SLIDE 1 — Compressible Aerodynamics (Title Slide)
What this slide says:
This is the title of the lecture series on compressible aerodynamics by Rajesh Ranjan.
Why this matters:
Compressible aerodynamics focuses on gas flows where density changes significantly.
This usually happens at Mach number greater than 0.3.
Key idea:
When density changes, all conservation equations become nonlinear and strongly coupled with thermodynamics.
That is what makes compressible flow fundamentally different from incompressible flow.
CFD connection:
Any CFD simulation involving high-speed gas flow must use a compressible solver.
Examples include:
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nozzles
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jet engines
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supersonic inlets
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shock waves
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rockets
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high-speed valves
SLIDE 2 — Overview: Theory, Computation, Problem Solving
What the slide says:
The course has 3 main pillars:
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Theory
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Computation (CFD)
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Problem-solving (numericals)
Why this matters:
To develop expertise in compressible flow, you must understand all three areas.
This PPT strengthens the theoretical foundation behind CFD models.
SLIDE 3 — What is Aerodynamics?
Simple explanation:
Aerodynamics = study of how gases flow, especially around bodies such as wings, aircraft, vehicles, and rockets.
Distinctions:
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Fluid dynamics: liquids + gases
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Gas dynamics: gas flows
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Aerodynamics: gas dynamics around bodies
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Hydrodynamics: water (liquid) flows
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Hemodynamics: blood flow
Why this matters:
Compressible aerodynamics is a branch of gas dynamics focusing on high-speed gases.
CFD relevance:
Choosing the right solver depends on understanding whether the flow is compressible or incompressible.
SLIDES 4 & 5 — Images of Aerodynamic Flow
These slides show visual examples:
Their purpose is to build intuition.
SLIDE 6 — Solids vs Fluids
Simple explanation:
Solids resist shear stress.
Fluids do not resist static shear — they continuously deform, which creates flow.
Key point:
Fluids respond to shear by changing shape (flow), not by maintaining a fixed deformation.
Equation (ASCII):
shear_stress = mu * (du/dy)
CFD relevance:
High-speed compressible flow requires including viscous stresses and thermodynamic coupling in Navier-Stokes equations.
SLIDE 7 — Solids vs Liquids vs Gases
Key idea:
Compressibility makes gases behave differently at high speeds → Mach effects.
SLIDES 8–12 — Gas Properties, Ideal Gas Law, R, gamma
Important equations (ASCII):
p = rho * R * T
R = Cp - Cv
gamma = Cp / Cv
a = sqrt(gamma * R * T)
Values of gamma:
CFD relevance:
Choosing the correct gamma in CFD is essential for accurate prediction of Mach number and shock strength.
SLIDE 13 — Types of Flows
Flow classification:
Why it matters:
Compressible aerodynamics involves almost all of these categories at once.
SLIDES 14–18 — Compressibility and Bulk Modulus
Compressibility (ASCII):
beta = (1/rho) * (d rho / d p)
Bulk modulus (ASCII):
K = - v * (d p / d v)
Interpretation:
High beta → highly compressible (gases)
High K → low compressibility (liquids)
Example values:
air: beta approx 1e-5
water: beta approx 5e-10
=> air is about 20000 times more compressible than water.
CFD relevance:
Compressibility controls density variations, Mach number, shock formation, etc.
SLIDES 19–21 — Mach Number and Compressible Flow
Mach number equation (ASCII):
M = u / a
Speed of sound (ASCII):
a = sqrt(gamma * R * T)
Why M > 0.3 matters:
Below M = 0.3, density changes are very small (< 5%).
Above M = 0.3, density changes start influencing momentum and energy equations.
CFD relevance:
When M > 0.3, a compressible solver is mandatory.
SLIDE 22 — Thermodynamics Review
Key thermodynamic quantities used in compressible flow:
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internal energy
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enthalpy
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heat
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work
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entropy
The compressible energy equation in CFD requires thermodynamics.
SLIDE 23 — Types of Gas Models
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Calorically Perfect Gas
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Thermally Perfect Gas
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Real Gas
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non-ideal equations of state (Peng-Robinson, Van der Waals)
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needed at high pressure or low temperature
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Chemically Reacting Gas
CFD relevance:
Wrong gas model → wrong temperature, density, shock prediction.
