To ensure a tight fit, cylindrical metal elements are sometimes made the same si

Question

To ensure a tight fit, cylindrical metal elements are sometimes made the same size (or even slightly larger) than the holes they will be placed into, and are cooled significantly just before being inserted into them. Suppose that on the Zebra z-pinch, at 20 degrees celcius, a cylindrical aluminum alloy cathode of diameter 5 cm needs to be cooled to fit into a hole in a holder. Suppose that the cathode needs to be made 115 microns smaller ( a micron, or micrometer, is 0.001 mm), in order to be inserted. To what temperature (in degrees celcius) must the aluminum alloy cathode be cooled? The expansion coefficient of the aluminum alloy is 2.4x10^-5.

Explanation / Answer

You have an aluminum cylinder, as you give it heat it shall expand, keep that in mind (unless they say otherwise)
* put alfa the linear coefficient of the given substance
now calculation .. V(final of the aluminum) = V(initial)*[3*alfa*(tf-ti)] - Vinitial
= 2*3*2.4*10^-5*(89-21) + 2 = 2.0097

Then i tell u that turpentine occupying the cylinder fully must have 2 L volume
now calculation .. V(final turpentine) = V(initial)*[3*alfa*(tf-ti)] + Vinitial
= 2*3*9*10^-4*(89-21) + 2 = 2.3672

Now you have to subtract the bigger volume of turpentine (the one calculated) by that of aluminum (the more capacity gained by the cylinder)
imagine it .. You'll get 0.3575 L overflowing (:

b) the volume remaining in the cylinder is that of the cylinder .. 2.0097
c) now if we changed the temp to reach the original one..
well here we have to use mental analysis i.e, the overflown amount over the initial volume will give us
0.3575/2.0097 = 0.1778
this ration also applies to the edges as well as the volume
this means that we multiply the edge (22 cm) by the ratio (0.1778) we get 3.9116 cm below the surface

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