Thermodynamics · Chapter 17

Compressible Flow

Compressible flow couples density, temperature, pressure, and velocity. Correct analysis starts by distinguishing static from stagnation states and by choosing the subsonic, choked, supersonic, or shock branch.

Learning objectives

Relate static and stagnation states through steady-flow energy.
Compute sound speed and classify flow by Mach number.
Select the correct isentropic area branch and identify choking.
Audit normal shocks with mass, momentum, energy, and entropy.
Recognize Rayleigh-flow and steam-nozzle property requirements.

17–1

Stagnation Properties

Energy carried by speed

Stagnation enthalpy \(h_0=h+V^2/2\) is the static enthalpy plus kinetic energy per mass. Stagnation pressure additionally requires an isentropic deceleration; friction or a shock leaves \(h_0\) unchanged when adiabatic but reduces \(P_0\).

Calorically perfect ideal gas

\[\frac{T_0}{T}=1+\frac{k-1}{2}Ma^2,\qquad \frac{P_0}{P}=\left(\frac{T_0}{T}\right)^{k/(k-1)}\]

Physical meaning

Static instruments move with the local fluid state; stagnation quantities describe the state after ideal recovery of speed. Confusing them can dominate a high-speed measurement error.

EXAMPLE 1Static and stagnation state

Air moves at 240 m/s with static temperature 285 K. Using supplied constant ideal-gas values \(c_p=1.005\ \mathrm{kJ/(kg\!\cdot\!K)}\), \(k=1.40\), and \(R=287\ \mathrm{J/(kg\!\cdot\!K)}\), find Mach number and stagnation temperature.

Recognition strategy

Velocity and static temperature with ideal-gas constants cue sound speed, Mach number, and the steady-flow kinetic-energy contribution to stagnation enthalpy.

Given information

Static temperature is absolute; pressure is absolute but not needed for these outputs.
Supplied ideal-gas constants apply over the stated narrow temperature range.

Model and interpretation

A streamline control volume conceptually decelerates the gas adiabatically and reversibly to define the stagnation state. Steady flow, negligible PE, no shaft work, and no heat transfer are assumed; KE is retained.

Step-by-step solution

1. Plan. Compute sound speed from \(a=\sqrt{kRT}=\sqrt{1.40(287)(285)}=338.5\ \mathrm{m/s}\).

2. Formulate. Divide velocity by sound speed to obtain \(Ma=240/338.5=0.709\), identifying a subsonic static state.

3. Calculate. Apply \(T_0=T+V^2/(2c_p)\), converting \(c_p\) to \(1005\ \mathrm{J/(kg\!\cdot\!K)}\).

4. Resolve. The kinetic-temperature increment is \(28.66\ \mathrm{K}\), hence \(T_0=313.7\ \mathrm{K}\).

Verification

Units: The quantity kRT has square metres per square second, so its square root is m/s; V squared divided by cp has kelvins after using cp in J/(kg K).

Signs and direction: Kinetic energy is positive, so ideal deceleration raises stagnation temperature above the 285 K static value; Mach number remains positive and subsonic.

Plausibility: The independent identity T0/T=1+0.2 Ma squared gives 1.1005, recovering about 313.7 K and agreeing with the direct energy calculation.

Final result: \(Ma=0.709\) and \(T_0=314\ \mathrm{K}\).

Engineering interpretation

A stationary probe that recovers velocity ideally reads the higher stagnation temperature. Real probes require recovery-factor, radiation, conduction, alignment, and calibration corrections.

17–2

Speed of Sound and Mach Number

The thermodynamic definition \(a^2=(\partial P/\partial\rho)_s\) describes propagation of a small, locally isentropic disturbance. For an ideal gas it becomes \(a=\sqrt{kRT}\). Mach number compares flow speed with information speed; absolute temperature and a state-appropriate \(k\) are essential.

17–3

One-Dimensional Isentropic Flow

Area–velocity relation

\[\frac{dA}{A}=(Ma^2-1)\frac{dV}{V}\]

A converging area accelerates subsonic flow but decelerates supersonic flow; a diverging area does the reverse. At \(Ma=1\), a smooth sonic transition requires an extremum of area. The area–Mach relation has two branches, so area ratio alone does not identify the state.

