Thermodynamics of Compressible Flows

In Chapter 2, we saw that a perfect gas has to be thermally as well as calorically perfect, satisfying the thermal state equation and at least two calorical state equations. Thus for a perfect gas:

pv = RT

or

p = pRT,

where p is the pressure and R is the gas constant, given by:

Theoretical Aerodynamics, First Edition. Ethirajan Rathakrishnan.

© 2013 John Wiley & Sons Singapore Pte. Ltd. Published 2013 by John Wiley & Sons Singapore Pte. Ltd.

where Ru is the universal gas constant equal to 8314 J/(kg K) and M is the molecular weight of the gas. Thus, of the four quantities p, p, T, R, in Equation (9.1i>), only two are independent.

Taking log on both sides and differentiating Equation (9.1i>), we get:

Let us assume unit mass of a gas receiving a small quantity of heat q. By the first law of thermodynamics, we know that heat is a form of energy [2]. Thus the quantity of heat q is equivalent to q units of mechanical energy. Hence addition of heat q will supply energy to the gas, resulting in the increase of its specific volume from v to (v + dv). Thus, the heat q added does a mechanical work of pdv. Let us assume that the expansion is taking place very slowly, so that no kinetic energy is developed. For this process, we can write:

q = du + pdv, (9.3)

where du is the increase in the internal energy of the gas. It is essential to note from Equation (9.3) that only a part of the heat q supplied is converted to mechanical work pdv and the rest of the heat is dumped into the internal energy of the gas mass. This demonstrates that the energy conversion is 100% efficient. The work pdv is referred to as flow work. Thus, for doing, say, 1 unit of work (pdv) we need to supply q/n amount of heat, where n is the efficiency of the work producing cycle and n is always less than 1. For example, the work producing devices, such as spark ignition (SI) engine, compression ignition (CI) engine and gas turbine (jet) engines has efficiencies of 40%, 60% and 30%, respectively.

For a perfect gas, the internal energy u is a function of the absolute temperature T alone. This hypothesis is a generalization for experimental results. It is known as Joule ‘slaw. Thus:

du = kdT. (9.4)

Substituting this into Equation (9.3), we have:

q = kdT + pdv. (9.5)

For a constant volume (isochoric) process, dv = 0. Thus for a constant volume process, Equation (9.5) reduces to:

q = kdT.

We can express this as:

q = cvdT,

where cv is called the specific heat at constant volume. It is the quantity of heat required to raise the temperature of the system by one unit while keeping the volume constant. Thus from Equation (9.5), with dv = 0, we get:

k = cv.

Similarly the specific heat at constant pressure, cp, defined as the quantity of heat required to raise the temperature of the system by one unit while keeping the pressure constant. Now, for p = constant, Equation (9.2) simplifies to:

dp dT

Y’

V P•

Р

There are three unknowns; the pressure p, density p and velocity V in Equations (9.10) and (9.11). Therefore these two equations alone are insufficient to determine the solution. To solve this motion, we can make use of the process Equation (9.9) as the third equation, presuming that the motion is isentropic.