Lecture32 - ChE 210: Meeting 32 April 6, 2011 Outline...

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Unformatted text preview: ChE 210: Meeting 32 April 6, 2011 Outline (F&R Chapters 7 & 8) Homework Due Friday 04/08/2011 - Exam Review 8.2, 8.9, 8.15, 8.25 Topics to be Covered - Energy Balance ° Types of Energy - Open vs. Closed System Shaft Work vs. Flow Work Steam Tables Reference States and Hypothetical Process Paths Calculating Enthalpy Changes - AP - AT 0 - Enthalpy of Mixing - Energy Balance Procedure - Table of Enthalpies - Psychrometric Chart Energy Balance Q—W=AEK+AEP+AU Q—WS=AEK+AEP+AH 1/6 ChE 210: Meeting 32 April 6, 2011 Steam Tables (8.5, 3.6, 8.7) 2/6 ChE 210: Meeting 32 April 6, 2011 Reference States and Hypothetical Process Paths - Cannot know U or H absolutely. - Can only know change in U or H relative to a reference state. - Define U or H of reference state to equal zero. - Calculate change in U or H from reference state to desired state with a hypothetical process path. - Change one condition at a time - T - P - ¢ - Mixing Example: Construct a hypothetical process path for heptane vapor at 200 C and 2 atm using a reference state of liquid heptane at 25 C and 1 atm. t 3/6 ChE 210: Meeting 32 April 6, 2011 Calculating Enthalpy Changes In general, AP at constant T and ¢ is easier to calculate for a liquid or solid than a gas. Liquids and solids at fixed T and ¢: A A A A A A AV~O and AUzO, so AH=AU+APV~VAP Ideal gases near 0 C and 1 atm: A ll O Nonideal gases require thermodynamic relationships that are beyond the scope of this course. [A .4 ‘ AT calculated using heat capacity data (Table 8.2) i l V 'l A .mk... AH = f CPlTldT L l l x 1.x pr: T1 Act) tabulated for normal melting and boiling point (Table 8.1) L—>v: AHV(Tb) VeL: -AHV(Tb) 89L: AHm(Tm) L931 -AHm(Tm) Enthalpy of Mixing A A AHS = H(solution) — H(solute) + H(solvent)] 4/6 ChE 210: Meeting 32 April 6, 2011 Given: Crystals are NaCzH3OZ*3HZO (Cp)all solutions = 3-5 kJ/kg—C (Cp)crysta|s = (C )H20,v = 32.4 kJ/kmoI—C P AHV(H20) = 4.39 x104 kJ/kmol AHS(25 C) = —1.71x1o4 kJ/kmol NaCZH302 AHHyd(25 C) = — 3.66 x 104 kJ/kmol NaC2H302 Find: Enthalpy of crystals at 50 C and 1atm Enthalpy of 15.4 mol% solution of NaczHgoz in water at 50 C 5/6 ChE 210: Meeting 32 April 6, 2011 6/6 ...
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This note was uploaded on 07/31/2011 for the course CHEM E 210 taught by Professor Shanks during the Spring '02 term at Iowa State.

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Lecture32 - ChE 210: Meeting 32 April 6, 2011 Outline...

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