Aerodynamics: GATE Aerospace Study Material Book-1

AEROSPACE ENGINEERING Aerodynamics

AERODYNAMICS CONTENTS

1 BASICS OF FLUIDS

2 AIRFOIL NOMENCLATURE

Basic Flow Analysis Techniques Computational Experimental Continuity Equation Types of Fluid flow Substantial Derivative Streamline, Streakline, Pathline Streamline Pathline Streakline Newton’s Law of viscosity Conservation of Momentum Equation (Differential form) Circulation Kelvin Theorem Bernoulli’s Equation Case 1: Steady irrotational flow Case 2: Steady but rotational flow Helmholtz Theorem Biot – Savart’s Law Energy Equation NACA Airfoils Nomenclature Four-digit series Five- digit series Modified 4 Digit and 5 Digit Series 1-series 6-series 7-series 8- series Objective Practice Question Stream function

3 POTENTIAL FUNCTION

 

Potential Function  Types of flows Uniform Flow Uniform Flow inclined at an angle Point Source    or Sink   

5 5 5 6 15 17 19 19 19 20 22 23 25 27 29 30 30 35 40 45 48 49 49 50 50 50 51 51 52 62 63



Vortex Flow Combination of Elementary flows Half Rankine Oval Full Rankine Oval Source– Sink Doublet Vortex Doublet Stationary or Non- rotating Cylinder (Uniform flow + Source-sink Doublet +Vortex Rotating Cylinder (Uniform flow + Source-sink Doublet + Vortex

64 64 65 66 68 69 69 71 73 74 76 81

4 LAMINAR FLOWS

VISCOUS

5

Corner Flows Spiral Flow Blasius Theorem Lift and Drag on Stationary cylinder Lift and Drag on Rotation cylinder Method of Images NACA Airfoils Nomenclature General 2d flow, between two infinite parallel plates Couette’s Flow Plane- Poiseuille’s Flow Different Boundary condition on plates Axisymmetric-Poisueilli’s Flow Expression for Velocity profile of fluid flowing between two moving drums in tangential direction

88 91 93 95 99 101 103 104 106 110 113 115

Boundary layer thickness Displacement Thickness Momentum Thickness Von Karmann’s Momentum Integral Equation For Boundary layer over a flat plate Turbulent Boundary Layer

122 125 127 130 133 136

Laws of Thermodynamics 0th Law 1st Law 2nd Law 3rd Law

138 139 139 139 139 142

119

BOUNDARY LAYER

6 GAS DYNAMICS

7 THIN AIRFOIL & FINITE WING THEORY

Compressibility

 

Speed of Sound Energy Conservation SHOCKS Static Normal Waveform Oblique Shocks Multiple Shock Waves The Detached Shock Wave Expansion Waves Nozzles and Nozzle Flows Diffusers Thin Airfoil & Finite Wing Theory Starting Vortex Kutta’s Condition Thin Airfoil Theory Thin Cambered Airfoil Prandtl’s Finite Wing Theory Special Case: Elliptical Lift distribution over finite wing Effect of Aspect Ratio High Lift Devices

143 143 147 148 154 158 163 165 167 172 182 182 183 184 189 194 196 200 203

8 LINEARIZED THEORY

9 WIND TESTING

TUNNEL

Linearized Theory Linearized Velocity Potential Equation Linearized Pressure coefficient Prandtl –Glauert Compressibility Correction Subsonic flow over a Wavy Wall Linearized Supersonic Pressure Coefficient Linearized Pressure Coefficient from oblique – shock relation

207 210 211 213 216 218 223

Pressure Measurement Pitot Tube Flow Rate Measurement Venturi meter Orifice meter Temperature Measurement Velocity Measurement (4) Dot Matrix Hot Wire Anemometry Laser Doppler Velocimetry Particle Image Velocimetry Density Measurement Schlieren Method Shadowgraph Technique The Interferometer

225 227 229 230 231 233 237 237 237 239 241 242 245 248 249

CHAPTER 1

BASICS OF FLUIDS 1.1

BASIC FLOW ANALYSIS TECHNIQUES

There are two ways to solve or analyze fluid problem.

1.1.1

1.1.2

COMPUTATIONAL 1.

Control–volume, or integral analysis

2.

Infinitesimal system or differential analysis

3.

Spectral methods

4.

Modelling using LES etc.

EXPERIMENTAL 1.

Experimental study on scaled model or dimensional analysis.

2.

Full practical problem experiments

In all cases, the flow must satisfy

1.2

1.

Conservation of mass (continuity)

2.

Momentum

3.

First law of thermodynamics (Conservation of energy)

4.

A state relation like

5.

Appropriate boundary condition at solid surfaces, interfaces, inlets and exits.

    P, T 

CONTINUITY EQUATION

Assume that there is an arbitrary domain of mass M, the mass inside the domain will change by adding or removing the mass from the domain with respect to time, hence

dM  mIn  mout dt R.H.S. of Equation (1.1) is mass flux through surface, which can be written as,

…(1.1)

dM     V. dA s dt

…(1.2)

From Gauss Divergence Theorem (a)

 F .ds   . Fd

(b)

 .ds     d

s

v

s

v

Therefore equation (1.2) becomes,

d dt

 d    V.dS  0 s

And using Guass divergence theorem, we get

  V.dS   .  Vd

…(1.3)

s

As time is independent of (x, y, z) coordinates hence, we can take time operator also inside the integral.

