JNTU B.Tech III HIGH SPEED AERODYNAMICS Previous Question Paper (September 2021 - R18)

Compressible flow is a branch of fluid mechanics where the density of the fluid varies significantly in response to changes in pressure or temperature. It is characterized by a Mach number M > 0.3, where the effects of fluid compressibility become non-negligible.

To derive the momentum equations for 3-D inviscid compressible flow, we consider a stationary infinitesimal control volume of dimensions dx, dy, and dz. Applying Newton's Second Law to this fluid element, the rate of change of momentum equals the sum of forces acting on it. For an inviscid fluid, the only surface force is pressure, and the body force is represented by gravity (g).

The momentum balance in the x-direction is given by:

Expanding the material derivative (Du/Dt = ∂u/∂t + u(∂u/∂x) + v(∂u/∂y) + w(∂u/∂z)), we obtain the x-momentum equation:

Similarly, for the y and z directions:

These equations represent the Euler equations for compressible flow, where u, v, and w are velocity components, ρ is density, p is pressure, and g_x, g_y, g_z are body force components per unit mass.

Given: Shock speed W₁ = 696 m/s, P₁ = 100 kPa, T₁ = 300 K. For air, γ = 1.4, R = 287 J/(kg·K), c_p = 1005 J/(kg·K). The speed of sound ahead is a₁ = √(γRT₁) = √(1.4 × 287 × 300) ≈ 347.2 m/s. The shock Mach number is M_s = W₁ / a₁ = 696 / 347.2 ≈ 2.005.

1. Ahead of the shock (Stationary frame): Static properties are P₁ = 100 kPa, T₁ = 300 K. Since the air is still, V₁ = 0, so stagnation properties are P₀₁ = 100 kPa and T₀₁ = 300 K.

2. Behind the shock (Stationary frame): Using normal shock relations for M_s ≈ 2.005: Pressure ratio P₂/P₁ = 1 + [2γ/(γ+1)] × (M_s² - 1) ≈ 4.53, so P₂ = 453 kPa. Temperature ratio T₂/T₁ = [1 + (2γ/(γ+1))(M_s² - 1)] × [(2 + (γ-1)M_s²) / ((γ+1)M_s²)] ≈ 1.69, so T₂ = 507 K. The velocity behind the shock is V₂ = [2a₁/(γ+1)] × (M_s - 1/M_s) ≈ 407.5 m/s. Stagnation properties: T₀₂ = T₂ + V₂²/(2c_p) = 507 + (407.5)²/2010 ≈ 590 K. P₀₂ = P₂(1 + ((γ-1)/2)M₂²)^γ/(γ-1) where M₂ = V₂ / √(γRT₂) ≈ 0.577, giving P₀₂ ≈ 563 kPa.

3. Moving frame (relative to shock): The shock is stationary. Ahead, M₁ = 2.005, P₁ = 100 kPa, T₁ = 300 K. P₀₁ = P₁(1 + 0.2 × 2.005²)^3.5 ≈ 782 kPa, T₀₁ = 300(1 + 0.2 × 2.005²) = 540.6 K. Behind, M₂ = 0.577, P₂ = 453 kPa, T₂ = 507 K. P₀₂ = 453(1 + 0.2 × 0.577²)^3.5 ≈ 563 kPa, T₀₂ = 507(1 + 0.2 × 0.577²) = 540.6 K.