Chapter 11

# Chapter 11 - Problem 11.2 Given Data on an air compressor...

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Hence the process is feasible! s0 > or for all real processes The second law of thermodynamics states that, for an adiabatic process Entropy s Temperature T s7 1 J kg K = (11.11b) sc p ln T 2 T 1 Rln p 2 p 1 = The governing equation for entropy is R 287 J kg K = c p 1004 J kg K = T 2 285 273 + () K = p 2 650 101 + ( ) kPa = T 1 20 273 + K = p 1 101 kPa = The data provided, or available in the Appendices, is: Solution Find: Whether or not the vendor claim is feasible Given: Data on an air compressor Problem 11.2
The process is Entropy s Temperature T T 1 247 C = T 2 520K = T 2 T 1 p 1 p 2 1k k = The governing equation for the lowest temperature is the isentropic relation Eq. 11.12b, or k 1.4 = p 2 650 101 + ( ) kPa = T 1 20 273 + () K = p 1 101 kPa = The data provided, or available in the Appendices, is: Solution Find: Lowest possible delivery temperature Given: Data on an air compressor Problem 11.3

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δ qT d s = But ds dh T = c p dT T = Hence, for p = const. Tds dh v dp = a) For a constant pressure process we start with Eq. 11.10b c v 717 J kg K = c v c p R = c p 1004 J kg K = R 287 J kg K = T 2 1200 273 + () K = T 1 100 273 + K = The data provided, or available in the Appendices, is: Solution Find: Heat to raise temperature to 1200 o C at a) constant pressure and b) constant volume Given: Air in a piston-cylinder Problem 11.5

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q u = Constant volume: q uw + = Constant pressure: From the first law: Heating to a higher temperature at constant pressure requires more heat than at constant volume: some of the heat is used to do work in expanding the gas; hence for constant pressure less of the heat is available for raising the temperature. q 789 kJ kg = qc v T 2 T 1 () = qT c v d = δ v dT = Hence δ d s = But ds du T = c v dT T = Hence, for v = const. Tds du p dv + = b) For a constant volume process we start with Eq. 11.10a q 1104 kJ kg = p T 2 T 1 = c p d = δ p dT = Hence
The two processes can be plotted using Eqs. 11.11b and 11.11a, simplified for the case of constant pressure and constant volume. a) For constant pressure s 2 s 1 c p ln T 2 T 1 Rln p 2 p 1 = (11.11b) so sc p ln T 2 T 1 = b) For constant volume s 2 s 1 c v ln T 2 T 1 v 2 v 1 + = (11.11a) so v ln T 2 T 1 = The processes are plotted in the associated Excel workbook

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Problem 11.5 (In Excel) Given: Air in a piston-cylinder Find: Heat to raise temperature to 1200 o C at a) constant pressure and b) constant volume; plot Solution The given or available data is: T 1 = 100 o C T 2 = 1200 o C R = 287 J/kg.K c p = 1004 J/kg.K c v = 717 J/kg.K The equations to be plotted are: T (K) a) s J/kg.K) b) s J/kg.K) 373 0 0 473 238 170 573 431 308 673 593 423 773 732 522 873 854 610 973 963 687 1073 1061 758 1173 1150 821 1273 1232 880 1373 1308 934 a) For constant pressure sc p ln T 2 T 1 = b) For constant volume v T 2 T 1 =
T-s

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Chapter 11 - Problem 11.2 Given Data on an air compressor...

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