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# Test Upload MANE 4070

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This page Test Prep (MCAT, SAT...) was uploaded by David Harris on Sunday February 7, 2016. The Test Prep (MCAT, SAT...) belongs to MANE 4070 at Rensselaer Polytechnic Institute taught by Professor A. Hirsa in Summer 2015. Since its upload, it has received 29 views. For similar materials see Aerodynamics in Business, management at Rensselaer Polytechnic Institute.

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Date Created: 02/07/16

MANE 4070 Aerodynamics 1 Final Exam December 12 2014 Instructor Professor Hicken Time 180 minutes Format Closedbook no aids no devices With internet access no cell phones Last Name First Name RIN Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Total UIOOOUIUI U 0 L r Question 1 7 marks NASA scientists are modeling the Orion spacecraft s reentry into the earth s atmosphere Orion has a radius of 25 m a If the scientists want to use a windtunnel experiment to predict the forces and moments on the spacecraft b If the scientists model the reentry ow at highaltitude using the NavierStokes partial differential equa c A windtunnel model is built with a radius of 015 m If the drag force experienced by the model is d Find the Mach number of the spacecraft for the conditions in part c ie V0 2 4000 ms and poo e If the lift on the capsule is negligible what is the dominant form of inviscid drag on the capsule list the requirements on the geometry and the ow that they must meet to achieve dynamic similarity tion what hypothesis do they risk violating and why D 45 N at sealevel conditions and V 28 m s what is the drag force on the actual spacecraft when its velocity is 4000 ms and the density is poo 0001 kg m3 Assume the two ows are dynamically similar 0001 kg m3 if the atmospheric temperature is To 2827 K You may assume air is an ideal gas at the given conditions Answer a To achieve dynamic similarity 1 the similarity parameters e g ReMa must be the same between the actual ow and the experiment and 2 the spacecraft and model must be geometrically similar 05 marks for similarity parameters must be the same 05 marks for geometries are similar The continuum hypothesis will be violated at high altitude because the gas is rare ed 1 mark for continuum hypothesis 1 mark for gas is rare ed or meanfree path is large or similar The coef cient of drag must be the same since the ows are dynamically similar For the model which has an area of S 7W2 2 00707 m2 D CD 01326 pmvgs Therefore the drag on Orion for the given conditions would be Orion has an area of S 1964 m2 1 1001on CDEpmvis 20800 N 1 mark for CD or correctly equating two CD if no value given 1 mark for drag on spacecraft For the given temperature am 2 xyRToo 337 ms Therefore the Mach number is Moo 2 119 1 mark for correct Mach number 05 partial mark if a0 correct e Wave drag is the dominant form of inviscid drag on the spacecraft 1 mark for wave drag drag due to shocks will also be accepted r Question 2 5 marks a Consider the following partial differential equation 8p v t o amp mm Identify the meaning of the two terms in the PDE Use the table provided below b Write the conservation of mass equation for steady ows c Write the conservation of mass equation for unsteady ows that are incompressible d The NavierStokes equations are am a a a as gt V39pa upI 5 pf 0 What conservation principle do these equations govern e A 3dimensional incompressible viscous ow can be modeled using the continuity equation and the Navier Stokes equations only How many scalar partial differential equations would you need to solve in this case table for answering question 2a ap at V p17 Answer a 8 p at is the rate of change ofdensity at a point V p i is the advective transport ofdensity away from the point 05 marks for meaning of 8 p at 05 marks for meaning of V p i For steady ows the time derivative vanishes from the conservation of mass equation continuity equation producing mm0 1 mark for correct equation For incompressible ow regardless of if they are steady or not the continuity equation becomes V a 0 1 mark for divergence of is