Question

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.

Short answer

Answer: The heat transfer coefficient is \(19.101 \frac{\text{W}}{\text{m}^2\cdot\text{K}}\).

Step-by-step solution

Calculate the mass transfer rate of naphthalene

First, we need to determine the mass transfer rate of naphthalene, which can be found using the mass loss and the time the experiment took: Mass loss of naphthalene: \(100\) g = \(0.1\) kg Time: \(45\) min = \(2700\) s Mass transfer rate: \(\dot{m} = \frac{\text{Mass loss}}{\text{Time}} = \frac{0.1 \text{ kg}}{2700 \text{ s}} = 3.7037 \times 10^{-5} \frac{\text{kg}}{\text{s}}\)

Find the mass transfer coefficient

The mass transfer coefficient, \(k_m\), can be found using the equation for mass transfer rate, \(\dot{m} = k_m A (p_A - p_{A,s})\), where \(A\) is the body surface area, and \(p_A\) and \(p_{A,s}\) are the vapor pressures at the free stream and at the surface, respectively. Since the air is kept at \(25^{\circ} \mathrm{C}\), the vapor pressure at the surface, \(p_{A,s}\), is the vapor pressure of naphthalene at this temperature, which is given as \(11 \mathrm{~Pa}\). The vapor pressure in the free stream, \(p_A\), is assumed to be zero because the air is dry. So we have: \(k_m = \frac{\dot{m}}{A (p_{A,s} - p_A)} = \frac{3.7037 \times 10^{-5} \frac{\text{kg}}{\text{s}}}{0.75 \text{ m}^2 (11 \text{ Pa})} = 4.493 \times 10^{-6} \frac{\text{kg}}{\text{m}\cdot \text{s}\cdot\text{Pa}}\)

Determine the Reynolds number

Now, we find the Reynolds number, \(Re\), using the given free-stream velocity, \(U_\infty = 2 \frac{\text{m}}{\text{s}}\), the mass diffusivity of naphthalene, \(D_{AB} = 0.61 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and assuming an air density of \(\rho = 1.2 \frac{\text{kg}}{\text{m}^3}\) (approximately the air density at standard conditions and \(25^{\circ} \mathrm{C}\)): \(Re = \frac{U_\infty \rho}{D_{AB}} = \frac{2 \frac{\text{m}}{\text{s}} \cdot 1.2 \frac{\text{kg}}{\text{m}^3}}{0.61 \times 10^{-5} \frac{\text{m}^2}{\text{s}}} = 3.9344 \times 10^5\)

Use the Chilton-Colburn analogy to find the heat transfer coefficient

Finally, we use the Chilton-Colburn analogy, which states that the ratio of heat transfer coefficient, \(h\), to mass transfer coefficient, \(k_m\), is equal to the ratio of thermal conductivity, \(k\), to mass diffusivity, \(D_{AB}\): \(h = \frac{k}{D_{AB}} k_m\) Assuming an air thermal conductivity of \(k = 0.026 \frac{\text{W}}{\text{m}\cdot\text{K}}\) (approximately the air thermal conductivity at \(25^{\circ} \mathrm{C}\)), we get: \(h = \frac{0.026 \frac{\text{W}}{\text{m}\cdot\text{K}}}{0.61 \times 10^{-5} \frac{\text{m}^2}{\text{s}}} (4.493 \times 10^{-6} \frac{\text{kg}}{\text{m}\cdot \text{s}\cdot\text{Pa}}) = 19.101 \frac{\text{W}}{\text{m}^2\cdot\text{K}}\) Thus, the heat transfer coefficient under the same flow conditions over the same geometry is \(19.101 \frac{\text{W}}{\text{m}^2\cdot\text{K}}\).