SLIDE 24 — First Law of Thermodynamics
ASCII form:
delta_q = delta_w + d e
delta_q = heat added
delta_w = work done
d e = change in internal energy
CFD energy equation (ASCII):
d/dt (rho * e) + div(rho * e * u) = - p * div(u) + viscous_work + heat_conduction
You are solving a PDE form of the first law.
SLIDE 25 — Second Law / Adiabatic / Reversible / Isentropic
Key relations (ASCII):
Adiabatic: delta_q = 0
Reversible: entropy_production = 0
Isentropic: ds = 0 = (delta_q / T) + entropy_production
Shocks are not isentropic because entropy increases.
SLIDE 26 — Entropy Equation
General form (ASCII):
T * ds = d e + p * d v
For ideal gas:
d e = Cv * dT
p = R * T / v
Hence:
ds = Cv * (dT/T) + R * (dv/v)
This is the basis for isentropic relations.
SLIDES 27–28 — Entropy Change Derivation
Integrating:
s2 - s1 = R * ln(v2/v1) + Cv * ln(T2/T1)
This expression is used to derive isentropic formulas.
SLIDES 29–30 — Isentropic Relations
From ds = 0:
p2/p1 = (T2/T1)^(gamma/(gamma-1))
rho2/rho1 = (T2/T1)^(1/(gamma-1))
p * v^gamma = constant
These relations are used for:
SLIDE 31 — 1D Flow Equations
Assumptions:
steady, 1D, inviscid, constant area
Continuity (ASCII):
rho1 * u1 = rho2 * u2
Momentum (ASCII):
p1 + 0.5 * rho1 * u1^2 = p2 + 0.5 * rho2 * u2^2
Energy (ASCII):
h1 + 0.5 * u1^2 = h2 + 0.5 * u2^2
where h = Cp * T
These form the basis for nozzle theory and shock jumps.
SLIDES 32–33 — Speed of Sound
Definition (ASCII):
a^2 = (d p / d rho) at constant entropy
For ideal gas:
a = sqrt(gamma * R * T)
Important insight:
For incompressible limit:
beta_s (isentropic compressibility) -> 0
=> speed of sound -> infinity
Pressure disturbances propagate instantaneously → incompressible assumption.
Problem 1 — Specific Gas Constant of Argon
Given:
M = 39.95 g/mol
Ru = 8.314 J/mol·K
Solution:
Convert molar mass:
M = 0.03995 kg/mol
R = Ru / M = 8.314 / 0.03995 approx 208 J/kg·K
Conclusion:
Argon has a lower specific gas constant than air, so it stores less energy per kg at the same temperature.
Problem 2 — Find Cp and Cv of Air
Given:
R = 287 J/kg·K
gamma = 1.4
Solution:
R = Cp - Cv
gamma = Cp / Cv
Cv = R / (gamma - 1) = 287 / 0.4 = 717.5
Cp = gamma * Cv = 1.4 * 717.5 approx 1004.5 J/kg·K
Conclusion:
Air behaves like a diatomic gas with Cp approx 1005 and Cv approx 718, giving gamma = 1.4.
Problem 3 — Choice of Gas Model
Given:
T = 300–800 K
p = approx 1 atm
Solution:
Low pressure and moderate temperature → ideal gas valid → Cp, Cv nearly constant.
Conclusion:
Air can be modeled as a calorically perfect gas.
Problem 4 — Density of Air
Given:
p = 200000 Pa
T = 300 K
R = 287
Solution:
rho = p / (R * T) = 200000 / (287 * 300) approx 2.32 kg/m^3
Conclusion:
Density increases proportionally with pressure under ideal gas behavior.
Problem 5 — Find Pressure
Given:
rho = 1.5
T = 320 K
R = 287
Solution:
p = rho * R * T
= 1.5 * 287 * 320 = 138 kPa
Conclusion:
Pressure rises directly with density and temperature in an ideal gas.
Problem 6 — Isothermal Compressibility
Given:
beta_T = 1 / p
p = 101325 Pa
Solution:
beta_T = 1 / 101325 approx 1e-5 m^2/N
Conclusion:
Air is highly compressible at atmospheric pressure.
Problem 7 — Compressibility: Air vs Water
Given:
beta_air = 1e-5
beta_water = 5e-10
Solution:
ratio = (1e-5) / (5e-10) = 2e4
Conclusion:
Air is almost 20,000 times more compressible than water.