zero The Navier Stokes equations govern conservation of linear momentum 1 mark for conservation of momentum The Navier Stokes equations are vector equations so they consist of 3 PDEs in 3dimensions Adding the continuity equation this gives 4 equations 1 mark for 4 equations L r Question 3 5 marks Consider the inviscid incompressible and irrotational 2dimensional ow described by the stream function Recall that in polar coordinates the stream function de nes the velocity via a What partial differential equation PDE does I satisfy You can give the name of the PDE or write down b Sketch the streamlines for the ow described by 11 and indicate the direction of the ow along the stream c Find the vorticity of the ow described by 11 for all points 36 7E 00 d For the ow described by 11 the circulation around any circle centered at the origin is l mZs Does this e A uniform freestream ow of u v 0 10 ms is added to the ow described by 1 If poo 1225 kg m3 13c ln mZs 3 8r 181 Mr and M9 2 the equation for the PDE lines Please include the x yaxes 1 8 l 8 r Hint If you need it the vorticity is given by a rug u r 8r r 89 contradict what you found in part c Explain nd the force magnitude and direction exerted by the ow on the surface of a circle centered at the origin J Answer 82 82 a The PDE satis ed by I is Laplace s equation V21 f f 0 8x 8y 1 mark for Laplace or its equation 05 marks for concentric circles about the origin 05 marks for clockwise ow direction indicated From the given equations for the velocity components we have ur 0 and we 2 2W Substituting these values into the vorticity equation we nd 18 l8url8 l 0 w r 9 ae m Alternatively we recognize that this is an irrotational ow except at the origin so a 0 by de nition 1 mark for a 0 05 partial credit if velocity components are correct No it does not contradict the result from part 0 because the velocity eld is singular at the origin 05 marks for No or It does not contradict 05 marks for velocity eld is singular or velocity blows up etc We also gave 05 partial marks for Yes because a 0 implies F 0 By the KuttaJoukowski theorem the force magnitude is F pooVooF 1225 N m Since the freestream ow is in the positive y direction and the vortex is clockwise the force direction is in the negative x direction 17quot 12250 Nm 05 marks for correct force magnitude 05 marks for correct force direction There was no carry over error if you found the vortex was counterclockwise r Question 4 8 marks a On the airfoil gure below clearly label 1 the thickness 2 the max camber 3 the mean camber line 4 the angle of attack 5 the chord length and 6 chord line b Place a checkmark beside the statements that are assumptions of thinairfoil theory D the angle of attack must be small and positive D the thicknesstochord ratio of the airfoil must be relatively small D the derivative of the mean camber line dzdx must be relatively small D the ow is inviscid irrotational and incompressible c Consider a symmetric thin airfoil Estimate its 6 at 06 50 and M00 06 d Consider a cambered thin airfoil with cmaC4 005 Estimate its cmaC4 at 06 30 and M 07 e According to thin airfoil theory what is signi cant about the point c 4 for symmetric airfoils Answer a See annotated Lecture 8 slide 4 13 mark for each correctly labeled term b All are assumptions of thinairfoil theory except for the rst the angle of attack does not need to be positive 05 marks for each correct 0 For a symmetric airfoil aL0 0 Therefore according to thinairfoil theory we have cm 2 27r5 7r 180 2 05483 Using this value with the PrandtlGlauret correction we nd c i 06854 1 M2 00 1 mark for incompressible cm 1 mark for corrected value C 1 According to thinairfoil theory c 4 is the aerodynamic center so Cm c4 005 regardless of a At Moo 2 07 we nd using the PrandtlGlauret correction Cmc40 Cmc4 m 007 1 mark for corrected value cmc4 e According to thinairfoil theory c 4 is the center of pressure and aerodynamic center for symmetric airfoils 05 for center of pressure 05 for aerodynamic center L r Question 5 7 marks Consider two wings with the following circulation distributions The two wings have rectangular planforms with spans of b 21 m and chord lengths