Question

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

Short answer

Question: Determine the values of the following quantities at a distance of 0.3 m from the leading edge of the heated flat plate: (a) hydrodynamic boundary layer thickness, (b) local friction coefficient, (c) average friction coefficient, (d) total drag force due to friction, (e) local convection heat transfer coefficient, (f) average convection heat transfer coefficient, and (g) rate of convective heat transfer. Answer: The required quantities are as follows: (a) Hydrodynamic boundary layer thickness: \(\delta = 0.0021\ \mathrm{m}\) (b) Local friction coefficient: \(C_{f} = 0.00293\) (c) Average friction coefficient: \(C_{f_{avg}} = 0.00584\) (d) Total drag force due to friction: \(F_{d} = 0.0474\ \mathrm{N}\) (e) Local convection heat transfer coefficient: \(h_{x} = 93.7\ \mathrm{W/ m^2 K}\) (f) Average convection heat transfer coefficient: \(h_{avg} = 189.4\ \mathrm{W/ m^2 K}\) (g) Rate of convective heat transfer: \(q_{conv} = 849.4\ \mathrm{W}\)

Step-by-step solution

Determine the properties of the air

First, let's find the film temperature by averaging the temperatures of the air and the heated plate: $$T_{film} = \frac{T_{air} + T_{plate}}{2}$$ Using the provided values, calculate the film temperature: $$T_{film} = \frac{15 + 65}{2} = 40^{\circ} \mathrm{C}$$ Now, look up the properties of air at the film temperature \(40^{\circ} \mathrm{C}\). Here are the necessary values: $$\rho = 1.127 \, \mathrm{kg/ m^3}$$ $$\mu = 1.97 × 10^{-5} \, \mathrm{kg/ m\cdot s}$$ $$k= 0.028 \, \mathrm{W/ m\cdot K}$$ $$C_{p}= 1007 \, \mathrm{J/ kg\cdot K}$$

Calculate the Reynolds number

Now, we need to find the Reynolds number at \(x = 0.3\, \mathrm{m}\). The Reynolds number can be calculated as: $$Re_{x} = \frac{\rho U x}{\mu}$$ Substituting the known values, we get: $$Re_{x} = \frac{1.127\, (3)(0.3)}{1.97 × 10^{-5}} = 5.14 \times 10^4$$ Since the Reynolds number is less than \(5 \times 10^5\), the flow is laminar.

Calculate part (a) - Hydrodynamic boundary layer thickness

To find the hydrodynamic boundary layer thickness, apply the laminar boundary layer correlation: $$\delta = \frac{5x}{\sqrt{Re_{x}}}$$ Substitute the values to get: $$\delta = \frac{5\, (0.3)}{\sqrt{5.14 × 10^4}} = 0.0021 \, \mathrm{m}$$

Calculate part (b) - Local friction coefficient

To find the local friction coefficient, apply the following formula: $$C_{f} = \frac{0.664}{\sqrt{Re_{x}}}$$ Using the calculated Reynolds number, we get: $$C_{f} = \frac{0.664}{\sqrt{5.14 \times 10^4}} = 0.00293$$

Calculate part (c) - Average friction coefficient

Calculate the average friction coefficient using the following formula: $$C_{f_{avg}} = \frac{1.328}{\sqrt{Re_{x}}}$$ Using the calculated Reynolds number, we get: $$C_{f_{avg}} = \frac{1.328}{\sqrt{5.14 \times 10^4}} = 0.00584$$

Calculate part (d) - Total drag force due to friction

To find the total drag force due to friction, apply this formula: $$F_{d} = C_{f_{avg}} \frac{1}{2} \rho U^{2} A$$ Here, \(A = x \, w\), since the plate is \(0.3\, \mathrm{m}\) wide. Substituting the known values: $$F_{d} = 0.00584 \frac{1}{2} (1.127) (3)^{2} (0.3) (0.3) = 0.0474 \, \mathrm{N}$$

Calculate part (e) - Local convection heat transfer coefficient

To find the local convection heat transfer coefficient, apply the following correlation for laminar flow: $$h_{x} = C_{hf}\left(\frac{k^3 \rho C_{p} Ux}{\mu}\right)^{1/2}$$ Here, \(C_{hf} = 0.664\). Substituting the values, we get: $$h_{x} = 0.664\left(\frac{0.028^3 \cdot 1.127 \cdot 1007 \cdot 3 \cdot 0.3}{1.97 \times 10^{-5}}\right)^{1/2} = 93.7 \, \mathrm{W/ m^2 K}$$

