Question

A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.

Answer 1

Here,
Coefficient of linear expansion of steel (a) = 1.2 × 10^-5 /K
outer diameter ( D1) = 8.70 cm
Inner diameter ( D2) = 8.69 cm
T1 = 27 + 273 = 300 K
T2 = ?
Change in due to cooling ,
D2 = D1 ( 1 + a ∆T )
8.69 = 8.70 [ 1 + 1.2 × 10^-5 × ( T2 - 300)]
(8.69 - 8.70) = 8.7 × 1.2 × 10^-5 × (T2 - 300)
T2 - 300 = -0.01/8.7 × 1.2 × 10^-5
T2 = 300 - 95.78
= 204.22 K
in clecius ,
C = 273 - 204.22 = 68.78°C ≈ 60°C

Answer 2

Answer:

Diameter of the shaft(L1)= 8.7×10^(-2)

Diameter of the central hole(L2)=8.69 ×10^(-2)

Temperature at which they form= 27°

In kelvin(T1)=273+27 =300 K

Let T2 be the temperature at which the wheel slip on to the shaft when it cooled.

Change in temperature=T1-T2 (here T1>T2)

Coefficient of linear expansion of steel(a) = 1.2×10^(-5) K^-1

we know,

a=(L1-L2)/L1×(T1-T2)

T1-T2 =(L1-L2)/L1×a

300-T2= (8.7-8.69)×10^(-2)/8.7×10^(-2)×1.2×10^(-5)

by solving it we get

T2 = 204.21

in celsius, T2= -68.78°

hence at temperature -68.78° the wheel slip on to the shaft.