Figure 17–1 A converging–diverging nozzle: subsonic flow accelerates in the converging part, reaches Mach 1 at the throat, and goes supersonic in the diverging part.

17–4

Isentropic Flow Through Nozzles

Choking occurs when the minimum area reaches Mach one. A converging nozzle then has a maximum ideal mass flow for fixed upstream stagnation state. A converging–diverging nozzle can use the supersonic branch only if back pressure is low enough; otherwise shocks or entirely subsonic flow appear.

EXAMPLE 2Choked converging nozzle

Air enters an adiabatic converging nozzle from a reservoir at \(P_0=600\ \mathrm{kPa\ abs}\) and \(T_0=360\ \mathrm{K}\). The throat area is 6.00 cm². For supplied \(k=1.40\), \(R=287\ \mathrm{J/(kg\!\cdot\!K)}\), determine critical pressure and maximum mass flow, explicitly assuming the back pressure satisfies \(P_b\le P^*\) so the nozzle chokes; otherwise report the result only as the maximum possible flow.

Recognition strategy

Reservoir conditions and a minimum area signal a choking branch. First compare back pressure with the critical pressure, then use the sonic mass-flux relation.

Given information

Reservoir stagnation properties are absolute; throat area is \(6.00\times10^{-4}\ \mathrm{m^2}\).
Air is stipulated ideal with constant properties and one-dimensional flow.
The nozzle chokes only when \(P_b\le P^*=317\ \mathrm{kPa\ abs}\); otherwise the calculated flow is merely an upper bound.

Model and interpretation

A steady open control volume encloses an isentropic nozzle: adiabatic, reversible, no shaft work, negligible PE, but changing KE. Choking sets \(Ma^*=1\) only for \(P_b\le317\ \mathrm{kPa\ abs}\); at higher back pressure the nozzle is unchoked and this approximate ideal result is only the maximum possible.

Step-by-step solution

1. Plan. Compute the critical ratio \(P^*/P_0=[2/(k+1)]^{k/(k-1)}=0.5283\), so \(P^*=317.0\ \mathrm{kPa\ abs}\).

2. Formulate. Use the choked ideal-gas flux \(\dot m/A=P_0\sqrt{k/(RT_0)}[2/(k+1)]^{(k+1)/(2(k-1))}\).

3. Calculate. Substitution gives mass flux \(1278.14\ \mathrm{kg/(m^2\!\cdot\!s)}\), with pressure entered in pascals.

4. Resolve. Multiply by throat area to obtain \(\dot m_{max}=0.7669\ \mathrm{kg/s}\); lower back pressure cannot increase this ideal mass flow.

Verification

Units: With pressure in pascals, P0 sqrt[k/(RT0)] has kg/(m squared s); multiplying by 6.00e-4 m squared gives kg/s.

Signs and direction: Flow is outward and mass rate positive; pressure decreases from 600 to 317 kPa absolute as the stream accelerates to the sonic throat.

Plausibility: The critical ratio is 0.5283, so choking requires Pb at or below 317 kPa absolute; the sonic mass and energy relations then bound the ideal flow at 0.767 kg/s.

Final result: \(P^*=317\ \mathrm{kPa\ abs}\) and \(\dot m_{max}=0.767\ \mathrm{kg/s}\).

Engineering interpretation

Choking fixes mass flow from upstream stagnation state and minimum area, not from back pressure alone. Vendor discharge coefficients, boundary layers, tolerances, and real-gas data are required for release sizing.

17–5

Shock Waves and Expansion Waves

A normal shock is adiabatic yet irreversible: mass, momentum, and stagnation enthalpy are conserved, while entropy rises and stagnation pressure falls. It always maps an upstream supersonic state to a downstream subsonic state. An oblique shock uses the normal Mach component; a smooth Prandtl–Meyer fan turns supersonic flow away from itself.

