EGR 334 ThermmodynamcisChapter 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:

Homework Assignment:

• Read Chap 4: Sections 1-3

From Chap 3: 138, 142,144,147

3Sec 3.10.2 : Incompressible Substance Model

or

What is a polytropic process?

12 1 1 1 2 1 2 2 1 2ln( / ) ln( / ) ln( / )W pdV C V V C V V pV V V p V p p

The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others.

N= 1: Isothermal process

constantNpv constantNpV

pv C

N= 0: Isobaric process p C

2 1( )W pdV p V V

4Sec 3.10.2 : Incompressible Substance Model

or

What is a polytropic process?

0Q

The exponent, N, may take on any value from -∞ to + ∞, but some values of N are more interesting than others.

N= k: Adiabatic process

constantNpv constantNpV

kpv C

1 12 1( )

1

N NN

N

C V VCW pdV dV C V dV

V N

N≠ 1: most polytropic processes

2 1

1 11 12 2 1 12 1 2 2 1 1

( ) ( )

1 1 1

N N N NN N p V V pV VCV CV p V pV

N N 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 3c) the work and heat transfer for each process.

7

P-v diagram

 

v

p

2

13State 1:

p1 = 0.5 MPa+0.1013MPa = 0.6013MPaT1 = 227 deg C.= 500 K

State 2: T2 = T1 = 227 deg C. = 500 KV2 = 2V1 State 3: p3 = p1 = 0.6013MPaV3 = V2

m = 1 kg of air

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

8

 

v

p

2

1 3

State 1: p1 = 0.6013 MPa T1 = 500K

m = 1 kg of air R= 0.2870 kJ/kg-KApply Ideal Gas Law: pV mRT

311 6 2

1

(1 )(0.2870 / )(500 ) 10000.2386

0.6013 10 /

mRT kg kJ kg K K MPa N mV m

p MPa N m kJ

State 2: T2 = 500 K V2 = 2V1 = 0.4773 m3

22 3 6 2

2

(1 )(0.2870 / )(500 ) 10000.3006

0.4773 10 /

mRT kg kJ kg K K MPa N mp MPa

V m N m kJ

State 3: p3 = p1 = 0.6013MPa V3 = V2 = 0.4773 m3

3 6 23 3

3

(0.6013 )(0.4773 ) 10 /1000

(1 )(0.2870 / ) 1000

p V MPa m N m kJT K

mR kg kJ kg K MPa N m

9

State 1: p1 = 0.6013 MPa T1 = 500K V1=0.2386 m3

m = 1 kg of air R= 0.2870 kJ/kg-K

2 1 1 2 1 2U U Q W

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: 2 1 2 1 2 1( ) ( ) 0vU U m u u mc T T

constantpV mRT

1 2 1 2 99.45Q W kJ

for constant T: and

2 21 2 1 1

1 1

ln lnV VC

W pdV dV C pVV V V

CpV

6 23 0.4773 10 /

(0.6013 )(0.2386 ) ln( ) 99.450.2386 1000

N m kJMPa m kJ

MPa N m

10

State 1: p1 = 0.6013 MPa T1 = 500K V1=0.2386 m3

m = 1 kg of air R= 0.2870 kJ/kg-K

State 2: p2= 0.3006 MPa T2 = 500 K V2 = 0.4773 m3 State 3: p3 = 0.6013MPa T3 = 1000 V3 = 0.4773 m3

---------------------------------------------------------------------------------

3 2 2 3 2 3U U Q W Process 2-3:

where:

so

3 2 3 2 3 2( ) ( )vU U m u u mc T T

onstantV c

2 3 3 2 358.75Q U U kJ

for constant T: 2 3 0W

0.28700.7175 /

1 1.4 1v

Rc kJ kg K

k

3 2 (1 )(0.7175 / )(1000 500) 358.75U U kg kJ kg K K kJ

11

End of slides for Lecture 11