Modern Engineering Thermodynamics

372 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.7 (Continued )
From Table C.13a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find R = 85.78 ft·lbf/(lbm·R) =
0.1102 Btu/(lbm·R). Then, at 212.0°F,
h fg ð212:0°FÞ = ½6289:78 R + ð1,827,997:8R 2 Þ/461:2 ð + 212:0RÞŠ
× ½0:1102 Btu/ ðlbm ⋅ RÞŠ = 992:37 Btu/lbm
Table C.1a gives h fg ð212:0°FÞ = 970:4 Btu/lbm: Thus, the value obtained from Rankine’s equation is in error by only +2.26%.
Exercises
19. Determine p sat from the Rankine equation given in Example 11.7 when T sat =212.0°F and compare it with the value of T sat
given in Table C.1a at 212.0°F. Answer: (p sat ) calc = 14.73 psia, and from Table C.1a, p sat (212.0°F) = 14.696 psia.
20. Using the relations given in Example 11.7, find the value of T sat for water when h fg = 1037 Btu/lbm and compare your
result with the value given in Table C.1a. Answer: (T sat ) calc = 124.6°F, and from Table C.1a, T sat = 100.0°F.
21. Determine h fg in Example 11.7 if the temperature is increased from 212.0°F to 500.0°F and compare your result with the
value given in Table C.1a. Answer: (h fg ) calc = 902.7 Btu/lbm, and from Table C.1a, h fg (500.0°F) = 714.8 Btu/lbm.
11.6 DETERMINING u, h, AND s FROM p, v, AND T
We are now ready to combine the previous results to produce u, h, ands relations from p, v, andT data. For a
simple substance, any two independent intensive properties fix its thermodynamic state. Consider the specific
internal energy described by a function of temperature and specific volume. We can write this as u = u(T, v).
Differentiating this function, we get
From Eq. (11.1), we can write
du =
∂u
∂T
∂u
∂v T
and using the Maxwell Eq. (11.15), this becomes
∂u
∂v T
dT +
v
∂u
∂v
dv
T
= T ∂s
− p
∂v T
= T ∂p
− p
∂T
v
In Chapter 3, we introduce the constant volume specific heat c v as
c v =
∂u
∂T v
(3.15)
and our equation for the total differential du then becomes
du = c v dT + T
∂p
− p dv (11.19)
∂T
v
Therefore, the change in specific internal energy for any simple substance can be determined by integrating
Eq. (11.19):
Z T2
Z v2
u 2 − u 1 = c v dT + T ∂p
− p dv (11.20)
T 1 v 1
∂T
Here, we achieved what we set out to do. Equation (11.20) has u cast completely in terms of the measurable
quantities p, v, T, and c v .
Similarly, we can consider the specific enthalpy to be given by a continuous function of temperature and pressure,
h = h(T, p). Then, its total differential is
dh =
∂h
dT +
∂h
dp
∂T p ∂p
T
In Chapter 3, we introduce the constant pressure specific heat c p as
c p =
∂h
(3.19)
∂T p
v