Step 1: Understand the Forces.
The force due to pressure and shear in the direction normal to the flow primarily contributes to the lift, particularly in aerodynamic contexts.
Step 2: Definition of Lift.
Lift is the component of the aerodynamic force that is perpendicular to the flow direction. It contrasts with drag, which is parallel to the flow direction.
An electrical wire of 2 mm diameter and 5 m length is insulated with a plastic layer of thickness 2 mm and thermal conductivity \( k = 0.1 \) W/(m·K). It is exposed to ambient air at 30°C. For a current of 5 A, the potential drop across the wire is 2 V. The air-side heat transfer coefficient is 20 W/(m²·K). Neglecting the thermal resistance of the wire, the steady-state temperature at the wire-insulation interface __________°C (rounded off to 1 decimal place).
GIVEN:
Kinematic viscosity: \( \nu = 1.0 \times 10^{-6} \, {m}^2/{s} \)
Prandtl number: \( {Pr} = 7.01 \)
Velocity boundary layer thickness: \[ \delta_H = \frac{4.91 x}{\sqrt{x \nu}} \]
Consider two identical tanks with a bottom hole of diameter \( d \). One tank is filled with water and the other tank is filled with engine oil. The height of the fluid column \( h \) is the same in both cases. The fluid exit velocity in the two tanks are \( V_1 \) and \( V_2 \). Neglecting all losses, which one of the following options is correct?
Choose the option that correctly matches the items in Group 1 with those in Group 2.
Consider a process with transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let \( \tau_m \) and \( \theta_m \) denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of \( \frac{\tau_m}{\theta_m} \) is __________ (rounded off to 2 decimal places).
Given: For \( G = \frac{1}{(\tau s + 1)^2} \), the unit step output response is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]
Is there any good show __________ television tonight? Select the most appropriate option to complete the above sentence.