Material 7
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The document discusses the composition and phases of a lead-tin alloy at eutectic composition after the eutectic reaction. It determines that the alpha phase comprises 45% of the alloy and is 19% tin, while the beta phase comprises 55% of the alloy and is 97% tin. It then calculates that for 200g of alloy, the alpha phase mass is 90g and the beta phase mass is 110g. It further calculates the masses of lead and tin in each phase. The document also contains information about determining the phases and amounts of cementite, austenite, and alpha in a steel alloy.
Material 7
- 1. SHEET 3 ENG. KAREEM. H. MOKHTAR
- 2. QUESTION 2 • Determine the amount and composition of each phase in 200 g of a lead-tin alloy of eutectic composition immediately after the eutectic reaction has been completed. (b) Calculate the mass of phases present. (c) Calculate the masses of lead and tin in each phase.
- 3. DETERMINE THE AMOUNT AND COMPOSITION OF EACH PHASE • Composition • ⍺ = 19% Sn • B = 97% Sn • Amount • ⍺ = 97−62 97−19 = 45 % • B = 62−19 97−19 = 55%
- 4. B) CALCULATE THE MASS OF PHASES PRESENT • Mass of ⍺ = total mass x percent of ⍺ =200 x 0.45= 90 g • Mass of B = total mass x percent of B =200 x 0.55= 110 g
- 5. CALCULATE THE MASSES OF LEAD AND TIN IN EACH PHASE. • Mass of Sn in ⍺ = 0.19 x 90 = 17.1 g • Mass of Sn in B = 0.97 x 110 = 106.7 g • Mass of Pb in ⍺ = 90 - 17.1 = 72.9 g • Mass of Pb in B = 110 – 106.7 =3.3 g What we have calculated in (a) Composition ⍺ = 19% Sn B = 97% Sn Amount ⍺ = 97−62 97−19 = 45 % B = 62−19 97−19 = 55% What we have calculated in (B) Mass of ⍺ = total mass x percent of ⍺ =200 x 0.45= 90 g Mass of B = total mass x percent of B =200 x 0.55= 110
- 6. Q3) DRAW SCHEMATIC PHASE DIAGRAMS FOR BINARY SYSTEMS WITH • a) complete liquid and solid solubility b) complete liquid but zero solid solubility, c) complete liquid and limited solid solubility.
- 7. Q4) FOR A 99.6 WT% FE-0.40 WT% C AT A TEMPERATURE JUST BELOW THE EUTECTOID, DETERMINE THE FOLLOWING a) composition of Fe3C (cementite) and alpha b) the amount of cementite in grams that forms per 100 g of steel c) the amount of austinite and alpha at proeutectoid Fe3C(cementite) 1600 1400 1200 1000 800 600 400 0 1 2 3 4 5 6 6.7 L g (austenite) g+L g + Fe3C a+ Fe3C L+Fe3C d Co, wt% C 1148°C T(°C) 727°C
- 8. a) composition of Fe3C and alpha CO = 0.40 wt% C Ca = 0.022 wt% C CFe C = 6.70 wt% C 3 Fe3C(cementite) 1600 1400 1200 1000 800 600 400 0 1 2 3 4 5 6 6.7 L g (austenite) g+L g + Fe3C a+ Fe3C L+Fe3C d Co, wt% C 1148°C T(°C) 727°C CO R S CFe C 3 Ca SOLUTION
- 9. b)THE AMOUNT OF CEMENTITE IN GRAMS THAT FORMS PER 100 G OF STEEL b) Amount of cementite= mass of alloy x percent of cementite Fe3C(cementite) 1600 1400 1200 1000 800 600 400 0 1 2 3 4 5 6 6.7 L g (austenite) g+L g + Fe3C a+ Fe3C L+Fe3C d Co, wt% C 1148°C T(°C) 727°C R S CFe C 3 Calpha
- 10. a) the amount of Austinite and alpha at proeutectoid Co = 0.40 wt% C Ca = 0.022 wt% C Caus = 0.76 wt% C austinite= 51.2 g alpha= 48.8 g Fe3C(cementite) 1600 1400 1200 1000 800 600 400 0 1 2 3 4 5 6 6.7 L g (austenite) g+L g + Fe3C a+ Fe3C L+Fe3C d Co, wt% C 1148°C T(°C) 727°C CO R S CausCalpha
- 11. TERM PAPER
- 12. TERM PAPER • Aim • To use the knowledge you have learned to solve real life problems • Search for certain information online and in books
- 13. TERM PAPER • Your duty • Submit the term paper before the due date (Hardcopy- 4/12) • Late submission but at the same week looses 30%
- 14. GRADING CRITERIA • Technical writing 3 marks • Correct procedure 1.5 marks • Correct assumptions and correct references 1.5 marks • Correct answers 4 marks