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8/10/2019 ti liu bo tr

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BO CO THC TP H 2014

I .Phn m u :

Thc tp k thut h l dp sinh vin b mn Hng khng khoa K thut Giao

thng tng hp li kin thc hc trong chng trnh i hc v tip xc vi quy

trnh sn xut thc t ca mt cng ty,nh my hoc thc hnh trong phng th

nghim ti trng. y ls chun b quan trng cho lun vnsau ny, v vy em

xin chn thnh cm n cc thy c v b mn to iu kin cho chng e m c

c tthc tp b ch nh vy.

Thi gian thc tp: t ngy 01/07/2014-15/08/2014 ti B mn Hng khng , khoa

K thut Giao thng.

I I .Mc lc :

A.Bo co ni dung cng vic c giao:

Bi ging Kh nglc hc2chng 1,2,3

1. Kin thc trng tm

2. Bi ging POWER POINTv Bi tpchng 1,2,3

3. Ti liu tham kho

B.Bo co thc tp ti phng th nghim ca b mn:

C.nh gi kt qu thctp:

D.Ph lc:Nht k thc tp


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A . BI GING KH NG LC HC II

1. KIN THC TRNG TM:

Ni dung kin thc:c tm tt y trong cc slide bi ging, ngoi ra sinh

vin cn nghin cu thm cc sch trong phn ti liu tham kho nm bt ccc phn trng tm ca mn Kh ng lc hc 2.

Sau y l mt scu hi n gin v kin thc c bn gip hc tt mn hc ny

Cnh my bay to ra lc nng nh th no?

Cnh my bay c thit k sao cho dng kh bn trn cnh chuyn ng nhanh

hn. Khi dng kh chuyn ng nhanh hn th p sut dng kh s gim xung. Do

p sut kh pha trn cnh my bay nh hn p sut kh pha di cnh. Schnh lch p sut tora mt lc gip nng cnh ln trong khng kh.

Cc nh lut c bn v chuyn ng:

Nh bc hc Isaac Newton a ra ba nh lut v chuyn ng vo nm 1665.

Ba nh lut ny gip gii thch chuyn ng ca my bay:

Nu mt vt th ang ng yn, n s khng th t chuyn ng. Nu mt vt

th ang chuyn ng, n s khng dng li hay chuyn hng tr khi c mt lc

tc dng ln n.

Mt vt th s chuyn ng xa hn v nhanh hn khi chng b tc dng mt lc


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mnh hn.

Khi mt vt th b tc dng theo mt hng, s lun c mt phn lc c cng

ln theo hng ngc li.

Cc lc tc dng ln my bay:

iu khin chuyn ng ca my bay:


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c th nghing my bay sang tri hoc phi, cc ailerons c nng ln

mt cnh v c h xung cnh cn li. Cnh c cc ailerons b h

xung, th s c nng ln cao. Cn cnh c cc ailerons c nng ln,th s b h xung.

tng hoc gim cao my bay, phi cng s iu chnh elevators ui

my bay. H thp elevators s khin cho mi ca my bay h xung, lm

gim cao ca my bay. Nng elevator ln lm tng cao ca my bay.

Khi Rudder b quay v mt pha, my bay s quay tri hoc phi. Mi camy bay s hng theo hng ca Rudder. Phi cng s phi hp chuyn

ng ca Rudder v ailerons chuyn hng bay ca my bay.

?

Chim bay thnh n theo hnh ch V v 2 l do sau:


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1. Khi ch chim bay theo hnh ch V so vi khi ch chim bay mt mnh ring ls:

- Li lc nng

- Gim lc cn

Nh vy chim s v cnh ( to lc y ging nh ng c my bay) t hn nns mt hn nn bay c xa hn.

Gii thch:

Ta xem ch chim nh 1 my bay.

- p sut pha di cnh ln hn pha trn cnh, nn ngay u mt cnh chng tas c wingtip vortices (finite wing m)

- Khi my bay bay s li pha sau n l trailing tip vortex (nhn hnh trn)

- Ta s c c vng upwash v downwash trn hnh

- Vng upwash s gip my bay li v lc nng, tc l ch chim theo bn nng st bay vo vng ny v do v tnh to thnh hnh ch V.

