What is the Common Temperature for Expansion of a Steel Ring and Aluminum Rod?

In summary, to fit the steel ring onto the aluminum rod, heat them up to a common temperature, and then cool them both down.
  • #1
BunDa4Th
188
0

Homework Statement



A steel ring with a hole having area of 3.96 cm2 is to be placed on an aluminum rod with cross-sectional area of 4.00 cm2. Both rod and ring are initially at a temperature of 36.0°C. At what common temperature can the steel ring be slipped onto one end of the aluminum rod?

Homework Equations



DeltaA = yA_0DeltaT
T = T_0 + DeltaT
DeltaT = A_a - A_s/(y_sA_s) - (y_aA_a)

The Attempt at a Solution



first i did 4 - 3.96 = .04 which is DeltaA

knowing that alluminum is 24x10^-6 (*C^-1) and steel is 11x10^-6 (*C^-1)
since the steel ring needs to be expand so it can fit in the alluminum rod i did
.04 = (11x10^-6)(3.96)(DeltaT) solve for DeltaT (i get this real huge number which is incorrect most likely)

T = T_0 + DeltaT when input its incorrect and when i try T = T_0 - DeltaT is also incorrect.
 
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  • #2
would i use this formula to solve for the temperature needed

DeltaA = A_i*alpha*(T2 - T1) solve for T2?
 
  • #3
Well, I gave this problem another try and still not understanding this, even after reading the book. Any help would really be helpful because this problem would really help me study for the final i have in 2 weeks. BTW, I tried 20 times and all incorrect...

would i have to find the temprature of both

DeltaA = (alpha_s)(A_s)(DeltaT) solve for deltaT
DeltaA = (alpha_a)(A_a)(DeltaT) solve for deltaT

with those two i would do DeltaT_a - DeltaT_s? then T = T_0 - DeltaT?
 
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  • #4
BunDa4Th said:
Well, I gave this problem another try and still not understanding this, even after reading the book. Any help would really be helpful because this problem would really help me study for the final i have in 2 weeks. BTW, I tried 20 times and all incorrect...

would i have to find the temprature of both

DeltaA = (alpha_s)(A_s)(DeltaT) solve for deltaT
DeltaA = (alpha_a)(A_a)(DeltaT) solve for deltaT

with those two i would do DeltaT_a - DeltaT_s? then T = T_0 - DeltaT?

If alpha represents a linear expansion coefficient then the your DeltaA equations are not correct. You might think about why, but my next comment will give you a big hint. Maybe you should try thinking in terms of the diameter changes needed to make the ring fit.
 
  • #5
Okay I don't understand why it is incorrect. Wouldn't it need a .04 cm^2 in change to allow the steel ring to fit? Also i got most of those formula and average coefficient of linear expansion from the book i read and from an example i looked at.
 
  • #6
BunDa4Th said:
Okay I don't understand why it is incorrect. Wouldn't it need a .04 cm^2 in change to allow the steel ring to fit? Also i got most of those formula and average coefficient of linear expansion from the book i read and from an example i looked at.

What you need is for the areas to become equal. You are heating both things, so they are both going to expand. So no, you are not looking for a .04 cm² change in one of the areas. You are looking for the changes to each area that will make them equal to each other.

The problem with using linear coefficients of expansion to find the change in area is that area is NOT a linear quantity. You need to either come up with the area coefficient of expansion (which you can do by thinking of the expansion of the area of a flat plate in terms of its linear expansions), or you need to compare a linear dimension of the two objects (the diameter or radius will do nicely).
 
  • #7
okay so what i need to do is heat up the steel ring to get the area of 4 cm^2 to fit at the end of the aluminum rod. Is that correct?

to get deltaA it is DeltaA = A - A_0. To get A it is A = L^2 where

L = L_0 + alpha*L_0DeltaT.

Now if that is the correct way to do it I am much more confuse. as to what L is and I don't understand what do you mean by area coefficient of expansion or copmare linear dimension.

(sorry i took to long to reply, I was working on my design project and trying to find a digital camera battery which only Samsung uses so it was hard to find.)
 
