# Linear expansion in a brass plug and steel ring

1. Aug 23, 2010

### rhombus

1. The problem statement, all variables and given/known data

"A round plug made of brass has a diameter of 87.53 mm at 20 C. The plug is to be fitted to a steel ring of inside diameter 87.43 mm. To what common temperature must they be brought in order to fit?"

The linear coefficient of expansion for brass:
$$\alpha_{b} = 18.7 \cdot 10^{-6} {\frac{\texttt{m}}{\texttt{m} \cdot \texttt{K}}$$

and steel:
$$\alpha_{s} = 13 \cdot 10^{-6} {\frac{\texttt{m}}{\texttt{m} \cdot \texttt{K}}$$

Unknowns: Delta T, L sub F (final L) for both brass and steel.

2. Relevant equations

As we are seeking a diameter, I assumed that the linear expansion equation would be sufficient:
$$\Delta L = \alpha \cdot \Delta T \cdot L_{0}$$

3. The attempt at a solution

I interpret this as a system of linear equations. Delta T is what is sought; L final is also unknown. So we have two equations with two unknowns.

Taking the original linear expansion equation for brass:
$$L_{b} - L_{f} = \alpha_{b} \cdot \Delta T \cdot L_{b}$$

I transform it algebraically to make L final the subject:

$$L_{f} = L_{b} - \alpha_{b} \cdot \Delta T \cdot L_{b}$$

I do the same for the steel ring and make the two equations equal each other, then I solve for delta T:

$$\Delta T = \frac{L_{s}-L_{b}}{\alpha_{s} \cdot L_{s} - \alpha_{b} \cdot L_{b}}$$

I substitute in my values, converting everything to base units (metres from millimetres), then cancelling units, and my result is

$$\Delta T = 199.91 \texttt{K}$$

Now, here is the rub - the answer provided in the assignment is -123 C. If I take 20 C and subtract 199, I get -179 C. These answers vary widely, but I cannot see what I have done wrong here. The coefficients were not the ones provided with the assignment, but came out of a table http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html" [Broken], which the instructor has used as a source in the past.

What have I done wrong here? If the work looks good, how likely is that this discrepancy would be caused by the linear expansion coefficients used?

PS: The LaTeX renderer in the post previewer is broken, so forgive me if the first example of the linear expansion equation is actually the linear expansion coefficient for steel. That is not what is in my source code.

Last edited by a moderator: May 4, 2017
2. Aug 24, 2010

### rl.bhat

For steel you have to use the linear coefficient of expansion (αs) where as for brass you have to use the volume coefficient of expansion which is equal to 3*(αb).

3. Aug 24, 2010

### rhombus

Why? We are measuring a diameter in both cases, which is linear.

4. Aug 25, 2010

### rhombus

Anything else to add? How does my work look?

5. Aug 25, 2010

### nasu

The result is quite sensitive to the values of the coefficients.
Using the values from this table,

http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion

(i.e. in units of 10^(-6), 19 for brass and 11 for steel) I find 122.6 degrees.

With your values (18.7 and 13), which may not be wrong after all, I've got your original result. It may be just a matter of parameters' values. Consider also the fact that for steel they provide a range of values.

6. Aug 25, 2010

### rhombus

Okay -- that is reassuring. It is probably just a question of which coefficients were used - your result is not far from result given. Thanks!