A radiometer is employed to monitor the temperature of manufactured parts \(\left(A_{1}=10 \mathrm{~cm}^{2}\right)\) on a conveyor. The radiometer is placed at a distance of \(1 \mathrm{~m}\) from and normal to the manufactured parts. When a part moves to the position normal to the radiometer, the sensor measures the radiation emitted from the part. In order to prevent thermal burn on people handling the manufactured parts at the end of the conveyor, the temperature of the parts should be below \(45^{\circ} \mathrm{C}\). An array of spray heads is programmed to discharge mist to cool the parts when the radiometer detects a temperature of \(45^{\circ} \mathrm{C}\) or higher on a part. If the manufactured parts can be approximated as blackbody, determine the irradiation on the radiometer that should trigger the spray heads to release cooling mist when the temperature is not below \(45^{\circ} \mathrm{C}\).

Step by step solution

01

Write down the known values

In order to solve the problem, let's first list down the given values: - Area of the manufactured parts: \(A_{1} = 10 \ \text{cm}^2\) - Distance between the radiometer and the parts: \(d = 1 \ \text{m}\) - Temperature threshold: \(T_{threshold} = 45^{\circ} \mathrm{C}\)

02

Convert the temperature to Kelvin

Since we will be working with the Stephan-Boltzmann Law, it is necessary to convert the temperature threshold to Kelvin. To do this, add 273.15 to the Celsius temperature: \(T_{threshold (K)} = T_{threshold} + 273.15 = 45 + 273.15 = 318.15 \ \mathrm{K}\)

03

Apply the Stephan-Boltzmann Law

The power emitted by a blackbody is given by the Stephan-Boltzmann Law: \(P_{emitted} = A \sigma T^4\) Where: - \(P_{emitted}\) is the power emitted by the blackbody, - \(A\) is the area of the blackbody (in this case, the area of the manufactured parts), - \(\sigma\) is the Stefan-Boltzmann constant: \(5.67 \times 10^{-8} \mathrm{W}\cdot\mathrm{m}^{-2}\cdot\mathrm{K}^{-4}\) - \(T\) is the temperature of the blackbody in Kelvin. Since we are interested in the irradiation detected by the radiometer, we need to calculate the power per unit area that reaches the radiometer at a distance \(d\). To do this, divide the emitted power by the surface area of a sphere with radius \(d\): \(I = \frac{P_{emitted}}{4 \pi d^2}\) Combining these two equations, we get: \(I = \frac{A \sigma T^4}{4 \pi d^2}\)

04

Calculate the irradiation

Now, we can plug in the known values into the equation and solve for the irradiation on the radiometer, \(I\): - Convert the area of the manufactured parts to square meters: \(A_{1} = 0.0010 \ \text{m}^2\) - Distance between radiometer and parts: \(d = 1 \ \text{m}\) - Temperature threshold in Kelvin: \(T_{threshold (K)} = 318.15 \ \mathrm{K}\) \(I = \frac{0.0010 \mathrm{m}^2 \times 5.67 \times 10^{-8} \mathrm{W}\cdot\mathrm{m}^{-2}\cdot\mathrm{K}^{-4} \times (318.15 \ \mathrm{K})^4}{4 \pi \times (1 \ \text{m})^2}\) \(I \approx 0.348 \ \mathrm{W/m^2}\) Thus, the irradiation on the radiometer that should trigger the spray heads to release cooling mist when the temperature is not below \(45^{\circ} \mathrm{C}\) is approximately \(0.348 \ \mathrm{W/m^2}\).

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