy, we have P2 = 180 kPa h2 = h1 = 270.18 kJ/kg Substituting, µ≅ T2 − T1 (19.51 − 20)K = = 0.0245 K/kPa P2 − P1 (180 − 200)kPa ...
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ME40 HW10 Solutions - 12-28 A substance is heated in a...

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