a cylinder of diameter 1cm at 30degree celcius is to slid into a hole on a steel plate. the hole has a diameter of 0.9997 cm at 30 degree celcius. to what temperature must be the plate be heated when steel has the alpha 1.1X10^-5/degree celcius
Dia.=0.9997 + 1.1*10^-5 (T-30) = 1.0cm,
0.9997 + 1.1*10^-5T - 33*10^-5 = 1,
1.1*10^-5*T=1 - 0.9997 + 0.00033
1.1*10^-5*T = 6.3*10^-4,
T = 57.3 deg.
2170.13
To determine the required temperature of the steel plate, we need to consider the expansion of both the cylinder and the hole.
1. Calculate the change in diameter of the cylinder:
ΔD = α × D × ΔT
where ΔD is the change in diameter, α is the coefficient of linear expansion (given as 1.1 × 10^-5/°C), D is the original diameter of the cylinder (1 cm), and ΔT is the change in temperature.
Plugging in the values:
ΔD = (1.1 × 10^-5/°C) × (1 cm) × ΔT
2. Calculate the new diameter of the cylinder at the given temperature:
D_new = D + ΔD
where D_new is the new diameter of the cylinder.
3. Calculate the change in diameter of the hole:
ΔD_hole = α × D_hole × ΔT
where ΔD_hole is the change in diameter of the hole, D_hole is the original diameter of the hole (0.9997 cm), and ΔT is the change in temperature.
Plugging in the values:
ΔD_hole = (1.1 × 10^-5/°C) × (0.9997 cm) × ΔT
4. Calculate the new diameter of the hole at the given temperature:
D_hole_new = D_hole + ΔD_hole
where D_hole_new is the new diameter of the hole.
5. Equate the new diameter of the cylinder to the new diameter of the hole and solve for ΔT:
D_new = D_hole_new
(1 cm) + (1.1 × 10^-5/°C) × (1 cm) × ΔT = (0.9997 cm) + (1.1 × 10^-5/°C) × (0.9997 cm) × ΔT
Solve this equation for ΔT to find the temperature change required.
Please note that this calculation assumes the thermal expansion of both the cylinder and the hole is uniform and isotropic, and there is no deformation or stress caused by the expansion.
To find the temperature to which the plate must be heated, we need to consider the thermal expansion of the steel plate.
The thermal expansion of a material is given by the equation:
ΔL = α * L * ΔT
Where:
ΔL is the change in length
α is the coefficient of linear expansion
L is the original length
ΔT is the change in temperature
In this case, since we are dealing with diameters, we can use the equation:
ΔD = α * D * ΔT
Where:
ΔD is the change in diameter
α is the coefficient of linear expansion
D is the original diameter
ΔT is the change in temperature
Given that the initial diameter of the hole in the steel plate is 0.9997 cm and the final diameter of the cylinder is 1 cm, we can calculate the change in diameter:
ΔD = 1 cm - 0.9997 cm = 0.0003 cm
Now, we can rearrange the equation to calculate the change in temperature:
ΔT = ΔD / (α * D)
Given that α is 1.1 x 10^-5 / degree Celsius (provided in the question) and the initial diameter D is 0.9997 cm, we can plug in the values:
ΔT = 0.0003 cm / (1.1 x 10^-5 / degree Celsius * 0.9997 cm)
Simplifying the expression:
ΔT = 0.0003 cm / (1.09967 x 10^-5 / degree Celsius)
ΔT ≈ 27.285 Celsius
Therefore, the steel plate must be heated to approximately 27.285 degrees Celsius for the cylinder to slide into the hole.