Modern Engineering Thermodynamics

11.5 The Clapeyron Equation 371
and, using Eq. (11.10), we obtain the Clapeyron equation as 2
dp
= h fg / T sat v fg
dT
sat
(11.17)
For most substances, v g ≫ v f , so we can approximate v fg ≈ v g . Also, for vapors at very low pressures, the saturated
vapor curve can be accurately approximated by the ideal gas equation of state, so we can write v g = RT sat /p sat .
Then Eq. (11.17) becomes
dp
= p sat h fg / RTsat
2 dT
sat
or
dp
p
sat
= h fg / RTsat
2 dTsat (11.18)
This equation is often called the Clapeyron-Clausius equation. For small pressure and temperature changes, h fg can
be assumed to be constant and Eq. (11.18) can be integrated from a reference state to any other state to give
or
ln ðp/p 0 Þ sat
= h fg /R T sat − T 0
T sat T 0
p sat = p 0 exp h fg /R
T sat − T 0
T sat T 0
where p 0 and T 0 are reference state values. An exponential relation between p sat and T sat fits experimental data
quite well for most substances at low pressure.
EXAMPLE 11.7
In 1849, William Rankine proposed the following relation between the saturation pressure and saturation temperature of
water:
ln p sat = 14:05 − 6289:78
T sat
− 913,998:92
T 2 sat
where p sat is in psia, and T sat is the temperature in °F + 461.2 (at that time –461.2°F was Rankine’s best estimate of absolute
zero temperature). Determine h fg at 212.0°F from the Rankine equation and compare the result with that listed in the steam
tables in Thermodynamic Tables to accompany Modern Engineering Thermodynamics.
Solution
Differentiating Rankine’s equation, we obtain
then, using Eq. (11.18), we get
1
p
dp
dT
sat
h fg =
RT2 dp
p dT
sat
= 6289:78
T 2 sat
+ 1,827,997:8
T 3 sat
= Rð6289:78 + 1,827,997:8/T sat Þ
(Continued )
2 The Clapeyron equation is valid for any type of phase change in a simple substance. For example, if we let the i subscript denote the
solid phase, then for melting we can write
and, for sublimation,
dp
= h
dT
solid
if / T sat v if
liquid
saturation
dp
= h
dT
solid
ig / T sat v ig
vapor
saturation