#### Transcript Polytropic Processes

EGR 334 Thermmodynamcis Chapter 3: Section 15 Lecture 11: Polytropic Processes Quiz Today? 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 Sec 3.10.2 : Incompressible Substance Model 3 What is a polytropic process? pv N constant or pV N constant The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others. N= 1: W Isothermal process N= 0: W pv C 1 pdV C V V C ln(V 2 / V1 ) p 1V1 ln(V 2 / V1 ) p 2V 2 ln( p 1 / p 2 ) Isobaric process pdV p (V 2 V1 ) pC Sec 3.10.2 : Incompressible Substance Model 4 What is a polytropic process? pv N constant or pV constant N The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others. N= k: pv C k Adiabatic process Q 0 N≠ 1: most polytropic processes W pdV CV 1 N 2 C V CV 1 N N 1 N dV C V 1 N 1 N N dV 1 N ( p 2V 2 )V 2 C (V 2 1 N V1 1 N 1 N ( p1V 1 )V1 1 N ) N p 2V 2 p1V1 1 N 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-v diagram p m = 1 kg of air State 1: p1 = 0.5 MPa+0.1013MPa = 0.6013MPa T1 = 227 deg C.= 500 K State 2: T2 = T1 = 227 deg C. = 500 K V2 = 2V1 3 2 v Apply 1st Law of Thermo: State 3: p3 = p1 = 0.6013MPa V3 = V2 1 Process 1-2: constant T ΔU1-2=Q1-2 - W1-2 Process 2-3: constant V ΔU2-3=Q2-3 - W2-3 8 m = 1 kg of air R= 0.2870 kJ/kg-K Apply Ideal Gas Law: State 1: p1 = 0.6013 MPa p pV m R T 1 3 2 T1 = 500K v V1 m R T1 (1kg )(0.2870 kJ / kg K )(500 K ) p1 State 2: T2 = 500 K p2 m R T2 (1kg )(0.2870 kJ / kg K )(500 K ) 0.4773 m State 3: p3 = p1 = 0.6013MPa T3 mR 10 N / m 2 0.2386 m 3 kJ V2 = 2V1 = 0.4773 m3 V2 p 3V 3 6 0.6013 M P a 1000 N m M Pa 3 6 2 2 kJ 10 N / m 0.3006 M P a kJ V3 = V2 = 0.4773 m3 3 1000 N m M Pa 6 (0.6013 M P a )(0.4773 m ) 10 N / m (1kg )(0.2870 kJ / kg K ) M Pa 1000 N m 1000 K 9 m = 1 kg of air R= 0.2870 kJ/kg-K State 1: p1 = 0.6013 MPa State 2: p2= 0.3006 MPa T1 = 500K T2 = 500 K V1=0.2386 m3 V2 = 0.4773 m3 State 3: p3 = 0.6013MPa T3 = 1000 V3 = 0.4773 m3 --------------------------------------------------------------------------------Process 1-2: U 2 U 1 Q1 2 W 1 2 where: U 2 U 1 m ( u 2 u 1 ) m c v (T 2 T1 ) 0 for constant T: W 1 2 p V m R T co n stan t p and C V V2 V2 pdV dV C ln p1V1 ln V V V 1 1 C 6 0.4773 10 N / m (0.6013 M P a )(0.2386 m ) ln( ) 0.2386 M Pa 3 Q1 2 W 1 2 99.45 kJ 2 kJ 1000 N m 99.45 kJ 10 m = 1 kg of air R= 0.2870 kJ/kg-K State 1: p1 = 0.6013 MPa State 2: p2= 0.3006 MPa T1 = 500K V1=0.2386 m3 T2 = 500 K V2 = 0.4773 m3 State 3: p3 = 0.6013MPa T3 = 1000 V3 = 0.4773 m3 --------------------------------------------------------------------------------- U 3 U 2 Q 23 W 23 Process 2-3: where: U 3 U 2 m ( u 3 u 2 ) m c v (T3 T 2 ) cv so R k 1 0.2870 1.4 1 0.7175 kJ / kg K U 3 U 2 (1kg )(0.7175 kJ / kg K )(1000 500) K 358.75 kJ for constant T: V c o n stan t W 23 0 Q 2 3 U 3 U 2 358.75 kJ 11 End of slides for Lecture 11