# Help wanted to understand the Scheil equation

1. Sep 21, 2010

### JDRF85

Hi guys

The Scheil equation for the segregation of impurities from a liquid to the solid during solidification is

Cs = C0Keff(1 - fs)^(Keff - 1)

Cs is the concentration of impurities in the solid

C0 is the initial impurity concentration in the liquid

Keff is the effective segregation coefficient (< or = 1)

fs is the mass fraction of the melt that is solidified.

I have trouble understanding what is going to happen as the solid fraction goes to 1. If anyone knows what's going on in this equation please help me out - I hate not understanding this!

Any further detail needed please let me know.

2. Oct 14, 2010

### MagnetDave

Try work the coefficient to its extremes.

In the case where k=0, the solid is pure A. (whatever A is)

In the case where k=1, the solid is the exact same as the liquid (i.e. there is no segregation).

What k tells, in essence, is the solid solubility of the two phases. If k=0 you are going to get a brutal eutectic where one phase is pure A and the other pure B (and you'll get a classic lamellar microstructure). If k=1 what precipitates out of the melt is in perfect harmony with the melt itself, so you see NO segregation, and you get a continuous, equiaxed microstructure. Everything else is somewhere in between.

Did this help?

3. Oct 14, 2010

### JDRF85

Thanks for the reply, MagnetDave. But I think I have not been clear with my question, so apologies. What I am intererested in is the evolution of the system with constant Keff somewhere between the extremes; as time goes on and solidfication proceeds the solid fraction fs is going to (by definition) go to unity, right? So what does this mean for the mathematics and the interpretation?

The bracket (1 - fs) goes to zero as fs goes to 1, and the exponent of that bracket which is (Keff - 1) will always be less than one. So surely this leads to a situation where you have approach or obtain a zero in the denominator on the right-hand side of the Scheil equation

Cs = C0(1 - fs)^(Keff - 1)

which seems to imply that Cs tends to infinity as the solidification completes. Whereas I would (naively?) expect it to tend to C0 if there is no volume change and no impurities are added or removed during the solidification.

Obviously I have not interpreted it correctly (nature abhors infinities or whatever) so I'd like to know where it "breaks down" and/or where I have gone wrong.

Cheers guys!

4. Oct 15, 2010

### MagnetDave

JDRF85,

What denominator?

-MD

5. Oct 17, 2010

### JDRF85

The exponent (Keff - 1) will always be between zero and -1 (not just less than 1 as I erroneously wrote in my last post). So in the way that 3^(-1/2) = 1/[3^(1/2)] and 3^(-1/7) = 1/[3^(1/7)], the Scheil equation can be expressed as either

Cs/C0 = (1 - fs)^(Keff - 1)

or

Cs/C0 = 1/[(1 - fs)^(1 - Keff)]

They are equivalent. As fs goes to unity, the RHS denominator goes to zero, bringing the LHS to infinity. I suppose this is not an accurate physical description (!) so I am interested in knowing when the validity of the Scheil equation breaks down and/or what has to be modified to keep it accurate.

Johnny

6. Oct 21, 2010

### MagnetDave

fs never gets there. What happens in a real system is that you run into the maximum solubility of the impurity, the drop out as a eutectic.

Scheil assumes that there is no diffusion into the solid (which is not entirely true) and that there is instant diffusion in the liquid (definitely not true). Large amounts of undercooling will hurt you, interstitial impurities in a high-solidus mixtures can hurt you. There have been a lot of refinements since the Scheil equation, but these tend to be very minor for most systems. Basically - Scheil does a nice enough job most the time, and is much easier to use, and needs fewer fudge factors.

7. Oct 21, 2010

### JDRF85

Thanks! Really, really, thankyou so much, because I was just at a loss here... It makes sense now that you've said it, but I just couldn't figure this one out. Cheers MagnetDave!