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2300_les19 - VI Entropy H The Tds Relations and Changes in...

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Unformatted text preview: VI. Entropy H. The Tds Relations and Changes in Entropy 4. Example Two pieces of copper, A and B, with masses 1 and 3 kg, and initial temperatures 0 and 200 C are brought together and allowed to equilibrate while insulated from the surroundings. Determine (a) the entropy change of A and B and (b) the entropy generation (Sgen) for the process, in kJ/K. Entropy change given by (7-28): ∆S = mCav ln T2 T1 Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 4. Example Final temperature, T2, is obtained from energy balance. [mCav (T2 − T1 )]A + [mCav (T2 − T1 )]B = 0 1(T2 − 0 ) + 3(T2 − 200 ) = 0 T2 = 150 C (151 C if use Cav for each ) Entropy changes for A and B are ∆U = Q − W = 0 ∆SA = 1(0.390 ) ∆SB = 3 (0.400 ) kJ 423 kJ ln = 0.171 kg 273 K kJ 423 kJ ln = −0.134 kg 473 K Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 1 VI. Entropy 4. Example Entropy generation from overall entropy balance on composite system (7-9) (S2 − S1 )sys = ∫ δQ T + Sgen There is no heat transfer (δQ = 0) so Sgen = ( S2 − S1 )sys = ∆SA + ∆SB = 0.171 + ( −0.134 ) = 0.037 Because Sgen > 0, the process is irreversible. kJ K Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 5. Isentropic processes with ideal gases a. approximate treatment assuming constant specific heats i. first isentropic relation (applies to open and closed systems), recall ∆s = cv ,av ln T2 v + R ln 2 T1 v1 (ideal gas, 7-33) Reversible and adiabatic ⇒ isentropic ⇒ ∆s = 0. (7-33) becomes (remember R = cp - cv, k = cp/cv) T v cv ,av ln 2 = R ln 1 T1 v2 or T2 ⎛ v1 ⎞ =⎜ ⎟ T1 ⎜ v 2 ⎟ ⎝⎠ R / Cv ⎛v ⎞ =⎜ 1⎟ ⎜v ⎟ ⎝ 2⎠ (k −1 ) (ideal gas, isentropic, 7-42) Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 2 VI. Entropy ii. second isentropic relation (applies to open and closed systems) ∆s = c p,av ln T2 P − R ln 2 T1 P1 (ideal gas, 7-34) Reversible and adiabatic ⇒ isentropic ⇒ ∆s = 0. (6.23) becomes (remember R = cp - cv, k = cp/cv) T P c p,av ln 2 = R ln 2 T1 P1 or T2 ⎛ p2 ⎞ =⎜ ⎟ T1 ⎝ p1 ⎠ R / cp ⎛p ⎞ =⎜ 2 ⎟ ⎝ p1 ⎠ ( k −1) / k (ideal gas, isentropic, 7-43) Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy iii. third isentropic relation (applies to open and closed systems) From (7-42) and (7-43), P2 ⎛ v1 ⎞ =⎜ ⎟ P1 ⎝ v 2 ⎠ k (ideal gas, isentropic, 7-44) Equations 7-42 – 7-44 can be written as Tv k −1 = C TP (1−k ) / k k =C (ideal gas, isentropic, where C is a constant.) Pv = C Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 3 VI. Entropy b. exact treatment using variable specific heats o s2 − s1 = s 2 − s1o − R ln P2 P1 (ideal gas, 7-39) Reversible and adiabatic ⇒ isentropic ⇒ ∆s = 0. (7-39) becomes o 0 s 2 − s1o exp( s2 / R ) Pr 2 P2 = exp = = 0 P1 R exp( s1 / R ) Pr1 (ideal gas, isentropic, 7-49) where Pr is called the relative pressure (Table A-17). We can also define vr, the relative specific volume (Table A-17). ⎛ v2 ⎞ v ⎜⎟ = r2 ⎜v ⎟ ⎝ 1 ⎠s =const . v r 1 (ideal gas, isentropic, 7-50) Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy I. Entropy Balance for a Closed System (Review) ⎛Time rate of change ⎞ ⎛ net rate of entropy ⎜ ⎟⎜ ⎜ of entropy within ⎟ = ⎜ transport to system ⎜ system at time t ⎟ ⎜ at time t ⎝ ⎠⎝ ⎞ ⎛ rate of entropy ⎟⎜ ⎟ + ⎜ generation within ⎟ ⎜ the system at time ⎠⎝ ⎞ ⎟ ⎟ t⎟ ⎠ dSsys dt =∑ j =1 n Qj Tj + Sgen dSsys = ∑ j =1 n δQj Tj + δ Sgen (S2 − S1 )sys = ∫ δQ T + Sgen Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 4 VI. Entropy J. Entropy Balance for an Open System nQ dScv = ∑ mi si − ∑ me se + ∑ j + Sgen cv dt i e j =1 T j (7-83) (S2 − S1 )CV = ∑ mi si − ∑ me se + ∫ i e δQ T + Sgen cv (7-82) This is the integrated form. Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy Reversible and Actual Work and the Generation of Entropy Consider two, steady flow processes, one reversible and the other irreversible (actual), with the same inlet and exit conditions. Q or Q in ,act in ,rev h1, s1 1 2 h2, s2 Wout ,act or Wout ,rev Energy balances: m(h1 − h2 ) = Wout ,act − Qin,act = Wout ,rev − Qin,rev Entropy balances: Qin,act = mT (s2 − s1 ) − TSgen and Qin,rev = mT (s2 − s1 ) Conclusion: ∴Wout ,rev = Wout ,act + TSgen TSgen represents the rate at which work is “lost” due to irreversibilities. Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 5 VI. Entropy K. Reversible, Steady-Flow Work 1. Steady-flow balance equations (applied to a compressor) Qin m out 2 1 0 = m1 − m2 ⎡ ⎤ V 2 − V12 + g (z2 − z1 )⎥ Qcv ,in + Wnonflow ,in = m ⎢(h2 − h1 ) + 2 2 ⎣ ⎦ m1s1 − m2 s2 + Q + Sgen = 0 T min Win Divide the energy and entropy balances by m : q + w in = ( h2 − h1 ) + s1 − s2 + V22 − V12 + g ( z2 − z1 ) 2 qrev =0 T where Q = mq, W = mw ,and Sgen = 0. Lesson 19, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy In differential form δq + δw in = dh + d ⎜ ⎜ ⎛ V2 ⎞ ⎟ + gdz ⎟ ⎝2⎠ and − ds + δqrev T =0 Rearrange the later to give δq rev = Tds Recall that Tds = dh − vdP (7-24) Then δ w rev ,in = vdP + d ( ke ) + d ( pe ) Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 6 VI. Entropy Integrating from 1 to 2 gives w rev ,in = ∫ vdP + ∆ke + ∆ pe 1 2 (7-51) For a compressor, wrev,in is a positive number. For a turbine, wrev,in is a negative number. Neglecting ∆ke and ∆pe gives w rev ,in = ∫ vdP 1 2 (7-52) For a turbine, it is convenient to rewrite (7-51) as w rev ,out = − ∫ vdP − ∆ke − ∆ pe 1 2 Lesson 19, Geof Silcox, Chemical Engineering, University of Utah 7 ...
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This note was uploaded on 09/02/2010 for the course PHYS 2300 taught by Professor Silcox during the Fall '09 term at Utah.

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