Therefore from equation (1.3), equation (1.2) can be written as



V

d   dt  .  V

  d  0

…(1.4)

Equation (1.4) is the integral form of conservation of mass, taking the volume to be very very small, we don’t required integration. Hence we get differential form of the equation as,

d  .  V  0 dt

 

…(1.5)

For Incompressible fluid

d 0 dt

 

.  V  0

.V  0 Assume (i)

If there is no source or sink

(ii)

Area and velocity vector are parallel

(iii)

All streamline has uniform velocity and density at inlet (1) and at outlet (2)

…(1.6)

1 A1V1  2 A2V2 If



…(1.7)

will be constant but then,

A1V1  A2 v2

…(1.8)

Continuity Equation in Polar coordinates Polar Coordinate:

 1  1    rur    u    uz   0 t r r r  z Also,

…(1.9)

u  ur i r  u i   uz i z

Question 1:

A plate is dipped inside honey bottle. Due to viscosity, honey sticks at the surface of the plate.

Keeping the plate vertical, honey starts dropping down due to earth’s gravity. Consider profile of honey as triangular and the variation of velocity of honey as linear. Find the equation of the thickness of honey at the bottom of the plate w.r.t time if honey is considered as incompressible fluid?

Solution:

dM . .  m in  m out dt Considering the control volume domain as triangular area per unit depth, in this domain no mass is coming in, and honey is going out.

Therefore,

. m In  0

dM  m.out dt



b

    u.dy 0

y b

 u  y   um

… (1.10)

b d  1    lb     0 u.dy dt  2  b 1 db  l     u.dy 0 2 dt b 1 db  l    u.dy 0 2 dt

Substituting the value of u from equation (1.10), we get



1 db l 2 dt

b

   um 0

y .dy b b



1 db l 2 dt

y2 1   um     2 b 0



l db 2 dt

b 2 1   um     2 b



l

db  umb dt b db t l  um  dt b0 b 0

 At

t  0,b  b0

t  t ,b  b When

t  ,b  0

Therefore,

  

l  In

In

b  umt b0

b t  um b0 l

b e b0

 um t l

b  b0e

 um t l

Question 1.2

There is a pump at the inlet of the tank, which is pumping the fluid of density,

in  1.4 kg / m 3 inside the tank with volume flow rate 2m 3 /sec. If another pump at the outlet of the tank is pumping fluid outside with flow rate expression for density and also calculate steady level density, initial density Solution:

0

1m 3 /sec. Then calculate

s inside the tank. Assume the

3

inside the tank and volume of tank is 1m .

At steady state, mass in

 mass out, hence

1Q1  2Q2 Also we know that,

d    dt  v  dv  0

If density is not changing

    Q entering  s  Q 1.4  2  s  1



s  2.8kg / m 3



dm . .  m In  m Out dt 

d    1  2.8   dt

 2.4    e t  2.4  0       r t  2.8  0  Question 1.3

There is a combustor tube of length I and uniform cross-section area (A) filled with fueloxidizer mixture, If the empirical relations given are: 

P  b  n 0    P0  where

  0.25

n  0.8 and 

P  u   0    P0  Where

ρ b and ρ u are the densities of the unburnt and burnt mixture and ρ 0 and P0 are

initial density and pressure. Calculate the following

Solution:

1.

What is the pressure inside the tube when total fuel- oxidizer mixture has burnt?

2.

How much fraction of fuel-oxidizer mixture is burnt when the flame reaches half distance?

3.

When flamer reached at half distance, what is the pressure inside?

1.

Using continuity equation

oQo  uQu  bQb Here, volume of the tube is same before and after burning, and mass is conserved, therefore

 u  o  b From, 

P  b  n 0   , we get  P0  1

1 P  P0 n 2.

We have to calculate the fraction of fuel- oxidizer burnt when the flame reaches at half of the distance, hence



b b

Q 2

Q Q  u 2 2

?



P  n 0   b  P0     b  u P  P  n 0    0    P0   P0 



b



b  u

n n 1

? e value of n  0.8

Substituting t





b

b  u



0.8  0.44 1.8

i.e. the fuel– oxidizer burnt is less than 50%.

3.

Pressure inside the tube when flame is at half way

 b 

Q Q  u  0Q 2 2

b 2



u

 0

2

substituting the value of

b and u





P  P  n 0    0    P0   P0     0 2 

n 1 P     1 2  P0  

P   2        n 1  P0  1

 P   2      P0   n  1 



Substituting the value of n and

 p   2      p0   1.8 

4

P  4     1.11  P0 



?

GATE Questions Question 5.7:

Consider a steady two– dimensional zero pressure gradient laminar flow of air over a flat plate as shown below. The free stream conditions are and

U   100m /sec,   1 atm

  1.8X 105 kg /m  sec. The ratio of displacement thickness to momentum

thickness of the boundary layer at a distance of 2 m from the leading edge is (GATE 2011)

Solution:

[a]

7.53

[b]

2.59

[c]

2.91

[d]

0.39

[b]

From Blasius solution of steady two – dimensional zero, pressure gradient laminar flow of air over a flat plate

* x  Question 5.8:

Solution:



1.721  0.664 ,  1 1 x 2 Re x Re 2x

 * 1.721   2.59  0.664

The laminar boundary layer over a large flat plate held parallel to the freestream is 5mm thick at a point 0.2 m downstream of the leading edge. The thickness of the boundary layer at a point 0.8 m downstream of the leading edge will be (GATE 2007) [a]

20 mm

[b]

5 mm

[c]

10 mm

[d]

2.5 mm

[a]

Blasius’s solution for laminar boundary layer over flat plate givens boundary layer thickness as

 

5.0x Rex

X is the distance of point from leading edge.

Assuming

Rex to be constant, we get:–

 0.2 5 0.2    0.8  0.8 0.8  0.8  20mm

Rex is local Reynold’s number.