Modern Engineering Thermodynamics

11.8 Thermodynamic Charts 381
so that
and
T sat
dp
dT
Z vg
v0
A
− p sat = T 2
sat
sat
Tsat
2
T sat
dp
dt
− p sat
sat
exp½AŠ − exp½AŠ = A
2
− 1 exp½AŠ
T sat
dv =
A 2
− 1 exp½AŠ ðv g − v 0 Þ
T sat
Since Eq. (b) must be valid to the saturated vapor line, it can be solved to find v g . For this problem, Eq. (b) is a quadratic
equation in v g with the following solution:
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v g = RT 2
sat RT
+ sat
− T sat
exp½BŠ
2p sat 2p sat p sat
From Eq. (c) and the preceding result for v g , we can now calculate v fg = v g – v f .
Then, with c 0 v equal to a constant (D 1), we get
Z Tsat
u g − u 0 + T
dp
− p dv = u 0 + c 0 v
dT
ðT A
sat − T 0 Þ + 2
− 1 exp½AŠ ðv g − v 0 Þ
T0
sat
T sat
then, with v g from the preceding equation, we can easily find h g = u g + p sat v g .
To find the saturated vapor entropy, we need to determine
Z vg
Z
dp
vg
A
dv = 2
dT
v0 sat
T 2 exp½AŠdv =
v0 sat
Then, with c 0 v equal to a constant, we get s g = s 0 +
Z Tsat
c 0 v
T0
= s 0 + c 0 v ln T sat
T 0
Now, we can determine the remaining saturated liquid properties as
dp
h f = h g − h fg = h g − T sat v fg
dT
and
A 2
Tsat
2 exp½AŠ ðv g − v 0 Þ
Z v
T dT + dp
dv
dT
v0 sat
A
+ 2
Tsat
2 exp½AŠ ðv g − v 0 Þ
sat
A
= h g − v 2
fg
T sat
exp½AŠ
A
u f = u g − u fg = u g − ðh fg − p sat v fg Þ = u g − v 2
fg exp½AŠ − p sat v fg
T sat
s f = s g − s fg = s g − h
fg
A
= s g − v fg exp½AŠ 2
T sat Tsat
2
Inserting the empirical values for the constants A 1 through D 1 into these equations provides the desired set of property
relations for this material in the saturated region.
Exercises
31. Determine an expression for T(∂p/∂T ) v – p for the material described in Example 11.11 in the superheated region. Answer:
∂p
= R
∂T v − 1
exp½BŠ − T
− B
2
exp½BŠ = R
v
v 2 v 2 T 2 v − exp½BŠ
1 − B
2
v 2 T
so that
T
∂p
− p = T
R ∂T
v − exp½BŠ
v
v 2 1 − B
2
−
RT T v − T
v 2 exp½BŠ
= B
2
v 2 exp½BŠ
where B = B 1 – B 2 /T.
32. Determine an expression for the specific internal energy of the material in Example 11.11 in the superheated region.
Answer:
where B = B 1 – B 2 /T.
u = u 0 +
Z T
T0
Z v
c 0 dT + T
o~
v0
∂p
∂T
− p
v
dv = u 0 + c 0 v ðT − T 0Þ − B 2 exp½BŠ
1
v − 1
v 0
(Continued )