HW_-_Casting_Solutions

HW_-_Casting_Solutions - 10.1. A disk 40 cm in diameter and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
10.1. A disk 40 cm in diameter and 5 cm thick is to be cast of pure aluminum in an open mold casting operation. The melting temperature of aluminum = 660 ° C, and the pouring temperature will be 800 ° C. Assume that the amount of aluminum heated will be 5% more than what is needed to fill the mold cavity. Compute the amount of heat that must be added to the metal to heat it to the pouring temperature, starting from a room temperature of 25 ° C. The heat of fusion of aluminum = 389.3 J/g. Other properties can be obtained from Tables 4.1 and 4.2 in the text. Assume the specific heat has the same value for solid and molten aluminum. Solution : Volume V = π D 2 h /4= π (40) 2 (5)/4 = 6283.2 cm 3 Volume of aluminum to be heated = 6283.2(1.05) = 6597.3 cm 3 From Table 4.1 and 4.2, density ρ = 2.70 g/cm 3 and specific heat C = 0.21 Cal/g- ° C = 0.88 J/g- ° C Heat required = 2.70(6597.3){0.88(660-25) + 389.3 + 0.88(800-660)} = 17,812.71{558.8 + 389.3 + 123.2} = 19,082,756 J 10.2. A sufficient amount of pure copper is to be heated for casting a large plate in an open mold. The plate has dimensions: length = 20 in, width = 10 in, and thickness = 3 in. Compute the amount
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/24/2011 for the course IE 215 taught by Professor Groover during the Fall '08 term at Lehigh University .

Page1 / 2

HW_-_Casting_Solutions - 10.1. A disk 40 cm in diameter and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online