Question

If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of the hole when the rivet is cooled to - 78.0 ∘C, the temperature of dry ice? Assume that the expansion coefficient of aluminum is 2.4×10−5(∘C)−1 and remains constant.

Answer 1

To fit a rivet into a 4.500 mm hole at -78.0℃, its diameter at 23.0℃ should be approximately 4.511 mm, considering aluminum's coefficient of linear expansion.

Step-by-step explanation:

To determine the diameter of a rivet at 23.0 ℃ that will equal the diameter of a hole when the rivet is cooled to -78.0 ℃, we use the linear expansion formula ΔL = αL₀ΔT, where:

• ΔL is the change in length (or diameter in this case),
• α is the coefficient of linear expansion,
• L₀ is the original length (or diameter), and
• ΔT is the temperature change.

Given the expansion coefficient for aluminum α = 2.4×10⁻⁵˚C⁻¹, the diameter of the hole d₀ = 4.500 mm, and the temperature change ΔT = 23.0 ℃ - (-78.0 ℃) = 101 ℃, we can calculate the required diameter of the rivet before cooling.

The change in diameter Δd is given by:

Δd = αd₀ΔT

Δd = (2.4×10⁻⁵)(4.500 mm)(101 ℃)

Δd = 0.010944 mm

The diameter of the rivet at 23.0 ℃ should be:

d₁ = d₀ + Δd

d₁ = 4.500 mm + 0.010944 mm

d₁ ≈ 4.511 mm

Thus, the rivet's diameter at 23.0°C should be approximately 4.511 mm to fit into a 4.500 mm hole when cooled to dry ice temperature.

Answer 2

Step-by-step explanation:

diameter at 23 degree celcius is 4.511 mm

GIVEN DATA:

diameter of hole = 4.500 mm

T_1 = 23.0 Degree celcius

T_2 = - 78.0 Degree celcius

expansion coeffcient of Al = 2.4*10^{-5} (degree celcius)^{-1}

diameter at 23 degree celcius is given as

`d = d_o (1+ \alpha \delta T)`

`= 4.5 (1+2.4*10^(-5) *(23-(-78)))`

= 4.511 mm

diameter of rivet after change in temperature is given as

`d= d_o + \Delta d`

`= d_o(1+ \alpha \delta T)`

`= 0.4500 *(1+2.4*10^(-5) *(23-(-78)))`

= 0.4511 cm