Coupled Calculation--Roughness Wall Boundary Conditon
Posted: Fri May 24, 2013 10:39 am
Hello,
I did coupled calculation with Syrthes, in my case, superheated steam flows in a tube, where the tube wall is heated. The heat conduction of the tube wall is calculated with Syrthes, and the inner face of the wall is set as interface between C_S and Syrthes. When in C_S, the boundary condition for the surface is set to be smooth, the calculation runs properly, with a reasonable pressure drop. However, when I changed the Smooth wall to Rougn wall with a roughness coef, the case can't run. So I suppose the coupling calculation doesn’t accept rough wall as a boundary condition.
So I want to add an additional head loss to model the friction pressure drop.
In the module of head loss, the head loss coefficient K should be defined by the user. The head loss term in the momentum conservation equation is that:
ρ*∂U/∂t=-grad P+⋯KρU(1)
K=1/2 αlpha |U| (2)
KρU^2=1/2 αlpha*ρ*U^2 (3)
I compare equation ∆P=0.5*f_D*L/D_h *ρU^2 and (3), it shows that when the head loss only dues to friction, αlpha is equivalent to f_D*L/D_h . While in the code, friction factor of smooth wall is computed, what we should define is the roughness part in addition to the smooth part, so I set αlpha=(f_D_rough-f_D_smooth)∙L/D_h.
the αlpha=4.2 in my case, I define the whole fluid volumical domaine as a head loss with αlphaXX=4.2, αlphaYY=4.2, αlphaZZ=4.2.
Finally, however the pressure drop I get is higher than I expected.
Could you help me find out a way to solve my case? Is my calculation with alpha right?
Best regards
Luyi
I did coupled calculation with Syrthes, in my case, superheated steam flows in a tube, where the tube wall is heated. The heat conduction of the tube wall is calculated with Syrthes, and the inner face of the wall is set as interface between C_S and Syrthes. When in C_S, the boundary condition for the surface is set to be smooth, the calculation runs properly, with a reasonable pressure drop. However, when I changed the Smooth wall to Rougn wall with a roughness coef, the case can't run. So I suppose the coupling calculation doesn’t accept rough wall as a boundary condition.
So I want to add an additional head loss to model the friction pressure drop.
In the module of head loss, the head loss coefficient K should be defined by the user. The head loss term in the momentum conservation equation is that:
ρ*∂U/∂t=-grad P+⋯KρU(1)
K=1/2 αlpha |U| (2)
KρU^2=1/2 αlpha*ρ*U^2 (3)
I compare equation ∆P=0.5*f_D*L/D_h *ρU^2 and (3), it shows that when the head loss only dues to friction, αlpha is equivalent to f_D*L/D_h . While in the code, friction factor of smooth wall is computed, what we should define is the roughness part in addition to the smooth part, so I set αlpha=(f_D_rough-f_D_smooth)∙L/D_h.
the αlpha=4.2 in my case, I define the whole fluid volumical domaine as a head loss with αlphaXX=4.2, αlphaYY=4.2, αlphaZZ=4.2.
Finally, however the pressure drop I get is higher than I expected.
Could you help me find out a way to solve my case? Is my calculation with alpha right?
Best regards
Luyi