Problem 8 — Bulk Modulus
Given:
beta_T = 4e-10
Solution:
K = 1 / beta_T = 2.5e9 Pa
Conclusion:
The fluid is extremely stiff and requires huge pressure for small compression.
Problem 9 — Speed of Sound at 300 K
Given:
gamma = 1.4
R = 287
T = 300
Solution:
a = sqrt(gamma * R * T) = sqrt(1.4 * 287 * 300) approx 347 m/s
Conclusion:
The computed sound speed matches the standard value for air at room temperature.
Problem 10 — Velocity From Mach Number
Given:
a = 340 m/s
M = 0.78
Solution:
U = M * a = 0.78 * 340 = 265 m/s
Convert to km/h: 265 * 3.6 approx 955 km/h
Conclusion:
Even subsonic Mach numbers correspond to very high true airspeeds.
Problem 11 — Mach Number From Velocity
Given:
U = 750 m/s
a = 300 m/s
Solution:
M = U / a = 2.5
Conclusion:
The projectile is deep in the supersonic regime.
Problem 12 — Constant Mach at Two Altitudes
Given:
T1 = 288 K
T2 = 217 K
Speed difference = 127 km/h
Solution:
Using U = M * a, temperature drop reduces a, giving M approx 0.78.
Conclusion:
Aircraft true airspeed reduces at altitude because the speed of sound decreases.
Problem 13 — T2/T1 From p2/p1
Given:
p2/p1 = 0.5
gamma = 1.4
Solution:
T2/T1 = 0.5^(1/3.5) approx 0.82
Conclusion:
A pressure drop of 50 percent lowers temperature to about 82 percent of its original value.
Problem 14 — Density Ratio
Given:
p2/p1 = 0.5
gamma = 1.4
Solution:
rho2/rho1 = 0.5^(1/1.4) approx 0.62
Conclusion:
Density drops significantly with pressure under isentropic expansion.
Problem 15 — Temperature After Compression
Given:
p2/p1 = 4
T1 = 300 K
gamma = 1.4
Solution:
T2/T1 = 4^0.286 approx 1.49
T2 approx 447 K
Conclusion:
Isentropic compression leads to a strong temperature rise.
Problem 16 — 1D Compressible Energy Equation
Given:
T1 = 300 K
u1 = 50
u2 = 250
Cp = 1005
Solution:
CpT1 + u1^2/2 = CpT2 + u2^2/2
T2 approx 270 K
Conclusion:
As velocity increases, temperature decreases because total energy is conserved.
Problem 17 — Continuity in 1D
Given:
rho1 = 1.2
A1 = 0.02
u1 = 100
rho2 = 0.8
A2 = 0.01
Solution:
u2 = (1.2 * 100 * 0.02) / (0.8 * 0.01) = 300 m/s
Conclusion:
Flow accelerates sharply when both density and area decrease.
Problem 18 — Check Compressibility
Given:
u1 = 80 m/s
a approx 347 m/s
Solution:
M approx 0.23
Conclusion:
Flow is safely in the incompressible regime (M < 0.3).
Problem 19 — Entropy Change (Isothermal)
Given:
p1 = 4 bar
p2 = 1 bar
T constant
R = 287
Solution:
v2/v1 = p1/p2 = 4
s2 - s1 = R * ln(4) approx 398 J/kg·K
Conclusion:
Isothermal expansion always increases entropy because volume increases.
Problem 20 — Isentropic Expansion
Given:
p1 = 5 bar
p2 = 1 bar
T1 = 400 K
gamma = 1.4
Solution:
T2 approx 252 K
s2 - s1 approx 0
Conclusion:
Entropy remains essentially constant, confirming an ideal isentropic process.
1. Why do we separate fluid dynamics into branches?
Because different physical regimes dominate different flows.
In gases, density changes matter; in liquids, they do not.
In low-speed flows, viscosity dominates; in high-speed flows, compressibility dominates.
Separating branches helps use simpler, more accurate models for each regime.
2. Why is aerodynamics usually compressible but hydrodynamics is not?
Because air is highly compressible (beta approx 1e-5), but water is extremely stiff (beta approx 5e-10).
So a high-speed air flow experiences density changes, but water does not.
