Materials Science and Engineering: An Introduction

$M = 4.08 \times 10^{-3} kg/h$
Given: 6-mm thick sheet of palladian with area of 0.25 $m^{2}$ at 600°C diffusion coefficient - $1.7 \times 10^{-8} m^{2}/s$ concentration at high and low pressure side - 2.0 0.4 kg of hydrogen per cubic meter of palladium Required: number of kilograms of hydrogen that pass per hour Solution: Using the combination of Equation 5.1 and 5.3, we obtain: $M = JAt = -DAt \frac{ΔC}{Δx} = -(1.7 \times 10^{-8} m^{2}/s)(0.25 m^{2})(3600 s/h)[\frac{(0.4 - 2.0 kg/m^{3})}{(6 \times 10^{-3} m) }] = 4.08 \times 10^{-3} kg/h$