-Nguyn nhn l do wing tip vortex ( xoy mi cnh ) gy ra. Xoy ny cchiu t di ln trn, t ngoi vo trong. Do vy, con chim pha sau phi bay mp ngoi cnh ca con chim bay trc n c xoy ny nng ln.Con no

bay pha trong hoc pha sau con bay trc s b xoy ny xung

- Li v lc nng tc l my bay c th bay gc tn nh, hay ni cch khcinduced drag s gim (gim lc cn).

2. Bay hnh ch V gip cc ch chim lin lc vi nhau tt hn, kim sot nhau

tt hn v ch chim dn u (ch chim mt nht) d dng bay li li ri v tr ngh mt, nhng li v tr cho ch chim bn cnh.


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2.BI GING POWER POINT V BI TPCHNG 1,2,3

Thi gian:t ngy 15/07/2014 n ngy 02/08/2014

a im:lm ti nh Gio vin ph trch:c L Th Hng Hiu

Bi ging power point:

Gii thiu cc mc chnh ca slide bi ging:

Chng 1: M u

1. What is aerodynamic?

2. Aerodynamics : classification and practical objectives

3. Definition of basic aerodynamic quantities : flow velocity, pressure,

density, temperature, viscosity, bulk elasticity, thermodynamic properties

4. Fluid statics and the atmosphere

5. Wing and airfoil geometry

6. Forces and moments on aircraft - Body fixed control coordinates

7. Force and moment coefficients

8. Pressure distribution on airfoil

9. Estimation of the lift, drag and pitching moment coefficients from the

pressure distribution

10.Center of pressure - Aerodynamic center

11.Dimensional Analysis and Flow similarity

12.Types of flow


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Chng 2: Mt s nguyn l v phng trnh c bn ca c hc thy kh

1. Nhc li mt s tnh cht ton hc ca cc i lng v hng v vect

2. Cc phng php m phng chuyn ng ca lu cht: phng php th

tch kim sot v phn t lu cht3. o hm ton phn ca vi phn th tch khi Transport theorem

4. Phng trnh lin tc

5. Phng trnh ng lng

6. Phng trnh nng lng

7. Cc phng trnh bo ton vit di dng vi phn o hm ton phn

8. Mt s khi nim v c tnh dng chuyn ng

1.Vn tc gc, vecto quay v bin dng2.Lu s

3.ng dng (lu tuyn) v qu o

4.Hm th vn tc-hm dng v mi lin h hm th-hm dng

9. Phng php gii cc phng trnh bo ton: phng php gii tch

v phng php s

Chng 3: Dng khng nn c qua cnh 2D

1. Gii thiu tng quan

2. Phng trnh Bernoulli - ng dng

3. Phng trnh cbn ca dng chuyn ng khng nn c, khng ma st

v khng quay

4. Cc chuyn ng th cbn

5. Chng nhp chuyn ng th

6. Dng chuyn ng qua c th bt k-khng c lc nng: phng php

mng ngun The numerical source panel method

7. Dng chuyn ng thc t qua hnh tr trn (tham kho)


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Chng 1: cc slide minh ho


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Chng 2:cc slide minh ho


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Chng 3:cc slide minh ho


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20/27

Bi tp sch Fundementals of Aerodynamics

Chng 1:

1.For most gases at standard or near standard conditions, the relationship among

pressure, density, and temperature is given by the perfect gas equation of state:

p = RT, where R is the specific gas constant. For air at near standard conditions,R = 287 J/ (kg K) in the International System of Units and R = 1716 ft lb/(slug

R) in the English Engineering System of Units. (More details onthe perfect gas

equation of state are given in Chapter 7.) Using the above information, consider

the following two cases:

(a) At a given point on the wing of a Boeing 727, the pressure and temperature

of the air are 1.9 N/ and 203 K, respectively. Calculate the density at

this point.

(b) At a point in the test section of a supersonic wind tunnel, the pressure and

density of the air are 1058 lb/ z and 1.23 slug/ft3, respectively. Calculate

the temperature at this point.

2. Starting with Equations (1.7), (l .8), and (1.11), derive in detail Equations

(1.15), (1.16), and (1.17).

3. Consider an infinitely thin flat plate of chord c at an angle of attack of in a

supersonic flow. The pressures on the upper and lower surfaces are different but

constant over each surface; i.e., (s) = and p,(s) = , where and , are

constants and , > . Ignoring the shear stress, calculate the location of the

center of pressure.

