Thermal Expansion of a circular steel plate

In summary: Having the equation in front of you makes it a lot easier to see where you might have made a mistake.
  • #1
Aoiumi
23
0

Homework Statement


A circular steel plate of radius 15 cm is cooled from 350 C to 20 C. By what percentage does the plates area decorate ?


Homework Equations



A=∏r^2
Af = Ai (1+2∂ΔT)
specific heat of steel = 12 x 10^-6

The Attempt at a Solution


r = 15 cm = .15 m
Ai = .070685 m^2

Af = Ai (1+2∂ΔT)
= (.070685m^2)(1 + 2 (12 x 10^-6)(330 C))
= 23.3 m^2?

What am I doing wrong? The final area does not look right at all. Thank you.
 
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  • #2
Af = Ai (1+2∂ΔT)

This would be an approximate form. Is the approximation valid for this problem?

Did you use the linear or volume coefficient for ∂?
Where did you get the value from?

If the temperature decreased, is ΔT positive or negative?
 
  • #3
= (.070685m^2)(1 + 2 (12 x 10^-6)(330 C))
It should be
= (.070685m^2)[(1 + 2 (12 x 10^-6)(330 C)]
Now try.
 
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  • #4
rl.bhat said:
= (.070685m^2) [ (1 + 2 (12 x 10^-6)(330 C) ]
... you've got one extra parenthesis.

"(.070685m^2) ( 1 + 2 (12 x 10^-6)(330 C) )" is fine as it is written.

But (.070685m^2)(1 + 2 (12 x 10^-6)(330 C)) ≠ 23.3m^2 ... well spotted.

... so I should add "check your arithmetic" to my list :)
 
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  • #5
Yes. You are right.
 
  • #6
Thank you for your help!

So, Af = .0712 m^2
Then percent decrease should be (.0712 - .070685)/.070685 or 72%. Does this make sense?
 
  • #7
Then percent decrease should be (.0712 - .070685)/.070685 or 72%. Does this make sense?
This is not correct.
1. (.0712 - .070685)/.070685 ≠ 0.72

2.
The percentage change in area A would be: $$p=100\frac{A_f-A_i}{A_i}$$ ... you swapped final and initial over.
If ##A_i > A_f## then the negative percentage tells you the change was a decrease.

Go back to your original calculation - is your final area bigger or smaller than the initial area?
Now compare with the question: should the plate be bigger or smaller after cooling?

Hint: ##\Delta T = T_f-T_i##
 
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  • #8
That helped. Thank you!
 
  • #9
No worries.

For the future: it is best practice to do all the algebra before you plug in the numbers.

Considering you have the same trouble with arithmetic that I do, I figure you'd do better to do what I do and avoid arithmetic as long as possible.

For example - the final answer you want is a percentage, so derive the equation for the percentage first: $$\begin{align}A_\% &= 100\frac{A_f-A_i}{A_i}\\ &=100\frac{A_i\big(1+2\alpha\Delta T\big)-A_i}{A_i}\\
&=200\alpha\Delta T\end{align}$$ .. so you could have avoided a bit of work.

I know the algebra looks more intimidating than the numbers, but it is much easier to troubleshoot.
 

1. What is thermal expansion?

Thermal expansion is the tendency of a material to increase in size or volume when exposed to an increase in temperature.

2. How does thermal expansion affect a circular steel plate?

When a circular steel plate is heated, it will expand in all directions, causing an increase in both diameter and thickness.

3. What causes thermal expansion in a circular steel plate?

The movement of atoms within the steel plate increases when the temperature rises, leading to an expansion of the material.

4. How is the amount of thermal expansion in a circular steel plate calculated?

The amount of thermal expansion in a circular steel plate can be calculated using the coefficient of thermal expansion, the original dimensions of the plate, and the change in temperature.

5. Can thermal expansion cause damage to a circular steel plate?

If the circular steel plate is not able to freely expand and contract, it may experience stress and potentially lead to damage such as warping or cracking. Proper design and installation can help prevent this.

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