  • #8
BunDa4Th said:
okay so what i need to do is heat up the steel ring to get the area of 4 cm^2 to fit at the end of the aluminum rod. Is that correct?
No. At the initial temperature, the steel ring has a smaller diameter than the aluminum rod. Heat them up, both will expand; Cool them down, both with contract. (Compare their expansion coefficients to determine which expands or contracts more.)

As OlderDan explained, you need to find what final temperature will make them the same size so that one can fit into the other. Find expressions for the size of each as a function of temperature; set those expressions equal and solve for the temperature difference.

Again, as OlderDan explained, since you have linear coefficients of expansion, it may be easier to work with linear dimensions like diameters instead of areas. (Treat the rod and hole as circles--find their initial diameters.)

... I don't understand what do you mean by area coefficient of expansion or copmare linear dimension.
This may help: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thexp.html"
 
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  • #9
Okay, I will give this a try again later tonight when i finish up my final project.

Thanks for all the explaining and Link.
 
  • #10
Well, I gave this problem another try and got lost on where to actually start first then i know i had to set two equation to equal each other so i thought

2alph_aDeltaT = 2alpha_sDeltaT but realize that the deltaT will cancel each other.

so I thought maybe i had to place the initial temperature into the equation but was not sure how since when solving it i will get the same result where deltaT is equal to 1 again.

I am to the point where i just want to give up on this problem and hope it never show up again...
 
  • #11
BunDa4Th said:
Well, I gave this problem another try and got lost on where to actually start first then i know i had to set two equation to equal each other so i thought

2alph_aDeltaT = 2alpha_sDeltaT but realize that the deltaT will cancel each other.
You need to compare the size, not just the change in size. Your equations should look like:
New Size = Original Size + Change in size

Get two equations like that and set them equal.
 
  • #12
okay if i did this correctly A =pir^2

steel ring diameter: 2.245
aluminum diameter: 2.257

L_0 +alphaL_0DeltaT = L_0 + alphaL_0DeltaT

Sorry if this is not correct or what you are trying to tell me but this chapter is just really confusing. I am just having a really hard time understanding this chapter.

steel = aluminum
 
  • #13
That looks good to me. Now you can solve for Delta T.
 
  • #14
Okay i solved for deltaT i got -454.80 which is incorrect because that is way to high. Did i do a mistake in the calculation? or is there a step i am missing?
 
  • #15
Okay, nevermind I figure out what i did wrong and got the answer. I also figure out how to solve it another way.

DeltaT = (A_s - A_a)/((A_a)(2alpha_a) - (A_s)(2alpha_s))

= -.04/(.000192 - .00008712)

DeltaT = -381
T = T_i + DeltaT T= -345

thanks for explaining everything and helping me figure out how to solve this equation.
 

1. What is thermal expansion of solids?

Thermal expansion of solids is the phenomenon where a solid material expands in size when it is heated and contracts when it is cooled. This is due to the increase or decrease in the average distance between the atoms or molecules that make up the solid.

2. Why do solids expand when heated?

When a solid is heated, the atoms or molecules within the material gain energy and vibrate faster. This increase in vibration causes the atoms to take up more space, resulting in the expansion of the solid.

3. What factors affect the amount of thermal expansion in a solid?

The amount of thermal expansion in a solid is affected by the type of material, its initial temperature, and the temperature change. Different materials have different coefficients of thermal expansion, which determine how much they will expand or contract when heated or cooled. Higher initial temperatures and larger changes in temperature also result in greater thermal expansion.

4. How is thermal expansion of solids measured?

Thermal expansion is typically measured using a device called a dilatometer, which measures the change in length or volume of a solid as it is heated or cooled. The coefficient of thermal expansion can then be calculated by dividing the change in length or volume by the initial length or volume and the change in temperature.

5. What are some practical applications of thermal expansion of solids?

Thermal expansion is an important phenomenon to consider in many practical applications, such as building construction, where materials need to be able to withstand changes in temperature without causing damage. It is also used in the design of engines and machines, as well as in the production of glass and ceramics. Understanding thermal expansion is also crucial in fields such as materials science and engineering.

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