6.4 Compressible and Non-Newtonian Flows

Key Takeaways

  • Compressible flow analysis is required when the Mach number (Ma) is 0.3 or higher.
  • Sonic velocity (c) is the speed at which pressure disturbances propagate, calculated as c = sqrt(k R T) for an ideal gas.
  • Stagnation properties represent fluid states when flow is decelerated isentropically to zero velocity.
  • Choked flow occurs at a throat when Ma = 1; further reduction in downstream pressure does not increase mass flow rate.
  • Non-Newtonian fluids exhibit shear-dependent viscosities (power-law model) or threshold yield stresses (Bingham plastics).
Last updated: July 2026

Fundamentals of Compressible Flow

In compressible flow, fluid density ($\rho$) changes significantly in response to pressure changes. While liquids are treated as incompressible, gases exhibit compressible behavior when the flow velocity ($v$) is high. The standard threshold for compressible flow analysis is a Mach number ($Ma$) of $0.3$ or higher:

Ma=vcMa = \frac{v}{c}

where $c$ is the local speed of sound (sonic velocity). For flows where $Ma < 0.3$, density variations are typically less than $5%$, allowing the use of incompressible equations.

Sonic Velocity: The speed of sound represents the velocity at which small pressure disturbances propagate through a medium. For an ideal gas undergoing an isentropic process, the sonic velocity is:

c=kRT=kRuTMc = \sqrt{k R T} = \sqrt{\frac{k R_u T}{M}}

where:

  • $k = C_p / C_v$ is the ratio of specific heats (isentropic expansion coefficient), which is approximately $1.4$ for diatomic gases like air and $1.3$ for triatomic gases like carbon dioxide.
  • $R$ is the specific gas constant ($R = R_u / M$, where $R_u = 8.314\text{ J/(mol}\cdot\text{K)}$ or $1545\text{ ft}\cdot\text{lb}_f\text{/(slug}\cdot^\circ\text{R)}$).
  • $T$ is the absolute temperature in Kelvin ($\text{K}$) or Rankine ($^\circ\text{R}$).
  • $M$ is the molecular weight of the gas.

Stagnation Properties: When a high-velocity gas stream is decelerated isentropically to zero velocity, its kinetic energy is converted into enthalpy, raising its temperature and pressure. The resulting states are called stagnation (or total) properties:

T0T=1+k12Ma2\frac{T_0}{T} = 1 + \frac{k-1}{2} Ma^2

P0P=(1+k12Ma2)kk1\frac{P_0}{P} = \left( 1 + \frac{k-1}{2} Ma^2 \right)^{\frac{k}{k-1}}

where $T_0$ and $P_0$ are the stagnation temperature and pressure, and $T$ and $P$ are the static temperature and pressure of the moving fluid.

Choked Flow: In a converging-diverging nozzle, if the pressure ratio across the nozzle is sufficiently large, the velocity at the throat (minimum area) reaches the speed of sound ($Ma = 1$). Once this condition is met, the flow is 'choked,' and the mass flow rate through the nozzle reaches a maximum value that cannot be increased by further lowering the downstream receiver pressure. The critical pressure ratio for choked flow of an ideal gas is:

PP0=(2k+1)kk1\frac{P^*}{P_0} = \left( \frac{2}{k+1} \right)^{\frac{k}{k-1}}

For air ($k = 1.4$), this ratio is $0.528$.

Rheology of Non-Newtonian Fluids

Non-Newtonian fluids do not obey Newton's law of viscosity; their shear stress ($\tau$) is not linearly proportional to the shear rate ($dv/dy$). Apparent viscosity depends on the applied shear rate and, in some cases, the duration of shearing.

Time-Independent Non-Newtonian Models:

  1. The Power-Law Model (Ostwald-de Waele): This is the most common model used in the FE exam to describe shear-dependent viscosity:

    τ=K(dvdy)n\tau = K \left( \frac{dv}{dy} \right)^n

    where $K$ is the flow consistency index ($\text{Pa}\cdot\text{s}^n$) and $n$ is the flow behavior index (dimensionless). The apparent viscosity ($\eta_{\text{eff}}$) is:

    ηeff=K(dvdy)n1\eta_{\text{eff}} = K \left( \frac{dv}{dy} \right)^{n-1}

    • Newtonian ($n = 1$): Viscosity is constant and equal to $K$.
    • Pseudoplastic / Shear-Thinning ($n < 1$): Viscosity decreases as shear rate increases. This behavior occurs because polymer chains or suspended particles align with the flow, reducing resistance. Examples include polymer solutions, ketchup, paper pulp, and paint.
    • Dilatant / Shear-Thickening ($n > 1$): Viscosity increases as shear rate increases. This occurs in highly concentrated suspensions where particles collide and interlock under high shear. Examples include starch-water suspensions (oobleck) and quicksand.
  2. Bingham Plastics: These fluids behave as solid bodies at low shear stress but flow as viscous fluids once the applied stress exceeds a threshold yield stress ($\tau_y$):

    τ=τy+μpdvdyforτ>τy\tau = \tau_y + \mu_p \frac{dv}{dy} \quad \text{for} \quad \tau > \tau_y

    where $\mu_p$ is the plastic viscosity. Examples include toothpaste, sewage sludge, drilling muds, and mayonnaise.

