I need help in this as soon as possible Thank you guys Unit 2: Homework / Homewo

Question

I need help in this as soon as possible
Thank you guys Unit 2: Homework / Homework: HW2 Of Ch178 20 Deadline: 100% until Tuesday, September 5 at 11:59 PM Tipler6 20.P.028. 1) A copper collar is to fit tightly about a steel shaft whose diameter is 6.0089 cm at 18 C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be ralsed so that it will just slip on the steel shaft, assuming the temperature of both the steel shaft and copper collar are raised simultaneously? c Submit You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question.

Explanation / Answer

Given

copper collar of diameter initially d1 = 5.9800 cm, and of steel shaft is d2 = 6.0089 cm

the circumferance of each one is C1 = 2pi*r1 = 2pi*d1/2 = pi*d1 = 3.14*5.9800 cm = 18.7772 cm

C2 = 2pi*r2 = 2pi*d2/2 = pi*d2 = 3.14*6.0089 cm = 18.867946 cm

we know the linear expansion of solids

the coefficient of linear expansion of copper is alpha_c = 17*10^-6 /0C , steel is alpha_s = 13*10^-6 /0C

temperature must the copper collar be raised so that it will just slip on the steel shaft is

from the relation Dl = l0*alpha*DT

where Dl is difference in length so that the copper collar will just slip on the steel shaft is 18.867946-18.7772 cm = 0.090746 cm

and DT = T2-T1 , here T1 = 18 0C, T2 = ?

substituting the values

0.090746 = 18.7772*17*10^-6(T2-18)

solving for T2 = 302.28 0C

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