Calculate part (f) - Average convection heat transfer coefficient

To find the average convection heat transfer coefficient, apply the following correlation for laminar flow: $$h_{avg} = C_{h_{avg}}\left(\frac{k^3 \rho C_{p} Ux}{\mu }\right)^{1/2}$$ Here, \(C_{h_{avg}} = 0.664\). Substituting the values, we get: $$h_{avg} = 1.128\left(\frac{0.028^3 \cdot 1.127 \cdot 1007 \cdot 3 \cdot 0.3}{1.97 \times 10^{-5}}\right)^{1/2} = 189.4 \, \mathrm{W/ m^2 K}$$

Calculate part (g) - Rate of convective heat transfer

Finally, to find the rate of convective heat transfer, use this formula: $$q_{conv} = h_{avg} A \Delta T$$ Calculate the heat transfer rate using the known values: $$q_{conv} = (189.4) (0.3) (0.3) (65 - 15) = 849.4 \, \mathrm{W}$$ In summary, the required quantities are as follows: (a) \(\delta = 0.0021\ \mathrm{m}\) (b) \(C_{f} = 0.00293\) (c) \(C_{f_{avg}} = 0.00584\) (d) \(F_{d} = 0.0474\ \mathrm{N}\) (e) \(h_{x} = 93.7\ \mathrm{W/ m^2 K}\) (f) \(h_{avg} = 189.4\ \mathrm{W/ m^2 K}\) (g) \(q_{conv} = 849.4\ \mathrm{W}\)

Key concepts

Hydrodynamic Boundary Layer

When a fluid, such as air, flows over a solid surface like a plate, a layer of fluid close to the surface begins to behave differently compared to the rest of the flow. This region is known as the hydrodynamic boundary layer. Within this layer, the velocity of the fluid changes from zero at the surface (due to the no-slip condition) to the freestream velocity farther away from the surface.
The thickness of this boundary layer, denoted as \( \delta \), indicates how far this effect reaches into the flow and is influenced by several factors including flow velocity, viscosity, and distance along the surface.
When the flow is laminar, as described in the exercise, the boundary layer thickness at a certain point \( x \) along the surface can be calculated using the formula \( \delta = \frac{5x}{\sqrt{Re_{x}}} \), where \( Re_{x} \) is the Reynolds number at that position. In this problem, calculating this value shows us the boundary layer is about \( 0.0021 \) meters thick at \( x = 0.3 \) meters, indicating a relatively thin boundary layer due to the laminar nature of the flow.

Reynolds Number

The Reynolds number \( (Re) \) is a dimensionless quantity in fluid mechanics that gives us insight into the flow regime of the fluid around a surface. It's an indicator of whether the flow will be laminar (smooth and orderly) or turbulent (chaotic and mixed).
The Reynolds number is calculated using the expression \( Re_{x} = \frac{\rho U x}{\mu} \), where \( \rho \) is the fluid density, \( U \) is the velocity of the fluid, \( x \) is the characteristic length, and \( \mu \) is the dynamic viscosity. A critical point is usually around \( 5 \times 10^5 \), above which flow tends to become turbulent.
In our air flow problem over a flat plate, the Reynolds number at \( x = 0.3 \) meters is about \( 5.14 \times 10^4 \), indicating that the flow is laminar. Understanding the Reynolds number is essential because it directs which formulas and approaches we can use to calculate other quantities, like boundary layer thickness and heat transfer coefficients.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient \( (h) \) is fundamental in determining how effectively heat is transferred from a surface to a fluid or vice versa. It is a measure of the convective heat transfer occurring between the surface and the fluid, and its value is influenced by factors such as fluid properties, flow velocity, and the temperature difference between the fluid and the surface.
There are two main types of convection heat transfer coefficients discussed in this exercise: **local** and **average**. The local convection heat transfer coefficient \( (h_x) \) gives the rate of heat transfer at a specific point on the surface, while the average coefficient \( (h_{avg}) \) provides a mean rate over the entire surface or significant portion of it.
For laminar flow over a flat plate, these coefficients can be computed from empirical correlations contingent on the Reynolds number and Prandtl number (another dimensionless number). For the local coefficient \( h_x \), the empirical formula is \( h_{x} = C_{hf}\left(\frac{k^3 \rho C_{p} Ux}{\mu}\right)^{1/2} \), where \( C_{hf} = 0.664 \). The average coefficient follows a similar structure but with its own correction factor. In the given exercise, these values were found to be \( 93.7 \ \mathrm{W/ m^2 K} \) and \( 189.4 \ \mathrm{W/ m^2 K} \), respectively, showing how the plate's temperature drives heat into the airflow.