Internal Energy and Enthalpy

Consider an individual molecule of a gas, say, an O2 molecule in air. This molecule is moving through space in a random fashion, occasionally colliding with a neighboring molecule. Because of its velocity through space, the molecule has translational kinetic energy. In addition, the molecule is made up of individual atoms which we can visualize as connected to each other along various axes; for example, we can visualize the O2 molecule as a “dumbbell” shape, with an О atom at each end of a connecting axis. In addition to its translational motion, such a molecule can execute a rotational motion in space; the kinetic energy of this rotation contributes to the net energy of the molecule. Also, the atoms of a given molecule can vibrate back and forth along and across the molecular axis, thus contributing a potential and kinetic energy of vibration to the molecule. Finally, the motion of the electrons around each of the nuclei of the molecule contributes an “electronic” energy to the molecule. Hence, the energy of a given molecule is the sum of its translational, rotational, vibrational, and electronic energies.

Now consider a finite volume of gas consisting of a large number of molecules. The sum of the energies of all the molecules in this volume is defined as the internal energy of the gas. The internal energy per unit mass of gas is defined as the specific internal energy, denoted by e. A related quantity is the specific enthalpy, denoted by h and defined as

h = e + pv [7.3]

For a perfect gas, both e and h are functions of temperature only:

e = e(T) [7.4a]

h=h(T) [7.4b]

Let de and dh represent differentials of e and h, respectively. Then, for a perfect gas,

de = cvdT [7.5a]

dh=cpdT [7.5b]

where cv and cp are the specific heats at constant volume and constant pressure, respectively. In Equations (7.5a and b), cv and cp can themselves be functions of T. However, for moderate temperatures (for air, for T < 1000 K), the specific heats are reasonably constant. A perfect gas where cv and cp are constants is defined as a calorically perfect gas, for which Equations (1.5a and b) becomes

[7.6 a] [7.6b]

For a large number of practical compressible flow problems, the temperatures are moderate; for this reason, in this book we always treat the gas as calorically perfect; that is, we assume that the specific heats are constant. For a discussion of compressible flow problems where the specific heats are not constant (such as the high-temperature chemically reacting flow over a high-speed atmospheric entry vehicle, that is, the space shuttle), see Reference 21.

Note that e and h in Equations (7.3) through (7.6) are thermodynamic state variables—they depend only on the state of the gas and are independent of any process. Although cv and cp appear in these equations, there is no restriction to just a constant volume or a constant pressure process. Rather, Equations (1.5a and b) and (1.6a and b) are relations for thermodynamic state variables, namely, e and h as functions of /’. and have nothing to do with the process that may be taking place.

For a specific gas, cp and cv are related through the equation

cp – cv = R [7.7]

Dividing Equation (7.7) by cp, we obtain

, cv R

Define у = cp/cv. For air at standard conditions, у = 1.4. Then Equation (7.8) becomes

R cp

yR

or

[7.9]

cp

У ~ 1

Similarly, dividing Equation (7.7) by c„, we obtain

Equations (7.9) and (7.10) are particularly useful in our subsequent discussion of compressible flow.