2300_les21 - VI. Entropy L. Adiabatic or Isentropic...

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Unformatted text preview: VI. Entropy L. Adiabatic or Isentropic Efficiencies of Steady Flow Devices 1. Turbines (ηT = 80-90%) ηT = w Actual turbine work =a Isentropic turbine work w s (7-60) Isentropic work, ws, is calculated assuming same inlet condition and same outlet pressure as actual turbine. Recall ⎛ V2 V2 ⎞ 0 = w in + (h1 − h2 ) + ⎜ 1 − 2 ⎟ + g (z1 − z2 ) ⎜2 2⎟ ⎝ ⎠ ηT = w a h1 − h2 a ≅ w s h1 − h2 s (7-61) Lesson 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 2. Compressors and Pumps (ηC = 75-85%) ηC = Isentropic compressor work w s = Actual compressor work wa Isentropic work, ws, is calculated assuming same inlet condition and same outlet pressure as actual compressor. Recall ⎛ V2 V2 ⎞ 0 = w in + (h1 − h2 ) + ⎜ 1 − 2 ⎟ + g (z1 − z2 ) ⎜2 2⎟ ⎝ ⎠ ηC = w s h2 s − h1 ≅ w a h2 a − h1 (7-63) Lesson 21, Geof Silcox, Chemical Engineering, University of Utah 1 VI. Entropy 3. Compressor Example (Ex. 7-15 of text) Given an adiabatic efficiency, ηC = 0.80, find required power input. Qin = 0 2 P2 = 800 kPa Air P1 = 100 kPa 1 T1 = 12 C = 285 K Win = ? m = 0 .2 kg s Lesson 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 3. Compressor Example (cont.) Approach: calculate the isentropic work, ws, and then find the actual work, wa, using the adiabatic efficiency. Do this (1) assuming ideal gas with constant specific heats and then (2) with value obtained using ideal gas with variable specific heats. First approach: ⎛T ⎞ w s = h2 s − h1 = c p (T2s − T1 ) = c pT1 ⎜ 2 s − 1 ⎟ ⎝ T1 ⎠ ⎡⎛ P ⎞( k −1) / k ⎤ − 1⎥ Ws = mCpT1 ⎢⎜ 2 ⎟ ⎢⎝ P1 ⎠ ⎥ ⎣ ⎦ ⎡⎛ 800 ⎞(1.4 −1) / 1.4 ⎤ − 1⎥ = 46.5 kW Ws = 0.2 (1.005 ) 285 ⎢⎜ ⎟ ⎢⎝ 100 ⎠ ⎥ ⎣ ⎦ Lesson 21, Geof Silcox, Chemical Engineering, University of Utah 2 VI. Entropy 3. Compressor Example (cont.) Second approach: w s = h2 s − h1 Find h1 from Table A-17 at 285 K. Find h2s by first finding pr2. s2 − s1 = ∫ c p T1 T2 P dT − R ln 2 = 0 T P1 o o s2 − s1 − R ln P2 =0 P1 P2 Pr 2 = P1 Pr1 Pr 2 = Pr1 P2 800 = 1.1584 = 9.2672 P1 100 Using Pr2, interpolate in Table A-17 to give h2s = 517.05 kJ/kg Ws = m(h2 s − h1 ) = 0.2 (517.05 − 285.14 ) = 46.4 kW Lesson 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 3. Compressor Example (cont.) ηC = Isentropic compressor work w s Ws = = Actual compressor work w a Wa Wa = Ws ηC = 46.4 kW = 58.0 kW 0. 8 Lesson 21, Geof Silcox, Chemical Engineering, University of Utah 3 VI. Entropy 4. Another Compressor Example On p. 565 of Metcalf & Eddy, Wastewater Engineering, 3rd ed., Chapter 10, on design of activated sludge process, recommends the use of the following equation to calculate the power requirements for blowers to aerate sludge. Pw = wRT1 [( p2 ) 29.7ne p1 0.283 −1 (SI units, typographical errors exactly as they appear) Our equation for a reversible process with an ideal gas is Win = mkRT1 k −1 ⎡⎛ P ⎞( k −1) / k ⎤ ⎢⎜ 2 ⎟ − 1⎥ ⎢⎝ P1 ⎠ ⎥ ⎣ ⎦ (7-57a) Is the Metcalf & Eddy equation correct? Lesson 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy 4. Another Compressor Example (cont.) Pw = Pw w R T1 p1 p2 n k 29.7 e wRT1 [( p2 ) 29.7ne p1 0.283 −1 Table provided by Metcalf & Eddy for their formula. = power requirement, kW = mass flow rate of air, kg/s = universal gas constant, 8.314 kJ/(kmol K) = absolute inlet temperature, K = absolute inlet pressure, atm = absolute outlet pressure, atm = (k-1)/k = 0.283 for air = 1.395 for air = constant for SI units conversion = efficiency (usual range for compressors is 0.70 to 0.90) Lesson 21, Geof Silcox, Chemical Engineering, University of Utah 4 VI. Entropy 4. Another Compressor Example (cont.) After correcting the obvious typographical errors, the Metcalf & Eddy equation becomes 0.283 ⎤ wRT1 ⎡⎛ p2 ⎞ ⎢⎜ ⎟ − 1⎥ 29.7ne ⎢⎝ p1 ⎠ ⎥ ⎣ ⎦ We need to check two things: Pw = 1) Based on Table A-1, is 0.283 = Yes, pretty close. 2) Is Ru/29.7 = R? Yes, pretty close. k − 1 1.400 − 1 = = 0.2857 k 1.400 R= Ru 8.314 kJ /( kmol K ) = MWair 28.97 kg / kmol Lesson 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy K. 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 21, Geof Silcox, Chemical Engineering, University of Utah 5 VI. Entropy H. Entropy Balance for an Open System (Review) 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 21, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy I. Closing Thoughts on Entropy and the Second Law 1. High temperature thermal energy is more available for doing work than low temperature thermal energy. W T ηth ,int rev = out = 1 − L TH QH 2. Entropy is not conserved. Sgen > 0 for all real processes. 3. Isolated systems will only change in a direction that results in an increase in entropy. 4. Irreversibilities result in the generation of entropy. Highly irreversible processes have high rates of entropy generation. High rates of entropy generation usually degrade the performance of a process. Lesson 21, Geof Silcox, Chemical Engineering, University of Utah 6 ...
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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 University of Utah.

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