Contents

2.1. CTE Calculation

The coefficient of thermal expansion is a material property that describes how much a material’s length changes when it’s temperature changes. The abreviation for coefficient of thermal expansion is CTE and the greek symbol alpha \(\alpha\) is often used for CTE in equations.

The formal definition of CTE is below:

\[ \frac{\Delta L}{L} = \alpha \Delta T \]

Below are two problems involving the coefficient of thermal expansion that are solved with code.

Problem 1: Cooling a Copper Bearing

Given:

A Copper bearing 80.10 mm at 20 *C needs to fit into a hole 80.00 mm. Assume the coefficient of thermal expansion of copper is 17 x 10^-6 /*C.

Find:

What temperature does the copper bearing need be cooled down to in order to fit inside the hole? If clearence is included, note it in your answer.

Solution:

Create a function called cl(). If 0.02 mm of clearence is assumed, then running cl(0.02) will calculate the temperature the copper bearing needs to be cooled to in order to fit in the 80.00 mm hole with 0.02 mm of clearence. The function definition is below. The fuction accepts an input argument for clearance. The default clearance is 0.

def cl(clnc=0):
    L0 = 80
    T2 = 20
    dL = 0.1 + clnc
    alpha = 17e-6
    T1 = T2 - ( (dL) / (alpha*L0))
    T1 = round(T1,2)
    return(T1)

We can call our function to answer the problem

Assuming no clearance, the copper bearing needs to be cooled to:

cl()

-53.53

Assuming 0.02 mm of clearance, the copper bearing needs to be cooled to:

cl(0.02)

-68.24

Assuming 0.01 mm of clearance, the copper bearing needs to be cooled to:

cl(0.01)

-60.88

Problem 2: Heating an aluminum case

Given:

At room temperature (20 °C), copper bearing with an outside diameter of 80.10 mm does not fit into an aluminum case with a hole diameter of 80.00 mm. In order for this to be accomplished, a furnace or hotplate can be used to heat the case.

Find:

What temperature must the case be heated up to in order for the bearing to fit in the case?

Assumptions:

Assume the bearing stays at room temperature. If assuming a clearance, note the clearance amount as part of your solution on your answer sheet below the answer.

Assume the coefficient of thermal expansion of aluminum is 24 × 10-6 °C-1 and the coefficient of thermal expasion of copper is 17 × 10-6 °C-1)

Solution

We’ll write a similar function to the function we created in Problem 1. Except for our Problem 2 function, we’ll solve for the higher temperature T2. The function definition is below.

def cl(clnc=0):
    L0 = 80
    #T2 = ?
    T1 = 20
    dL = 0.1 + clnc
    alpha = 24e-6
    T2 = T1 + ( (dL) / (alpha*L0))
    T2 = round(T2,2)
    return(T2)

Assuming no clearence, the case needs to be heated up to:

cl()

72.08

Assuming 0.05 mm of clearence, the case needs to be heated up to:

cl(0.05)

98.13
2. Atomic Structure and Bonding 2.2. Bond Energy vs. Distance