Thermal Expansion of a circular steel plate

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SUMMARY

The discussion focuses on calculating the percentage decrease in the area of a circular steel plate when cooled from 350°C to 20°C. The initial area (Ai) is calculated using the formula A=∏r², resulting in Ai = 0.070685 m². The final area (Af) is derived using the thermal expansion formula Af = Ai (1 + 2∂ΔT), where ∂ is the coefficient of linear expansion for steel (12 x 10^-6). The correct percentage decrease in area is determined to be approximately 0.72%, following the formula p = 100(Af - Ai)/Ai.

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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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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?
 
= (.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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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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Yes. You are right.
 
Thank you for your help!

So, Af = .0712 m^2
Then percent decrease should be (.0712 - .070685)/.070685 or 72%. Does this make sense?
 
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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That helped. Thank you!
 
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.
 

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