Estimate the maximum and minimum thermal conductivity values for a cermet that contains 85 vol \(\%\) titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and \(69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for \(\mathrm{TiC}\) and \(\mathrm{Co}\), respectively.

Short Answer

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Question: Estimate the maximum and minimum thermal conductivity values for a cermet containing 85 vol% TiC particles in a cobalt matrix given the thermal conductivities of TiC and Co as 27 and 69 W/m·K, respectively. Answer: The estimated maximum and minimum thermal conductivity values for the cermet containing 85 vol% TiC particles in a cobalt matrix are 33.3 W/m·K and 29.3 W/m·K, respectively.

Step by step solution

01

Understand the Rule of Mixtures

The Rule of Mixtures is a widely used model which estimates the properties of a composite material by taking weighted averages of the individual properties of the constituents. In this case, it will be used to calculate the thermal conductivity values for the cermet. The Rule of Mixtures expression for finding the thermal conductivity (K) of the cermet is given by: \(K = p \cdot K_\text{TiC} + (1-p) \cdot K_\text{Co}\), where p is the volume fraction of TiC and \(K_\text{TiC}\) and \(K_\text{Co}\) are the thermal conductivities of TiC and Co, respectively.
02

Calculate the volume fraction

The problem statement tells us that the cermet contains 85 vol% TiC particles. That means the volume fraction of TiC (p) is 0.85. The volume fraction of the cobalt matrix (1-p) is 1 - 0.85 = 0.15.
03

Calculate the maximum and minimum thermal conductivity values

Using the Rule of Mixtures formula, we can now calculate the maximum and minimum thermal conductivity values for the cermet. - For maximum thermal conductivity (assuming Co matrix provides better thermal conductivity due to its higher value): \(K_\text{max} = 0.85 \cdot K_\text{TiC} + 0.15 \cdot K_\text{Co} = 0.85 \cdot 27 \mathrm{~W/m \cdot K} + 0.15 \cdot 69 \mathrm{~W/m \cdot K} = 22.95 \mathrm{~W/m \cdot K} + 10.35 \mathrm{~W/m \cdot K} = 33.3 \mathrm{~W/m \cdot K} \) - For minimum thermal conductivity (assuming TiC particles provide a hindrance to thermal conductivity of the matrix): \(K_\text{min} = \dfrac{0.85}{27 \mathrm{~W/m \cdot K}} + \dfrac{0.15}{69 \mathrm{~W/m \cdot K}} = \dfrac{0.85}{27} + \dfrac{0.15}{69} = 0.03148 + 0.00217\) Now, we need to take the reciprocal of the result to get the minimum thermal conductivity value: \(K_\text{min} = \dfrac{1}{(0.03148 + 0.00217)} = 29.3 \mathrm{~W/m \cdot K}\)
04

Present the results

The estimated maximum and minimum thermal conductivity values for the cermet containing 85 vol% TiC particles in a cobalt matrix are 33.3 W/m·K and 29.3 W/m·K, respectively.

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