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## Q. 5.1

Determining the amount of phases in carbon steel
Determine the amount of gamma and alpha phases in a 10-kg, 1040 steel casting as it is being cooled slowly to the following temperatures: (a) 1173 K, (b) 1001 K, and (c) 999 K.

## Verified Solution

(a) Referring to Fig. 5.4b, we draw a vertical line at 0.40% C at 1173 K.
We are in the single-phase austenite region, so percent gamma is 100 (10 kg) and percent alpha is zero. (b) At 1001 K, the alloy is in the two-phase gamma-alpha field. When we draw the phase diagram in greater detail, we can find the weight percentages of each phase by the lever rule:

Percent alpha = $\left\lgroup\frac{C_{\gamma }-C_{0}}{C_{\gamma }-C_{\alpha }} \right\rgroup 100=\left\lgroup\frac{0. 77-0.40}{0.77-0.022} \right\rgroup100 = 50$ %, or 5 kg.

Percent gamma = $\left\lgroup\frac{C_{0 }-C_{\alpha }}{C_{\gamma }-C_{\alpha }} \right\rgroup100=\left\lgroup\frac{0. 40-0.022}{0.77-0.022} \right\rgroup100 = 50$ %, or 5 kg.

(c) At 999 K, the alloy will be in the two-phase alpha and $Fe_{3}$C region. No gamma phase will be present. Again, we use the lever rule to find the amount of alpha present:

Percent alpha =$\left\lgroup\frac{6.67 – 0.40}{6.67 – 0.022} \right\rgroup$×100% =94% , or 9.4 kg.