Question

Air \((k=0.028 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7)\) at \(50^{\circ} \mathrm{C}\) flows along a 1 -m-long flat plate whose temperature is maintained at \(20^{\circ} \mathrm{C}\) with a velocity such that the Reynolds number at the end of the plate is 10,000 . The heat transfer per unit width between the plate and air is (a) \(20 \mathrm{~W} / \mathrm{m}\) (b) \(30 \mathrm{~W} / \mathrm{m}\) (c) \(40 \mathrm{~W} / \mathrm{m}\) (d) \(50 \mathrm{~W} / \mathrm{m}\) (e) \(60 \mathrm{~W} / \mathrm{m}\)

Short answer

Based on the given information and calculations, the heat transfer per unit width between the plate and air is approximately 92.01 W/m. However, from the given options (a, b, c, d, e), the closest value would be 60 W/m (option e).

Step-by-step solution

Find the dynamic viscosity of the air

Using Sutherland's formula for the dynamic viscosity of gases, we can find the dynamic viscosity (µ) of air at 50°C: $$\mu = 1.458 \times 10^{-6} \frac{T^{3/2}}{T + 110.4}$$ Here, T is the absolute temperature in Kelvin, which is \(273.15+50=323.15K\). Plug in the value, and we have: $$\mu = 1.458 \times 10^{-6} \times \frac{323.15^{3/2}}{323.15 + 110.4} = 2.092 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}$$.

Obtain the velocity and kinematic viscosity of the air

Using the Reynolds number definition, we can find the velocity (U) and kinematic viscosity (ν) of the air: $$\operatorname{Re} = \frac{UL}{\nu}$$ Rearranging the equation, we have: $$\nu = \frac{UL}{\operatorname{Re}}$$ Plug in the given values (U = velocity, L = 1m, Re = 10,000), and we get: $$\nu = \frac{U}{10,000}$$ However, we also know that: $$\nu = \frac{\mu}{\rho}$$ We need to find the density of air at the given temperature. Let's use the ideal gas law to estimate the air density (ρ) at 50°C: $$\rho = \frac{P}{RT} $$ Assuming atmospheric pressure, P = \(101325 \mathrm{Pa}\), the gas constant for air, R = \(287 \mathrm{J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\), and T = \(323.15 \mathrm{K}\). Calculate the density: $$\rho = \frac{101325}{287 \cdot 323.15} = 1.122 \mathrm{kg} \cdot \mathrm{m}^{-3}$$ Now, calculate the kinematic viscosity, ν: $$\nu = \frac{2.092 \times 10^{-5}}{1.122} = 1.866 \times 10^{-5} \mathrm{m}^2 \cdot \mathrm{s}^{-1}$$ And finally, we can find the velocity (U): $$U = 10,000 \cdot 1.866 \times 10^{-5} = 0.1866 \mathrm{m} \cdot \mathrm{s}^{-1}$$

Calculate hydrodynamic boundary layer thickness(delta)

Using the Blasius solution, find the thickness of the boundary layer (δ) at the end of the plate: $$\delta = 0.37x\operatorname{Re}_{x}^{-1/5}$$ Where x is the distance along the plate, and \(\operatorname{Re}_x = \frac{Ux}{\nu}\), substitute the necessary values, we have: $$\delta = 0.37(1)\left(\frac{0.1866 \cdot 1}{1.866 \times 10^{-5}}\right)^{-1/5} = 7.66 \times 10^{-3}\mathrm{m}$$

Determine thermal boundary layer thickness(delta_t)

Now, we calculate the thermal boundary layer thickness (δ_t), knowing that: $$\delta_t = \frac{\delta}{(\operatorname{Pr})^{1/3}}$$ Substitute the necessary values (Pr = 0.7), and we have: $$\delta_t = \frac{7.66 \times 10^{-3}}{(0.7)^{1/3}} = 9.13 \times 10^{-3}\mathrm{m}$$

Compute the heat transfer coefficient (h)

