Solution to Homework of Chapter 11

26 gas porosity can be eliminated by preventing the

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stronger the part. 26. Gas porosity can be eliminated by preventing the gas from initially dissolving in the molten metal, using such techniques as vacuum melting, controlled atmospheres, flux blankets, low superheats, and careful handling and pouring. In addition, dissolved gases can be removed by vacuum 29. As molten material temperature increases, the metal fluidity increases, allowing the material possibly to flow into small spaces between sand grains. Upon solidification, sand grains embedded in the surface of the casting. Chemical activity rate also increases and mold reactions are accelerated leading to reactions which may alter casting material structure and properties. 45. Casting patterns generally incorporate several types of modifications or allowances. These include shrinkage allowances to compensate for thermal contraction, draft to permit pattern removal, machining allowances, distortion allowance, and compensation for thermal changes in mold dimensions. Problems: 1. Plate dimension: 2” x 4” x 6”; H/D = 1.5; n =2 t riser = 1.25 t casting ( V/A ) 2 riser = 1.25(V/A ) 2 casting ( V/A )riser = 1.12 ( V/A )casting { ( π /4) D 2 H } / { 2( π D 2 /4 ) + π DH } = { ( 1.15 )(2 x 4 x 6 ) } / {2(2x4) + 2(2x6) + 2(4x6)} substituting H = 1.5 D 1.5 D 3 / 8 D 2 = 1.12 ( 0.545 ) 3 D / 16 = 0.6104
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D = 3.26 in H = 1.5 D = 4.88 in Vriser = (π/4) D 2 H = 40.73 in3 Yield = Vol casting / ( Volcasting + Vriser ) = 48 / ( 48 + 40.73 ) = 54% 3. For the 3" x 5" x 10" solid t s = B (V/A) 2 11.5= B( 3 x 5 x 10) 2 / { 2 (3x5) + 2 (3x10) + 2 (5x10)] 2 11.5 = B (150) 2 / (30 + 60 + 100) 2 = B (150 / 190) 2 = B (0.789) 2 = 0.623 B B = 11.5 / 0.623 = 18.46 For a casting of 0.5" x 8" x 8" cast under the same conditions: t s = 18.46 x (.5x8x8) 2 / [2 (.5x8) + 2(.5x8) + 2(8x8)] 2 = 18.46 x (32) 2 / [ 8 + 8 + 128] 2 = 0.91 min
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