ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

[...]... lbf in–2 0 .14 50 BTU 0.9478 kJ kJ kg 1 BTU BTU lb 1 0.9478 0.4299 kJ kg 1 K 1 BTU lb 1 R 1 0.2388 kJ kmole 1 BTU lbmole 1 0.2388 K 1 R 1 1 kJ K BTU R 1 0.5266 1 1 kJ kg K BTU lb 1 R 1 0.2388 kJ kmole 1 K 1 BTU lbmole 1 R 1 0.2388 °C, K °F, °R (9/5)T+32 °C, K s kJ kJ kg 1 kJ kmole 1 m3 m3 m s 1 m3 kg 1 m3 kmole 1 kJ kJ kJ kg 1 °R 1. 8 s BTU 0.9478 BTU lb 1 0.4299 BTU lbmole 1 0.4299 ft3 35. 315 gallon... 35. 315 gallon 264.2 ft s 1 3.2 81 ft3 lbm 1 16. 018 ft3 lbmole 1 16. 018 BTU 0.9478 ft lbf 737.5 BTU lb 1 0.4299 kg kg 1 1bm lbm 1 m ft 3.2 81 f k activity of component k, βP, βT, kmole lbmole molecules molecules kmole 1 lbmole -1 2.2046 compressibility βs /fk K 1, atm 1 R 1, bar 1 1 atm γk ˆ φ k /φk thermal conductivity η η First Law efficiency relative efficiency bar 0.555, 1. 013 1. 013 ˆ ˆ activity coefficient,... stream availability or exergy Subscripts a b c chem c.m c.v e f f f fg g H I inv id iso L max m min net p p,o R rev r s sf sh Th TM TMC wwet kJ kg 1 1 BTU lb 0.4299 K bar 1 ºR atm 1 1.824 1 kJ kmole BTU lbmole 1 0.4299 kJ K 1 kJ kg 1 BTU R 1 BTU lb 1 0.2388 0.2388 air boundary critical chemical control mass control volume exit flow saturated liquid (or fluid) formation saturated liquid (fluid) to... weight can be calculated using Eq 5, i.e., M = 0.78×28 + 0. 21 32 + 0. 01 39.95 = 28.975 kg per kmole of mixture The total mass m = 3 .12 ×28.02 + 0.84×32 + 0.04×39.95 = 11 5.9 kg, and mass fractions are: YN2 = mN2 /m = 3 .12 ×28.02 /11 5.9 = 0.754 Similarly YO2 = 0.232, and YAr = 0. 013 8 Remark The mixture of N2, O2, and Ar in the molal proportion of 78 :1: 21 is representative of the composition of air (see the... Properties C James Clark Maxwell (18 31 18 79) Relations 1 First Maxwell Relation a Remarks 2 Second Maxwell Relation a Remarks 3 Third Maxwell Relation a Remarks 4 Fourth Maxwell Relation a Remarks 5 Summary of Relations D Generalized Relations 1 Entropy ds Relation a Remarks 2 a 3 a 4 a E 1 2 3 4 a 5 a 6 7 F 1 G H 1 a b 2 a b c I J 1 a b 2 a b 3 4 a 5 a b 6 a b K 1 a b 2 a b c 3 Internal Energy (du)... Problems G Chapter 7 Problems H Chapter 8 Problems I Chapter 9 Problems J Chapter 10 Problems K Chapter 11 Problems L Chapter 12 Problems M Chapter 13 Problems Appendix A Tables Appendix B Charts Appendix C Formulae Appendix D References Chapter 1 1 INTRODUCTION A IMPORTANCE, SIGNIFICANCE AND LIMITATIONS Thermodynamics is an engineering science topic,which deals with the science of “motion” (dynamics)... a Deformable Control Volume 3 Second law and Entropy A Introduction 1 Thermal and Mechanical Energy Reservoirs a Heat Engine b Heat Pump and Refrigeration Cycle B Statements of the Second Law 1 Informal Statements a Kelvin (18 24 -19 07) – Planck (18 58 -19 47) Statement b Clausius (18 22 -18 88) Statement C Consequences of the Second Law 1 Reversible and Irreversible Processes 2 Cyclical Integral for a Reversible... of 6.023 10 26 molecules (or Avogadro number of molecules) of a species is called one kmole of that substance The total mass of those molecules (i.e., the mass of 1 kmole of the matter) equals the molecular mass of the species in kg Likewise, 1 lb mole of a species contains its molecular mass in lb For instance, 18 .02 kg of water corresponds to 1 kmole, 18 .02 g of water contains 1 gmole, while 18 .02 lb... k id λ 1 r Gruneisen constant kW m 1 K 1 BTU ft 1 R 1 0 .16 05 ω ρ φ φ Φ specific humidity density equivalence ratio, fugacity coefficient relative humidity, absolute availability(closed system) Φ' relative availability or exergy φ JT µ ν σ Ψ fugacity coefficient Joule Thomson Coefficient chemical potential stoichiometric coefficient entropy generation absolute stream availability Ψ' kg m–3 1bm ft–3... case of thin metal walls Heat transfer cannot occur across the adiabatic boundary In this case the boundary is impermeable to heat flux, e.g., as in the case of a Dewar flask P1 Q=0 P 2>P 1 T 2>T 1 To Combustion P1 Q P 2>P 1, T 2=T 1 Storage tanks Chamber Figure 2: (a) Compression of natural gas for gas turbine applications; (b) Compression of natural gas for residential applications System Boundary Control . K 1 , atm 1 R 1 , bar 1 0.555, 1. 013 β s atm 1 bar 1 1. 013 γ k activity coefficient, ˆ α k / ˆ α k id ˆ φ k /φ k Gruneisen constant λ thermal conductivity kW m 1 K 1 BTU ft 1 R 1 0 .16 05 ηFirst. kJ kg 1 BTU lb 1 0.4299 q c charge R gas constant kJ kg 1 K 1 BTU lb 1 R 1 0.2388 R universal gas constant kJ kmole 1 BTU lbmole 1 0.2388 K 1 R 1 S entropy kJ K 1 BTU R 1 0.5266 s. m 3 kg 1 ft 3 lb m 1 16. 018 v specific volume (mole basis) m 3 kmole 1 ft 3 lbmole 1 16. 018 W work kJ BTU 0.9478 W work kJ ft lb f 737.5 w work per unit mass kJ kg 1 BTU lb 1 0.4299 w