One-dimensional isentropic flow relations

Why it matters

Nozzles, diffusers, wind-tunnel test sections and engine intakes are all designed first with one-dimensional isentropic relations. With one Mach number you get every static-to-stagnation ratio and the area needed, so these relations (and the isentropic table built from them) are the most used tool in gas dynamics and in the GATE AE paper.

Key ideas

Assumptions. The quasi-one-dimensional model treats properties as uniform across each section of a duct whose area A(x) varies slowly. The gas is calorically perfect (constant cp, cv, γ). The flow is steady, adiabatic, with no shaft work and no friction — that is, isentropic. Shocks, wall friction (Fanno flow) and heat transfer (Rayleigh flow) break these assumptions and are treated separately.

Energy equation. With no heat or work, h + V²/2 = h₀ is constant along the duct. For a perfect gas h = cp·T, and using cp = γR/(γ−1) and a² = γRT this becomes T₀/T = 1 + (γ−1)/2 · M². This temperature relation needs only adiabatic flow — it holds even across a shock.

Isentropic link. Because the flow is also reversible, p/p₀ = (T/T₀)^(γ/(γ−1)) and ρ/ρ₀ = (T/T₀)^(1/(γ−1)). So once you know M, every static property follows from the stagnation values, which stay constant along an isentropic duct.

Area–velocity relation. Combining continuity (ρ·V·A = constant), the Euler equation (dp = −ρ·V·dV) and dp/dρ = a² gives dA/A = (M² − 1)·dV/V.

• M < 1: area decrease → velocity increase (a subsonic nozzle converges).
• M > 1: area increase → velocity increase (a supersonic nozzle diverges).
• M = 1 can occur only where dA = 0, i.e. at a minimum-area section (throat).

So to accelerate gas from rest to supersonic speed you need a converging–diverging (de Laval) duct with sonic flow at the throat.

Area–Mach relation. Writing continuity between any section and the (real or imaginary) sonic section of area A* gives A/A* as a function of M only. Every value A/A* > 1 has two solutions, one subsonic and one supersonic; which one occurs depends on the pressure ratio imposed on the duct. A/A* = 1 only at M = 1.

Mass flow. For given p₀, T₀ and throat area, the mass flow is greatest when the throat is sonic. Once that happens the nozzle is "choked": lowering the back pressure further cannot increase the mass flow.

Trends with M. As M rises, T, p and ρ all fall and V rises; p falls fastest (exponent 3.5 for air), ρ next (2.5), then T (1).

Formulas

T₀/T = 1 + (γ−1)/2 · M² — adiabatic flow (no isentropic assumption needed).

p₀/p = (1 + (γ−1)/2 · M²)^(γ/(γ−1)) — isentropic; for air the exponent is 3.5.

ρ₀/ρ = (1 + (γ−1)/2 · M²)^(1/(γ−1)) — isentropic; for air the exponent is 2.5.

dA/A = (M² − 1)·dV/V — area–velocity relation.

A/A* = (1/M)·[(2/(γ+1))·(1 + (γ−1)/2 · M²)]^((γ+1)/(2(γ−1))) — area–Mach relation; exponent 3 for air.

ṁ = (p₀·A*/√T₀)·√(γ/R)·(2/(γ+1))^((γ+1)/(2(γ−1))) — choked mass flow; for air ṁ = 0.0404·p₀·A*/√T₀ (SI units).

Symbols: T, p, ρ static temperature (K), pressure (Pa), density (kg/m³); subscript 0 = stagnation (reservoir) value; M Mach number; γ ratio of specific heats; R specific gas constant (J/(kg·K)); A area (m²); A* sonic (throat) area (m²); ṁ mass flow (kg/s).

Worked examples

Example 1 (standard). Air expands isentropically from a reservoir at p₀ = 500 kPa, T₀ = 400 K to p = 100 kPa. Find M, T and V. γ = 1.4, R = 287 J/(kg·K).

1. p₀/p = 5 = (1 + 0.2·M²)^3.5.
2. 1 + 0.2·M² = 5^(1/3.5) = 1.5838, so M² = 0.5838/0.2 = 2.919.
3. M = 1.709.
4. T = T₀/(1 + 0.2·M²) = 400/1.5838 = 252.6 K.
5. a = √(1.4 × 287 × 252.6) = 318.6 m/s; V = M·a = 1.709 × 318.6 = 544 m/s.

Answer: M = 1.71, T = 252.6 K, V ≈ 544 m/s.

Example 2 (GATE level). A section of a convergent–divergent nozzle has twice the throat area, and the throat is sonic. Reservoir: p₀ = 1.0 MPa, T₀ = 500 K. Find the two possible Mach numbers there and, for the supersonic one, the static pressure and temperature.

1. Solve A/A* = 2.0 from the area–Mach relation (iteration or table): subsonic M = 0.306, supersonic M = 2.197.
2. Supersonic branch: 1 + 0.2·M² = 1 + 0.2 × 4.827 = 1.9655.
3. T = 500/1.9655 = 254.4 K.
4. p₀/p = 1.9655^3.5 = 10.646, so p = 1.0 × 10⁶ / 10.646 = 93.9 kPa.
5. (For the subsonic branch: p = 937 kPa, T = 490.8 K.)

Answer: M = 0.306 or 2.197; supersonic branch p ≈ 93.9 kPa, T ≈ 254.4 K.

Common mistakes

• Using static temperature in place of T₀ (or vice versa) in the ratios.
• Inverting ratios: p/p₀ is less than 1; p₀/p is greater than 1.
• Applying isentropic pressure or density relations across a shock. Only T₀ survives a shock; p₀ drops.
• Picking the wrong branch of A/A* — check whether the flow downstream of the throat is supersonic.
• Assuming the sonic area A* is always a physical throat; it is a reference area and changes across a shock.
• Taking 1/3.5 as the exponent for the temperature step: T₀/T has exponent 1, p₀/p has 3.5.

For GATE AE

Expect: static/stagnation ratios at a given M, M from a pressure ratio, the two roots of A/A*, choked mass flow, and statements about converging and diverging ducts in subsonic and supersonic flow. Practise inverting p₀/p for M quickly (take the 1/3.5 power, subtract 1, divide by 0.2), and remember key table values: at M = 2, p/p₀ = 0.1278, T/T₀ = 0.5556, A/A* = 1.6875.

Quick check

1. Find p/p₀ at M = 2 for air.
2. What Mach number gives p₀/p = 2 in air?
3. In supersonic flow, does a converging duct speed the flow up or slow it down?
4. Which ratio, T₀/T or p₀/p, still holds across a normal shock with the upstream stagnation value?
5. What is A/A* at M = 1?

Answers: 1. 0.1278; 2. M ≈ 1.05; 3. slows it down; 4. T₀/T (T₀ is unchanged by an adiabatic shock); 5. 1.