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- - **A question in Hrvojebs PHD thesis**
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hi Hrvoje Jasak,
I have a quehi Hrvoje Jasak,
I have a question in your phd thesis. In page 123, about Non-orthogonality error, it reads "For the complete diffusion term, the non-orthogonality error has the following form: E_d = \ Sigma _f S \ cdot [ ( \ rho U)_f K \ cdot (\ Nambla \ phi)_f] =\ Nambla \ cdot (\ Gamma_D \ cdot \ Nambla \ phi), where \ Gamma _D =(\ rho U)_f k. " I think in expresion $ E_d $ shoud get rid of "(\ rho U)_f" item, maybe "(\ rho \Gamma _\ phi)_f" which refers to Equation (3.24) in Page 83. Am I right? |

You are right - thank you veryYou are right - thank you very much.
I will try to rebuild the Thesis and put an updated version on my web site, but this might be quite painful. Thanks for the effort, I really appreciate it. Hrv |

Hi Hrv,
I have another questiHi Hrv,
I have another question. In your PHD thesis page 145,equation(3.136) is $a_P U_P = H(U) -\ Nambla p $, it is a semi-discretised form, it reads "It has been consequently divided through by the volume in order to enable face interpolation of the coefficients." So the expression of equation (3.136) is right. It means the item $\ Nambla p$ is derived from \ frac{\ int _V \ Nambla p dV}{V_P}=\ frac{(\ Nambla p)_P V_P}{V_P}. then refer equation(3.10) in page 79 and equation(3.26)in page 84, " $(\ Namble \ phi)_P = \ frac {1}{V_P} \ Sigma _f S \ phi_f$" but in page 146, equation(3.142) --final form of discretised incompressible NS system for momentum equation is $a_P U_P = H(U) - \ Sigma_f S (p)_f $, it means the item $\ Nambla p$ in equation 145 is equal with $\ Sigma_f S (p)_f$, and this conflicts with equation(3.26), right? Maybe my understanding is wrong! Leosding |

I don't see what your problem I don't see what your problem is:
grad P discretises as: \grad \phi = \ frac {1}{V_P} \ Sigma _f S \ phi_f In order to discretise the momentum equation, I have to integrate over the CV, which brings in the volumes. So, if I want to interpolate the coefficients, I have to make them intensive and I divide the equation by the volume. Nothing tricky there... Hrv |

I know your mean.
My qustion I know your mean.
My qustion is(again): In order to discretise the momentum eqution, with integrate over the CV, you will gain the equation: A<sub>P</sub> U<sub>P</sub> = H<sup>'</sup>(U)+Ñp V<sub>P</sub> ----Equ(A); then wanting to interpolate the coefficients, you divide the equation Equ(A) by the volume V<sub>P</sub>, and get (3.136) in your thesis: a<sub>P</sub>U<sub>P</sub> = H(U)+Ñp; but the equation (3.142) is a<sub>P</sub>U<sub>P</sub> = H(U)+SS(p)<sub>f</sub>; Does it imply the Ñp = SS(p)<sub>f</sub>?( I think it's wrong, as of (3.26) in page 84,Ñf<sub>P</sub>=1/V<sub>P</sub>SSf<sub>f</sub>.) Of course, if equation (3.142) is from directly integrating (2.24) over the CV, it is right, then it imply the a<sub>P</sub> in (3.136) and (3.142) are different. If (3.142) is from (3.136), the item "SS(p)<sub>f</sub>" should be divided by "V<sub>P</sub>", right? Thank for your kindness to my stupid question. Leo |

Before you divide by the volumBefore you divide by the volume you've got the volume integral of grad p which is equal to the sum_f S_f p_f.
After the division you have: aP UP = H(U)+ grad p; So: grad p = 1/V sum_f S_f p_f Hrv |

yes, the equation:
Ñp=1/V SS<yes, the equation:
Ñp=1/V SS<sub>f</sub> p<sub>f</sub> is RIGHT! but how did you derive to (3.142) from (3.136) directly? They are smae "a<sub>P</sub>" token in two equations. It's self-contradicting from your thesis. |

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