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# Principles of Aerodynamics - Coursework Example

## Extract of sample Principles of Aerodynamics

2: Coefficient of Drag = CD = 0.054 Area = S = 15 m2 Thrust = T= 1500 N Density= A = 0.5 kg/m3 For a steady and level flight, drag force is equal to the thrust produced by engines, D = T = 1500 N D = (? A V?2) S CD = (? (0.5) (V?2)) (15) (0.054) = 1500 => V? = 86.06 m/s = 8.6 E +1 m/s Question No. 3: Question No. 4: The sketches shown below illustrate the trend of variation in CL, CD, and L/D ratios with increasing angle of attack. Question No. 5: Critical Mach number corresponds to that value of Mach number for free stream flow for which a localized mach number of ‘1’ is obtained at any point around the airfoil. When this condition arises, a shock wave is created at the point where the flow reaches the sonic speed. As the speed increases, regions of very low pressure are created. This causes the flow to separate from the airfoil thereby substantially increasing the drag forces on it. The figure illustrates this phenomenon. Some of the important design features incorporated in the aircrafts in order to contain the effects of this situation are using thin airfoil and / or super critical airfoil (Anderson Introduction 763). Making an airfoil thinner increases the value of Critical Mach Number and hence the airplane can fly at very high speeds without a significant increase in drag forces on it. ...
Their unique design limits the rise of drag forces even after the critical mach number is reached. Such airfoils have successfully been utilized in TACT aircraft program run by NASA Dryden Flight Research Center (Cury). Question No. 6: In the above illustration, the triangle represents the fuselage of an aircraft as seen from front. ‘’ is the angle of bank for the turn. LV = L cos  LH = L sin  The centripetal force required by the aircraft to take the turn is provided by horizontal component of lift force given by LH, equating the two; L sin  =  … (a) the component LV balances the weight of the aircraft, hence = L cos  = mg From the above equation, L = mg / cos  Putting values in (a) and simplifying; g tan  =  putting values for  = 15o and r = 1500 m gives v = 62.79 m/s Load Factor = L/W = L/L cos  = 1/ cos  = 1/cos 15o = 1.035 Question No. 7: Following are the control surfaces used to control the motion of an aircraft along different axes: (1) The longitudinal axis: Ailerons (2) The Vertical or Normal Axis: Rudder (3) The lateral Axis: Elevator The figure shows the above mentioned control surfaces and the functions they perform. All the control surfaces work on the principle of creating drag for the incoming wind thereby changing the direction of the wind. Due to this change in velocity, a momentum change occurs which causes a force to act on the control surface and the desired movement of the aircraft is achieved this way. The ailerons tilt the aircraft around the longitudinal axis. They are always installed in pairs. The opposite motion of the two ailerons creates a couple which acts about the longitudinal axis to cause the desired motion. Rudder rotates the ...Show more

## Summary

Task 1: Question No. 1: Area of the aircraft = S = 29 m2 Coefficient of Lift = CL = 0.67 Altitude = A = 29000 ft Velocity = V? = 385 km/hr = 106.94 m/s Lift = L = ? Solution: Density at given altitude ‘A’ = A = 0.4671 kg/m3 (Anderson Introduction 763) Lift on an airfoil is given as L = (?…
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