Question

What is the relationship between \(\vec{E_f} \,and\, \vec{S_f}\) using the over-relaxed approach?

(a) \(\vec{E_f}=(\vec{S_f} ).\vec{e}\)

(b) \(\vec{E_f}=(\frac{S_f}{cos ⁡\theta}) \vec{e}\)

(c) \(\vec{E_f}=(\vec{S_f})×\vec{e}\)

(d) \(\vec{E_f}=((\vec{S_f}).\vec{e} ) \vec{e}\)

I got this question during an online exam.

This is a very interesting question from Diffusion Problem in chapter Diffusion Problem of Computational Fluid Dynamics

Answer 1

The correct answer is (b) \(\vec{E_f}=(\frac{S_f}{cos ⁡\theta}) \vec{e}\)

Explanation: The relationship is given by \((\frac{S_f}{cos ⁡θ})\vec{e}\). This is calculated in CFD packages as \(\vec{E_f}=\frac{\vec{S_f}.\vec{S_f}}{\vec{e}.\vec{S_f}}\vec{e}\). The derivation is given as

\(\vec{E_f}=(\frac{S_f}{cos⁡\theta})\vec{e}=(\frac{S_f^2}{S_f cos⁡\theta})\vec{e} =\frac{\vec{S_f}.\vec{S_f}}{\vec{e}.\vec{S_f}}\vec{e}\).