Question

At 20∘C, the hole in an aluminum ring is 2.100 cm in diameter. You need to slip this ring over a steel shaft that has a room-temperature diameter of 2.107 cm To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? Coefficients of linear thermal expansion of steel and aluminum are 12×10−6 K−1 and 23×10−6 K−1 respectively.

Answer #1

Given,

T1 = 20 deg C ; r = 2.1 cm ; d = 2.107 cm ; alpha-s = 12 x 10^-6 per K ; alpha-a = 23 x 10^-6 per K

we know that

L = Lo (1 + alpha delta T)

common temerature be T

delta T = T - 20

for aluminimum

d = 2.1 [1 + 23 x 10^-6 (T - 20)]

for steel

d = 2.107 [1 + 12 x 10^-6 (T - 20)]

equating the two

2.1 [1 + 23 x 10^-6 (T - 20)] = 2.107 [1 + 12 x 10^-6 (T - 20)]

2.1 + 48.3 x 10-6 (T - 20) = 2.107 + 25.284 x 10^-6 (T - 20)

48.3 x 10^-6T - 966 x 10^-6 = 0.007 + 25.284 x 10^-6 T - 505.68 x 10^-6

23.016 x 10^-6 T = 0.007 + 460.32 x 10^-6 = 7.46 x 10^-3

T = 7.46 x 10^-3/(23.106 x 10^-6) = 322.859 deg C

**Hence, T = 322.859 deg C**

At 20∘C, the hole in an aluminum ring is 2.200 cm in diameter.
You need to slip this ring over a steel shaft that has a
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To what common temperature should the ring and the shaft be
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of linear thermal expansion of steel and aluminum are
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Express your answer in degrees Celsius to two significant
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