Knowing the thermal conductivity (k) of the air and the thickness of the thermal boundary layer, we can find the heat transfer coefficient (h) : $$h = \frac{k}{\delta_t}$$ Plug in the values (k = 0.028W/m.K), and we have: $$h = \frac{0.028}{9.13 \times 10^{-3}} = 3.067 \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}$$

Calculate the heat transfer per unit width (Q)

We now compute the required heat transfer (Q) per unit width for the plate: $$Q = h \cdot \Delta T \cdot W$$ Here, \(\Delta T\) is the temperature difference between the air and the plate (50°C - 20°C = 30°C), and W = 1 m (since we are considering the heat transfer per unit width). Calculate the heat transfer per unit width, Q: $$Q = 3.067 \cdot 30 \cdot 1 = 92.01 \mathrm{W} \cdot \mathrm{m}^{-1}$$ The heat transfer per unit width between the plate and air is approximately 92.01 W/m, which is not among the given options (a, b, c, d, e). However, from these options, the closest value would be 60 W/m (option e).

Key concepts

Reynolds number

The Reynolds number is a fundamental concept in fluid dynamics, which helps predict the flow pattern in different fluid flow situations. It is a dimensionless number used to determine whether a flow is laminar or turbulent. The Reynolds number, often denoted as \( ext{Re} \), is calculated using the equation:

• \( ext{Re} = \frac{UL}{u} \)
• \( U \) is the fluid flow velocity
• \( L \) is the characteristic linear dimension (such as the length of a plate)
• \( u \) is the kinematic viscosity of the fluid

In this scenario, the Reynolds number of 10,000 indicates a combination of moderate velocity, a plate length of 1 meter, and air properties which suggest likely laminar flow across the plate. Understanding when the flow switches between laminar to turbulent, around a Reynolds number approximately between 2,300 to 4,000, is crucial in determining the flow's behavior, which directly affects heat transfer rates.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. In fluids like air, it tells us how easily the heat energy is transmitted through the particles. It is denoted by \( k \) and has units of \( ext{W/m} \, ext{K} \), reflecting how much heat (in watts) is transferred per meter distance for each degree of temperature difference. In this exercise, air has a thermal conductivity of \( 0.028 \, \text{W/m} \, \text{K} \).
This value plays a key role in calculating the heat transfer coefficient, which ultimately helps determine how efficiently heat moves from the air to the plate. Thus, understanding the material's thermal conductivity is essential for designing and analyzing heat exchangers, insulations, and any system where heat transfer is a critical factor.

Boundary Layer Thickness

Boundary layers are thin zones at the fluid-solid interfaces where the velocity of the fluid changes from zero at the solid surface (due to the no-slip condition) to the free stream velocity. For heat transfer calculations, there are two types to consider: the hydrodynamic boundary layer and the thermal boundary layer.

• Hydrodynamic Boundary Layer: This pertains to the flow velocity transition. Its thickness \( \delta \) is vital in determining how fluid properties vary near the surface.
• Thermal Boundary Layer: This defines how temperature changes through the layer and affects heat transfer. It's usually denoted as \( \delta_t \).

The thermal boundary layer thickness was calculated using the given Prandtl number \( \text{Pr} \) as:\[ \delta_t = \frac{\delta}{(\text{Pr})^{1/3}} \]In heat transfer applications, the thickness of these layers is crucial as thinner layers typically mean more efficient heat transfer due to less resistance.

Heat Transfer Coefficient

The heat transfer coefficient \( h \) is a parameter describing the heat transfer between a solid surface and a fluid flowing over it. It is typically expressed in units of \( ext{W} \, \text{m}^{-2} \, \text{K}^{-1} \). This coefficient incorporates surface characteristics, fluid properties, and flow conditions in determining the effectiveness of heat transfer processes.
In this exercise, the heat transfer coefficient was calculated by dividing the thermal conductivity by the thermal boundary layer thickness:\[ h = \frac{k}{\delta_t} \]This means, for practical applications, if we can reduce the thickness of the thermal boundary layer (by increasing flow or surface texture), we can enhance the heat transfer rate. A higher heat transfer coefficient indicates more effective energy exchange, crucial in thermal management systems like HVAC units, radiator designs, and packaging for electronics.