diagrama fe3c - probleme

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Calculati cantitatea de faze si constituenti din aliajul Fe-C dat: Def.: cantitatea procentuala a uneia dintre faze se obtine raportand segmentul opus (in raport cu compozitia fazei respective) la suma segmentelor. OTELURI 0.002 – 0.77 %C – otel hipoeutectoid ⇒ A ∈ (0.002 – 0.77) – Fα + Ce + P 0.77 %C – otel eutectoid 0.77 – 2.11 %C – otel hipereutectoid ⇒ B ∈ (0.77 – 2.11) – P + Ce FONTE 2.11 – 4.3 %C – fonta hipoeutectica ⇒ C ∈ (2.11 – 4.3) – P + Ce + Le 4.3 %C – fonta eutectica

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Page 1: Diagrama Fe3C - Probleme

Calculati cantitatea de faze si constituenti din aliajul Fe-C dat:

Def.: cantitatea procentuala a uneia dintre faze se obtine raportand segmentul opus (in raport cu compozitia fazei respective) la suma segmentelor.

OTELURI0.002 – 0.77 %C – otel hipoeutectoid ⇒ A ∈ (0.002 – 0.77) – Fα + Ce + P0.77 %C – otel eutectoid0.77 – 2.11 %C – otel hipereutectoid ⇒ B ∈ (0.77 – 2.11) – P + CeFONTE2.11 – 4.3 %C – fonta hipoeutectica ⇒ C ∈ (2.11 – 4.3) – P + Ce + Le4.3 %C – fonta eutectica4.3 – 6.67 %C – fonta hiperutectica ⇒ D ∈ (4.3 – 6.67) – Le + Ce

1. Otel hipoeutectoidA ∈ (0.002 – 0.77) – Fα + Ce + P

Q Fα = 6.67−A

6.67−0.002 · 100%

QCe = A−0.002

6.67−0.002 · 100%

Page 2: Diagrama Fe3C - Probleme

QP = A−0.002

0.77−0.002 · 100%

2. Otel hipereutectoidB ∈ (0.77 – 2.11) – P + Ce

QP = 6.67−B

6 .6 7−0.77 · 100%

QCe = B−0.77

6.67−0.77 · 100%

3. Fonta hipoeutecticaC ∈ (2.11 – 4.3) – P + Ce + Le

QA = 4.3−C

4.3 –2 .11 · 100%

100 % A ………………….21% CeQA ………………………… X = QCe

QP = 4.3−C

4.3−0.77 · 100%

QLe = C−0. 77

4.3−0.77 · 100%

4. Fonta hiperutecticaD ∈ (4.3 – 6.67) – Le + Ce

QCe = D – 4.3

6.67−4.3 · 100%

QLe = 6.67−D

6.67−4.3 · 100%