An aluminum plate thickness 2 mm and 3 \times 3 cm, has a 1 cm diameter hole in the center. The...

Question:

An aluminum plate thickness 2 mm and 3 {eq}\times {/eq} 3 cm, has a 1 cm diameter hole in the center. The plate is heated such that the thickness increases to 3 mm. What is the new diameter of the hole in cm?

Thermal Expansion in Length and Area

Materials expand with increase of temperature and contract with decrease of temperature. The expansion and contraction of the materials are almost linear over a wide range of temperature and the change in length of a material with temperature is given by the equation {eq}\Delta L = L \alpha \Delta T {/eq}. In this equation {eq}L, \ \ \alpha, \ \ \Delta T {/eq} are the length at some reference temperature, coefficient of thermal expansion, and change in temperature. The area or volume of the material also will expand or contract with temperature. The coefficient of expansion for area is twice that of the coefficient of linear expansion. Volume coefficient of expansion is three times of linear expansion coefficient.

Given data

• Thickness of the aluminum plate {eq}x = 2.0 \times 10^{-3} \ m {/eq}
• Diameter of the hole at the center of the sheet {eq}d = 1.0 \times 10^{-2 } \ m {/eq}
• On heating the thickness of the sheet increases to {eq}x_1 = 3.0 \times 10^{-3 } \ m {/eq}
• Coefficient of linear thermal expansion of aluminum {eq}\alpha = 23.6 \times 10^{-6} \ / ^o C {/eq}

Let {eq}\Delta T {/eq} be the increase in temperature that caused the increase in thickness.

Then the increased thickness can be expresses as {eq}x_1 = x \times [ 1 + \alpha \Delta T ] {/eq}

Therefore the increase in temperature {eq}\Delta T = \dfrac { (\dfrac { x_1 } { x } - 1 ) } { \alpha } \\ \Delta T = \dfrac { ( \dfrac { 3.0 \times 10^{-3 } } { 2.0 \times 10^{-3 } } - 1 ) } { 23.6 \times 10^{-6} } \\ \Delta T = 2.12 \times 10^4 \ ^o C {/eq}

Initial area of the hole {eq}A = \pi ( d/2)^2 \\ A = \pi \times ( \dfrac { 1.0 \times 10^{-2 } } { 2 } )^2 \\ A = 7.85 \times 10^{-5} \ m^2 {/eq}

Area of the hole after the given increase in temperature {eq}A_1 = A \times ( 1 + 2 \alpha \Delta T ) \\ A_1 = 7.85 \times 10^{-5 } \times ( 1 + 2 \times 23.6 \times 10^{-6 } \times 2.12 \times 10^4 ) \\ A_1 = 1.571 \times 10^{-4} \ m^2 {/eq}

If {eq}d_1 {/eq} is the new diameter we have the equation {eq}A = \pi ( d_1/2 )^2 {/eq}

Therefore the new diameter {eq}d_1 = \sqrt { \dfrac { 4 A_1 } { \pi } } \\ d_1 = \sqrt { \dfrac { 4 \times 1.571 \times 10^{-4 } } { \pi } } \\ d_1 = 1.414 \times 10^{-2 } \ m \\ d_1 = 1.41 \ cm {/eq}