Question

Consider a flow over a surface with the velocity and temperature profiles given as $$ \begin{aligned} &u(y)=C_{1}\left(y+y^{2}-y^{3}\right) \\ &T(y)=C_{2}-e^{-2 C_{2} y} \end{aligned} $$ where the coefficients \(C_{1}\) and \(C_{2}\) are constants. Determine the expressions for the friction coefficient \(\left(C_{f}\right)\) and the convection heat transfer coefficient \((h)\).

Short answer

Question: Determine the expressions for the friction coefficient (Cf) and the convection heat transfer coefficient (h) using the given velocity and temperature profiles. Answer: The expressions for the friction coefficient (Cf) and the convection heat transfer coefficient (h) are: $$ C_f = \frac{2\mu C_1}{ρu^2_\infty} $$ $$ h = - k (2 C_{2}^2) $$

Step-by-step solution

Calculate wall shear stress (τw) using the velocity profile

From the given velocity profile \(u(y)=C_{1}\left(y+y^{2}-y^{3}\right)\), we can determine the wall shear stress \(τ_w\) by differentiating the velocity profile with respect to y and then evaluating it at y = 0. To find \(τ_w\), we first find the velocity gradient: $$\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{\mathrm{d}(C_{1}(y + y^{2} - y^{3}))}{\mathrm{d}y} = C_{1}(1 + 2y - 3y^{2})$$ Now, to find the wall shear stress \(τ_w\), we take the velocity gradient at y = 0, $$ τ_w = \mu \left.\frac{\mathrm{d}u}{\mathrm{d}y}\right|_{y=0} = \mu C_{1} (1 + 2\cdot0 - 3\cdot0^{2}) = \mu C_{1}$$ where \(\mu\) is the dynamic viscosity.

Calculate the friction coefficient (Cf)

The friction coefficient \(C_f\) is defined as: $$ C_f = \frac{2τ_w}{ρu^2_\infty} $$ We can use the definition of the Reynolds number using the displacement thickness \(\delta\) to relate \(u_\infty\) to \(C_1\): $$ Re_\delta = \frac{\rho u_\infty \delta}{\mu} = \frac{u_\infty \delta}{\nu} $$ where \(ρ\) is the fluid density, \(u_\infty\) is the free stream velocity, \(\nu\) is the kinematic viscosity, and \(\delta\) represents the displacement thickness. From the given velocity profile and our calculation of \(τ_w\), we have \(τ_w = \mu C_1\), and we can express \(C_1\) in terms of \(u_\infty\) as: $$ C_1 = \frac{τ_w}{\mu} = \frac{ρu^2_\infty C_f}{2} $$ So the relationship between Cf and \(C_1\) is: $$ C_f = \frac{2\mu C_1}{ρu^2_\infty} $$

Calculate the temperature gradient at the surface (dT/dy)

From the given temperature profile \(T(y)=C_{2}-e^{-2 C_{2} y}\), we can determine the temperature gradient at the surface by differentiating the temperature profile with respect to y and evaluating it at y = 0. $$\frac{\mathrm{d}T}{\mathrm{d}y} = \frac{\mathrm{d}(C_{2}-e^{-2 C_{2} y})}{\mathrm{d}y} = 2C_{2}^2e^{-2 C_{2} y}$$ Now, evaluating the temperature gradient at the surface (y = 0): $$ \left.\frac{\mathrm{d}T}{\mathrm{d}y}\right|_{y=0} = 2C_{2}^2e^{-2 C_{2} \cdot0} = 2C_{2}^2 $$

Calculate the convection heat transfer coefficient (h)

To find the convection heat transfer coefficient (h), we use the relationship involving the temperature gradient: $$ h = -k \left.\frac{\mathrm{d}T}{\mathrm{d}y}\right|_{y=0} $$ where k is the thermal conductivity. Using the temperature gradient expression from Step 3: $$ h = - k (2 C_{2}^2) $$ Thus, we have determined the expressions for the friction coefficient \(\left(C_{f}\right)\) and the convection heat transfer coefficient \((h)\) as: $$ C_f = \frac{2\mu C_1}{ρu^2_\infty} $$ $$ h = - k (2 C_{2}^2) $$