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1. EGR 334 ThermmodynamcisChapter 3: Section 15

   Lecture 11: Polytropic Processes Quiz Today?
2. Main concepts for today’s lecture: Polytropic Process is defined and explained Let’s do some example problems. Reading Assignment: Read Chap 4: Sections 1-3 Homework Assignment: From Chap 3: 138, 142,144,147
3. Sec 3.10.2 : Incompressible Substance Model What is a polytropic process? or The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others. N= 1: Isothermal process N= 0: Isobaric process
4. Sec 3.10.2 : Incompressible Substance Model What is a polytropic process? or The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others. N= k: Adiabatic process N≠ 1: most polytropic processes
5. Problem 3.148 T or F. a) T or F: The change in specific volume from saturated liquid to saturated vapor (vg - vf) at a specified saturation pressure increases as the pressure decreases. T b) T or F: A two phase liquid-vapor mixture with equal volumes of saturated liquid and saturated vapor has a quality of 50%. F c) T or F: The following assumptions apply for a liquid modeled as incompressible: the specific volume is constant and the specific internal energy is a function only of temperature. T d) T or F: Carbon dioxide (CO2) at 320 K and 55 bar can be modeled as an ideal gas. ( pr = 0.75 Tr = 1.05 Z = 0.74 ) F e) T or F: When an ideal gas undergoes a polytropic process with n=1, the gas temperature remains constant. T
6. Example Problem: (3.139) One kilogram of air in a piston cylinder assembly undergoes two processes in series from an initial state where p1_gage= 0.5 MPa, and T1 = 227 deg C. Process 1: Constant temperature expansion until the volume is twice the initial volume. Process 2: Constant volume heating until the pressure is again 0.5 MPa. Sketch the two processes in series on a p-v diagram. Assuming ideal gas behavior, determine a) the pressure at state 2. b) the temperature at state 3 c) the work and heat transfer for each process.
7. p P-v diagram m = 1 kg of air State 1: p1 = 0.5 MPa+0.1013MPa = 0.6013MPa T1= 227 deg C.= 500 K v 1 3 State 2: T2 = T1= 227 deg C. = 500 K V2 = 2V1 2 Apply 1st Law of Thermo: Process 1-2: constant T ΔU1-2=Q1-2 - W1-2 Process 2-3: constant V ΔU2-3=Q2-3- W2-3 State 3: p3 =p1 = 0.6013MPa V3 = V2
8. p m = 1 kg of air R= 0.2870 kJ/kg-K Apply Ideal Gas Law: State 1: p1 = 0.6013 MPa T1= 500K v 1 3 State 2: T2 = 500 K V2 = 2V1 = 0.4773 m3 2 State 3: p3 =p1 = 0.6013MPa V3 = V2 = 0.4773 m3
9. m = 1 kg of air R= 0.2870 kJ/kg-K State 1: p1= 0.6013 MPa T1= 500K V1=0.2386 m3 State 2: p2= 0.3006 MPa T2 = 500 K V2 = 0.4773 m3 State 3: p3 = 0.6013MPa T3 = 1000 V3 = 0.4773 m3 --------------------------------------------------------------------------------- Process 1-2: where: for constant T: and
10. m = 1 kg of air R= 0.2870 kJ/kg-K State 1: p1= 0.6013 MPa T1= 500K V1=0.2386 m3 State 2: p2= 0.3006 MPa T2 = 500 K V2 = 0.4773 m3 State 3: p3 = 0.6013MPa T3 = 1000 V3 = 0.4773 m3 --------------------------------------------------------------------------------- Process 2-3: where: so for constant T:
11. End of slides for Lecture 11