open TELEMAC-MASCARET The mathematically superior suite of solvers

Hello,

Yes, I refer to dimensional analysis, and I just remark that in your formula :

(dT/dt)+(u.grad(T))-(1/h.div(h.vt.grad(T)))=s*T

s*T must have the same dimension (in the sense of units) than dT/dt, and as dT has the same dimension as T it means that s should have the same dimension as 1/dt, which is the inverse of a time. Elsewhere in your formulas you have 10*s+1, which means that s should have no dimension, so there is a problem here, in the sense that your formulas are not independent of the choice of units. This would be the first thing to sort out before going further, you seem to be lacking a coefficient in your formula.

If we check PRIVE%ADR(1)%P%R(I), it is compared to CS%ADR(1)%P%R(I), and has thus no dimension, assuming that CS is cubic meters of sediment per cubic meters of water. PRIVE%ADR(1)%P%R(I) is equal to:

0.0273D0*2650.D0*(U%R(I)**2+V%R(I)**2)/(GRAV*H%R(I))

and (U%R(I)**2+V%R(I)**2)/(GRAV*H%R(I)) has no dimension (it is the Froude number squared). I guess that 2650.D0 is kg/m**3, so it is dimensionly correct if 0.0273D0 is in m**3/kg ?

With best regards,

Jean-Michel Hervouet

Hello,,

Sorry,sir. My formula I need to solve is:
(dT/dt)+(u.grad(T))-(1/h.div(h.vt.grad(T)))=β*XWC*（s'-s)*(k/(k*s+1))*T/H

βhas no dimension, XWC is the settling velocity of sediment (m/s),
s'=0.0273D0*2650.D0(g/L)*(U%R(I)**2+V%R(I)**2)/(GRAV*H%R(I)), s is the suspended sediment concentration, the unit is g/L,so s' has the same dimension with s. The unit of k is L/g, so k*s has no dimension. H is depth, the unit is m. T has the same dimension with dT.

So the dimension of the right part of this formula is: (m/s)*g/L*(L/g/(g/L+1))*T/m.It is the same with the dT/dt (T/s). Am I right?

Thank you sir.

zqzuoan

Hello,

I am not at work and will be able to test your case only tomorrow, but if you have dry zones in your model, there is a risk of division by 0 because under a test IF(ABS(H%R(I)).LE.0.2D0) you do a division by H%R(I).

You should first write IF(H%R(I).LE.0.2D0) and the division by H%R(I) should be a division by MAX(H%R(I),1.D-3) for example. If this gives a difference of behaviour we would have to think more of the values of those coefficients when the depth tends to 0, because the term (U%R(I)**2+V%R(I)**2)/(GRAV*H%R(I)), which is F**2 may tend to infinity on dry zones.

With best regards,

Jean-Michel Hervouet

Hello,JMH

Yes, I have thought this problem. And I will modify my program. But I am not sure that if it is right to add the right part of the formula to TEXP.I modified the difsou.f as follows: TEXP%ADR(1)%P=β*XWC*（s'-s)*(k/(k*s+1))*T%ADR(1)%P/H

With best regards,
zqzuoan