A metal rod made of Aluminum for 1.5 meters of length, and then Nickle for the r

Question

A metal rod made of Aluminum for 1.5 meters of length, and then Nickle for the remaining 4.5 meters of its length (measurements made at 20. C degree). The linear thermal expansion coefficients are: alpha_AI = 23 times 10^-6/C degree and alpha_Ni = 13 times 10^-6/C degree. The temperature of the rod is then increased to 520 degree C. a. How much does the Aluminum expand by (in length)? b. How much does the Nickle expand by (in length)? c. How much does the entire rod expand by (in length)? d. Calculate the "effective" thermal expansion coefficient for this composite rod.

Explanation / Answer

We use formula,

L= L**T

a) LAl= LAl*Al*T = 1.5*23*10^-6*(520-20) = 0.01725 m

b) LNi= LNi*Ni*T = 4.5*13*10^-6*(520-20) = 0.02925 m

c) Lrod = LAl + LNi = 0.01725 + 0.02925 = 0.0465 m

d) eff = Lrod/(Lrod*T) = 0.0465/(6.0*500) = 15.5*10^-6 / Co

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