- #1
ronin777
- 3
- 0
Hi all,
I want to use SPH discretization for my cooling rate but i don't know if its valid.
So, this is the equation i want to discretize.
L = (1/dT/dt) . ∫v(T) dT
So,
dT/dt = (1/L) . ∫v(T) dT
dT/dt = (1/L). Ʃv(T) dT (with dT →0)
The limits of the above integral are
Tini: Initial temperature at which columnar grain nucleates
Tnuc: Nucleation temp. of equiaxed grain
L: Average length of the appearing columnar grain
during solidification
dT/dt: The cooling rate caused by diff. in mould and melt temperature
The above transition is columnar to equiaxed in grain structures which takes place during solidification of the melt.
Background of the problem: I have a situation in the form where i have a molten Aluminium Copper alloy melt poured in a mould to be solidified. So, the mould temperature is lower than than than the poured melt. I am thinking about a relation which associates temperature or temperature gradient across the fluid in the mould or the cooling rate because of the temperature gradient with the viscosity situation in the melt.
Cheers
Ronin
I want to use SPH discretization for my cooling rate but i don't know if its valid.
So, this is the equation i want to discretize.
L = (1/dT/dt) . ∫v(T) dT
So,
dT/dt = (1/L) . ∫v(T) dT
dT/dt = (1/L). Ʃv(T) dT (with dT →0)
The limits of the above integral are
Tini: Initial temperature at which columnar grain nucleates
Tnuc: Nucleation temp. of equiaxed grain
L: Average length of the appearing columnar grain
during solidification
dT/dt: The cooling rate caused by diff. in mould and melt temperature
The above transition is columnar to equiaxed in grain structures which takes place during solidification of the melt.
Background of the problem: I have a situation in the form where i have a molten Aluminium Copper alloy melt poured in a mould to be solidified. So, the mould temperature is lower than than than the poured melt. I am thinking about a relation which associates temperature or temperature gradient across the fluid in the mould or the cooling rate because of the temperature gradient with the viscosity situation in the melt.
Cheers
Ronin