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The temperature at the point (x, y, z) in a substance with conductivity K = 8.5 is u(x, y, z) = 5y2 + 5z2. Find the rate of heat flow inward across the cylindrical surface y2 + 2+ = 7,0 5 x 5 5. X

User HoverHell
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Answer: There is no net heat flow across the cylindrical surface.

Explanation:

To find the rate of heat flow inward across the cylindrical surface, we need to calculate the heat flux vector and then integrate it over the surface.

The heat flux vector can be calculated using Fourier's law of heat conduction:

q = -K * ∇u

where q is the heat flux vector, K is the thermal conductivity, and ∇u is the gradient of the temperature function u(x, y, z).

Let's first calculate the gradient ∇u:

∇u = (∂u/∂x, ∂u/∂y, ∂u/∂z)

∂u/∂x = 0 (no x-dependence in the temperature function)

∂u/∂y = 10y

∂u/∂z = 10z

Therefore, the gradient ∇u is (0, 10y, 10z).

Next, we calculate the heat flux vector:

q = -K * ∇u = -8.5 * (0, 10y, 10z) = (0, -85y, -85z)

The heat flux vector is (0, -85y, -85z).

To calculate the rate of heat flow inward across the cylindrical surface, we need to integrate the dot product of the heat flux vector and the outward unit normal vector over the surface.

The outward unit normal vector for the cylindrical surface is (1, 0, 0) since the surface is perpendicular to the x-axis.

The dot product of the heat flux vector and the outward unit normal vector is:

q_dot_n = (0, -85y, -85z) dot (1, 0, 0) = 0

Integrating this dot product over the cylindrical surface will give us the rate of heat flow inward. However, since the dot product is zero, it indicates that there is no net heat flow across the cylindrical surface.

User Andrey Langovoy
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