4. Consider an infinitely thin flat plate with a 1 m chord at an angle of attack of

10 in a supersonic flow. upper and lower surfaces are given by = 4 x

+ 5.4 x , pl = 2 x - 1.73 x , u = 288 , and =

731 , respectively, where x is the distance from the leading edge in meters and

p and I are in newtons per square meter. Calculate the normal and axial forces, the

lift and drag, moments about the leading edge, and moments about the quarter

chord, all per unit span. Also, calculate the location of the center of pressure.

5.Consider an airfoil at 12 angle of attack. The normal and axial forcecoefficients are 1.2 and 0.03, respectively. Calculate the lift and drag coefficients.

6. Consider an NACA 2412 airfoil (the meaning of the number designations for

standard NACA airfoil shapes is discussed in Chapter 4). The following is a

tabulation of the lift, drag, and moment coefficients about the quarter chord for

this airfoil, as a function of angle of attack.


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From this table, plot on graph paper the variation of xcp/C as a function of cz.

7.The drag on the hull of a ship depends in part on the height of the water wavesproduced by the hull. The potential energy associated with these waves therefore

depends on the acceleration of gravity g. Hence, we can state that the wave drag

on the hull is D = f ( ) where c is a length scale associated with

the hull, say, the maximum width of the hull. Define the drag coefficient as

. Also, define a similarity parameter called the Fmude number,

Fr = V/ Using Buckinghams pi theorem, prove that = f (Fr).

8.The shock waves on a vehicle in supersonic flight cause a component of drag

called supersonic wave drag . Deiine the wave-drag coefficient as

, where S is a suitable reference area for the body. In supersonicHight,the flow is governed in part by its thermodynamic properties, given by the

specific heats at constant pressure c 1, and at constant volume . Define the ratio

Ey. Using Buckinghams pi theorem, show that . Neglect theinfluence of friction.

9.Consider two different flows over geometrically similar airfoil shapes, one

airfoil being twice the size of the other. The fiow over the smaller airfoil has

freestream properties given by = 200 K, = 1.23 kg/m3, and = 100 m/s. The

flow over the larger airfoil is described by = 800 K, = 1.739 kg/m3, and

= 200 m/s. Assume that both /at and a are proportional to T1/2. Are the twoflows dynamically similar?

10.Consider a Lear jet fiying at a velocity of 250 m/s at an altitude of 10 km,

where the density and temperature are 0.414 kg/ and 223 K, respectively.

Consider also a one-fifth scale model of the Lear jet being tested in a wind tunnel

in the laboratory. The pressure in the test section of the wind tunnel is 1 atm =

1.01 105 N/ . Calculate the necessary velocity, temperature, and density of the


22/27

airflow in the wind-tunnel test section such that the lift and drag coefficients are

the same for the wind-tunnel model and the actual airplane in flight. Note: The

relation among pressure, density, and temperature is given by the equation of state

described in

11.A U-tube mercury manometer is used to measure the pressure at a point onthe wing of a wind-tunnel model. One side of the manometer is connected to

the model, and the other side is open to the atmosphere. Atmospheric pressure

and the density of liquid mercury are 1.01 105 N/ and l.36 104 kg/ ,

respectively. When the displacement of the two columns of mercury is 20 cm,

with the high column on the model side, what is the pressure on the wing?

12.The German Zeppelins of World War I were dirigibles with the following typ-

ical characteristics: volume = 15,000 , and maximum diameter = 14.0 m.

Consider a Zeppelin flying at a velocity of 30 m/s at a standard altitude of 1000

m (look up the corresponding density in any standard altitude table). The Zep-pelin is at a small angle of attack such that its lift coefficient is 0.05 (based on

the maximum cross-sectional area). The Zeppelin is flying in straight-and-level

flight with no acceleration. Calculate the total weight of the Zeppelin.

13.Consider a circular cylinder in a hypersonic How, with its axis perpendicular

to the How. Let Q5 be the angle measured between radii drawn to the leading edge

(the stagnation point) and to any arbitrary point on the cylinder. The pressure

coefficient distribution along the cylindrical surface is given by

Calculate the drag coefficient for

the cylinder, based on projected frontal area of the cylinder.

14.Derive Archimedes principle using a body of general shape.

