Polytropic Processes

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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 )
pC
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 23  W 23
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 23  0
Q 2  3  U 3  U 2  358.75 kJ
11
End of slides for Lecture 11

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