A cylinder is made up of two coaxial layers, one of radius R and the other of inner radius R and outer radius 2R. The inner and the outer portions are respectively made up of substances of thermal conductivities K and 2K respectively. If the effective thermal conductivity between the flat ends of the cylinder is xk/4 , then the value of x is ________
Question
A cylinder is made up of two coaxial layers, one of radius R and the other of inner radius R and outer radius 2R. The inner and the outer portions are respectively made up of substances of thermal conductivities K and 2K respectively. If the effective thermal conductivity between the flat ends of the cylinder is xk/4 , then the value of x is ________
Solution
The thermal resistance of a cylindrical layer is given by the formula R = ln(r2/r1) / (2πkL), where r1 and r2 are the inner and outer radii of the layer, k is the thermal conductivity of the material, and L is the length of the cylinder.
For the inner layer, we have R1 = ln(R/R) / (2πKL) = 0, because ln(1) = 0.
For the outer layer, we have R2 = ln(2R/R) / (2π2KL) = ln(2) / (4πKL).
The total thermal resistance of the system is the sum of the resistances of the layers, so R_total = R1 + R2 = 0 + ln(2) / (4πKL) = ln(2) / (4πKL).
The effective thermal conductivity of the system is given by the formula k_eff = L / (R_total * A), where A is the cross-sectional area of the cylinder. Since the cylinder is coaxial, A = πR^2.
Substituting the given values, we get k_eff = L / [ln(2) / (4πKL) * πR^2] = 4KL / ln(2).
Comparing this with the given effective thermal conductivity xk/4, we get 4KL / ln(2) = xk/4.
Solving for x, we get x = 16L / ln(2).
Since the length L of the cylinder does not affect the value of x, we can set L = 1 for simplicity, giving x = 16 / ln(2) ≈ 23.03.
Therefore, the value of x is approximately 23.03.
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