Question

Reconsider Prob. 14-82. In order to reduce the migration of water vapor through the wall, it is proposed to use a \(0.051-\mathrm{mm}\)-thick polyethylene film with a permeance of $9.1 \times 10^{-12} \mathrm{~kg} / \mathrm{s}^{2} \mathrm{~m}^{2}$.Pa. Determine the amount of water vapor that will diffuse through the wall in this case during a \(24-\mathrm{h}\) period. Answer: \(26.4 \mathrm{~g}\)

Short answer

Answer: 26.4 g

Step-by-step solution

Identify the given information

We are given the following information: - Thickness of polyethylene film: \(d = 0.051 \, \text{mm}\) - Permeance of polyethylene film: \(P = 9.1 \times 10^{-12} \, \frac{\text{kg}}{\text{s}^{2} \, \text{m}^{2}\text{.Pa}}\) - Time period: \(t = 24 \, \text{h}\)

Convert units

First, we must convert the thickness of the polyethylene film to meters and the time period to seconds: - Thickness of polyethylene film: \(d = 0.051 \times 10^{-3} \, \text{m}\) - Time period: \(t = 24 \times 3600 \, \text{s}\)

Calculate the mass flow rate

The mass flow rate of water vapor through the polyethylene film, \(\dot{m}\), can be calculated using the formula: \(\dot{m} = P \cdot \Delta p \cdot A\) We can assume that the pressure difference, \(\Delta p\), across the film is constant and equal to the atmospheric pressure, which is approximately \(1 \, \text{atm} = 101325 \, \text{Pa}\). The area, \(A\), through which the water vapor is diffusing is not given, but since we are interested in the amount of water vapor that will diffuse through the wall, we can just assume an area of \(1 \, \text{m}^2\). Now, we can calculate the mass flow rate: \(\dot{m} = (9.1 \times 10^{-12} \, \frac{\text{kg}}{\text{s}^{2} \, \text{m}^{2}\text{.Pa}}) \cdot (101325 \, \text{Pa}) \cdot (1 \, \text{m}^2)\) \(\dot{m} = 9.22096 \times 10^{-7} \, \frac{\text{kg}}{\text{s}}\)

Calculate the total amount of water vapor diffused

To find the total amount of water vapor that will diffuse through the wall during the 24-hour period, we simply multiply the mass flow rate by the time period and convert the result to grams: Total amount of water vapor \(= \dot{m} \cdot t\) Total amount of water vapor \(= (9.22096 \times 10^{-7} \, \frac{\text{kg}}{\text{s}}) \cdot (24 \times 3600 \, \text{s})\) Total amount of water vapor \(= 0.0263517 \, \text{kg}\) Total amount of water vapor \(= 26.4 \, \text{g}\) (rounded to one decimal place) So, the amount of water vapor that will diffuse through the wall in this case during a 24-hour period is \(26.4 \, \text{g}\).