Answer:
An examination is conducted to assess how basic thin airfoils perform in slightly supersonic flow conditions, utilizing the nonlinear transonic theory initially proposed by von Kármán[1]. Formulas for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are devised based on a transonic similarity variable. Aerodynamic coefficients are computed in similarity form for flat plates and asymmetric wedge airfoils, and their graphical representations are created. Sample plots are provided for a flat plate and a particular asymmetric wedge, shown on conventional coordinate axes of Cl, Cd, and Cmc/4 in relation to angle of attack and Cl against Mach Number to showcase distinct characteristics of nonlinear flow.
Explanation:
Factors influencing friction
The magnitude of friction is contingent on the following elements: i) The surface area in contact. ii) The applied pressure on the surfaces. Force is determined by Pressure multiplied by Area; thus, if the contact area increases or if the pressure applied rises, the frictional force will also escalate.
Methods for reducing friction
i) Smooth the contact surface. ii) Apply oil or grease to fill small gaps in flat surfaces. iii) Use ball bearings to minimize contact area among rotating components.
Lubrication
To minimize friction, various methods may be employed: Oil can be either thin or viscous, which depends on its SAE number (SAE indicating Society of Automotive Engineers). Highly viscous oils may not reach all components effectively. In contrast, very thin oils may drain away quickly, resulting in wastage. Grease is preferable in such situations, particularly around ball-bearings. Regular grease or oil should not be utilized under high speed, high pressure, and high temperature conditions—specialized lubricants are required then. The consistency of oil varies with temperature; it thickens in the cold and thins in the heat. Therefore, the choice of lubricant should be seasonally appropriate, and it's always wise to consult the equipment's operating manual prior to making a selection.[[TAG_11]]
Answer: c. 4.56 × 105 J
Explanation:
Given the mass of the lead brick, m = 7.25 kg
Starting temperature T1 = 18.0 °C
Ending temperature T2 = 328 °C
The specific heat capacity for lead, c = 128 J/(kg∙°C)
And the latent heat of fusion Lfusion = 23,200 J/kg
The required energy Q =?
Using the following equations
Energy required, Q = mc (T2 - T1) + mLfusion
Substituting in the values we have: 7.25 kg * 128 J/(kg∙°C) * (328 - 18°C) + 7.25 kg * 23200 J/kg
= 455880 J
= 4.56 x 10^5 J
