It is desired to slip an aluminum ring over a steel rod. At 10.0^C the inside di

Question

It is desired to slip an aluminum ring over a steel rod. At 10.0^C the inside diameter of the ring is 4.00 cm and the diameter of the steel rod is 4.040 cm.

a) In order for the ring to slip over the rod, should the ring be heated or cooled?
b) Find the temperature of the ring at which it fits over the rod. The rod remains at 10.0^C

Once the aluminum ring is slipped over the steel rod, the ring and rod are allowed to equilibrate at a temperature of 22^C. The ring is now stuck on the rod.

c) If the temperature of both the ring and the rod are changed together, should the system be heated or cooled to remove the ring?
d) Find the temperature at which the ring can be removed.

Use the Coefficient of linear expansion for steel as 1.2 x 10 -5 K-1 and for aluminum as 2.4 x 10 -5 K-1

Explanation / Answer

   the ring should be heated    for simplicity let us assume that the ringis cut and unrolled then it would be a rectangle    if the rectangle is heated it would expandalong its length and width    its length is the circumference of thering    as the length of the rectangle increases thecircumference also increases and so does its diameter (b) Diameter of the ring do=4 cm Diameter of the steel rod d=4.04 cm Increase in length d=d- do=0.04 cm Temperature  To=273+10=373    as L = C = d    L = C          = d     d = (do) T    then the temperature will be    T = To + (d / do)       = 373+(0.04 /  2.4 x 10 -5*4)       =789.6 K      =789.6-273     =  516.6.oC C) system is cooled to remove AL ring from the steel rod d) then the temperature will be    T = To- (d / do)       = 373-0.04 /  2.4 x 10 -5*4)       =-43.6 K      =-43.6-273     =  -316.6oC       =789.6 K      =789.6-273     =  516.6.oC C) system is cooled to remove AL ring from the steel rod d) then the temperature will be    T = To- (d / do)       = 373-0.04 /  2.4 x 10 -5*4)       =-43.6 K      =-43.6-273     =  -316.6oC    T = To- (d / do)       = 373-0.04 /  2.4 x 10 -5*4)       =-43.6 K      =-43.6-273     =  -316.6oC       =-43.6 K      =-43.6-273     =  -316.6oC

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