Question

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. Density ρ = 2.70 g/cm3 and specific heat C = 0.88 J/g-°C. Assume the specific heat has the same value for solid and molten aluminium.

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. Density p = 2.70 g/cm3 and specific heat C = 0.88 J/g-°C. Assume the specific heat has the same value for solid and molten aluminium.

Answer 1

Answer:- Page no o Objective 1910 compute the amount of heat that must be added to the melal to heat it to the pouring temperature starting from a room lemperature of 25°C given data, Diameter o dist D = uocm thickness of disk CH = 5cm melling temperature of alluminium Powring temperature (TS)= 800°c (tm)- 660°C wroom temperature (To)- 25°C Heal of fusion of alluminium (L) = 389.3 Ila Density (O) = 2-70 g/cm3. Specific heat CC) -0.88 5/9°c. Assume specifre heat has the same value for solid and molten alluminium Total amount of heat added to alluminium Q = m [c Com-tv) +++c CTS-Tm)] -0 calculating min volume of alluminium (v)= 7 xdext. v= 4 x 40² xs V= 6283.2 CM3

page no actual value will be 5% more than vi, so V = 1.05% V. Vi= 1.05 X 6288.2 cm3 Vi= 6597.34 cm3 .. mass (m)= evi m=(2.70 91cm3) (6597.34 cm3) ma 17812.839 substitute this m value and given values in ean o .::Q = 17812.83 (0.88 (660-25)+ 384.3 + 0.88 (800-600] 8 = 12312.83 0.88 (695) + 384.3 + 0.88 (140)] Q = 17819.83 (558.8 + 389.3 + 123.2] Q = 17812.83 (1071.33 Q19082884.78 J.