Casting a metal (Enthelpy of Fusion Question)

AI Thread Summary
The discussion focuses on calculating the final temperatures of aluminum and a steel mold during the cooling process of molten aluminum poured into the mold. The user initially set up equations for heat transfer but realized that the cooling of aluminum should be calculated from its melting point of 660°C rather than from its initial temperature of 700°C. After reworking the equations, the user found that the final temperature (Tf) for both the aluminum and steel mold is approximately 94.77°C when using Excel's goal seek function. Additionally, there is a cautionary note about the potential reactions between molten aluminum and stainless steel, even at lower temperatures. The conversation emphasizes the importance of accurate heat transfer calculations in metal casting.
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I want to take a liquid metal, Aluminum at 700oC and pour it into a steel mold starting at 20oC. For the purpose of the question let’s say that the heat transfer is 100% with no losses to the surroundings.

Here are some numbers:

Al
Mass = 200 g = 0.2 Kg
Ti, Al = 700oC
Cp = 900 J/Kg-K
ΔHf = 10.67 kJ/mol
Melting point of Al = 660oC

Steel
Mass = 3Kg
Ti, st = 20oC

So the liquid aluminum will cool once it hits 660, but the steel mold should never change state due to its higher melting point.

At this point I want to find the final temperatures of the Aluminum and the Steel mold.

Here's what I have so far can someone please check my theory/number placement:

(900 J/Kg-K)(0.2Kg)(700oC-660oC)+(200g)(1 mol/26.96g) + (900 J/Kg-K)(0.2Kg)(Tf, Al - 20oC)

= (486 J/Kg-K)(3Kg)(Tf, St - 20oC)

So my guess is both final temperatures will be the same, I just wanted to know if this work looks correct or if I need to change/move some numbers.
 
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This part (900 J/Kg-K)(0.2Kg)(Tf, Al - 20oC) is not correct. One is still cooling from Tmelt to Tf. The Al decreases in temperature while the Fe heats up.

At equilibrium Tf(Al) = Tf(Fe)
 
That's actually why I posted this, I had a feeling that using the 20oC on the Aluminum side didn't make sense, so I think the left hand side should be changed to (900 J/Kg-K)(0.2Kg)(Tf, Al - 660oC).

Is that correct, or am I still missing something else?
 
Try (660°C-Tf(Al)). The Al is cooling from 660°C to Tf, 660°C > Tf.
 
Ok, I reworked it again let me know what you link:

LHS (Aluminum)
(900 J/Kg-K)(0.2Kg)(700oC-660oC)+(200g)(1 mol/26.96g)(10.67 kJ/mol)+(900J/Kg-K)(0.2Kg)(660oC - Tf)

RHS (Steel)
(486 J/Kg-K)(3Kg)(Tf - 20oC)

I took the sum of the LHS and the RHS and put them both into excel

I set Tf as my variable on both sides of the equation and used goal seek to find Tf = 94.77oC

With that LHS=RHS=109019.9
 
It is my experience that molten Al can do nasty things to Stainless. Even though you are well below SS melting point reactions can still occur. The fact that your steel is cool should help.

Good luck and dry molds.
 
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