Time-Dependent Non-Newtonian Models:

  • Thixotropic Fluids: Apparent viscosity decreases over time under constant shear stress (e.g., non-drip paints, honey).
  • Rheopectic Fluids: Apparent viscosity increases over time under constant shear stress (e.g., gypsum pastes).

Summary of Flow Models

Fluid TypeGoverning Stress EquationViscosity Behavior
Newtonian$\tau = \mu (dv/dy)$Constant Apparent Viscosity
Power-Law ($n < 1$)$\tau = K (dv/dy)^n$Shear-Thinning (Apparent Viscosity Decreases)
Power-Law ($n > 1$)$\tau = K (dv/dy)^n$Shear-Thickening (Apparent Viscosity Increases)
Bingham Plastic$\tau = \tau_y + \mu_p (dv/dy)$Yield Stress Required Before Flow

Worked Example 1: Sonic Velocity and Mach Number

Problem: Carbon dioxide ($M = 44.01\text{ g/mol}$, $k = 1.30$) flows through a pipeline at a temperature of $150^\circ\text{C}$ and a velocity of $180\text{ m/s}$. Determine the speed of sound and the Mach number of the gas.

Solution:

First, convert the temperature to Kelvin:

T=150+273.15=423.15 KT = 150 + 273.15 = 423.15\text{ K}

Calculate the specific gas constant ($R$):

R=RuM=8.314 J/(molK)0.04401 kg/mol188.9 J/(kgK)R = \frac{R_u}{M} = \frac{8.314\text{ J/(mol}\cdot\text{K)}}{0.04401\text{ kg/mol}} \approx 188.9\text{ J/(kg}\cdot\text{K)}

Calculate the speed of sound ($c$):

c=kRT=1.30188.9 J/(kgK)423.15 K=103,912 m2/s2322.4 m/sc = \sqrt{k R T} = \sqrt{1.30 \cdot 188.9\text{ J/(kg}\cdot\text{K)} \cdot 423.15\text{ K}} = \sqrt{103,912\text{ m}^2\text{/s}^2} \approx 322.4\text{ m/s}

Calculate the Mach number ($Ma$):

Ma=vc=180 m/s322.4 m/s0.558Ma = \frac{v}{c} = \frac{180\text{ m/s}}{322.4\text{ m/s}} \approx 0.558

Since $Ma = 0.558 > 0.3$, the flow must be modeled as compressible subsonic flow.

Worked Example 2: Effective Viscosity of a Power-Law Fluid

Problem: A polymer solution ($K = 2.5\text{ Pa}\cdot\text{s}^{0.6}$, $n = 0.6$) is sheared in a rheometer at a shear rate of $150\text{ s}^{-1}$. Calculate the shear stress and the effective viscosity of the solution.

Solution:

Using the power-law equation, calculate the shear stress ($\tau$):

τ=K(dvdy)n=2.5 Pas0.6(150 s1)0.6\tau = K \left( \frac{dv}{dy} \right)^n = 2.5\text{ Pa}\cdot\text{s}^{0.6} \cdot (150\text{ s}^{-1})^{0.6}

τ=2.520.3150.8 Pa\tau = 2.5 \cdot 20.31 \approx 50.8\text{ Pa}

Calculate the effective (apparent) viscosity ($\eta_{\text{eff}}$):

ηeff=τdv/dy=K(dvdy)n1=2.5(150)0.4=2.50.13540.339 Pas\eta_{\text{eff}} = \frac{\tau}{dv/dy} = K \left( \frac{dv}{dy} \right)^{n-1} = 2.5\cdot (150)^{-0.4} = 2.5 \cdot 0.1354 \approx 0.339\text{ Pa}\cdot\text{s}

Note that the effective viscosity ($0.339\text{ Pa}\cdot\text{s}$) is significantly lower than the consistency index $K$ ($2.5$), demonstrating the shear-thinning behavior of the fluid.

Test Your Knowledge

Which of the following conditions characterizes the flow at the throat of a converging-diverging nozzle when it has reached choked flow, and how can the mass flow rate be increased past this point?

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B
C
D
Test Your Knowledge

A fluid has a flow behavior index (n) of 1.4 in the power-law model. What classification of fluid is this, and how does its apparent viscosity change with increasing shear rate?

A
B
C
D
Test Your Knowledge

Which of the following fluids behaves as a solid under low shear stresses but flows like a viscous liquid once a threshold yield stress is exceeded?

A
B
C
D