Thermal Expansion of Aluminium Sleeve over Steel shaft

Click For Summary
SUMMARY

The discussion centers on calculating the necessary temperature for an aluminum sleeve to fit over a steel rod, given their coefficients of thermal expansion. The aluminum sleeve, with an internal diameter of 4.000 cm at 10 degrees Celsius, must be heated to 410 degrees Celsius (683.8 K) to expand sufficiently to fit over the steel rod, which has a diameter of 4.040 cm. The thermal expansion equation used is ΔL = α × L_initial × ΔT, where α for aluminum is 25 x 10^-6. The participants also clarify that temperature differences can be calculated in either Celsius or Kelvin without conversion.

PREREQUISITES
  • Understanding of thermal expansion principles
  • Familiarity with the thermal expansion equation ΔL = α × L_initial × ΔT
  • Knowledge of coefficients of thermal expansion for aluminum and steel
  • Basic algebra for solving equations
NEXT STEPS
  • Research the coefficients of thermal expansion for various materials
  • Learn about the applications of thermal expansion in engineering
  • Explore methods for calculating thermal expansion in composite materials
  • Investigate temperature control techniques for material fitting processes
USEFUL FOR

Mechanical engineers, materials scientists, and students studying thermal dynamics who are interested in the practical applications of thermal expansion in material fitting and assembly processes.

hemetite
Messages
50
Reaction score
0

Homework Statement



An aluminium sleeve of internal diameter 4.000cm at 10 degrees Celsius is heated and slipped over a steel rod of diameter 4.040cm (at 10 degrees Celsius. When the aluminium sleeve and steel rod return to room temp they are impossible to separate. Assume that the coeffiecients of thermal expansion are constant with temperature.

1) To what temperature must the aluminium sleeve be heated before it can slipped over the steel rod?

2) Can the aluminium sleeve and steel rod be separated by heating or cooling form together to some common temperature? If so, what temperature?

Homework Equations



delta L = alpha x L initial x delta temperature


The Attempt at a Solution



This problem should be quite straightforward and simple...but i am not confident of my answer i got for part 1)...care to spot my mistake?

Let the room temperature be 295 K.
At 10 degrees = 273.15 K + 10 = 283.15 K

Firstly i check the Diameter for both aluminum and steel at room temperature.

The aluminum ring have expanded by
delta L = (25x10^-6)(4x10^-2)(295-283.15)
= 1.185 x 10 ^ (-5) m

The diameter of aluminium at room temp
= (4x10^-2) + = 1.185 x 10 ^ (-5)
= 0.04001185 m

Okay i cut short...i just do the same for the steel rod and it is 0.040405744m.

To get the temperature of the aluiminium so that it can be slipped over the rod ..
Diameter of steel - Diameter aluminium = 0.000393894 m

so i put back in the thermal expansion equation

0.000393894 = (25x10^-6)(0.04001185)(Tempearture - 295 K)

solving the equation:
The Temperature the aluminium need to be heated = 683.8 K = 410 degree celsius

I felt that the number is a bit big...any mistakes?
 
Physics news on Phys.org
It looks like you are making this problem unnecessarily complicated. Assume that this is done in a relatively cold room at 10 degrees Celsius. Also, since a difference in temperature is the same regardless of whether it is in Kelvin or Celsius, you do not need to convert back and forth. Just use the thermal expansion equation

ΔL = αAl L ΔT

once to find the temperature change at which the inner diameter of the sleeve matches the diameter of the rod.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
849
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K