Lecture 11:
Polytropic Processes Quiz Today?

• Polytropic Process is defined and explained
• Let’s do some example problems.
• Read Chap 4: Sections 1-3
From Chap 3: 138, 142,144,147

Sec 3.10.2 : Incompressible Substance Model
What is a polytropic process?
3 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: Isothermal process pv

C
W
  pdV
  
1 C V V

C ln( /
2 1
)
 p V
1 1 ln( /
2 1
)
 p V
2 2 ln( p
1
/ p
2
)
N= 0: Isobaric process
W
  pdV

(
2

V
1
)

Sec 3.10.2 : Incompressible Substance Model
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= k: Adiabatic process
Q

0 pv k 
C
N ≠ 1 : most polytropic processes
W
  pdV
 
V
C
N
  
N dV C V dV

(
2
1

N
1


V
1
1

N
N
)

CV
2
1

N
1


CV
N
1
1

N

(
N p V
2
2
) V
2
1

N
1


(
N
N p V
1
1
) V
1
1

N
 p V
2 2
 p V
1 1
1

N
4

Problem 3.148 T or F.
a) T or F : The change in specific volume from saturated liquid to saturated vapor (v g
- v f
) 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 (CO
2 as an ideal gas. ( p r
) at 320 K and 55 bar can be modeled
= 0.75 T r
= 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 constan t. T
5

Example Problem: (3.139) One kilogram of air in a piston cylinder assembly undergoes two processes in series from an initial state where p
1_gage
= 0.5 MPa, and T
1
= 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.
6
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.

P-v diagram m = 1 kg of air p
State 1: p
T
1
1
= 0.5 MPa+0.1013MPa = 0.6013MPa
= 227 deg C.= 500 K
State 2:
T
2
V
2
= T
1
= 2V
1
= 227 deg C. = 500 K
State 3: p
3
V
3
= p
1
= V
2
= 0.6013MPa
Apply 1 st Law of Thermo:
Process 1-2: constant T
ΔU
1-2
=Q
1-2
- W
1-2
Process 2-3: constant V
ΔU
2-3
=Q
2-3
- W
2-3
1 v
3
2
7

m = 1 kg of air R= 0.2870 kJ/kg-K
Apply Ideal Gas Law: pV
 mRT
State 1: p
1
= 0.6013 MPa T
1
= 500K p
1
3
2 v
V
1
 mRT p
1
1

(1 kg )(0.2870
0.6013
State 2:
T
2
= 500 K V
2
= 2V
1
/
MPa
)(500 )
= 0.4773 m 3
10
6
MPa
N m
2
1000 kJ p
2
 mRT
2
V
2

(1 kg )(0.2870
/
0.4773
m 3
)(500 ) MPa
10 6 N m 2
1000 kJ
State 3: p
3
= p
1
= 0.6013MPa V
3
= V
2
= 0.4773 m 3
T
3
 p V
3 3 mR

(0.6013
MPa )(0.4773
m
3
) 10
6
N m
2
(1 kg )(0.2870
/ ) MPa kJ
1000



0.2386
1000 K m
3
0.3006
MPa
8

m = 1 kg of air R= 0.2870 kJ/kg-K
State 1: p
1
= 0.6013 MPa T
1
= 500K V
1
=0.2386 m 3
State 2: p
2
= 0.3006 MPa T
2
= 500 K V
2
= 0.4773 m 3
State 3: p
3
= 0.6013MPa T
3
= 1000 V
3
= 0.4773 m 3
---------------------------------------------------------------------------------
Process 1-2:
U
2

U
1

Q

W where: U
2

U
1

(
2
 u
1
)
 v
(
2

T
1
)

0
W
  pdV
 
V
C pV
 mRT
 ln


V
V
1
2


 ln


V
V
1
2



MPa m
3
(0.6013
)(0.2386
) ln(
6
0.4773 10 /
)
N m
0.2386
MPa
2 p

C
V kJ
1000

99.45
kJ
Q

W

99.45
kJ
9

m = 1 kg of air R= 0.2870 kJ/kg-K
State 1: p
1
= 0.6013 MPa T
1
= 500K V1=0.2386 m 3
State 2: p
2
= 0.3006 MPa T
2
= 500 K V
2
= 0.4773 m 3
State 3: p
3
= 0.6013MPa T
3
= 1000 V
3
= 0.4773 m 3
---------------------------------------------------------------------------------
Process 2-3:
U
3

U
2

Q

W where: so
U
3

U
2

(
3
 u
2
)
 v
(
3

T
2
) c v
 k
R

1

0.2870

U
3

U
2

(1 kg )(0.7175
/

0.7175
/


K

358.75
kJ
10
W

0
V
 c onstant
Q

U
3

U
2

358.75
kJ

11