Exercício - Revisão Prova - 2023-2

quinta-feira, 16 de novembro de 2023 10:19 Página 1 de Nova Seção 1 c) ρ (\frac{∂v}{∂t} + V_x \frac{∂v}{∂x} + V_y \frac{∂v}{∂y} + V_z \frac{∂v}{∂z}) = ρg_x - \frac{∂P}{∂x} + μ (\frac{∂^2v_x}{∂y^2} + \frac{∂^2v_x}{∂z^2}) ρ (\frac{∂v}{∂t} + V_x \frac{∂v}{∂x} + V_y \frac{∂v}{∂y} + V_z \frac{∂v}{∂z}) = ρg_y - \frac{∂P}{∂y} + μ (\frac{∂^2v_y}{∂x^2} + \frac{∂^2v_y}{∂z^2}) ρ (\frac{∂v}{∂t} + V_x \frac{∂v}{∂x} + V_y \frac{∂v}{∂y} + V_z \frac{∂v}{∂z}) = ρg_z - \frac{∂P}{∂z} + μ (\frac{∂^2v_z}{∂x^2} + \frac{∂^2v_z}{∂y^2}) 0 = ρg_x - \frac{∂P}{∂x} + μ \frac{∂^2V_x}{∂y^2} 0 = - ρg_senoθ - \frac{∂P}{∂x} + μ \frac{∂^2V_x}{∂y^2} \frac{∂^2V_x}{∂y^2} = \frac{ρg_senoθ}{μ} + \frac{1}{μ} \frac{∂P}{∂x} \int \frac{∂^2V_x}{∂y^2} dy = \int (\frac{ρg_senoθ}{μ} + \frac{1}{μ} \frac{∂P}{∂x}) dy \frac{∂V_x}{∂y} = \frac{ρg_senoθ}{μ} y + \frac{1}{μ} \frac{∂P}{∂x} y + C_1 y \to \int dy V_x = \frac{ρg_senoθ.y^2}{2μ} + \frac{1}{2μ} \frac{∂P}{∂x} y^2 + C_1.y + C_2 \hspace{1cm} (I) d) aplicando \hspace{0.2cm} CC_1 \hspace{0.2cm} a \hspace{0.2cm} (I)... 0 = C_2 aplicando \hspace{0.2cm} CC_2 \hspace{0.2cm} a \hspace{0.2cm} (I)... 0 = \frac{ρg_senoθ h^2}{2μ} + \frac{1}{2μ} \frac{∂P}{∂x} h^2 + C_1 \frac{h}{h} \int aplicando 0 = \frac{ρg_senoθ.h^2}{2μ} + \frac{1}{2μ} \frac{∂P}{∂x} h^2 + C_1 \frac{h}{h} C_1 = - \frac{ρg h \hspace{0.1cm}senoθ}{2μ} - \frac{h}{2μ} \frac{∂P}{∂x} V_x = (\frac{ρg \hspace{0.1cm}senoθ}{2μ} + \frac{1}{2μ} \frac{∂P}{∂x}) y^2 - (\frac{ρg \hspace{0.1cm}senoθ}{2μ} + \frac{h}{2μ} \frac{∂P}{∂x})(y) P_{rel} = P_{abs} - P_{atm} \hspace{2cm} P_{rel} = P_{2_{abs}} - P_{atm} \hspace{2cm} P_{atm} - P_{atm} = 0 P_{2_{abs}} = P_{atm} = 101325 Pa \frac{∂P}{∂x} \approx \frac{ΔP}{Δx} = \frac{P_{final} - P_{inicial}}{x_{final} - x_{inicial}} = \frac{P_2 - P_1}{L} \frac{ΔP}{Δx} = \frac{0 - 5400}{1.3} = -41 538.85 \frac{Pa}{m} V_x = -3982.43 y^2 + 318.6 y e) V_{max} -> onde \tau é menor L_d \hspace{0.2cm} no \hspace{0.2cm} centro \hspace{0.2cm} -> \hspace{0.2cm} y = \frac{h}{2} = 0.04 m L_d \hspace{0.2cm} V_x = 6.372 \frac{m}{s} f) \space τ \space quando \space y_1 = 0 τ = μ \frac{∂V_x}{∂y} \rightarrow Lei \space de \space visc \space de \space Newton \frac{∂V_x}{∂y} = -7964.86 y + 318.6 quando \space y_1 = 0 \rightarrow \frac{∂V_x}{∂y} = 318.6 .s quando \space y = 0 \rightarrow \frac{∂V_x}{∂y} = 318.6 \cdot \frac{1}{5} γ = 0.001 \cdot 318.6 = 0.32 Pa r Página 4 de Nova Seção 1