Question

In a manufacturing facility, \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\) wet brass plates coming out of a water bath are to be dried by passing them through a section where dry air at 1 atm and \(25^{\circ} \mathrm{C}\) is blown parallel to their surfaces at \(4 \mathrm{~m} / \mathrm{s}\). If the plates are at \(15^{\circ} \mathrm{C}\) and there are no dry spots, determine the rate of evaporation from both sides of a plate.

Short answer

Answer: The rate of evaporation from both sides of the wet brass plate is approximately 0.0732 kg/s.

Step-by-step solution

Calculate the Mass Transfer Coefficient

To calculate the mass transfer coefficient (k), we will use the following equation derived from heat and mass transfer principles: \[k = \frac{h_{fg}}{C_{pl}(T_s - T_\infty)}\] where \(h_{fg}\) is the latent heat of vaporization of water, \(C_{pl}\) is the specific heat of liquid water, \(T_s\) is the surface temperature of the plate, and \(T_\infty\) is the air temperature. We are given that \(T_s = 15^{\circ}\mathrm{C}\) and \(T_\infty = 25^{\circ}\mathrm{C}\). The latent heat of vaporization for water at room temperature is approximately \(h_{fg} = 2.4 \times 10^6 \mathrm{~J/kg}\). The specific heat of liquid water is \(C_{pl} = 4186 \mathrm{~J/(kg \cdot K)}\). Let's calculate k: \[k = \frac{2.4 \times 10^6 \mathrm{~J/kg}}{4186 \mathrm{~J/(kg \cdot K)} (15^{\circ}\mathrm{C} - 25^{\circ} \mathrm{C})} \approx 11.43 \mathrm{~m/s}\]

Calculate the Evaporation Rate

Now that we have the mass transfer coefficient, we can determine the rate of evaporation from the plate. The evaporation rate equation can be given as: \[q = A \times k \times \rho_{vapor}(T_\infty - T_s)\] where q is the evaporation rate, A is the area of the plate, and \(\rho_{vapor}\) is the vapor density. The given dimensions of the plate are \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\), which is equal to \(0.4 \mathrm{~m} \times 0.4 \mathrm{~m}\), so the surface area is \(A = (0.4 \mathrm{~m} \times 0.4 \mathrm{~m}) = 0.16 \mathrm{~m^2}\). Assuming the vapor density at these conditions is \(\rho_{vapor} = 0.02 \mathrm{~kg/m^3}\), we can calculate the evaporation rate: \[q = 0.16 \mathrm{~m^2} \times 11.43 \mathrm{~m/s} \times 0.02 \mathrm{~kg/m^3} (25^{\circ}\mathrm{C} - 15^{\circ} \mathrm{C}) \approx 0.0366 \mathrm{~kg/s}\]

Calculate the Evaporation Rate from Both Sides

Since air is blown parallel to both sides of the plate, the evaporation rate from both sides should be the same. Therefore, the evaporation rate from both sides of the plate would be double the value we obtained in Step 2: \[q_{Total} = 2 \times 0.0366 \mathrm{~kg/s} = 0.0732 \mathrm{~kg/s}\] So, the rate of evaporation from both sides of the wet brass plate is approximately \(0.0732 \mathrm{~kg/s}\).

Key concepts

Evaporation Rate Calculation

The evaporation rate tells us how quickly a liquid is turning into vapor from a surface. To calculate it, we use the formula:

• \[ q = A \times k \times \rho_{vapor}(T_{\infty} - T_s) \]

Here, \( q \) is the rate of evaporation, \( A \) is the area of the plate, \( k \) is the mass transfer coefficient, \( \rho_{vapor} \) is the vapor density of the substance, and \( T_{\infty} - T_s \) represents the temperature difference between the air and the surface.
The area of the plate is calculated by multiplying its length and width, which for a plate measuring 40 cm by 40 cm, translates to an area of 0.16 m².
Using the given values for \( k \), \( \rho_{vapor} \), and the respective temperatures, we can solve for \( q \) to find the individual evaporation rate for one side of the plate. The total evaporation rate from both sides is twice this value, since air flows over both sides.

Latent Heat of Vaporization

Latent heat of vaporization is the energy required to turn a liquid into a vapor without changing its temperature. It's crucial in processes where liquid water evaporates to form steam or vapor.
For water at room temperature, the latent heat is approximately \( 2.4 \times 10^6 \text{ J/kg} \).
This value illustrates the amount of energy needed to break the intermolecular bonds in water, allowing it to pass into the vapor phase.
Understanding latent heat is key for calculating the mass transfer coefficient, as seen in the formula:

• \[ k = \frac{h_{fg}}{C_{pl}(T_{s} - T_{\infty})} \]

In this formula, \( h_{fg} \) represents the latent heat of vaporization, showing its integral role in mass transfer calculations.

Mass Transfer Coefficient

The mass transfer coefficient (\( k \)) quantifies the rate at which mass is transferred from one phase to another.
It depends on factors like the flow conditions and the physical properties of the substances involved.
In our calculation:

• \[ k = \frac{2.4 \times 10^6 \text{ J/kg}}{4186 \text{ J/(kg·K)} (15^{\circ}\text{C} - 25^{\circ}\text{C})} \approx 11.43 \text{ m/s} \]

We use the specific heat capacity \( C_{pl} \) along with the latent heat of vaporization \( h_{fg} \).
The calculated \( k \) tells us how efficiently mass is being transferred into the air from the brass plates.
A higher mass transfer coefficient implies faster evaporation, critical for improving drying processes in industrial applications.

Vapor Density

Vapor density \( (\rho_{vapor}) \) indicates the mass of vapor per unit volume. It's pivotal in calculating the evaporation rate.
For our brass plates, an assumed value of \( \rho_{vapor} = 0.02 \text{ kg/m}^3 \) was used.
This value affects how much vapor is generated over time.

• Evaporation increases as vapor density rises.
• It's used in the equation \[ q = A \times k \times \rho_{vapor}(T_{\infty} - T_s) \], impacting the overall rate \( q \) of evaporation.

Understanding vapor density helps predict the evaporation process, crucial for drying operations.
Thus, it plays a significant role in setting conditions for efficient mass transfer and evaporation control.