Question

The average heat transfer coefficient for air flow 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

Question: Determine the average heat transfer coefficient for air flow over an odd-shaped body using the Chilton-Colburn analogy. The sublimation rate of naphthalene is 0.003 kg/s, the surface area is 2 m², time is 3600 s, vapor pressure is 500 Pa, and mass diffusivity 0.25 × 10⁻⁵ m²/s. Assume air properties at 1 atm pressure and 300 K temperature, with R = 287 J/kg·K, and Cp = 1005 J/kg·K. Solution: Step 1: Calculate mass flux at the surface Calculate the mass flux (ṁ) using the given data: $$ \dot{m}=\frac{m_{\text{naphthalene}}}{A t} = \frac{0.003 kg/s}{2 m^2 \cdot 3600s} = 4.17 \times 10^{-7}\, kg/m^2·s $$ Step 2: Calculate mass transfer coefficient First, calculate the molar concentration (C) of the air using the ideal gas law: $$ C=\frac{P}{R T} = \frac{101325 \,Pa}{287 \,J/kg \cdot K \cdot 300 \,K} = 1.18 \, kg/m^3 $$ Then, calculate the mass transfer coefficient (k_M) using the given equation: $$ k_{M}=\frac{\dot{m}}{C D_{A B}\left(1-\frac{P_{A}}{P}\right)} = \frac{4.17 \times 10^{-7} \,kg/m^2·s}{1.18 \,kg/m^3 \cdot 0.25 \times 10^{-5} \, m^2/s \cdot \left(1- \frac{500 \,Pa}{101325 \,Pa}\right)} = 1.19 \times 10^{-2} \,m/s $$ Step 3: Use the Chilton-Colburn analogy to calculate the heat transfer coefficient To use the Chilton-Colburn analogy, we assume Re and Pr are constant. Based on this assumption, we can rewrite the analogy as follows: $$ h=k_{M} \cdot\left(\frac{Re_{D} \operatorname{Pr}}{Re_{D} \operatorname{Sc}}\right) = k_{M} \cdot \frac{Pr}{Sc} $$ Now, we can calculate the heat transfer coefficient (h) using the mass transfer coefficient (k_M) and Pr/Sc ratio. We will assume the Pr/Sc ratio to be 0.71, which is typical for air at atmospheric pressure and room temperature: $$ h = k_{M} \cdot \frac{Pr}{Sc} = 1.19 \times 10^{-2} \,m/s \cdot 0.71 = 8.45 \times 10^{-3} \,W/m^2·K $$ FINAL STEP: Report the heat transfer coefficient The average heat transfer coefficient for air flow over the odd-shaped body under the given conditions is 8.45 × 10⁻³ W/m²·K.

Step-by-step solution

Calculate mass flux at the surface

Calculate the mass flux of naphthalene sublimated from the surface of the body using the mass loss data given in the problem statement. The mass flux (ṁ) can be calculated using the mass sublimated (m_naphthalene), surface area (A), and time (t). $$ \dot{m}=\frac{m_{\text{naphthalene}}}{A t} $$

Calculate mass transfer coefficient

The mass transfer coefficient (k_M) can be calculated using the mass flux (ṁ), vapor pressure (P_A), mass diffusivity (D_AB), and air properties at the given temperature and pressure. The molar concentration (C) of the air can be determined using the ideal gas law: $$ C=\frac{P}{R T} $$ The mass transfer coefficient is then calculated using the following equation, which is derived from Fick's law of diffusion: $$ k_{M}=\frac{\dot{m}}{C D_{A B}\left(1-\frac{P_{A}}{P}\right)} $$

Use the Chilton-Colburn analogy to calculate heat transfer coefficient

Based on the Chilton-Colburn analogy, the heat transfer coefficient (h) can be related to the mass transfer coefficient (k_M) using the Reynolds number (Re) and Prandtl (Pr) and Schmidt (Sc) numbers. We can also approximate air as an ideal gas with a constant heat capacity ratio (C_p/C_v). The relation is given as: $$ h=k_{M} \cdot\left(\frac{Re_{D} \operatorname{Pr}}{Re_{D} \operatorname{Sc}}\right) $$ Re and Pr can be calculated from given gas properties at the free stream velocity. Calculate the heat transfer coefficient based on mass transfer coefficient and gas properties.