EXAMPLE 3Normal-shock loss audit

Ideal air with \(k=1.40\) encounters a normal shock at \(Ma_1=2.00\), \(P_1=90.0\ \mathrm{kPa\ abs}\), and \(T_1=250\ \mathrm{K}\). Determine downstream Mach number, pressure, temperature, and entropy rise using \(R=0.287\ \mathrm{kJ/(kg\!\cdot\!K)}\), \(c_p=1.005\ \mathrm{kJ/(kg\!\cdot\!K)}\).

Recognition strategy

A sudden stationary transition from supersonic flow identifies the normal-shock branch; isentropic area relations cannot cross this irreversible discontinuity.

Given information

All static states use absolute pressure and temperature; constant ideal-gas properties are supplied.
The shock is stationary, normal, adiabatic, and one-dimensional.

Model and interpretation

A thin steady control volume surrounds the shock. Mass, momentum, and total enthalpy are conserved; there is no heat or shaft work and PE is negligible. The process is irreversible, not isentropic, so stagnation pressure falls.

Step-by-step solution

1. Plan. Apply the normal-shock relation to obtain \(Ma_2^2=[1+0.2Ma_1^2]/[1.4Ma_1^2-0.2]=0.3333\), hence \(Ma_2=0.577\).

2. Formulate. Compute \(P_2/P_1=1+2k(Ma_1^2-1)/(k+1)=4.50\), giving \(P_2=405\ \mathrm{kPa\ abs}\).

3. Calculate. Compute density ratio \(\rho_2/\rho_1=2.667\), then \(T_2/T_1=(P_2/P_1)/(\rho_2/\rho_1)=1.6875\), so \(T_2=421.9\ \mathrm{K}\).

4. Resolve. Audit entropy: \(s_2-s_1=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)=0.0942\ \mathrm{kJ/(kg\!\cdot\!K)}>0\).

Verification

Units: Pressure ratios, density ratios, and Mach numbers are dimensionless; cp ln(T2/T1) and R ln(P2/P1) both have kJ/(kg K).

Signs and direction: Across the stationary normal shock, velocity and Mach number fall while static pressure, temperature, density, and entropy all rise in the downstream direction.

Plausibility: Mass and momentum jump ratios give P2/P1=4.50 and rho2/rho1=2.667; stagnation enthalpy is conserved and Delta s=0.0942 kJ/(kg K) is positive.

Final result: \(Ma_2=0.577\), \(P_2=405\ \mathrm{kPa\ abs}\), \(T_2=422\ \mathrm{K}\), and \(\Delta s=0.0942\ \mathrm{kJ/(kg\!\cdot\!K)}\).

Engineering interpretation

The shock forces a supersonic stream onto the subsonic branch while conserving stagnation temperature and destroying stagnation pressure. Installation work requires shock-position, unsteadiness, thermal, structural, and CFD validation.

17–6

Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow)

Rayleigh flow models steady, one-dimensional, constant-area flow with heat transfer and negligible wall friction. Heating drives either branch toward Mach one, but property trends differ on subsonic and supersonic branches. State tables or equations must use the correct branch; heat into the control volume is positive and entropy generation cannot be negative.

17–7

Steam Nozzles

Steam may approach saturation or enter the two-phase region, where ideal-gas formulas and constant \(c_p\) are invalid. Identify each state from absolute pressure and temperature, then use authoritative steam tables or a validated equation of state. Quality applies only inside the equilibrium dome; delayed condensation requires a specialist nonequilibrium model.

Problem-solving strategy

Recognize the compressible-flow problem family and identify the requested property or performance measure.
Draw a closed-system or control-volume boundary and label every state with independent properties.
Choose a property source and model before selecting equations; interpolate only between valid tabulated states.
Write conservation and constitutive relations symbolically, count unknowns, and iterate when properties depend on the answer.
Audit dimensions, signs, limiting behavior, conservation, entropy generation, and model applicability.

Chapter recap

Distinguish static and stagnation properties before using measured data.
Use Mach number to choose area-response and choking branches.
Treat shocks as adiabatic but irreversible and verify positive entropy generation.
Before moving on, be able to audit mass, momentum, energy, property validity, and back-pressure response.

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