Question

The thermal conductivity of stainless steel has been characterized experimentally to vary with temperature as \(k(T)=9.14+0.021 T\) for \(273

Short answer

The average thermal conductivity of stainless steel between 300 and 1200 K is 164.418 W/m·K. The rewritten expression in the desired form is \(k(T) = 14.873(1 + 6.12 \times 10^{-4} T)\), where \(k_0 = 14.873 \;\mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) and \(\beta = 6.12 \times 10^{-4} \;\mathrm{K}^{-1}\).

Step-by-step solution

Integrate the given expression

First, we need to integrate the given function over the temperature range from 300 to 1200 K. The integral will give us the cumulative thermal conductivity, and dividing by the temperature range will give us the average thermal conductivity. Integrate \(k(T)\) with respect to temperature \(T\): $$\int_{300}^{1200} (9.14 + 0.021T) dT$$

Calculate the integral

Evaluate the integral: $$\int_{300}^{1200} (9.14 + 0.021T) dT = \left[9.14T + 0.0105T^2\right]_{300}^{1200}$$ Plug in the limits of the integration: $$(9.14(1200) + 0.0105(1200)^2) - (9.14(300) + 0.0105(300)^2)$$

Evaluate the expression

Calculate the value: $$ (10968 + 151200) - (2742 + 9450) = 160168 - 12192 = 147976$$

Calculate the average thermal conductivity

Divide the result by the temperature range (1200 - 300) to find the average thermal conductivity: $$\frac{147976}{(1200-300)} = \frac{147976}{900} = 164.418$$ So the average thermal conductivity between 300 and 1200 K is \(164.418 \;\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)

Rewrite the expression

Now we need to rewrite the expression for the thermal conductivity in the form \(k(T) = k_{0}(1 + \beta T)\). Since \(k(273) = k_0\), we can find \(k_0\) by plugging T = 273 K into the given expression: $$k_0 = 9.14 + 0.021(273) = 9.14 + 5.733 = 14.873$$ Now we have \(k(T) = 14.873(1 + \beta T)\). To find \(\beta\), we can equate this expression with the original one and solve for \(\beta\): $$9.14 + 0.021T = 14.873(1 + \beta T)$$ From this, we can determine that \(\beta = \frac{0.021 - \frac{9.14}{14.873}}{14.873} = 6.12 \times 10^{-4}\) Now we have the desired expression: $$k(T) = 14.873(1 + 6.12 \times 10^{-4} T)$$ where \(k_0 = 14.873 \;\mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) and \(\beta = 6.12 \times 10^{-4} \;\mathrm{K}^{-1}\).

Key concepts

Stainless Steel

Stainless steel is a versatile and widely used material known for its resistance to corrosion and high durability. It is composed of iron, mixed with other elements such as chromium, which enhances its corrosion resistance. This material is often used in construction, cookware, and industrial equipment.
One important property of stainless steel is its ability to conduct heat, also known as thermal conductivity. Thermal conductivity is crucial in applications like heat exchangers and cooking utensils, where efficient heat transfer is needed. The thermal property of stainless steel can change with temperature, affecting its performance in various applications.

Temperature Dependence

The thermal conductivity of materials, including stainless steel, often depends on temperature. This means as the temperature changes, the material's ability to conduct heat will also change. In the given exercise, the thermal conductivity of stainless steel is expressed as a function of temperature: \[ k(T) = 9.14 + 0.021T \] This expression shows that the thermal conductivity increases linearly with temperature from 273 K to 1500 K. Here, 9.14 is the base thermal conductivity at 0 degrees Celsius (273 K), and 0.021 is the rate at which thermal conductivity increases per degree Kelvin increase in temperature.
This kind of relationship helps engineers and scientists to predict how well stainless steel will perform in different thermal environments and to design systems accordingly.

Average Thermal Conductivity

To find the average thermal conductivity over a specified temperature range, we integrate the function of temperature-dependent thermal conductivity over this range. This requires using the concept of definite integration, which sums all values of the function within the limits. For the temperature range 300 K to 1200 K in our solution, it involves solving the integral:\[ \int_{300}^{1200} (9.14 + 0.021T) \, dT \] We then divide this result by the temperature range (which is 1200 K - 300 K = 900 K) to find the average value. This gives us the average thermal conductivity, reflecting how well the material conducts heat on average across that temperature span. Achieving this involves calculating:\[ \frac{147976}{900} = 164.418 \] Thus, between these temperatures, the average thermal conductivity of stainless steel is approximately 164.418 W/m·K.

Mathematical Integration

Mathematical integration is a fundamental process in calculus used to find the total effect of a variable as it changes over a certain range. In the context of thermal conductivity, it helps calculate quantities like cumulative total thermal conductivity over a temperature span.Let's consider the given function of thermal conductivity: \[ k(T) = 9.14 + 0.021T \] To integrate this between 300 K and 1200 K, we calculate:\[ \int_{300}^{1200} (9.14 + 0.021T) \, dT = \left[9.14T + 0.0105T^2\right]_{300}^{1200} \]Plugging in these limits provides the cumulative thermal conductivity over the desired range. Integration helps resolve how the stainless steel's heat conduction capability builds up over the increasing temperatures. Successfully solving this integral provides accurate measures for engineering problems, ensuring efficient material usage and safety.