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 \(x=x_{\mathrm{cr}}\) : (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, \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (f) Average convection heat transfer coefficient, W/m² \(\cdot \mathrm{K}\) (g) Rate of convective heat transfer, W

From the given data, we have the air properties as temperature \(T_\infty = 15^\circ C\), pressure \(P_\infty = 1 atm\), and velocity \(U_\infty = 3.0 m/s\). The plate's width is \(w = 0.3 m\) and its temperature is \(T_w = 65^\circ C\). First, we need to determine the critical distance \(x_{cr}\) using the Blasius boundary layer equations. \(x_{\mathrm{cr}}\) can be determined using the Reynolds number, \(Re_{x_{\mathrm{cr}}}\), at which the transition from laminar to turbulent flow occurs. This value is usually taken as \(Re_{x_{\mathrm{cr}}} = 5 \times 10^5\). To find the critical distance \(x_{cr}\), we first find the kinematic viscosity, \(\nu\), of air at the film temperature, \(T_f = (\frac{T_w + T_\infty}{2})\). The film temperature is a simplification where we assume the temperature distribution within the boundary layer to be linear. Using air property table, at \(T_f = 40^\circ C\), we get the kinematic viscosity \(\nu = 1.96 \times 10^{-5} m^2/s\). Now, we can determine the critical distance, \(x_{\mathrm{cr}}\): \(x_{\mathrm{cr}} = \dfrac{Re_{x_{\mathrm{cr}}} \cdot \nu}{U_\infty} = \dfrac{5 \times 10^5 \cdot 1.96 \times 10^{-5}}{3} = 0.327 m\)

The hydrodynamic boundary layer thickness \(δ\) can be calculated using the Blasius solution for laminar boundary layer: \(δ = \dfrac{5}{\sqrt{Re_x}} x\) Therefore, at \(x_{cr}\), we have: \(δ_{cr} = \dfrac{5}{\sqrt{Re_{x_{cr}}}} x_{cr} = \dfrac{5}{\sqrt{5 \times 10^5}} \cdot 0.327 = 7.32 \times 10^{-3} m\)

The local friction coefficient \(C_f\) can be determined using the Blasius solution for laminar boundary layers: \(C_f = \dfrac{0.664}{\sqrt{Re_x}}\) So, at \(x_{cr}\), we have: \(C_{f_{cr}} = \dfrac{0.664}{\sqrt{5 \times 10^5}} = 0.000938\)

The average friction coefficient can be found using a similar formula: \(C_{f_{ave}} = \dfrac{1.328}{\sqrt{Re_x}}\) So, at \(x_{cr}\): \(C_{f_{ave_{cr}}} = \dfrac{1.328}{\sqrt{5 \times 10^5}} = 0.00188\)

The total drag force due to friction can be found using the formula: \(D_{fr} = \dfrac{1}{2} ρ U_\infty^2 w C_{f_{ave}} x_{\mathrm{cr}}\) Using the air properties at the initial condition, \(ρ = 1.225 kg/m^3\). We substitute the values into the formula: \(D_{fr} = \dfrac{1}{2} \cdot 1.225 \cdot 3^2 \cdot 0.3 \cdot 0.00188 \cdot 0.327 = 0.517 N\)

The local convection heat transfer coefficient \(h_x\) can be found using the Reynolds and Prandtl numbers: \(h_x = \dfrac{0.332 k}{x} Re_x^{1/2} Pr^{1/3}\) At the film temperature, \(T_f = 40^\circ C\), we have \(k = 0.0285 W/m\cdot K\) and \(Pr = 0.709\). Then, at \(x_{cr}\), we get: \(h_{x_{cr}} = \dfrac{0.332 \cdot 0.0285}{0.327} \cdot (5 \times 10^5)^{1/2} \cdot 0.709^{1/3} = 153.7 W/m^2 \cdot K\)

The average convection heat transfer coefficient can be found using a similar formula: \(h_{ave} = \dfrac{0.664 k}{x} Re_x^{1/2} Pr^{1/3}\) So, at \(x_{cr}\): \(h_{ave_{cr}} = \dfrac{0.664 \cdot 0.0285}{0.327} \cdot (5 \times 10^5)^{1/2} \cdot 0.709^{1/3} = 306.7 W/m^2 \cdot K\)

The rate of convective heat transfer \(Q\) can be calculated using the average convection heat transfer coefficient and temperature difference: \(Q = h_{ave} \cdot w \cdot x_{cr} \cdot (T_w - T_\infty)\) Substituting the values: \(Q = 306.7 \cdot 0.3 \cdot 0.327 \cdot (65 - 15) = 755.5 W\) In conclusion, the following quantities are computed at \(x = x_{cr}\): (a) Hydrodynamic boundary layer thickness, \(δ_{cr} = 7.32 \times 10^{-3} m\); (b) Local friction coefficient, \(C_{f_{cr}} = 0.000938\); (c) Average friction coefficient, \(C_{f_{ave_{cr}}} = 0.00188\); (d) Total drag force due to friction, \(D_{fr} = 0.517 N\); (e) Local convection heat transfer coefficient, \(h_{x_{cr}} = 153.7 W/m^2 \cdot K\); (f) Average convection heat transfer coefficient, \(h_{ave_{cr}} = 306.7 W/m^2 \cdot K\); (g) Rate of convective heat transfer, \(Q = 755.5 W\).