Thermal Expansion of a circular steel plate

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Homework Help Overview

The discussion revolves around the thermal expansion of a circular steel plate, specifically examining how its area changes when cooled from 350°C to 20°C. The original poster seeks to determine the percentage decrease in the plate's area based on the given temperature change and the coefficient of thermal expansion for steel.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for calculating the final area of the plate and question the validity of the approximation used. There are inquiries about the correct interpretation of the temperature change and whether the linear or volume coefficient of expansion should be applied.

Discussion Status

Some participants have provided guidance on checking arithmetic and clarifying the formula used for percentage change. There is an ongoing exploration of the calculations, with multiple interpretations of the results being considered. The original poster has acknowledged the feedback and is working through the calculations.

Contextual Notes

There is a noted confusion regarding the signs of temperature change and the implications for the area calculation. The original poster's initial arithmetic appears to have led to an incorrect conclusion about the area change.

Aoiumi
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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##
 
Last edited:
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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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