Fluid Mechanics and Heat Transfer: Flow Over a Flat Plate

Drag Coefficient and Shear Stress Calculations for Flat Plates

• Flow Over a Flat Plate Parameters:     * The ambient temperature is defined as $T_{\infty} = 10^{\circ}C$.     * The wall temperature is initially defined as $T_w = 10^{\circ}C$.     * The local drag coefficient ($C_{f,x}$) for laminar flow over a flat plate is given by the correlation:         * $C_{f,x} = 0.664 \times Re_x^{-1/2}$     

• The average drag coefficient ($C_{f,L}$) for the entire length of the plate is given by:         * $C_{f,L} = 1.328 \times Re_L^{-1/2}$

• Shear Stress and Force Formulation:     * The shear stress ($\tau_w$) exerted on the fluid by the wall is calculated using the average drag coefficient:         * $\tau_w = \frac{1}{2} \times \rho \times u^2 \times C_{f,L}$     * Where $u$ represents the free stream velocity.     * Substituting the average drag coefficient formula into the shear stress equation yields:         * $\tau_w = \frac{1}{2} \times \rho \times u^2 \times 1.328 \times Re_L^{-1/2}$     * The total force ($F$) is the product of wall shear stress and the surface area:         * $\text{Force} = \text{Stress} \times \text{Area}$     * Assuming a unit length for the width of the plate ($W = 1$), the area is $L \times 1$.     * The total force is derived as:         * $F = \frac{1}{2} \times \rho \times u^2 \times 1.328 \times Re_L^{-1/2} \times L \times 1$

Power Consumption and Velocity Relationships

• Power Derivation:     * Power is defined as the product of force and velocity:         * $\text{Power} = \text{Force} \times \text{Velocity}$     * Substituting the force expression:         * $\text{Power} = \frac{1}{2} \times \rho \times u^2 \times 1.328 \times Re_L^{-1/2} \times L \times 1 \times u$     * Recalling the definition of the Reynolds number ($Re_L = \frac{u \times L}{\nu}$), the power expression becomes:         * $P = 0.664 \times \rho \times u^3 \times \left(\frac{u \times L}{\nu}\right)^{-1/2} \times L$     * Simplifying the powers of velocity ($u$) and length ($L$):         * $P = 0.664 \times \rho \times u^3 \times u^{-1/2} \times L^{-1/2} \times \nu^{1/2} \times L$         * $P = 0.664 \times \rho \times u^{5/2} \times \nu^{1/2} \times L^{1/2}$

• Proportionality:     * From the derived formula, Power ($P$) is proportional to the kinematic viscosity ($\nu$) raised to the power of one-half:         * $P \propto \nu^{1/2}$

Comparative Analysis of Temperature Conditions on Fluid Properties

• Condition Changes (Question 1a):     * Scenario: The wall temperature is increased to $T_w = 90^{\circ}C$.     * The film temperature ($T_f$) is calculated as the average of the wall temperature and the free stream temperature:         * $T_f = \frac{90 + 10}{2} = 50^{\circ}C$     * The properties of water vary between the cold condition ($10^{\circ}C$) and the film condition ($50^{\circ}C$).

• Viscosity Data:     * Dynamic viscosity at $10^{\circ}C$ ($\mu_{10^{\circ}C}$): $0.013\,g/am.s$     * Dynamic viscosity at $50^{\circ}C$ ($\mu_{50^{\circ}C}$): $0.00548\,g/amis$     * Comparing the power requirements between the two states using the proportionality $P \propto \mu^{1/2}$ (assuming constant density):         * $\frac{P_{50^{\circ}C}}{P_{10^{\circ}C}} = \left(\frac{\mu_{50^{\circ}C}}{\mu_{10^{\circ}C}}\right)^{1/2}$         * $\frac{P_{50^{\circ}C}}{P_{10^{\circ}C}} = \left(\frac{0.00548}{0.013}\right)^{1/2} \approx 0.65$

