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The steady-state temperature distribution in a one-dimensional wall of thermal conductivity 50 W/m K and thickness 50 mm is observed to be T(degree C) = a + bx^2, where a = 200 degree C, b = -2000 degree C/m2, and x is in meters. A. What is the heat generation rate q in the wall? b. Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate?
The steady-state temperature distribution in a one-dimensional wall of thermal conductivity 50 W/m K and thickness 50 mm is observed to be T(degree C) = a + bx^2, where a = 200 degree C, b = -2000 degree C/m2, and x is in meters. A. What is the heat generation rate q in the wall? b. Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate?
Final answer:To address the question about the steady-state temperature distribution in a wall and calculate the heat generation rate and heat fluxes, we apply the principles of thermal conductivity. The problem requires applying Fourier's Law to the provided temperature distribution, but exact calculation of internal heat generation rate q and precise heat fluxes at the wall faces necessitates more specific data or a detailed derivation based on boundary conditions.Explanation:To understand the heat generation rate and heat fluxes in a one-dimensional wall based on the given temperature distributionT(degrees C) = 200 - 2000x^2, wherexis in meters, we need to apply principles of thermal conductivity and heat transfer. Given the thermal conductivity (k) of the material is 50 W/mK, and the wall thickness is 50 mm, or 0.05 meters.Firstly, we can determine the heat generation rate within the wall, q, by recognizing that the temperature distribution's format suggests an internal heat source due to the x^2 dependence. Since the exact source of heat within the material isn't external thermal gradients but rather internal generation, it's essential to use a modified form of Fourier's law, which incorporates an internal generation term. However, with the given equation and lack of additional specifics, it's complicated to directly calculateqwithout deriving or knowing the internal energy generation equations or boundary conditions for temperatures.Secondly, to determine the heat fluxes at the two wall faces, we use Fourier's Law in the formqh = -kA(dT/dx), whereqhis the heat flux (W/m^2), and-(dT/dx)is the temperature gradient. At each wall face (x=0 and x=0.05m), the gradient and thus the heat flux can be calculated. It's essential to note, the heat generated inside the wall must equal the heat exiting through the wall faces, aligning with the law ofenergy conservation. Without specific computations here, typically, one would differentiate the temperature equation with respect toxto find the gradient at each face, then apply the values into Fourier's expression to find the fluxes....