15.Consider a light, single~engine, propeller-driven airplane similar to a Cessna

Skylane. The airplane weight is 2950 lb and the wing reference area is 174 ftz.

The drag coefficient of the airplane is a function of the lift coefficient CL

for reasons that are given in Chapter 5; this function for the given airplane is

= 0.025 -4- 0.054 .

(a) For steady, level flight at sea level, where the ambient atmospheric density is

0.002377 slug/ft3, plot on a graph the variation of , , and the lift-to-dragratio L / D with flight velocity ranging between 70 ft/s and 250 ft/s.

(b) Make some observations about the variation of these quantities with velocity.


23/27

Chng 2 v 3: nh dng PDFfile word

(hnh minh ha)

3.TI LIU THAM KHO:

1. 1.John D. Anderson Jr, Fundementals of Aerodynamics, McGraw-Hill,2001

2. Houghton and Carpenter, Aerodynamics for Engineering Students, 5thEdition, Edward Arnold, 2003 (E-book)

3. Aerodynamics of Flights, W. Phillips (E-book)

Chapter 1- Overview ofAerodynamics


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B. BO CO THC TP TI PHNG TH NGHIM CA B MN

Thi gian:t ngy 19/07/2014 n ngy 10/08/2014

a im:phng th nghim b mn K thut Hng khng ti C5, C1 v B4 Gio vin ph trch:thy Trn Tin Anh

Ngi h tr, gim st:ch Hng Nht

Sinh vin thc tp: Quang Thnh, Tin t, Thin T (K10) v cc bn K11

Ni dung cng vic:

1. Di chuyn cc thit b th nghim v dng c ca b mn t hai

phng th nghim B4 v C1 v phng th nghim mi ti C5

2. Dn dp v sinh v h tr thy c sp xp cc thit b vo phng thnghim mi

3. Sa cha cc thit b v dng c b h hng

4. Phn loi v sp xp dng c, sch v, thit b, vv theo yu cu

ca ngi ph trch

5. Phn cng thi gian trc ti phng th nghim mi h tr thy c

v trng coi cc ti snca b


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C. NH GI KT QU THCTP:

Nhng kt qu thu c:

+ y c hi em n tp v tng hp cc kin thc chuyn ngnh trng tm gip

cho vic lm lun vn+ Thng qua vic son v chnh sa slide bi ging gip em c kh nng t duy

tng hp v la chn thng tin mt cch hiu qa

+ c thc tp ti trng, lm vic chung vi cc thy c v anh ch em trong b

mn Hng khng, gip gn cht mi quan h thy tr -bn b

+ Hc thm c nhiu kin thc ngoi chng trnh ging dy tcth ly cho

cuc sng

Phn nh gi c nhn:+ Xt v mt tng th, em hon thnh c chng trnh thc tp v cng vic

c giao

+ Tuy nhin em vn cha hon thnh mt cch hiu qu nht do mt s nguyn

nhn ch quan v khch quan

+ Mt s cng vic c giao khng hon thnh ng hn v ng yu cu

+ Knh mong qu thy c b qua v phn hi cho em em c th chnh sa, lm

cho ng yu cu v tt hn.


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D. PH LC: NHT K THC TP 2014

Thi gian Ni dung cng vic a im Cn b hng dn

Tun 1Th 315/07/2014

Th 416/07/2014

Th 517/07/2014

Th 618/07/2014

Th 719/07/2014

Phn cng cng vic1.Son slide2. Dn phng TN3.Trc phng TN

Chnh sa slide biging Kh ng 2 chng 1

Dn phng th nghimb mn B4

P.102A5

Ti nh

B4

C Hiu

C Hiu

Thy Tin Anh

Tun 2T 21/07 n 26/07

Trc phng TN vph dn dp

Chnh sa slide bi

ging Kh ng 2 chng 2 v chng 3

C5

Ti nh

Thy Tin Anh

C Hiu

Tun 3T 28/07 n 02/08

Th 7 02/08/2014

Trc phng TN vph dn dp

Np bi ging, bi tp

KLH2

C5 Thy Tin AnhCh Nht

C Hiu


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Tun 4T 04/08 n 09/08 1.Trc phng TN v

ph dn dp

2.Dn dp phng TNC1

3.Sa cha cc thit bb h hng

C5

C1

C5

Thy Tin AnhC HiuCh Nht

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