of c 3 m N wing 1 me 219v 2A sinn9 219v 001sin9 0005 sin29 00003 sin39 nl F9 219 iAn sinn9 219 001sin9 nl wing 2 a Find the coef cient of lift for both wings b Find the coef cient of induced drag for both wings c What is unusual about wing l Hint A2 is nonzero d What is special about wing 2 e List a disadvantage of wing 2 f Suppose an aircraft is designed to operate at standard sealevel conditions How would you have to change its span to get the same induced drag if it operated at an altitude where the density is half the value of the sealevel value You can assume the velocity lift and span ef ciency remain unchanged Answer a The coef cient of lift is de ned by CL A17rAR The two wings have the same aspect ratio of AR 2 b2 S bc 7 and the same A1 001 coef cient therefore the coef cient of lift for both wings is CL 022 1 mark for correct CL 05 partial credit if AR is correct No carry over if AR was incorrect To nd the coef cient of induced drag for wing l we rst need to compute its span ef ciency factor Using the given formula we nd 1 3 2 An 1 n A1 For the second wing we can either compute its span ef ciency factor explicitly or recognize that it is an elliptical distribution since An 0Vn gt 1 Either way we nd 62 l Adapting the given formula for I I o I 2 o 1 mark for correct Cal1 05 partial credit if e1 correct 1 mark for correct CD2 No carry over if CL or AR were incorrect 0 Wing l is unusual because A2 7S 0 which means that its lift distribution is nonsymmetric about the root 1 mark for nonsymmetric lift distribution or similar d Wing 2 has an elliptical lift distribution andor has a minimum induced drag for all planar wings with the same parameters 1 mark for elliptical lift or minimum induced drag e A disadvantage of wing 2 is that it increases the root bending moment which will likely increase the necessary structural weight Wing 2 may also be more dif cult to manufacture 1 mark for increased root moment or increased structural weight or dif cult to manufacture f Recall that induced drag behaves as D olt L2 pooVOEbZe this proportionality can also be derived from the given C13 expression if necessary If the two designs have the same induced drag DH 2 Dig then L2 L2 pooVOEl e poo2V be gt 192 219 Therefore the span would need to be increased by a factor of 1 mark for increase by 2 r L Question 6 7 marks A Pitot tube is mounted on an aircraft that is ying at an altitude where p 5530 Pa p 00888 kgm3 and T 217 K a List 2 of the 3 assumptions necessary in order to apply the Bernoulli equation along a streamline b Assume the conditions in part a hold Use the Bernoulli equation to nd the velocity of the aircraft if the Pitot tube measures 0 5814 Pa c What is the aircraft s Mach number for the velocity found in part b d The aircraft increases velocity until 0 7055 Pa The assumptions of the Bernoulli equation no longer hold Find the aircraft s velocity assuming the ow is isentropic e The aircraft increases its velocity further to V 370 ms For this velocity what total pressure does the Pitot tube measure Answer a Flow must be 1 steady 2 inviscid and 3 incompressible along the streamline 1 mark partial credit of 05 for each correct assumption b Rearrangng the Bernoulli equation we nd 2190 poo poo v m 80 ms 1 mark for correct velocity xyRT 295 ms therefore Mo Vooaoo 0271 1 mark for correct Mach number 05 partial credit for a c The freestream speed of sound is a 1 Since the ow is isentropic we can use the totaltostatic pressure ratio to nd the Mach number y l 2 a 7 1 p The freestream speed of sound is still a0 295 ms therefore V aeoMoo 177 ms Moo 1 06 1 mark for correct freestream Mach number 1 mark for correct velocity We also accepted correct answers that used total enthalpy e Here the Mach number is Vooaoo and pressure downstream of the shock are given by the normalshock relations 1 w 1gt21M3 ng r 12 Finally using the totaltostatic pressure ratio between the shock and the Pitot tube gives 11 y1 081 Mi 1 9210 Pa PZZPoo 1 1 17143 14200 Pa 05 marks each for correct M2 and p2 after shock I mark for correct p0 for Pitot tube P0P2 1 125 so there is a normal shock ahead of the Pitot tube The Mach number L Question 7 6 marks