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### AA311.Midterm1.Solution

Course: AA 311, Fall 2011
School: Washington
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Word Count: 898

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Atmospheric AA311 Flight Mechanics Exam #1 Solutions Problem 1. (10 points) An aircraft pressure sensor reads 54.0 kPA. On this day, the sea-level pressure is 100.0 kPA, the sea-level temperature is 10 C, and the lapse rate is -7 C per kilometer. Compute the geometric altitude of the aircraft. Solution. Given the pressure P (h) = 54.0 kPA, we must first find h in the given non-standard conditions: P (h) = PSL TSL...

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Atmospheric AA311 Flight Mechanics Exam #1 Solutions Problem 1. (10 points) An aircraft pressure sensor reads 54.0 kPA. On this day, the sea-level pressure is 100.0 kPA, the sea-level temperature is 10 C, and the lapse rate is -7 C per kilometer. Compute the geometric altitude of the aircraft. Solution. Given the pressure P (h) = 54.0 kPA, we must first find h in the given non-standard conditions: P (h) = PSL TSL + Lh TSL 0 -( LR ) g We know every value in this expression except for h. Solving for h, we find: P (h) - LR g0 TSL h = -1 L PSL (-0.007 K/m)(287.0368 9.81 m/s2 283.16 K 54.0 kPa - = - 1- -0.007 K/m 100.0 kPa = 4796 m Converting to geometric altitude: h = 4796 m hG = rearth rearth - h h = 4800 m Nm/kgK) Problem 2. (10 points) An aircraft flies at an altitude of 5000 m. The aircraft's rectangular wing (8 m span, 1.5 m chord) is made up of uniform airfoil elements along the span. Suppose that the pressure distribution at some distance along the span takes the form as shown in Figure 1. Calculate the lift coefficient of that airfoil section. Assume that the flow is uniform along the span, and is not affected by wing-tip vortices. The Pitot tube pressure transducer reads a total pressure of ptotal = 54640 P a. Calculate the total lift force produced by the wing. How fast is the aircraft flying? Assume the aircraft is flying straight and level. What is the gross mass of the aircraft (including fuel, and passengers)? Solution. The lift coefficient is equal to the area enclosed by the pressure coefficient curves. The area is readily found as the sum: cl = 1.5 0.9 + 0.1 0.4 + 0.6 0.1 + 0.9 0.4 = 1.085. 2 Remark: The cartoon is misleading: the upper surface Cp,u cannot cross the abscissa at a constant slope with the given parameters. If you assumed a constant slope, and computed the area of the two triangles you obtained cl = 0.98. That solution is also worth full credit. The total pressure is the sum of dynamic pressure and static pressure. The static pressure is found from the table: pstatic = 54048 P a. hence the dynamic pressure is q = ptotal - pstatic = 592 P a. 2 From q = 1 V , and using = 0.74 kg/m3 we have 2 V = The lift force is 2 q = 40 m/s. L = CL Sq = 1.085 8 m 1.5 m 592 P a = 7708 N. From mg = W = L, we have m= L 7708 N = = 786 kg. g 9.81 m/s2 Figure 1: Pressure distribution over the wing in Problem 2. Problem 3. (10 points) Consider a wing, modeled as an infinitely thin flat plate, with a chord of 2 m and span of 30 m. The plane is flying at a standard altitude of 5000 m at 80 m/s with zero incidence angle to the flow. Assume all incompressible, turbulent flow over the wing. Part a. Calculate the boundary layer thickness at the trailing edge of the wing. Part b. Sketch boundary the layer thickness profile over the wing. On the same plot sketch a more realistic boundary layer thickness profile over the wing. Part c. Calculate the drag due to skin friction on the wing. Part d. What is the drag due to flow separation on the wing? Solution. Part a. We know that the absolute viscosity coefficient is a function of temperature. From the attached graph we see that at T = 255 K, is approximately m = 1.54 10-5 kg . ms We can calculate the Reynold's number at the trailing edge (x = c): Rec = V c 0.74 kg/m3 80 m/s 2 m = = 7.6 106 . -5 kg 1.54 10 ms The boundary layer thickness for turbulent flow at the trailing edge is given by = Part b. A general sketch of the all turbulent profile and a more realistic boundary layer plot is shown below. In a realistic scenario, the flow will start laminar and then transition to turbulent at a point xcr on the wing. Since the laminar boundary layer is not as thick as the turbulent boundary layer, the estimate in part a is too high. 0.37 2 m 0.37 2 0.37c = = m = 0.031 m = 3.1 cm. 0.2 6 )0.2 Rec (7.6 10 23.77 Figure 2: Turbulent and realistic boundary layer above a flat plate. Part c. The total skin friction coefficient for the plate is Cf = 0.074 0.074 = = 0.0031 Re0.2 23.77 c So the total drag (on both sides of the plate) is given by Df = 2Cf q S, where q = 1 2 1 V = 0.74 kg/m3 802 m/s = 2368 P a. 2 2 S = cb = 60 m2 . Hence the total skin friction drag is Df = 2Cf q S = 2 0.0031 2368 P a 60 m2 = 880.9 N. Problem 4. (10 points) Consider an airspeed measurement device as shown in Figure 3. The air density is = 1.1901 kg/m3 . The pressure transducer measures p1 = 9.7773 104 P a and p2 = 9.8665 104 P a. Calculate V2 . Calculate V . Calculate p3 (assume p3 is connected to a tube facing the airflow). Figure 3: Airspeed measurement setup in Problem 4. Solution. From the continuity equation, we have V1 = Recall Bernoulli's equation: 1 1 p1 + V12 = p2 + V22 . 2 2 Rearranging Bernoulli's equation, we get: 2 V22 = (p1 - p2 ) + V12 . Substituting equation (1) into Bernoulli's equation, we have V22 or V2 = Substituting values, V2 = 2(9.773 - 9.8665) 104 P a = 10 m/s. 1 1.1901 kg/m3 . 1 - ( 0.25 )2 2(p1 - p2 ) . [1 - (A2 /A1 )2 ] 2 = (p1 - p2 ) + A2 A1 2 A2 V2 . A1 (1) V22 , From the continuity equation we have A2 V2 = 4 10 m/s = 40m/s, A1 which is the same as the free-stream velocity: V1 = V = 40 m/s. From Bernoulli's equation, and from the fact that V = V1 we have that p = p1 , hence the instrument, p3 , measures the sum 1 2 1 1 p3 = p + V = p1 + V12 = 9.7773 104 P a + 1.1901 kg/m3 402 m2 /s4 = 98, 725.1 P a. 2 2 2 Figure 4: Table showing standard properties in US Customary and SI units. Figure 5: Atmospheric conditions.
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