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Crocco relation

Assumptions

  • \(\Pr = \frac{\mu c_p}{k} = 1\)
  • adiabatic wall \(\frac{\partial T}{\partial y} = 0\)
  • zero pressure gradient: \(\beta = 0\), therefore \(m = 0\) and \(u_e\) is constant

Relation

\[ \tau(\eta) = 1 + \frac{\gamma - 1}{2} M_e^2 \left(1 - f'^{\,2}\right) \]
Derivation

Step 1: Start from the 2D compressible boundary layer energy equation

The 2D compressible boundary layer equations give the energy equation:

\[ \rho c_p \left( u \frac{\partial T}{\partial x} + v \frac{\partial T}{\partial y} \right) = u \frac{dp}{dx} + \frac{\partial}{\partial y}\!\left( k \frac{\partial T}{\partial y} \right) + \mu \left(\frac{\partial u}{\partial y}\right)^{\!2} \]

Step 2: Eliminate the pressure term

Define \(H = c_p T + u^2/2\). The pressure term \(u\,dp/dx\) in the energy equation has no counterpart in a clean transport form, so we eliminate it by adding the \(x\)-momentum equation multiplied by \(u\):

Multiply the \(x\)-momentum equation by \(u\):

\[ \rho \left( u \frac{\partial (u^2/2)}{\partial x} + v \frac{\partial (u^2/2)}{\partial y} \right) = -u\frac{dp}{dx} + u\frac{\partial}{\partial y}\!\left( \mu \frac{\partial u}{\partial y} \right) \]

Add this to the energy equation:

\[ \begin{aligned} \rho \left( u \frac{\partial (c_p T + u^2/2)}{\partial x} + v \frac{\partial (c_p T + u^2/2)}{\partial y} \right) &= \underbrace{u\frac{dp}{dx} - u\frac{dp}{dx}}_{= \, 0} \\ &\quad + \frac{\partial}{\partial y}\!\left( k \frac{\partial T}{\partial y} \right) + u\frac{\partial}{\partial y}\!\left( \mu \frac{\partial u}{\partial y} \right) + \mu \left(\frac{\partial u}{\partial y}\right)^{\!2} \end{aligned} \]

The pressure terms cancel exactly. The left-hand side is simply \(\rho(u\,\partial H/\partial x + v\,\partial H/\partial y)\).

For the right-hand side, the product rule gives:

\[ u\frac{\partial}{\partial y}\!\left( \mu \frac{\partial u}{\partial y} \right) + \mu \left(\frac{\partial u}{\partial y}\right)^{\!2} = \frac{\partial}{\partial y}\!\left( \mu u \frac{\partial u}{\partial y} \right) = \frac{\partial}{\partial y}\!\left( \mu \frac{\partial (u^2/2)}{\partial y} \right) \]

After cancellation the equation reads:

\[ \rho \left( u \frac{\partial H}{\partial x} + v \frac{\partial H}{\partial y} \right) = \frac{\partial}{\partial y}\!\left( k \frac{\partial T}{\partial y} \right) + \frac{\partial}{\partial y}\!\left( \mu \frac{\partial (u^2/2)}{\partial y} \right) \]

Step 3: Require Pr = 1

With \(\Pr = 1\), thermal conductivity satisfies \(k = \mu c_p\). The two remaining terms on the right combine into a single flux of \(H\):

\[ \frac{\partial}{\partial y}\!\left( k \frac{\partial T}{\partial y} \right) + \frac{\partial}{\partial y}\!\left( \mu \frac{\partial (u^2/2)}{\partial y} \right) = \frac{\partial}{\partial y}\!\left( \mu c_p \frac{\partial T}{\partial y} + \mu \frac{\partial (u^2/2)}{\partial y} \right) = \frac{\partial}{\partial y}\!\left( \mu \frac{\partial H}{\partial y} \right) \]

The result is a source-free transport equation for \(H\):

\[ \rho u \frac{\partial H}{\partial x} + \rho v \frac{\partial H}{\partial y} = \frac{\partial}{\partial y}\!\left(\mu \frac{\partial H}{\partial y}\right) \]

Because the right-hand side contains no forcing term, a spatially uniform \(H\) is always a valid solution, provided the boundary conditions allow it.

Step 4: Apply the boundary conditions

The wall-normal derivative of \(H\) is:

\[ \frac{\partial H}{\partial y} = c_p \frac{\partial T}{\partial y} + u \frac{\partial u}{\partial y} \]

At the wall both terms vanish independently:

  • No-slip: \(u_w = 0\), so \(u_w\,\partial u/\partial y|_w = 0\) regardless of the velocity gradient.
  • Adiabatic: \(\partial T/\partial y|_w = 0\) (zero heat flux by definition).

Therefore \(\partial H/\partial y|_w = 0\), a homogeneous Neumann condition. At the boundary-layer edge, \(H = H_e = c_p T_e + u_e^2/2\) (Dirichlet).

Step 5: Identify the exact solution

The \(H\)-equation has no source term, and the zero-pressure-gradient case has a constant edge state. Since \(m = 0\), \(u_e\) is constant, so \(H_e = c_p T_e + u_e^2/2\) is constant as well. If \(H\) is constant, all derivatives of \(H\) vanish, so the transport equation is satisfied. The edge condition fixes that constant:

\[H = H_e\]

Substituting the definitions of \(H\) and \(H_e\) gives total enthalpy conservation across the boundary layer:

\[c_p T + \tfrac{1}{2}u^2 = c_p T_e + \tfrac{1}{2}u_e^2\]

This is the key result: the total enthalpy is the edge total enthalpy at every \(\eta\).

Step 6: Non-dimensionalise

Divide by \(c_p T_e\), use \(u_e^2/(2 c_p T_e) = (\gamma - 1)M_e^2/2\), and identify \(T/T_e = \tau\) and \(u/u_e = f'\):

\[ \overbrace{\frac{T}{T_e}}^{\tau} = 1 + \frac{\gamma - 1}{2} M_e^2 \left(1 - \overbrace{\frac{u^2}{u_e^2}}^{f'^{\,2}}\right) \]

Therefore:

\[ \boxed{ \tau(\eta) = 1 + \frac{\gamma - 1}{2} M_e^2 \left(1 - f'^{\,2}\right) } \]

This is the Crocco relation used in the verification. It requires \(\Pr = 1\) and the zero-pressure-gradient case tested here.

Results

This holds for any \(M_e\) when \(\beta = 0\). Two cases are tested:

Case \(M_e\) \(m\) \(\beta\)
A 1.5 0 0.0
B 3.0 0 0.0

Crocco relation verification at Me=1.5

Crocco relation verification at Me=3.0

The similarity solution and Crocco relation agree well for both Mach numbers.

Run

The verification script is vnv/verification/falkner_skan/crocco/verification_crocco.py.

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python verification_crocco.py