Modern Engineering Thermodynamics

11.6 Determining u, h, and s from p, v, and T 373
Introducing the definition of specific enthalpy into Eq. (11.1) gives
or
and, from this equation, we can deduce that
Using the Maxwell Eq. (11.16), we get
and our total differential dh becomes
du = dh − pdv− vdp= Tds− pdv
dh = Tds+ vdp (11.21)
∂h
∂p
T
∂h
∂p
T
= T ∂s
+ v
∂p
T
= −T ∂v
+ v
∂T p
The change in specific enthalpy for any simple substance is then given by
dh = c p dT + v − T ∂v
dp (11.22)
∂T p
Z T2
h 2 − h 1 = c p dT +
T 1
Z p2
p 1
v − T ∂v
dp (11.23)
∂T p
Again, we are successful. Equation (11.23) has h cast completely in terms of the measurable quantities, p, v,
T, and c p . Also note that Eqs. (11.20) and (11.23) are related by the fact that
h 2 − h 1 = u 2 − u 1 + p 2 v 2 − p 1 v 1
Finally, we can carry out the same type of analysis for the specific entropy of a simple substance. If we let
s = s (T, v), then
ds =
∂s
dT +
∂s
dv
∂T v ∂v T
From Eqs. (11.1) and (3.15), we can deduce that
∂s
= 1
∂u
∂T v T ∂T v
and, using the Maxwell Eq. (11.15), we can write the total differential ds as
ds =
c
v
dT +
∂p
dv (11.24)
T ∂T
Integrating this gives a relation for the change in specific entropy of a pure substance based completely on
measurable quantities:
s 2 − s 1 =
Z T2
T 1
c v
T dT + Z v2
v 1
= c v
T
v
∂p
∂T
dv (11.25)
v
By assuming s = s(T, p), we can also show that (see Problem 27 at the end of this chapter)
ds = c
p ∂v
dT − dp (11.26)
T ∂T p
and
s 2 − s 1 =
Z T2
T 1
Z
c p2
p
T dT − ∂v
dp (11.27)
∂T p
This completes the process of discovering relations for the unmeasurable u, h, ands properties in terms of the
measurable properties p, v, and T. The following example illustrates the use of these results.
p 1