Heat Transfer Nuance and Power Savings Analysis

• Power Savings (Question 1b):     * The relative power at the higher temperature is $65\%$ of the power at the lower temperature ($P_f = 0.65 \times P_c$).     * The power saving is calculated as:         * $P_{saving} = P_c - P_f = P_c - 0.65 \times P_c = 0.35 \times P_c$     * The literal expression for power saving is:         * $P_{saving} = 0.35 \times 0.664 \times \rho \times \nu_c^{1/2} \times u^{5/2} \times L^{1/2}$

• Heating Power via Convection:     * The convection correlation for the Nusselt number ($Nu$) is:         * $Nu = \frac{\bar{h} \times L}{k} = 0.664 \times Re_L^{1/2} \times Pr^{1/3}$     * The average heat transfer coefficient ($\bar{h}$) is:         * $\bar{h} = 0.664 \times \frac{k}{L} \times \left(\frac{u \times L}{\nu}\right)^{1/2} \times Pr^{1/3}$     * The power required for heating is given by:         * $\text{Power (heating)} = \bar{h} \times A \times \Delta T$         * $\text{Power (heating)} = \left(0.664 \times \frac{k}{L} \times Re_L^{1/2} \times Pr^{1/3}\right) \times (L \times 1) \times \Delta T$         * $\text{Power (heating)} = 0.664 \times k \times Re_L^{1/2} \times Pr^{1/3} \times \Delta T$

• Determining Velocity Threshold:     * For the power saved by heating the oil to exceed the power required to do the heating (P_{saving} > P_{heating}):         * 0.35 \times 0.664 \times \rho \times \nu_c^{1/2} \times u^{5/2} \times L^{1/2} > 0.664 \times u^{1/2} \times L^{1/2} \times \nu_f^{-1/2} \times Pr^{1/3} \times k \times \Delta T     * Simplifying common terms:         * 0.35 \times \rho \times \nu_c^{1/2} \times u^2 > \nu_f^{-1/2} \times Pr^{1/3} \times k \times \Delta T     * Given values for the fluid at film temperature:         * $k = 0.64\,W/mK$, $Pr_f = 3.57$, $\Delta T = 80$, $\rho = 1000$.         * $\nu_f = 3.16 \times 10^{-7}\,m^2/s$, $\nu_c = 1.2 \times 10^{-5}\,m^2/s$.     * The inequality becomes:         * u^2 > \frac{(3.16 \times 10^{-7})^{-1/2} \times 3.57^{1/3} \times 0.64 \times 80}{0.35 \times 1000 \times (1.2 \times 10^{-5})^{1/2}}         * u^2 > 3630         * $u \approx 60.3\,m/s$

• Laminar Flow Stability Check:     * To ensure flow remains laminar (Re_L < 5 \times 10^5):         * \frac{u \times L}{\nu} < 5 \times 10^5         * L < \frac{5 \times 10^5 \times 3.16 \times 10^{-7}}{60.3}         * L < 2.62 \times 10^{-3}\,m

Case Study: Convective Heat Transfer over a Surface

• Input Parameters (Question 2):     * $T_{wall} = 50^{\circ}C$     * $T_{\infty} = 20^{\circ}C$     * Film Temperature: $T_{film} = \frac{50 + 20}{2} = 35^{\circ}C$     * Distance along plate: $x = 100\,cm = 1.0\,m$     * Velocity: $V_{\infty} = 0.05\,m/s$     * Dynamic/Kinematic viscosity at $35^{\circ}C$ (interpolated/given): $\nu = 0.00725$

• Reynolds Number Calculation:     * $Re_x = \frac{u \times x}{\nu} = \frac{0.5 \times 100}{0.00725} = 68965.5172$     * Since Re_x < 5 \times 10^5, the flow is confirmed as laminar.