Therefore airflows become compressible at moderate speeds; water flows rarely do.
3. Why is Prandtl important in aerodynamics?
Prandtl introduced boundary layer theory, explaining why real flows have skin friction drag.
His work bridges inviscid outer flow with viscous wall effects, enabling modern aerodynamics and CFD.
4. Why can solids resist static shear but fluids cannot?
In solids, molecules are locked in fixed positions, so they can sustain shear deformation without moving.
In fluids, molecules rearrange easily, so even small shear forces cause continuous deformation (flow).
5. Why are liquids treated as incompressible while gases are not?
Liquid molecules are tightly packed; their volume decreases extremely little under pressure.
Gas molecules have large separation, so pressure significantly changes density.
6. What physically causes compressibility in gases?
Large spacing between molecules allows them to move closer under pressure.
Therefore, density depends strongly on pressure and temperature.
7. Why do gases diffuse faster than liquids?
Because gas molecules move faster, have larger mean free paths, and lower intermolecular attraction.
This leads to rapid mixing and diffusion.
8. Why is the ideal gas law valid only at low pressures?
At low pressure, molecules are far apart, so:
9. Why is gamma higher for monoatomic gases than diatomic gases?
Monoatomic gases have only 3 translational degrees of freedom.
Diatomic gases also have rotational degrees of freedom.
More degrees of freedom mean larger Cv, reducing gamma = Cp/Cv.
10. Why do specific heats vary with temperature?
At higher temperatures, vibrational and rotational modes become active.
These require additional energy, increasing Cp and Cv.
11. Why do real gases deviate from ideal behavior at high pressure?
Because:
12. Why do we assume constant gamma in compressible flow?
Because for air below approx 1000 K, Cp and Cv vary very little.
Thus gamma is nearly constant, simplifying analysis.
13. Why does the continuum assumption fail at high altitude?
Because density becomes extremely low and mean free path becomes comparable to flow length scales.
The fluid can no longer be treated as continuous; kinetic theory is required.
14. Why consider viscosity even in "inviscid" flows?
Inviscid solvers ignore viscosity in the bulk flow but viscous effects near walls still determine drag and separation.
Also shocks generate entropy through irreversible viscous dissipation.
15. Why do compressible effects appear only at high Mach numbers?
Density changes scale approximately with M^2.
At M < 0.3, density change is less than about 5 percent; negligible.
Above M = 0.3, compressibility becomes important.
16. Why is turbulence modeling more difficult in compressible flow?
Because density, temperature, and viscosity all vary.
Turbulence interacts with shocks, expansions, and heating, making the flow highly nonlinear.
17. What does bulk modulus measure?
It measures a fluid's resistance to compression.
High bulk modulus = fluid is stiff (like water).
Low bulk modulus = fluid compresses easily (like air).
18. Why is water nearly incompressible but not perfectly incompressible?
Even water can compress under extremely high pressure, but the effect is tiny.
Engineering flows rarely reach pressures strong enough to significantly compress water.
19. Why is air 20000 times more compressible than water?
Because air's bulk modulus is approx 1e5 Pa, while water's is approx 2e9 Pa.
Bulk modulus ratio determines compressibility ratio.
20. Why does density depend on both compressibility and pressure change?
Density change = rho * beta * dp.
Even a highly compressible fluid shows small density change if pressure change is small.
21. Why distinguish between isothermal and isentropic compressibility?
Isothermal compressibility is for slow processes that allow heat transfer.
Isentropic compressibility is for fast, adiabatic processes like sound waves.
22. Why are isothermal and isentropic compressibility different?
Because isentropic processes include no heat transfer and involve reversible compression, which stiffens the gas compared to isothermal compression.
23. Why is Mach number the correct measure of compressibility?
Mach = u / a compares flow speed to disturbance speed.
If flow is much slower than pressure waves (u << a), density stays nearly constant.
If u becomes comparable to a, density changes strongly.
24. Why does compressible flow usually mean Mach > 0.3?
Below M = 0.3, density changes less than approx 5 percent.
Above M = 0.3, density changes meaningfully affect momentum and energy equations.
25. Why does speed of sound depend on temperature?
Higher temperature means more molecular kinetic energy, increasing the rate at which pressure disturbances propagate.
26. Why do aircraft fly at constant Mach, not constant speed?
Shock formation, drag rise, lift characteristics, and engine performance all depend on Mach, not true airspeed.
