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A 0.6 percent Carbon hypoeutectoid plain-carbon steel is slowly cooled from 950 oC to a temperature just slightly below 723 oC. Calculate the weight percent proeutectoid ferrite in the steel. Please do not include any numbers after the decimal and give the percentage as "xx" and not "0.xx"

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zuwairahdahir

Answer:

Wt of austenit=80%

Wt of preoceutectoid ferrite=19%

Explanation:

The weight percent austenite is calculated from the ratio of the segment of the tie line to the left of the 0.65 percent C to the entire length of the tie line.

Wt of austenit=(0.65 -0.02) / (0.08 -0.02 ) x 100% =0.63 /0.78 x 100% =80%

The weight percent proeutectoid ferrite is calculated from the ratio of the segment of the tie line to the right of the 0.65 percent C to the entire length of the tie line.

0.80 0.65 0.15 Wt % proeutectoid ferrite 100% 100%

Wt of preoceutectoid ferrite=(0.85 -0.65 ) / (0.08 -0.02 ) x 100% =0.15 /0.78 x 100% =19%

lazicah

Answer:

19%

Explanation:

0.2C-----------------0.6C-------------0.8C

weight percent proeutectoid ferrite is calculated from the ratio of the segment of the tie line to the right of the 0.6 percent Carbon to the entire length of the tie line.

Wt % of proeutectoid ferrite = (0.8 - 0.6)/(0.8 - 0.2) ×100% = 19%

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