FINAL STEP: Report the heat transfer coefficient

After completing the above steps, report the calculated heat transfer coefficient (h) as the average heat transfer coefficient for air flow over the odd-shaped body under the given conditions.

Key concepts

Chilton-Colburn Analogy

In the study of fluid flow and heat transfer, the Chilton-Colburn analogy plays a fundamental role. This analogy provides a powerful way to relate heat and mass transfer processes. It is particularly useful when one type of transfer is easy to measure, but the other is not. This happens frequently in experimental setups where conditions perfectly suited for measuring either heat or mass transfer are present, but not both.

The analogy ties together dimensionless numbers such as the Nusselt (Nu), Reynolds (Re), and Prandtl (Pr) numbers for heat transfer, with a similar set of numbers for mass transfer, which include the Sherwood (Sh), Reynolds, and Schmidt (Sc) numbers. The main expression derived from this analogy is:

• The heat transfer coefficient (h) is proportional to the mass transfer coefficient (k_M)
• Multiplied by the ratio of Reynolds and Prandtl numbers: \( Re \cdot \frac{Pr}{Sc} \).

This relationship is advantageous because it allows engineers and scientists to substitute difficult heat transfer measurements with more feasible mass transfer experiments.

Mass Transfer Coefficient

The mass transfer coefficient is a critical parameter in the analysis of mass transfer processes. It represents the rate at which a substance moves from one phase to another. In the case of the exercise, naphthalene is sublimating from a solid to a gas phase.

To calculate the mass transfer coefficient (\(k_{M}\)), we rely on Fick's Law of Diffusion, which considers:

• Mass flux of the sublimating naphthalene (determined experimentally)
• Molar concentration of air, derived using the ideal gas law
• The diffusivity of naphthalene vapor in air

The formula is given by:\[k_{M} = \frac{\dot{m}}{C \cdot D_{AB} \cdot \left(1 - \frac{P_A}{P}\right)}\]This coefficient is crucial as it directly influences the conversion of mass transfer data into heat transfer coefficients using the Chilton-Colburn analogy.

Reynolds Number

The Reynolds number is a dimensionless quantity that helps predict flow patterns in different fluid flow situations. It's an essential element when analyzing both heat and mass transfer, especially when using the Chilton-Colburn analogy.

Calculated as the ratio of inertial forces to viscous forces, it can be expressed as:\[Re = \frac{\rho u L}{\mu}\]Where:

• \(\rho\) is the fluid density,
• \(u\) is the velocity,
• \(L\) is the characteristic length,
• \(\mu\) is the dynamic viscosity.

In this exercise scenario, the Reynolds number is calculated based on known properties of air and flow conditions over the body. It influences both the heat and mass transfer characteristics by indicating whether the flow is laminar or turbulent.

Prandtl Number

The Prandtl number is a dimensionless number that relates the momentum diffusivity (viscosity) and the thermal diffusivity of a fluid. It offers insight into how easily heat can be transferred in fluid flows and thus is crucial in the context of heat transfer.

This number is given by the formula:\[Pr = \frac{\mu \cdot c_p}{k}\]Where:

• \(\mu\) is the dynamic viscosity,
• \(c_p\) is the specific heat capacity at constant pressure,
• \(k\) is the thermal conductivity.

A high Prandtl number indicates that the fluid's thermal diffusivity is low compared to its momentum diffusivity, meaning the substance retains heat effectively. It's used to relate mass and heat transfer processes alongside the Reynolds number in the Chilton-Colburn analogy.

Schmidt Number

The Schmidt number is a dimensionless number that describes the ratio of momentum diffusivity (viscosity) to mass diffusivity. It plays an analogous role to the Prandtl number, but for mass transfer instead of heat transfer.

The Schmidt number is calculated by:\[Sc = \frac{\mu}{\rho \cdot D_{AB}}\]Where:

• \(\mu\) is the dynamic viscosity,
• \(\rho\) is the density,
• \(D_{AB}\) is the diffusivity.

In the exercise, the Schmidt number characterizes the mass transfer process of naphthalene sublimation. When combined with other dimensionless numbers like the Reynolds number, it helps link mass transfer to heat transfer via the Chilton-Colburn analogy.