# Regarding the ITTC Guidelines for Ship CFD applications

I was reading through the ITTC – Recommended Procedures and Guidelines - Practical Guidelines for Ship CFD Applications (https://ittc.info/media/4196/75-03-02-03.pdf) document and this little thing kept me thinking. In page 9 the distance y of the first point from the wall can be computed is said to be computed as $$y = \frac { y ^ { + } L } { \operatorname { Re } _ { L } \sqrt { \frac { C _ { f } } { 2 } } }$$. And right below $$C _ { f } = \frac { 1.328 } { \sqrt { R e _ { L } } }$$.

Now let's try to deduce the formula for $$y$$. $$\begin{cases} \tau _ { w } = \frac { 1 } { 2 } \rho U _ { e } ^ { 2 } C _ { f } \\ u ^ { * } = \sqrt { \frac { \tau _ { w } } { \rho } } \end{cases}$$ $$\Rightarrow u ^ { * } = \sqrt { \frac { \frac { 1 } { 2 } \rho U _ { e } ^ { 2 } C _ { f } } { \rho } } \Rightarrow u ^ { * } = U_e\sqrt{\frac{C_f}{2}}$$

$$\begin{cases} y ^ { + } = \frac { yu ^ { * } } { \nu } \Rightarrow y=y ^ { + }\frac{\nu}{u^{*}}= y^{+}\frac{\nu}{U_e\sqrt{\frac{C_f}{2}}}\\ Re_L= \frac{U_e L}{\nu} \Rightarrow \frac{Re_L}{L} = \frac{U_e}{\nu} \Rightarrow \frac{\nu}{U_e}=\frac{L}{Re_L} \end{cases}$$

$$\Rightarrow \boxed{y =\frac{y ^ { + }L}{Re_L\sqrt{\frac{C_f}{2}}} }$$

So in the deduction $$C_f$$ is the local friction coefficient, but in the guidelines, $$C_f$$ is obtained using a formula for an average friction coefficient. So in my opinion $$C _ { f }$$ should be calculated by $$\frac { 0.6642 } { \sqrt { Re _ { x } } }$$.

What do you think?

• The entire purpose of the ITTC guidelines is to produce test data that is comparable with other tests in other places. (This is still more of a dream than a reality). It's not like the ITTC 57 line has any real validity or love among hydro people, but it's what is used for comparison purposes. In the end, you still have form factors to bring the full size ship numbers into agreement with real world experience. – Phil Sweet Jan 31 '19 at 12:37