How Do You Solve Thermal Expansion Equations Under Time Pressure?

  • Thread starter Thread starter Jayhawk1
  • Start date Start date
  • Tags Tags
    Confused
AI Thread Summary
To solve the thermal expansion problem, first establish the equations for the diameters of the brass plug and iron ring as functions of temperature. The diameter of the brass plug increases according to D_b = 8.755cm(1 + 10x10^-6°C^-1 * ΔT), while the diameter of the iron ring is given by D_i = 8.745cm(1 + 12x10^-6°C^-1 * ΔT). Set these two equations equal to find the temperature at which the diameters match. After simplifying, you can isolate ΔT and calculate the common temperature needed for the plug to fit into the ring. This approach will yield the required temperature change to ensure a proper fit.
Jayhawk1
Messages
44
Reaction score
0
https://www.physicsforums.com/showthread.php?p=543646#post543646

I asked this question earlier and I didn't get any significant help, atleast beyond what I already knew. I do not know how to calculate this, nor do my classmates who I have seen also posed this question to the forum. Please help! I know the concept, but I can not calculate it. I have worked on it for a few days now, and it is due in about an hour.
 
Physics news on Phys.org
You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.
 
Jayhawk1 said:
A brass plug is to be placed in a ring made of iron. At room temperature (20oC), the diameter of the plug is 8.755 cm and that of the inside of the ring is 8.745 cm. They must be brought to what common temperature (in Co) in order to fit? The coefficient of linear expansion for brass is 10x10-6(Co)-1 and that for iron is 12x10-6(Co)-1.

I don't know what to do with this... I know I have to subtract and divide some of these numbers, but I have no clue as to how to do it exactly. I need help please! Thank you.,

Diameter of brass plug as a function of the temperature change is

D_b = D_{b0}(1+10*10^{-6}/C^o*{\Delta}T)

Diameter of iron ring as a function of the temperature change is

D_i = D_{i0}(1+12*10^{-6}/C^o*{\Delta}T)

Set the two diameters equal and solve for the temperature change. Add the change to the starting temperature.
 
HallsofIvy said:
You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.


The diameter of the inside of the ring actually gets larger. If it were a disk, all the linear dimensions would increase. The diameter of the "missing" part of the disk that makes it a ring will expand in proportion to the outer diameter.
 
Thank you but...

Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.
 
Jayhawk1 said:
Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.

D_b = 8.755cm(1+10*10^{-6}/C^o*{\Delta}T)

D_i = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)

8.755cm(1+10*10^{-6}/C^o*{\Delta}T) = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)

8.755cm-8.745cm = 8.745cm(12*10^{-6}/C^o*{\Delta}T) - 8.755cm(10*10^{-6}/C^o*{\Delta}T)

(8.755cm-8.745cm)*10^6C^o = (8.745cm*12 - 8.755cm*10){\Delta}T

Can you finish this?
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

How do you Calculate Ampacity of a Conductor Geometry?
2
Replies
64
Views
5K
Solar Panels and Solar Radiation Flux Density Help - Very Confused
Replies
35
Views
5K
Is the Thermal/Mechanics Problem incoherent?
Replies
8
Views
2K
How Do You Solve Equations with Variables V, U, and T?
Replies
10
Views
1K
Where Should I Ask About Running a Thermal Conductivity Test for Soils?
Replies
2
Views
1K
How Do You Solve Adiabatic Expansion Problems in Thermodynamics?
Replies
5
Views
3K
Equilibrium and Pressure Distribution in a Spherical Body Under Gravity
Replies
5
Views
4K
How do I convert 1.2atm to N/m^2 when solving for gas work?
Replies
8
Views
3K
Positioning Device Based On Thermal Expansion
Replies
2
Views
3K
Archived Finding the thermal energy and Q for an isobaric process.
Replies
1
Views
4K

Hot Threads

Recent Insights

Back
Top