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Lec27 - (6.2a(2.17 1 st Tds equation is in term of du The 2...

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ME 200: Thermodynamics I Lecture 27: T-ds relations, Entropy change for incompressible substances Sections 6.3-6.4 Problems: SP27; 6.11, 6.14 Professor Hukam Mongia Office Hours: MWF 2:30 to 3:30 pm in ME 83 (Other times email for appointment) Email: [email protected] Phone: 765-494-5640 Course Website: https://engineering.purdue.edu/ME200 Course Secretaries: Diana Akers (ME 84) and Marilyn Morrison (ME 100) See Kul Inn (ME 189) about grading Homework Assignments 10/18/2009 1
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Introducing the T dS Equations How easy is to compute Or TdS =d Q ? Consider a pure, simple compressible system undergoing an internally reversible process. Negligible impact of motion and gravity:
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Unformatted text preview: (6.2a) (2.17) 1 st Tds equation is in term of du; The 2 nd TdS Equation in terms of dh. Why do we need in term of dh? Closed versus Open System H = U + pV dH = d(U + pV) = dU + d(pV) = (dU + pdV) + Vdp =TdS + Vdp Therefore, TdS = dH – Vdp the 2 nd TdS equation Mass basis: Tds = dh – vdp In a constant pressure process: Tds = dh e.g., during phase change, ds = (h g – h f )/T For an incompressible Substance, apply the 1 st Tds equation: Tds = du = c(T) dT Heating water from ambient cold day (32 o F) to 200 o F : s 1 = -0.0003 Btu/lb. o R s 2 = 0.294 Btu/lb. o R Saturated vapor s g = 1.7762 Btu/lb. o R...
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