Section 7.2 · A brief review of thermodynamics
Perfect gas, internal energy and enthalpy
A perfect gas is one in which intermolecular forces are neglected. In compressible-flow applications the molecules are on average more than ten molecular diameters apart, far enough to justify the assumption. Pressure, density and temperature are then related by the equation of state
where R is the specific gas constant (R = 287 J/(kg·K) for air) and v is the specific volume, the volume per unit mass. The symbol v also denotes the y component of velocity; the meaning is always clear from context.
The energy of a molecule is the sum of its translational, rotational, vibrational and electronic energies. Summed over all molecules in a volume of gas and divided by its mass, it is the specific internal energy e. The specific enthalpy is
For a perfect gas both depend on temperature only, e = e(T) and h = h(T) (7.4), with de = cvdT and dh = cpdT (7.5). For air below about 1000 K the specific heats are nearly constant; a perfect gas with constant cv and cp is calorically perfect, and this course always treats the gas as such:
e and h are state variables: they depend only on the state of the gas, not on the process that brought it there. Although cv and cp appear in (7.6), the relations are not restricted to constant-volume or constant-pressure processes.
For a given gas, cp − cv = R (7.7). Dividing by cp and by cv, with γ = cp/cv (1.4 for air at standard conditions):
Worked examples 7.1–7.3
A room has a rectangular floor m by m and a ceiling m high. The air pressure and temperature in the room are atm and °C. Calculate (a) the internal energy and (b) the enthalpy of the air in the room.
The energy and enthalpy per unit mass follow directly from the temperature; the mass follows from the equation of state. The check H/E = h/e = cp/cv = γ holds for any calorically perfect gas. The notes use 1 atm = 1.01 × 105 N/m² and 25 °C = 298 K; the exact column uses 101 325 N/m² and 298.15 K, a 0.3% difference in E and H. Use consistent units throughout.
(7.2) A blow-down wind-tunnel reservoir has an internal volume of m³. Air is pumped in at kg/s, and heat exchangers hold the air at K. Calculate the rate of increase of pressure at the instant the pressure is 10 atm. (7.3) At this constant pumping rate, how long does it take to raise the pressure from to atm?
With ρ = M/V and p = ρRT at constant T, dp/dt = (RT/V) dM/dt, independent of the pressure itself: the pressure rises linearly in time. With R = 287 J/(kg·K), RT/V = 287 × 300/30 = 2870 exactly; the notes write 2867.13 N/(m²·s) and, with 1 atm = 1.01 × 105 N/m², find 352.3 s. The exact values are 2870 N/(m²·s) and 353.0 s.
Section 7.2 · A brief review of thermodynamics
First and second laws, and the entropy change
First law. For a system of fixed mass, an increment of heat δq added to the system and work δw done on it by the surroundings change its internal energy:
e is a state variable, so de is an exact differential that depends only on the end states. δq and δw depend on the process. Three kinds of process matter here:
- Adiabatic: no heat is added to or taken from the system.
- Reversible: no dissipative phenomena occur: no viscosity, thermal conduction or mass diffusion.
- Isentropic: both adiabatic and reversible.
For a reversible process δw = −p dv, so
Second law. The first law says nothing about the direction of a process, for example whether heat flows from a hot plate to a cold one or the reverse. The entropy, a new state variable, decides it:
Here δq is the actual heat added and dsirrev the entropy generated by viscosity, thermal conduction and mass diffusion inside the system. Hence
A process proceeds in the direction in which the entropy of the system plus its surroundings increases or, at best, stays the same.
Calculating the entropy change. With heat added reversibly, δqrev = T ds in (7.12), and with dh = de + p dv + v dp:
These are the first law written in terms of entropy. With dh = cpdT, de = cvdT and the equation of state, integrating between states 1 and 2 of a calorically perfect gas gives
Entropy is a function of two state variables, s = s(p, T) or s = s(v, T).
Section 7.2 · A brief review of thermodynamics
Isentropic relations
In an isentropic process δq = 0 and dsirrev = 0, so ds = 0. Setting s2 − s1 = 0 in (7.25), with cp/R = γ/(γ − 1):
and in (7.26), with cv/R = 1/(γ − 1) and v2/v1 = ρ1/ρ2:
Together:
Worked examples 7.4–7.5
A Boeing 747 flies at a standard altitude of 10.8 km, where Appendix D gives p∞ = N/m² and T∞ = K. The pressure at a point on the wing is N/m². Assuming isentropic flow over the wing, calculate the temperature at this point.
The notes state p = 1.916 × 104 N/m² but evaluate the ratio with 1.914 × 104, and report T = 207.3 K. Either pressure gives T ≈ 205.9 K: (1.916/2.3422)0.2857 = 0.9442.
Air in the reservoir of Examples 7.2–7.3 is at atm and K. It expands through the tunnel duct to a location where the pressure is atm. Calculate the temperature there if (a) the expansion is isentropic and (b) the expansion is nonisentropic, with an entropy increase of J/(kg·K).
The dissipation in the nonisentropic expansion leaves the gas hotter, 175.3 K instead of 127.5 K, at the same pressure: part of the energy that would have gone into kinetic energy stays as internal energy.