Therefore pilots and autopilots maintain Mach, not velocity.
27. Why are liquids considered incompressible even at high velocities?
Because their speed of sound is extremely high.
Even at high velocities, Mach number remains very small (M << 0.1).
Thus density variations remain negligible.
28. Why must compressible flow involve thermodynamics?
Because density depends on temperature and pressure, and temperature changes require knowledge of heat, work, and internal energy.
29. Why is internal energy mainly a function of temperature for ideal gases?
Because ideal gases assume no intermolecular forces.
Thus internal energy depends entirely on kinetic energy of molecules.
30. Why is calorically perfect gas assumption valid below approx 1000 K?
Below this temperature, vibrational modes are not yet active.
Thus Cp and Cv stay nearly constant.
31. Why do high-temperature flows require thermally perfect gas models?
Because Cp and Cv vary significantly when vibrational modes activate, especially above 1000 K.
Using constant Cp and Cv becomes inaccurate.
32. Why use real-gas equations at high pressure?
Because ideal gas law fails when molecule spacing becomes small and intermolecular forces become strong.
Real gas equations (Peng-Robinson, Van der Waals) correct this.
33. Why include both heat and work in the first law for fluids?
Fluids transport energy through:
34. Why isentropic = adiabatic + reversible?
Adiabatic removes heat transfer; reversible removes internal friction.
Only when both occur is entropy change zero.
35. Why does entropy increase across a shock?
Shocks are irreversible processes involving viscous dissipation.
Kinetic energy converts to internal energy, increasing disorder.
36. Why is an adiabatic process not always isentropic?
Adiabatic means no heat transfer, but friction or shocks can still generate entropy.
Thus ds > 0 even though dq = 0.
37. Why does ds include both dv/v and dT/T terms?
For gases, entropy depends on both microscopic disorder (T) and macroscopic volume (v).
Temperature changes affect molecular energy; volume changes affect molecular spacing.
38. Why can entropy change be written in terms of p, v, and T for ideal gases?
Because ideal gases are simple compressible substances.
Two independent variables define their state.
Thus ds can be expressed using any convenient pair like (T, v) or (p, T).
39. Why does p * v^gamma = constant only in isentropic flow?
Because this relation comes directly from ds = 0 and constant Cp, Cv.
If entropy changes, this form breaks.
40. Why do pressure, temperature, and density scale via gamma?
Because gamma expresses how energy is partitioned between pressure-volume work and internal energy.
Thus all isentropic relations contain gamma.
41. Why do isentropic relations fail in shocks?
Because entropy increases across shocks (irreversible).
Isentropic equations assume ds = 0, which is violated.
42. Why is 1D flow a valid assumption in many problems?
When area changes slowly and boundary layers are thin, variations occur mainly along the flow direction.
Thus a 1D approximation is accurate and simplifies analysis.
43. Why does momentum equation include dynamic pressure?
Because dynamic pressure (0.5 * rho * u^2) represents kinetic energy per unit volume.
It quantifies how much momentum is carried by the fluid.
44. Why does the energy equation use enthalpy instead of internal energy?
Because enthalpy = internal energy + flow work.
Using enthalpy automatically accounts for pressure work in flowing systems.
45. Why require constant area for simplified 1D equations?
If area varies, acceleration is affected by area change, requiring area derivatives.
Constant area simplifies the equations drastically.
46. Why is speed of sound defined as sqrt( dp / d rho ) at constant entropy?
Because sound is a small, rapid, adiabatic pressure disturbance.
Taking derivative at constant entropy captures this physical behavior.
47. Why are sound waves isentropic?
They are rapid and of small amplitude, so no significant heat transfer or irreversibility occurs.
Thus ds approx 0.
48. Why does incompressibility imply infinite speed of sound?
If density does not change (beta = 0), then any pressure change would require an infinite dp/d rho.
Thus mathematically, sound speed goes to infinity.
49. Why does higher gamma increase speed of sound?
Higher gamma means gas is stiffer (more pressure for a given density change).
Stiffer gas carries disturbances faster.
50. Why does temperature affect speed of sound but pressure does not (for ideal gas)?
Because a = sqrt(gamma * R * T).
In ideal gases, pressure and density scale together, so their ratio cancels.
Only temperature remains.
