Question

Air at \(52^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), and 10 percent relative humidity enters a 5 -cm-diameter tube with an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube inner surface is wetted uniformly with water, whose vapor pressure at \(52^{\circ} \mathrm{C}\) is \(13.6 \mathrm{kPa}\). While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to \(10 \mathrm{kPa}\). Detemine \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the log-mean driving force for mass transfer in molar concentration units, \((c)\) the water evaporation rate in \(\mathrm{kg} / \mathrm{h}\), and \((d)\) the length of the tube.

Short answer

Question: Calculate the average mass transfer coefficient, the log-mean driving force for mass transfer, the water evaporation rate, and the length of the tube in a process where air containing water vapor flows through a tube with wetted inner surfaces, given the temperature, pressure, humidity, and properties of air and water vapor. Answer: To calculate the required variables, follow these steps: 1. Calculate the molar and mass flow rate of the inlet air 2. Calculate the molar fractions of dry air and water vapor at the inlet 3. Calculate the molar fractions of dry air and water vapor at the outlet 4. Determine the average mass transfer coefficient 5. Determine the log-mean driving force for mass transfer 6. Calculate the water evaporation rate 7. Determine the length of the tube Use the given information and appropriate formulas for each step to solve the problem.

Step-by-step solution

Calculate the molar and mass flow rate of the inlet air

First, we will need to obtain the molar flow rate and mass flow rate of dry air and water vapor. To do this, we will use the ideal gas law, which states that \(PV = nRT\). We can modify it to find the molar flow rate and mass flow rate: For dry air: Molar flow rate = \(\dot{n}_{dry air} = \frac{PV}{RT} - \frac{10\%}{100\%}\frac{PV}{RT}\) For water vapor: Molar flow rate = \(\dot{n}_{water vapor} = \frac{10\%PV}{RT}\) We will then convert these molar flow rates to mass flow rates using the molecular weights of dry air and water vapor: Mass flow rate = \(\dot{m} = \dot{n} \cdot M\)

Calculate the molar fractions of dry air and water vapor at the inlet

Next, we need to find the molar fractions of dry air and water vapor at the beginning of the tube. The molar fractions will be the ratio of the individual molar rates to the total molar flow rate: \(y_{dry air} = \frac{\dot{n}_{dry air}}{\dot{n}_{dry air} + \dot{n}_{water vapor}}\) \(y_{water vapor} = \frac{\dot{n}_{water vapor}}{\dot{n}_{dry air} + \dot{n}_{water vapor}}\)

Calculate the molar fractions of dry air and water vapor at the outlet

Now we will find the molar fractions of dry air and water vapor at the end of the tube: \(y'_{dry air} = \frac{\dot{n}_{dry air}}{\dot{n}_{dry air} + \dot{n}'_{water vapor}}\) \(y'_{water vapor} = \frac{\dot{n}'_{water vapor}}{\dot{n}_{dry air} + \dot{n}'_{water vapor}}\) Where: Partial pressure of vapor at the outlet, \(P'_{water vapor} = 10 kPa\) \(\dot{n}'_{water vapor} = \frac{P'_{water vapor}V}{RT}\)

Determine the average mass transfer coefficient

Using the molar fractions, we can now calculate the average mass transfer coefficient, which is given by: \(K_{mass transfer} = \frac{W_{water vapor}}{A(V \cdot t)}\frac{1}{ln(\frac{y_{dry air}' - y_{water vapor}'}{y_{dry air} - y_{water vapor}})}\) Where: \(W_{water vapor}\) is the molecular weight of water vapor (18 g/mol) \(A\) is the cross-sectional area of the tube \(V \cdot t\) is the volume flow rate of the inlet air

Determine the log-mean driving force for mass transfer

Now we will find the log-mean driving force for mass transfer in molar concentration units. The log-mean driving force is calculated as follows: \(\Delta C_{log-mean} = \frac{C_{water vapor} - C_{water vapor}'}{ln(\frac{C_{water vapor}'}{C_{water vapor}})}\) Where: \(C_{water vapor} = \frac{P_{water vapor}RT}{(P-P_{water vapor})(Z^2)}\) \(C_{water vapor}' = \frac{P_{water vapor}'RT}{(P-P_{water vapor}')(Z^2)}\) \(Z\) is the compressibility factor (assumed to be 1)

Calculate the water evaporation rate

We can now find the water evaporation rate using the mass transfer coefficient and the log-mean driving force: \(W_{evaporation} = K_{mass transfer} \cdot \Delta C_{log-mean}\) To convert this to \(\frac{kg}{hr}\), multiply by the appropriate conversion factors.

Determine the length of the tube

Finally, we will find the length of the tube required for these conditions using the mass transfer coefficient and the water evaporation rate: \(Length = \frac{W_{evaporation}}{K_{mass transfer} \cdot A(V \cdot t)}\) Now we have determined (a) the average mass transfer coefficient, (b) the log-mean driving force for mass transfer, (c) the water evaporation rate, and (d) the length of the tube.

Key concepts

Partial Pressure

Partial pressure helps understand how much of a particular gas is present in a mixture of different gases. In simple terms, partial pressure is the pressure that a gas would have if it occupied the entire volume on its own. When dealing with humid air, like in this exercise, the partial pressure of water vapor is essential because it tells us how close the air is to becoming saturated.
This is especially important for processes such as evaporation. In the given exercise, we're told that the water vapor pressure at 52°C is 13.6 kPa, but the partial pressure of the vapor leaving the outlet reaches 10 kPa. This change in partial pressure from the inlet to the outlet reflects the evaporation that has occurred.
The remaining pressure is contributed by the dry air. Knowing the partial pressures allows us to understand the driving force behind the mass transfer and how much vapor can still be absorbed by the air.

Molar Flow Rate

Molar flow rate refers to the quantity of a specific component moving through a surface per unit time, measured in moles per time period (e.g., moles per second). It is a key factor in mass transfer calculations because it helps us understand how much of each gas is present in the flow. Molar flow rate is often derived from the ideal gas law, where the ideal conditions allow us to interrelate pressure, volume, and temperature.
In this exercise, we calculate the molar flow rate to determine how much dry air and water vapor are entering and exiting the tube. This information helps in computing the mass flow rates and the changes in vapor composition. By knowing the molar flow rates, engineers can adjust the system to better control the evaporation process or any mass transfer actions happening in real-world applications.

• The molar flow rate formula for a component is: \( \dot{n} = \frac{PV}{RT} \)
• Compute separately for dry air and water vapor to understand their contributions.
• Conversions to mass flow rate require multiplying by the molecular weight of the component.

Evaporation Rate

The evaporation rate describes how quickly a liquid changes into vapor. It depends on several factors, including temperature, surface area, and the partial pressure above the liquid. In mass transfer studies, the evaporation rate can be determined using the mass transfer coefficient and the driving force for mass transfer.
In this exercise, the water evaporation rate helps determine how much water turns into vapor as air moves through the tube. The computation involves understanding the balance of forces that encourage water molecules to escape from the liquid surface. Engineers and scientists use this data to design systems like humidifiers or drying systems effectively.
Calculate the evaporation rate using:

• The mass transfer coefficient, which is a measure of how readily molecules transfer from one phase to another.
• The log-mean driving force, which accounts for, the difference in concentrations driving the mass transfer.

By understanding these elements, we can optimize processes for efficiency and ensure controlled evaporation rates suitable for industrial applications.