# Blasius Friction factor

1. Nov 7, 2009

### Libertiene

"blasius" Friction factor

I am trying to find out if the formula below can be used to estimate the friction factor for turbulent flows with Reynolds number around 100,000, in an air system.

f = 0.079*RE^-0.25

I have not been able to find this exact formula anywhere other than on this website http://www.seykota.com/rm/friction/friction.htm which states that it is valid for 4000<Re<100000.

The above formula has been called the "blasius equation" so maybe it has been derived from it in some way? I think that if this formula works it cant be that accurate.

Does anyone know where this comes from? Or prove it?

2. Nov 7, 2009

3. Nov 9, 2009

### minger

Re: "blasius" Friction factor

The Blasius and Prandlt formulas differ quite a bit at large Reynolds numbers, so for completeness, here is the Prandlt derivation.

With a profile known (assuming pipe flow here), the average pipe velocity can be calculated:
$$u_{av} = \frac{Q}{A} = \frac{1}{\pia^2}\int^a_0\bar{u}2\pi r\,dr = \frac{1}{a^2}\int^a_0 \bar{u}2(a-y)\,dy$$
Turbulent pipe flow has very little wake. Therefore, the law of the wall is accurate all the way across the pipe. If we then neglect the (very thin) viscous sublayer and substitute the simple log-law, then:
$$u_{av} = v^*\left(\frac{1}{k}ln\frac{av^*}{\nu} + B - \frac{3}{2k}\right)$$
From the definition of pipe-friction factor
$$C_f = 2\tau_w / \rho u^2_{av}$$
The following hold:
$$\begin{split} \frac{u_{av}}{v^*} &= \left(\frac{2}{C_f}\right)^{1/2} \\ \frac{av^*}{\nu} &= Re_D\left(\frac{C_f}{8}\right)^{1/2} \\ Re_D &= \frac{2au_{av}}{\nu} \end{split}$$
Realizing that the previosu equation for average velocity is actually a friction factor relation, we introduce base-10 logs, and clean up for:
$$\frac{1}{\Lambda^{1/2}} = 1.99\log_{10}(Re_D\Lambda^{1/2}) - 1.02$$
Where
$$\Lambda = 4C_f$$
Is the Darcy-Friction Factor. Since Prandtl neglected the sublayer and wake, he slightly adjusted the cosntants to better fir the pipe-friction data, particualrly at low Reynolds. The final form is:
$$\frac{1}{\Lambda^{1/2}} = 2.0\log_{10}(Re_D\Lambda^{1/2})-0.8$$