Local and Average Heat Transfer Coefficient Determination

• Nusselt Number Calculations:     * Local Nusselt Number ($Nu_x$):         * $Nu_x = 0.332 \times Pr^{1/3} \times Re_x^{1/2}$         * Using $Pr = 4.87$:         * $Nu_x = 0.332 \times (4.87)^{1/3} \times (68965.5)^{1/2} = 147.78$     * Local Convection Coefficient ($h_x$):         * $h_x = \frac{Nu_x \times k}{x} = \frac{147.78 \times 0.62}{1} = 91.623\,W/m^2K$

• Average Coefficients and Heat Flux:     * Average Heat Transfer Coefficient ($\bar{h}$):         * $\bar{h} = 2 \times h_x = 2 \times 91.623 = 183.246\,W/m^2K$     * Average Heat Flux ($q''$):         * $q'' = \bar{h} \times \Delta T = 183.246 \times (50 - 20) = 5497.4\,W/m^2$     * Total Heat ($q$):         * $q = q'' \times \text{Area} = 5497.4 \times (0.2 \times 1) = 4398\,W$

Boundary Layer Characteristics

• Thickness Calculation:     * The boundary layer thickness ($\delta$) for laminar flow is given by:         * $\delta = 4.92 \times \frac{x}{\sqrt{Re_x}}$         * $\delta = 4.92 \times (68965)^{-1/2} \approx 0.01873\,m = 1.873\,cm$     * Since $\delta \ll L$, the thin boundary layer assumption and theoretical results are justified.

Shear Stress and Heat Flux at Specified Conditions

• Scenario (Question 3):     * $u_{\infty} = 0.5\,m/s$     * $L = 1\,cm = 0.01\,m$     * $\Delta T = 1^{\circ}C$     * $T_{\infty} = 20^{\circ}C$     * Properties at $50^{\circ}C$ (specified for computation):         * $\rho = 0.997\,g/cm^3$         * $\mu = 0.01\,cm^2/s$

• Computation of Average Shear Stress (\tau):     * $\tau = \frac{1}{2} \times \rho \times u^2 \times C_{f,L}$     * $C_{f,L} = 1.328 \times Re_L^{-1/2}$     * $Re_L = \frac{0.5 \times 1}{0.01} \times 100 = 5000$     * $\tau = 0.5 \times 997 \times 0.5^2 \times 1.328 \times (5000)^{-1/2}$     * $\tau = 124.625 \times 1.328 \times 0.01414 = 2.34\,N/m^2$

• Computation of Average Heat Flux:     * Using properties at $20^{\circ}C$: $k = 0.59\,W/mK$, $Pr = 7.07$.     * $q'' = \bar{h} \times \Delta T = \left(\frac{Nu_L \times k}{L}\right) \times \Delta T$     * $q'' = \frac{0.59}{0.01} \times 0.664 \times (7.07)^{1/3} \times (5000)^{1/2}$     * $q'' = 5316.7\,W/m^2$

Derivation of the Similarity Solution for Boundary Layer Flow

• Governing Equations:     * Continuity Equation: $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$     * Momentum Equation: $u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2}$

• Similarity Variable and Stream Function:     * Definitions for Blasius derivation:         * $\eta = y \times \sqrt{\frac{u_{\infty}}{\nu x}}$         * $u = u_{\infty} f'(\eta)$         * $\psi = (u_{\infty} \nu x)^{1/2} f(\eta)$

• Goal: Show that the Blasius equation can be generalized as:     * $2f''' + (n+1)ff'' + 2m[1 - (f')^2] = 0$     * This incorporates the case where free stream velocity follows a power law $u_{\infty} = C x^m$.

• Derivation Steps:     * Substitution for stream function gradients:         * $u = \frac{\partial \psi}{\partial y}$         * $v = -\frac{\partial \psi}{\partial x}$     * The momentum equation is transformed into similarity coordinates by calculating the derivatives of $f$ with respect to $x$ and $y$:         * $\psi_y \psi_{yx} - \psi_x \psi_{yy} = m u \frac{du}{dx} + \nu \psi_{yyy}$     * Applying the power law $u \propto x^m$ leads to the coefficients involving $m$ and $n$ in the final ODE.

• Energy Equation Analogy:     * The governing energy equation is:         * $u \frac{\partial T}{\partial x} + v \frac{\partial T}{\partial y} = \alpha \frac{\partial^2 T}{\partial y^2}$     * Defining non-dimensional temperature $\theta(\eta) = \frac{T - T_w}{T_{\infty} - T_w}$, the energy equation transforms into an ODE similar to the momentum equation, depending on the Prandtl number ($Pr$).