Ideal Gas Entropy State Relations -1 Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
Entropic State Eqns. – Ideal Gases
• Recall Gibbs eqn.
– for ideal gas
– integrate
AE3450
dv
T
p
T
du
ds
dv
v
R
T
dT
c
v
2
1
2
1
2
1
v
v
T
T
v
s
s
v
dv
R
T
dT
c
ds
2
1
2
1
1
2
v
v
T
T
v
v
dv
R
T
dT
c
s
s
1
2
1
2
12
ln
2
1
v
v
R
T
dT
c
s
s
s
T
T
v
1
2
1
2
12
ln
2
1
R
T
dT
c
s
s
s
T
T
v
Ideal Gas Entropy State Relations -2
School of Aerospace Engineering
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
Entropic State Eqns. – Ideal Gases
• Using
h
version
– for ideal gas
– integrate
–
s
(
T
,
p
)
behavior?
AE3450
dp
T
v
T
dh
ds
dp
p
R
T
dT
c
p
2
1
2
1
2
1
v
v
T
T
p
s
s
p
dp
R
T
dT
c
ds
1
2
1
2
12
ln
2
1
p
p
R
T
dT
c
s
s
s
T
T
p
T dependence,
s
due to T
p dependence,
s
due to p
and dependence is separable!!
Ideal Gas Entropy State Relations -3 Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
Entropic State Eqns. – Ideal Gases
• So can generally write
– with
– e.g.,
with
AE3450
1
2
12
12
ln
p
p
R
T
s
s
o
can calculate
and/or find
in tables
2
1
12
T
T
p
o
T
dT
c
T
s
easy to
calculate
2
1
12
T
s
T
s
T
s
o
o
o
T
T
p
o
ref
T
dT
c
T
s
T
ref
typically
0
K
(
or
R)
School of Aerospace Engineering
CPG (+Ideal) Equations
• Simplifying
for
c
p
=const.
– normalizing
AE3450
1
2
12
ln
2
1
p
p
R
T
dT
c
s
T
T
p
entropy more sensitive to T changes
1
2
1
2
12
ln
ln
p
p
R
T
T
c
s
p
1
2
1
2
ln
ln
R
T
T
c
v
1
2
ln
2
1
R
T
dT
c
T
T
v
s
(
T
,
p
)
s
(
T
,
)
1
2
1
2
12
ln
ln
p
p
T
T
R
c
R
s
p
1
2
1
2
ln
ln
p
p
T
T
c
p
R
1
2
1
1
2
12
ln
ln
p
p
T
T
R
s
Ideal Gas Entropy State Relations -5
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Air Example
•
Given:
Air initially @ 27 C, 1 atm; later
same air compressed to 327 C, 50 atm
•
Find:
Entropy change for cpg and non-cpg cases
•
Assume:
tpg
air
Ideal Gas Entropy State Relations -6
School of Aerospace Engineering
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Air Example: Solution
•
Analysis (cpg):
State Eqn.
kgK
J
R
air
287
need values for
R
,
kgK
kJ
426
.
0
factor of 50 increase in
p
only about twice the effect
of 2
increase in
T
1
2
1
1
2
12
ln
ln
p
p
T
T
R
s
air
air
27C
1 atm
327C
50 atm
1
2
4
.
1
~
600
300
,
K
air
1
50
ln
300
600
ln
287
5
.
3
12
K
K
kgK
J
s
kgK
kJ
122
.
1
696
.
0
must use absolute
T
Ideal Gas Entropy State Relations -7
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Air Example: Solution
•
Analysis (non-cpg):
State Eqn.
kgK
kJ
415
.
0
<2% difference in
s
o
(
T
)
term for cpg air calculation
over 300-600K
1
2
1
2
12
ln
p
p
R
T
s
T
s
s
o
air
air
o
air
air
air
300K
1 atm
600K
50 atm
1
2
kgK
kJ
122
.
1
707
.
0
from Table D.1
kgK
kJ
kgK
kJ
s
12
6
.
409
5
.
702
1
.
122
p
term same,
even if not cpg
was 0.696
s
< 0 ???
air had to have been cooled during process
1
2
if isothermal p
?
T
s
600K
300K
School of Aerospace Engineering
T-s Diagrams
• Often drawing a process on a T-s state diagram will
be very helpful to understanding the process
– isothermal process?
– isentropic process?
– isobaric process?
• what do lines of
constant
p
look like?
• can show for
tpg
• lower pressure lines?
