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In an experiment, a sphere of crystalline sodium chloride \((\mathrm{NaCl})\) was suspended in a stirred tank filled with water at \(20^{\circ} \mathrm{C}\). Its initial mass was \(100 \mathrm{~g}\). In 10 minutes, the mass of the sphere was found to have decreased by 10 percent. The density of \(\mathrm{NaCl}\) is \(2160 \mathrm{~kg} / \mathrm{m}^{3}\). Its solubility in water at $20^{\circ} \mathrm{C}\( is \)320 \mathrm{~kg} / \mathrm{m}^{3}$. Use these results to obatin an average value for the mass transfer coefficient.

Short Answer

Expert verified
Answer: The average mass transfer coefficient (k) can be expressed as a function of the mass transfer area (A) using the following formula: k = (1/60000 kg/s) / (A *((ρ_s / M_s) - C_w)). To find the exact value of k, more information about the geometry of the sodium chloride sphere and the mass transfer area (A) is needed.

Step by step solution

01

Find the mass transfer rate

In 10 minutes, the mass of the sodium chloride sphere decreased by 10 percent, so the mass transfer rate is the amount of mass transferred per minute. Since the initial mass of the sphere is 100 g, the mass transferred in 10 minutes is 10% of 100 g, or 10 g. Therefore, the mass transfer rate R_m is: R_m = 10 g / 10 min = 1 g/min
02

Find the driving force

The driving force for mass transfer is the difference between the sodium chloride concentration in the sphere and the solubility of sodium chloride in water. The concentration of sodium chloride in the sphere is given by its density ρ_s and molar mass M_s: C_s = ρ_s / M_s The solubility of sodium chloride in water is given as 320 kg/m³: C_w = 320 kg/m³ The driving force ΔC is then the difference between the concentration in the sphere and the solubility in water: ΔC = C_s - C_w
03

Convert units

To find the average mass transfer coefficient, we need to convert the mass transfer rate, R_m, and the density of sodium chloride, ρ_s, to the same units. We will convert grams to kilograms and minutes to seconds: 1 g/min × (1 kg / 1000 g) × (1 min / 60 s) = 1/60000 kg/s Now, we have the mass transfer rate in kg/s.
04

Calculate the mass transfer coefficient

To find the average mass transfer coefficient, k, we will divide the mass transfer rate by the driving force. Since we don't have the information about the mass transfer area (A), we will express k as a function of A: k = R_m / (A * ΔC) Replacing R_m and ΔC with their values: k = (1/60000 kg/s) / (A *((ρ_s / M_s) - C_w)) Now we have the average mass transfer coefficient as a function of the mass transfer area. To find the exact value of k, we would need more information about the geometry of the sodium chloride sphere and the mass transfer area (A).

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Most popular questions from this chapter

The pressure in a pipeline that transports helium gas at a rate of $7 \mathrm{lbm} / \mathrm{s}\( is maintained at \)14.5$ psia by venting helium to the atmosphere through a \(0.4\)-in-internal-diameter tube that extends $30 \mathrm{ft}$ into the air. Assuming both the helium and the atmospheric air to be at \(80^{\circ} \mathrm{F}\), determine \((a)\) the mass flow rate of helium lost to the atmosphere through the tube, (b) the mass flow rate of air that infiltrates into the pipeline, and (c) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation.

The average heat transfer coefficient for airflow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton-Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at \(1 \mathrm{~atm}\) at a free-stream velocity of $2 \mathrm{~m} / \mathrm{s}$ over a body covered with a layer of naphthalene. The surface area of the body is \(0.75 \mathrm{~m}^{2}\), and it is observed that $100 \mathrm{~g}\( of naphthalene has sublimated in \)45 \mathrm{~min}$. During the experiment, both the body and the air were kept at \(25^{\circ} \mathrm{C}\), at which the vapor pressure and mass diffusivity of naphthalene are $11 \mathrm{~Pa}\( and \)D_{A B}=0.61 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$, respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry.

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