A brass plug is to be placed in a ring made of iron. A 20 oC the diameter of the

Question

A brass plug is to be placed in a ring made of iron. A 20oC the diameter of the plug is 8.753 cm and that of the inside of the ring is 8.753 cm. They must both be brought to what common temp in order to fit?

brass = 19E-6 C0-1

iron = 12E-6 C0-1

The example states that since this is a linear expansion problem, to fit the plug in, the d needs to be the same for both. SO in class we said

dplug = dring and used

(BRASS) Lo(1+T)=Lo(1+T) (IRON)

The answer we got was -163 via this problem. Then T new is -143.

I tried solving the problem Lo(1+T)=Lo(1+T) and cannot get a proper answer. Can someone show me the step by step algebra? thank you!

how do you go from the first line to the second*

Explanation / Answer

D = Do*(1 + c*dT)

Do = Initial diameter
dD = change in diameter
c = coefficient of thermal expansion
dT = change in temperature w.r.t. the initial 20o C


Diameter of ring: Dr = Dro*(1 + ci*dT)
Diameter of plug: Dp = Dpo*(1 + cb*dT)

Dro = initial diameter of ring = 8.743 cm
Dpo = initial diameter of plug = 8.753 cm
ci = coefficient for iron = 12 x 10-6
cb = coefficient for brass = 19 x 10-6

so,

Dr = Dp
Dro*(1 + ci**dT) = Dpo*(1 + cb*dT)

dT = (Dro - Dpo) / ( Dpo*cb - Dro*ci )

Dpo*cb = 8.753*19*10-6 = 1.663 x 10-4
Dro* ci = 8.743*12*10-6 = 1.049 x 10-4

dT = (8.743 - 8.753)/[1.663*10-4 - 1.049*10-4]
dT = -0.01/6.139x10-5
dT = -162.9

So the temperature drops by 163o C. Since it started at 20oC the final temperature will be. 20o - 163o = -143o C

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