AE3450
T
s
T
s
p
1
p
2
p
3
p
1
>
p
2
>
p
3
Note: p lines diverge to the right
slope increases with s,T
p
s
Ideal Gas Entropy State Relations -9 Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
Isentropic Relations - TPG
• What can we say for
s
=0?
• For tpg
AE3450
1
2
ln
0
2
1
p
p
R
T
dT
c
T
T
p
2
1
1
ln
1
2
T
T
p
T
dT
c
R
p
p
R
s
R
s
T
dT
T
c
R
isen
o
o
T
T
p
e
e
p
p
2
1
2
1
1
1
2
e
s
o
(
T
)/
R
available in book tables as p
r
1
2
1
2
r
r
isentropic
p
p
p
p
T
v
T
v
v
v
r
r
isentropic
1
2
1
2
similarly
Ideal Gas Entropy State Relations -10
School of Aerospace Engineering
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
Isentropic Relations – TPG and CPG
• For isen. & cpg
– using
p
=
RT
AE3450
1
2
1
2
ln
ln
0
p
p
R
T
T
c
p
R
c
isen
p
T
T
p
p
1
2
1
2
isentropic process
1
1
2
T
T
1
2
1
2
T
T
p
p
isen
1
1
2
1
1
1
2
1
2
p
p
T
T
p
pv
const
p
p
s
1
since
>1
RT
p
Ideal Gas Entropy State Relations -11
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Automobile Engine Example
•
Given:
Air compressed in auto engine cylinder
with compression ratio of 10 (=V
1
/V
2
),
a typical value
•
Find:
T
2
, p
2
•
Assume:
tpg, cpg (
=1.4)
compression fast enough that no heat transfer to
cylinder walls BUT slow enough to be reversible
T
1
=300K
p
1
=1 atm
air
School of Aerospace Engineering
AE3450
Auto Engine Example: Solution
•
Analysis:
Mass Conserv.
1
2
2
1
2
1
2
1
m
m
V
V
m
m
2
nd
Law
rev+adiab
isentropic
1
1
2
1
2
T
T
T
2
=754K
p
2
=25atm
T
1
=300K
p
1
=1 atm
air
1
2
1
V
V
51
.
2
10
0
.
4
1
2
T
T
1
.
25
10
1
.
4
1
2
1
2
p
p
Ideal Gas Entropy State Relations -13
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Auto Engine Example: Solution
•
What if don’t assume cpg:
2
1
2
1
V
V
v
v
Use Table D.1
vs. 754
K
vs.
25
atm
T
1
=300K
p
1
=1 atm
air
p
atm
T
p
T
p
p
p
r
r
isen
3
.
24
38
.
1
6
.
33
2
1
2
1
2
v
T
T
K
T
v
K
v
v
v
r
r
r
isen
730
~
10
4
.
62
300
10
2
2
2
2
1
still have
T
(K)
C
p
kJ/kgK
p
r
v
r
300
1.0000
1.3800
62.396
700
1.0700
28.682
7.0051
720
1.0758
31.882
6.4822
740
1.0819
35.351
6.0087
Ideal Gas Entropy State Relations -14
School of Aerospace Engineering
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Stagnation Example
•
Given:
Air @ 300 K, 1 atm,
700 m/s
•
Find:
Pressure (
p
o
) if flow is slowed reversibly and adiabatically to
zero velocity (stagnation pressure)
•
Assume:
tpg, cpg (
=1.4) – and no external work
T
=300K
p
=1 atm
v=700 m/s
T
o
p
o
Ideal Gas Entropy State Relations -15
Copyright © 2013 by Jerry M. Seitzman. All rights reserved.
AE3450
Stagnation Example: Solution
•
Analysis:
1
st
Law
already showed
if
Q
=0,
W
=0
and if cpg
2
v
2
h
h
o
T
=300K
p
=1 atm
v=700 m/s
T
o
p
o
R
T
c
T
T
p
o
2
2
v
2
1
2
v
stagnation temperature
2
nd
Law
rev+adiab
isentropic
T
s
T
o
T
1
T
T
p
p
o
o
kgK
J
R
air
287
4
.
1
~
air
287
300
1
.
81
4
.
1
700
2
.
0
1
v
2
1
1
2
2
2
2
kg
J
s
m
RT
T
T
o
0
.
8
81
.
1
3
.
5
1
T
T
p
p
o
o
T
o
=544K
p
o
=8.0atm
What if
process
irrev.?
p
o
p
