2300_les18 - VI. Entropy G. Entropy Change of a Pure...

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Unformatted text preview: VI. Entropy G. Entropy Change of a Pure Substance 1. T-s diagram for a pure substance Sat. vap. line Sat. liq. line P = constant T sg sf s sfg = sg − sf Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 2. Tabular entropy data a. Superheated vapor: s from table b. Saturated liquid: s = sf c. Saturated mixture: s = sf + xsfg = (1 - x)sf + xsg d. Compressed liquid: s(T,p) ≅ sf at T (7-12) 3. Entropy change: ∆S = m∆s = m(s2 - s1) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 1 VI. Entropy 4. Example. The piston-cylinder device shown below contains 1.00 kg of water at 150 C and 100 kPa. The piston is frictionless and weightless. We now add 50 kJ of heat to the water. Find the increase in u, h, and s for the water. How much work is done by the water? Sketch the process on a T-s diagram. 100 kPa 1.00 kg H2O (system) T1 = 150 C T2 = ? C P1 = P2 = 100 kPa Qin = 50 kJ 1 kg H2O Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy a. assume heat capacity of piston and cylinder are negligible, ∆KE and ∆PE are negligible, and that expansion is quasi-equilibrium. b. data from Table A-6 (superheated water) at 150 C and 0.10 MPa. u1 = 2582.8 kJ/kg h1 = 2776.4 kJ/kg s1 = 7.6134 kJ/(kg K) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 2 VI. Entropy c. first law for a closed system ∆U = Qin − Wout ∆U = Q − P ∆V ( for a reversible process ) ∆H = Q m( h2 − h1 ) = Q h2 = Q + h1 = 50.0 kJ + 2776 .4 kJ = 2826 .4 kJ Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy d. interpolate in Table A-6, using h2, to find T2, u2, and s2 T2 = 175.3 C u2 = 2620.9 kJ/kg s2 = 7.725 kJ/(kg K) ∆u = u2 - u1 = 38.1 kJ/kg ∆h = h2 - h1 = 50.0 kJ/kg ∆s = s2 - s1 = 0.1117 kJ/(kg K) W = Q - ∆u = 50.0 - 38.1 = 11.9 kJ Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 3 VI. Entropy e. Sketch on T-s diagram Sat. vap. line 2 1 sg sf s p = 100 kPa = constant T Sat. liq. line Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy H. Isentropic processes ∆S = 0 or s2 = s1 (7-13) T P1 > P2 1 2 P1 P2 Good approximation for processes that are nearly adiabatic and reversible. Example: isentropic expansion of sat. vapor in a piston-cylinder device or turbine. s Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 4 VI. Entropy I. The TdS Relations and Changes in Entropy 1. First and second laws for a closed system dU = δQ − δW dS = ∑ j =1 n δQj Tj + δ Sgen a. First Tds equation for simple, compressible substance dU = δQrev − δWrev dS = δQrev T δ Wrev = PdV Tds = du + Pdv or ds = du Pdv + T T (7-23, 25) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy b. Second Tds equation for simple, compressible substance Recall that h = u + Pv or dh = du + Pdv + vdP Tds = du + Pdv Tds = dh − vdP or ds = dh vdP − T T (7-24, 26) 7-25 and 7-26 are valid for reversible and irreversible processes. Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 5 VI. Entropy 2. ∆S for solids and liquids (incompressible substances) ds = du pdv + T T du cdT = T T But dv = 0 Then ds = and c = cp = cv. Integration gives ∆s = s2 − s1 = ∫ c(T ) 1 2 dT T ≅ cav ln 2 T T1 (7-28) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 3. ∆S for ideal gases ds = ds = du Pdv dT dv + = cv +R T T T v dh vdP dT dP − = cp −R T T T P ∆s = ∫ cv (T ) 1 2 v dT + R ln 2 T v1 dT P − R ln 2 T P1 (ideal gas, 7-31) ∆s = ∫ c p (T ) 1 2 (ideal gas, 7-32) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 6 VI. Entropy a. Approximate analysis using averaged specific heats ∆s = cv ,av ln T2 v + R ln 2 T1 v1 T2 P − R ln 2 T1 P1 (ideal gas, 7-33) ∆s = c p,av ln (ideal gas, 7-34) where Tav = T1 + T2 and cav = c(Tav ) 2 Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy b. Analysis for ideal gases using integrated specific heats Let s o = ∫ c p (T ) 0 T dT (see Equation 7-37 and Table A-17) T Then ∫ T2 T1 c p (T ) T2 T1 dT dT dT o o = ∫ c p (T ) − ∫ c p (T ) = s2 − s1 0 0 T T T o s2 − s1 = s 2 − s1o − R ln P2 P1 P2 p1 (ideal gas, 7-39) s2 − s1 = s 2o − s1o − R ln (ideal gas, 7-40) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 7 VI. Entropy c. Example - Is it possible? Is ∆S > 0? Compressed air 2 kg at 4 atm, 300 K Cold air 1 kg at 1 atm 273 K Hot air 1 kg at 1 atm, 327 K Magic Vortex Tube Assume ideal gas with constant specific heat evaluated at 300 K. From (7-34), ∆s = c p ln T2 P − R ln 2 T1 P1 From Table A-1, 2, cp = 1.005 kJ/(kg K) and R = 0.2870 kJ/(kg K) Lesson 18, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy c. Example - Is ∆S > 0? For the hot stream ∆shot = c p ln T2 P 327 1 kJ − R ln 2 = 1.005 ln − 0.2870 ln = 0.4845 T1 P1 300 4 K For the cold stream ∆scold = 1.005ln 273 1 kJ − 0.2870ln = 0.3031 300 4 K kJ K The total change in entropy is ∆Stot = 0.4845 + 0.3031 = 0.7876 We conclude that the process is possible (although we don’t know how it works) because ∆S > 0. Lesson 18, Geof Silcox, Chemical Engineering, University of Utah 8 ...
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