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Math example

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Math example

Lift(LL) can be determined by Lift Coefficient (CLC_L) like the following equation.

L=12ρv2SCLL = \frac{1}{2} \rho v^2 S C_L
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Lift($$L$$) can be determined by Lift Coefficient ($$C_L$$) like the following
equation.


$$
L = \frac{1}{2} \rho v^2 S C_L
$$

Calculates Darcy friction factor using the Colebrook equation originally published in [1]_. Normally, this function uses an exact solution to the Colebrook equation, derived with a CAS. A numerical can also be used.

1f=−2log⁡10(ϵ/D3.7+2.51Ref)\frac{1}{\sqrt{f}}=-2\log_{10}\left(\frac{\epsilon/D}{3.7} +\frac{2.51}{\text{Re}\sqrt{f}}\right)

Parameters

  • Re : float Reynolds number, [-]
  • eD : float Relative roughness, [-]
  • tol : float, optional None for analytical solution (default); user specified value to use the numerical solution; 0 to use mpmath and provide a bit-correct exact solution to the maximum fidelity of the system’s float; -1 to apply the Clamond solution where appropriate for greater speed (Re > 10), [-]

Returns

  • fd : float Darcy friction factor [-]

Notes

The solution is as follows:

fd=ln⁡(10)2⋅3.72⋅2.512(ln⁡(10)ϵ/D⋅Re−2⋅2.51⋅3.7⋅lambertW[ln⁡(10)10(ϵRe2.51⋅3.7D)⋅Re2/2.512])f_d = \frac{\ln(10)^2\cdot {3.7}^2\cdot{2.51}^2} {\left(\ln(10)\epsilon/D\cdot\text{Re} - 2\cdot 2.51\cdot 3.7\cdot \text{lambertW}\left[\ln(\sqrt{10})\sqrt{ 10^{\left(\frac{\epsilon \text{Re}}{2.51\cdot 3.7D}\right)} \cdot \text{Re}^2/{2.51}^2}\right]\right)}

Some effort to optimize this function has been made. The lambertw function from scipy is used, and is defined to solve the specific function:

y=xexp⁡(x)lambertW(y)=xy = x\exp(x) \text{lambertW}(y) = x

This is relatively slow despite its explicit form as it uses the mathematical function lambertw which is expensive to compute.

For high relative roughness and Reynolds numbers, an OverflowError can be encountered in the solution of this equation. The numerical solution is then used.

The numerical solution provides values which are generally within an rtol of 1E-12 to the analytical solution; however, due to the different rounding order, it is possible for them to be as different as rtol 1E-5 or higher. The 1E-5 accuracy regime has been tested and confirmed numerically for hundreds of thousand of points within the region 1E-12 < Re < 1E12 and 0 < eD < 0.1.

The numerical solution attempts the secant method using scipy’s newton solver, and in the event of nonconvergence, attempts the fsolve solver as well. An initial guess is provided via the Clamond function.

The numerical and analytical solution take similar amounts of time; the mpmath solution used when tol=0 is approximately 45 times slower. This function takes approximately 8 us normally.

Examples

Colebrook(1E5, 1E-4) 0.018513866077471

References

.. [1] Colebrook, C F.”Turbulent Flow in Pipes, with Particular Reference to the Transition Region Between the Smooth and Rough Pipe Laws.” Journal of the ICE 11, no. 4 (February 1, 1939): 133-156. doi:10.1680/ijoti.1939.13150.