Consider the M00 19 ow around a wedge with halfangle 90 and angle of attack 11quot see the gure below The freestream pressure is poo 1 atm gt A450 21 9 poo 1 atm When answering the questions below give your answers to 2 signi cant digits unless indicated otherwise a Find the shock angle 3 for the shock that forms on the lower surface b Find the Mach number and pressure along the lower surface of the wedge c Find the Mach number and pressure along the upper surface of the wedge d For M 19 at what angle of attack will the oblique shock detach Give your answer to the nearest whole number Answer a The ow de ection angle is 9 110 90 200 From the given 93 M curves we nd that 3 580 1 mark for correct shock angle b The normal component of the incoming Mach number is Mn 2 0 sin 161 Therefore using the normal shock relations M 2 I 1lty 1gt2M rMoo r 12 The Mach number after the oblique shock is M2 Mnaz sin 3 9 m 11 1 mark for pressure pg 6 28 29 05 partial credit for equation 1 mark for Mach number M2 6 108 11 05 partial credit for equation We gave 05 partial credit if Mn was correct No carry over for incorrect 3 andor 9 27 M2 r1 moo 1 m 29 atm 0665 c The ow is de ected by 92 20 on the upper surface Using the PrandtlMeyer table we nd vMoo 23590 Denoting the region after the expansion with 3 we nd VM3 VMoo l 92 25590 Interpolating between the data points in the table we nd M3 m 197 Using this value and the totaltostatic pressure ratio the expansion is isentropic we nd Y 1 1y 1M 2 y 1y1M322 gtlt poo 0897 atm IDs1903 poo poopgoo 1 mark for Mach number M3 6 197 20 1 mark for pressure 3 E 088 09 05 for correct equation No carry over if M3 was incorrect d For M 19 the oblique shock becomes detached if the de ection angle is greater than QmaX 210 found from the 93M curve From the geometry this corresponds to an angle of attack of a w 120 1 mark for 120 05 partial credit for QmaX m 210 10 Question 8 5 marks Consider the supersonic wind tunnel depicted below The eXit area of the nozzle which is the same as the cross sectional area of the test section is A6 16 m2 In the following you can assume there are no shocks in the convergingdiverging nozzle or the test section test section A M22 gt v a To achieve M 2 in the test section what area does the throat need to have Use the closest entry in the appropriate table b If the reservoir to the left of the nozzle above is at 0 5 atm what is the static pressure in the test section when the test section s Mach number is M 2 c If the reservoir temperature is To 288 K what is the ow velocity in the test section when M 2 d When the Mach number is M 2 in the test section the mass ow rate is 0109 kg s What is the mass L ow rate when the Mach number in the test section is increased to M 3 J Answer a For supersonic ow the throat must be sonic so At A At M 2 the area ratio is A A m 175 Therefore ifAe 16 m2 then A A 914 1 mark for correct A value b The ow is isentropic so we can use the totaltostatic pressure ratio 1 1 F p 2 p0 1 YTMZ 0639 atm 1 mark for correct pressure 0 Using the totaltostatic temperature ratio we nd r l 1 T T01TM2 160 K Therefore the speed of sound in the testsection is a xyRT 254 ms and the velocity is V aM 507 ms 1 mark for correct temperature in test section 1 mark for correct velocity 1 Two possible solutions Solution 1 A6 changes but At A stays the same The ow is choked the throat is sonic so the mass ow rate is the same m 0109 kg s 11 Solution 2 At A changes but Ae stays the same First we need to know the stagnation density in the reservoir Using the mass ow rate for the M 2 case we have peVeAe 0109 kgs gt p 134 x 10 5 kgm3 Using the totaltostatic ratio for the density the ow is isentropic we nd the stagnationtotal density in the reservoir to be 1 1 p0p 1YTM2 584gtlt105kgm3 Next we use the totaltostatic ratios to nd the density and temperature in the test section at M 3 1 A 11 pe p0 1 YTMZ 445 x 10 6 kgm3 7 1 1 Te To 1TM2 103 K Finally we can use the temperature and Mach number to nd the velocity and then with the density and area the mass ow rate m peVeAe 445 x 10 6 3 were 16 kgs 00434 kgs 1 mark for correct value of m 12

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