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ENGR 2213 Thermodynamics. F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma. Reversible Steady-Flow Work. For a steady-flow device undergoing an internally reversible process,. ( δ q) rev - ( δ w) rev = dh + d(ke) + d(pe). - PowerPoint PPT Presentation

### Transcript of ENGR 2213 Thermodynamics

• ENGR 2213 Thermodynamics

F. C. LaiSchool of Aerospace and Mechanical EngineeringUniversity of Oklahoma

• Reversible Steady-Flow Work(q)rev - (w)rev = dh + d(ke) + d(pe)(q)rev = T ds = dh - v dpFor a steady-flow device undergoing an internally reversible process,Neglect the changes in kinetic and potential energies,(q)rev - (w)rev = dh(w)rev = - v dp

• Workreversible work in closed systemsreversible work associated withan internally reversible processan steady-flow device The larger the specific volume, the larger the reversible work produced or consumed by the steady-flow device.

• WorkTo minimize the work input during a compression process Keep the specific volume of the working fluid as small as possible.To maximize the work output during an expansion process Keep the specific volume of the working fluid as large as possible.

• WorkSteam Power Plant Pump, which handles liquid water that has a small specific volume, requires less work.Gas Power PlantWhy does a steam power plant usually have a better efficiency than a gas power plant? Compressor, which handles air that has a large specific volume, requires more work.

• Work1. To approach an internally reversible process as much as possible by minimizing the irreversibilities such as friction, turbulence, and non-quasi-equilibrium compression.Minimizing the Compressor Work To keep the specific volume of the gas as small as possible by maintaining the gas temperature as low as possible during the compression process.This requires that the gas be cooled as it is compressed.

• Steady-Flow WorkPolytropic Processes (pvn = constant)

• Steady-Flow WorkPolytropic Processes (pvn = constant)Isentropic Processes (pvk = constant)

• Steady-Flow WorkIsothermal Processes (pv = constant)1. n = k12. 1 < n < k23. n = 13

• Example 1Air entering a compressor at p1 = 100 kPa and T1 = 20 C and exiting at p2 = 500 kPa. If the airundergoes a polytropic process with n = 1.3, determine the work and heat transfer per unit mass of flow rate.

• Example 1 (continued)= 425 K= - 164.15 kJ/kgPolytropic processes

• Example 1 (continued)= - 164.15 + (426.35 293.17)Table A-17, T1 = 293 K, h1 = 293.17 kJ/kg T2 = 425 K, h2 = 426.35 kJ/kg = - 30.97 kJ/kg

• Isentropic Efficiency for Turbines Isentropic efficiency for a turbine is defined as the ratio of the actual performance of a turbine to the performance that would be achieved by undergoing an isentropic process for the sameinlet state and the same exit pressure.

• Isentropic EfficiencyTurbines

• Example 2Air enters a turbine at p1 = 300 kPa and T1 = 390 K and exits at p2 = 100 kPa. Given that the actual work output from the turbine is 74 kJ/kg and if the turbine operates adiabatically, determine the isentropic efficiency for the turbine.

• Example 2 (continued)Table A-17 T1 = 390 Kpr1 = 3.481, h1 = 390.88 kJ/kg= 3.481 (100/300)= 1.1603Table A-17 pr2 = 1.1603, h2s = 285.27 kJ/kg= 390.88 285.27 = 105.6 kJ/kg= 0.7

• Isentropic Efficiency for Compressors Isentropic efficiency for a compressor is defined as the ratio of the performance of a compressor that would be achieved by undergoing an isentropic process to the actual performance for the same inlet state and the same exit pressure.

• Isentropic EfficiencyCompressors

• Example 3Air enters an insulated compressor at p1 = 95 kPa and T1 = 22 C. Given that p2/p1 = 6 and c = 0.82, determine the exit temperature for the air.

• Example 3 (continued)Table A-17 T1 = 295 Kpr1 = 1.3068, h1 = 295.17 kJ/kg= 1.3068 (6)= 7.841Table A-17 pr2 = 7.841, T2s = 490.29 K h2s = 493.0 kJ/kgTable A-17 h2 = 536.4 kJ/kg, T2 = 532 K

• Example 40.5 kilogram of water executes a Carnot powercycle. During the isothermal expansion, the wateris heated until it is a saturated vapor from an initialstate where the pressure is 1.5 MPa and the qualityis 25%. The vapor then expands adiabatically topressure of 100 kPa. Find the heat addition and rejection from this cycle. the cycle efficiency.

• Example 4 (continued)Given:p1 = p2 = 1.5 MPap3 = p4 = 100 kPax1 = 0.25W23 = 403.8 kJ/kgFind:Q12 = ?Q34 = ? = ?

• Example 4 (continued)Q12 = m(u2 u1) + mp(v2 v1)Table A-5 p1 = p2 = 1.5 MPa, hf = 844.84 kJ/kg, hfg = 1947.3 kJ/kg, hg = 2792.2 kJ/kgsf = 2.315 kJ/kg K, sfg = 4.1298 kJ/kg K, sg = 6.4448 kJ/kg Kh1 = hf + x1 hfg = m(h2 h1)s1 = sf + x1 sfg = 844.84 + 0.25(1947.3) = 1331.67 kJ/kg= 2.315 + 0.25(4.1298) = 3.3474 kJ/kg Kh2 = hg = 2792.2 kJ/kgs2 = sg = 6.4448 kJ/kg K

• Example 4 (continued)Q12 = m(h2 h1)Table A-5 p3 = p4 = 100 kPa, hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg, hg = 2675.5 kJ/kgsf = 1.3026 kJ/kg K, sfg = 6.0568 kJ/kg K, sg = 7.3594 kJ/kg K= 0.5(2792.2 1331.67) = 730.27 kJQ34 = m(h4 h3)s3 = s2 = 6.4448 kJ/kg K s4 = s1 = 3.3474 kJ/kg K

• Example 4 (continued)Q34 = m(h4 h3)h3 = hf + x3 hfg = 417.46 + 0.849(2258) = 2334.5 kJ/kgh4 = hf + x4 hfg = 417.46 + 0.338(2258) = 1180.66 kJ/kg= 0.5(1180.66 2334.5) = -576.92 kJ