Chapter 7  Section B  NonNumerical Solutions
7.2
(
a
) Apply the general equation given in the footnote on page 266 to the particular derivative of interest
here:
µ
∂
T
∂
P
¶
S
=−
µ
∂
T
∂
S
¶
P
µ
∂
S
∂
P
¶
T
The two partial derivatives on the right are found from Eqs. (6.17) and (6.16); thus,
µ
∂
T
∂
P
¶
S
=
T
C
P
µ
∂
V
∂
T
¶
P
For gases, this derivative is positive. It applies to reversible adiabatic expansions and compressions
in turbines and compressors.
(
b
) Application of the same general relation (page 266) yields:
µ
∂
T
∂
V
¶
U
µ
∂
T
∂
U
¶
V
µ
∂
U
∂
V
¶
T
The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus,
µ
∂
T
∂
V
¶
U
=
1
C
V
·
P
−
T
µ
∂
P
∂
T
¶
V
¸
For gases, this may be positive or negative, depending on conditions. Note that it is zero for an
ideal gas. It applies directly to the
Joule expansion
, an adiabatic expansion of gas confined in a
portion of a container to fill the entire container.
7.3
The equation giving the thermodynamic sound speed appears in the middle of page 257. As written,
it implicitly requires that
V
represent
specific
volume. This is easily confirmed by a dimensional
analysis. If
V
is to be
molar
volume, then the right side must be divided by molar mass:
c
2
V
2
M
µ
∂
P
∂
V
¶
S
(
A
)
Applying the equation given in the footnote on page 266 to the derivative yields:
µ
∂
P
∂
V
¶
S
µ
∂
P
∂
S
¶
V
µ
∂
S
∂
V
¶
P
This can also be written:
µ
∂
P
∂
V
¶
S
·µ
∂
P
∂
T
¶
V
µ
∂
T
∂
S
¶
V
¸·
µ
∂
S
∂
T
¶
P
µ
∂
T
∂
V
¶
P
¸
·µ
∂
T
∂
S
¶
V
µ
∂
S
∂
T
¶
P
µ
∂
P
∂
T
¶
V
µ
∂
T
∂
V
¶
P
¸
Division of Eq. (6.17) by Eq. (6.30) shows that the first product in square brackets on the far right is
the ratio
C
P
/
C
V
. Reference again to the equation of the footnote on page 266 shows that the second
product in square brackets on the far right is
−
(∂
P
/∂
V
)
T
, which is given by Eq. (3.3).
Therefore,
µ
∂
P
∂
V
¶
S
=
C
P
C
V
µ
∂
P
∂
V
¶
T
=
C
P
C
V
µ
−
1
κ
V
¶
662
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View Full DocumentSubstitute into Eq. (
A
):
c
2
=
VC
P
M
C
V
κ
or
c
=
±
P
M
C
V
κ
(
a
) For an ideal gas,
V
=
RT
/
P
and
κ
=
1
/
P
. Therefore,
c
ig
=
±
M
C
P
C
V
(
b
) For an incompressible liquid,
V
is constant, and
κ
=
0, leading to the result:
c
=∞
. This of
course leads to the conclusion that the sound speed in liquids is much greater than in gases.
7.6
As
P
2
decreases from an initial value of
P
2
=
P
1
, both
u
2
and
˙
m
steadily increase until the critical
pressure ratio is reached. At this value of
P
2
,
u
2
equals the speed of sound in the gas, and further
reduction in
P
2
does not affect
u
2
or
˙
m
.
7.7
The massﬂow rate
˙
m
is of course constant throughout the nozzle from entrance to exit.
The velocity
u
rises monotonically from nozzle entrance
(
P
/
P
1
=
1
)
to nozzle exit as
P
and
P
/
P
1
decrease.
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 Spring '11
 Ku
 Thermodynamics, Trigraph, Adiabatic process, ∂z ∂T

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