Section 7.3
Definition of compressibility
All real substances are compressible to some extent. Squeeze an element of fluid of specific volume v with a pressure increase dp; its volume changes by dv. The compressibility is the fractional change in volume per unit change in pressure:
The value depends on the process. If heat transfer holds the temperature constant, τ is the isothermal compressibility; if there is no heat transfer and friction is ignored, the compression is isentropic and τ is the isentropic compressibility:
With v = 1/ρ, dv = −dρ/ρ², so
Whenever the fluid experiences a pressure change dp, its density changes by ρτ dp.
Typical values at 1 atm. For a perfect gas, τT = 1/p and τs = 1/(γp) (Section 8.3), so air is roughly 20 000 times more compressible than water:
| Fluid at 1 atm | ρ (kg/m³) | τT (m²/N) | τs (m²/N) |
|---|---|---|---|
| Air, 288 K | 1.225 | 9.87 × 10−6 | 7.05 × 10−6 |
| Water, 20 °C | 998 | 4.6 × 10−10 | ≈ 4.6 × 10−10 |
Why low-speed flow can be treated as incompressible. In a flow, pressure changes are of the order of the dynamic pressure, dp ≈ ½ρV². For a gas, (7.37) then gives dρ/ρ ≈ τs·½ρV² = ½ρV²/(γp) = M²/2. The density change is small only while the Mach number is small: about 5% at M ≈ 0.3, the usual limit of incompressible aerodynamics. In water the same speeds change the density by a negligible amount.
Sea-level air, p = 101 325 N/m², T = 288.16 K; water ρ = 998 kg/m³, τ = 4.6 × 10−10 m²/N. The exact air curve is the density rise from the moving gas to its stagnation state, ρ0/ρ − 1 (Section 7.5).
Section 7.4
Governing equations for inviscid, compressible flow
| Flow type | Primary dependent variables | Equations to solve |
|---|---|---|
| Inviscid, incompressible (low-speed aerodynamics) | p, | continuity (1), momentum (3): 4 equations |
| Inviscid, compressible (high-speed aerodynamics) | p, | continuity (1), momentum (3), energy (1), equation of state (1), internal energy (1): 7 equations |
Continuity, integral and differential form:
Momentum:
Energy:
Equation of state p = ρRT (7.1) and internal energy e = cvT (7.6) close the system.
Section 7.5
Definition of total (stagnation) conditions
The static pressure p, temperature T and density ρ are the values you would feel riding along with the gas at the local flow velocity; static pressure measures the purely random motion of the molecules. Now imagine taking a fluid element passing through a point and slowing it adiabatically to zero velocity. The temperature it reaches is the total temperature T0, and the corresponding enthalpy is the total enthalpy h0 = cpT0. These are defined quantities: every point of a flow has a T0 and h0, whether or not the gas is actually brought to rest there.
For adiabatic flow (
Derivation of (7.51) (four steps)
In steady flow the right side vanishes, so for steady, adiabatic, inviscid flow
At any point the total enthalpy is the static enthalpy plus the kinetic energy, both per unit mass. If all streamlines come from a common uniform freestream, h0 is the same on every streamline, and in a steady adiabatic flow
throughout the flow, the latter for a calorically perfect gas. This simple statement is a powerful tool.
Total pressure and density. Now slow the element to rest both adiabatically and reversibly, that is isentropically. The pressure and density it reaches are the total pressure p0 and total density ρ0; its temperature is the same T0, since an isentropic process is also adiabatic. The isentropic assumption enters only the definition: p0 and ρ0 can be assigned at every point of a general nonisentropic flow.
Sonic conditions. Accelerating (or decelerating) the element adiabatically to sonic speed defines T* and a* = √(γRT*) in the same way.
Worked examples 7.6–7.7
At a point in an airflow the pressure, temperature and velocity are atm, K and m/s. Calculate the total temperature and total pressure at this point.
T0 needs only an adiabatic deceleration, (7.54); p0 needs an isentropic one, (7.32) with T0. Bernoulli’s equation, which assumes constant density, would give a total pressure of only a few atmospheres here (last row).
An airplane flies at a standard altitude of 3 km, where Appendix D gives T∞ = K and p∞ = N/m². A Pitot tube at the nose measures N/m². The airplane flies at a high subsonic speed, faster than 480 km/h. Calculate its velocity.
480 km/h = 133.3 m/s is already M ≈ 0.41, so the flow is compressible and Bernoulli’s equation does not apply. The Pitot tube compresses the flow isentropically to rest, so it reads p0; (7.32) gives T0, and the energy equation (7.54) the velocity. The last rows show the 8% error Bernoulli would make.
Section 7.6
Some aspects of supersonic flow: shock waves
Calculating the shape and strength of shock waves is an essential part of supersonic aerodynamics and the main thrust of Chapters 8 and 9. A shock wave is an extremely thin region, typically of the order of 10−5 cm, across which the flow properties change drastically. Across it:
- pressure, density, temperature and entropy increase;
- Mach number, velocity and total pressure decrease;
- total enthalpy, and so total temperature, stays the same.
Behind a normal shock the flow is always subsonic. Behind an oblique shock the Mach number falls but usually stays supersonic. Continue with Chapter 8, Normal Shock Waves, and Chapter 9, Oblique Shock and Expansion Waves.
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Summary