lecture 6

Elements ofAerospace

Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.1

Lecture 06Elements of Aerospace EngineeringBasic Aerodynamics

AE602 Elements of Aerospace Engineering

Manoj T. NairIIST


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.2

Agenda

1 Basic Aerodynamics

2 Continuity Equation

3 Momentum Equation


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.3

Basic Aerodynamics I

The 3 fundamental physical principles1 Mass in conserved2 Newton’s second law (F=ma) holds3 Energy is conserved

Mass in conserved - Continuity Equation

Newton’s second law - Momentum Equation

Energy is conserved - Energy Equation


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.4

Continuity Equation I

Physical principle Mass can be neither created nordestroyed

Consider the streamlines that go through thecircumference of the circleThe streamlines form a tube - stream tubeThe cross sectional area of the tube may change - frompoint 1 to 2


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.5

Continuity Equation II

However, as long as the flow is steady, the mass that flowsthrough the section 1 should be same as the mass thatflows through section 2Let at section 1, the cross sectional area be A1 and thevelocity be V1

Consider all the fluid element at section 1 at a given instantAfter a lapse of time dt , these fluid elements move adistance V1dtThe elements have swept out a volume A1V1dtdownstream of point 1


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.6

Continuity Equation III

The mass of the gas in this volume is

dm = ρ1(A1 V1 dt)

The mass flow m through the area A is the mass crossingA1 during time interval dtTherefore,

Mass flow =dmdt

= m1 = ρ1A1V1 kg/s

Also the mass through section 2 is

m2 = ρ2A2V2

Mass conservation means

ρ1A1V1 = ρ2A2V2

This is the continuity equation for steady fluid flow


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.7

Continuity Equation IV

We assumed that the velocity V and density ρ wereuniform over the cross sectional area A

In real life, both V and ρ would vary across thecross-sectional area A

When using the equation, we can assume V and ρ to bemean values of velocity and density at the cross section


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.8

Continuity Equation V

The stream tube need not be bounded by a solid wall

Consider the flow over an airfoil

The space between the two streamlines is the stream tube


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.9

Continuity Equation VI

Incompressible and Compressible FlowIn real life, fluid is compressible to some extentIf we take a volume of fluid and apply enough pressureson the boundary of the volume, the volume of the elementwill decreaseHowever, the mass would remain the same

As a result, the density ρ would changeThe amount by which ρ changes depends on the nature ofthe fluid and the applied pressure


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.10

Continuity Equation VII

If the fluid is a liquid, e.g. water, then the change in volumeis very small and ρ is essentially constantIf the material is a gas, the volume can readily change andρ can be a variableSo we can have incompressible and compressible fluidAerodynamic flows can also be classified into two

1 Compressible flowdensity of the fluid elements can change from point to point,i.e. ρ1 6= ρ2this variability of the density is important at high speeds

2 Incompressible flowdensity of the fluid element remains constant, i.e. ρ1 = ρ2

Incompressible flow is a myth - it never occurs in natureHowever for flow where the variation of ρ is negligiblysmall, it is convenient to make the assumption that ρ isconstantIncompressible flow assumption can be made for flow inliquids


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.11

Continuity Equation VIII

For low-speed flow of air, V < 100m/s, flow can beassumed to be incompressible

All the airplanes from Wright Flyer (1903) to the late 1930shad velocities <100m/s

Hence early development of aerodynamics always dealtwith incompressible flows

For incompressible flow, continuity equation becomes

A1V1 = A2V2


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.12

Momentum Equation I

Physical principleF = m a

Consider a infinitesimally small fluid element moving alonga streamline with velocity V

At some instant the fluid element is located at PThe x-axis is parallel to the streamlineThe y and z-axis is perpendicular to the streamlineWhat is the force on this element?


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.13

Momentum Equation II

The force is a combination of three phenomena1 Pressure acting in a normal direction on all the six faces of

the element2 Frictional shear acting tangentially on all six faces of the

element3 Gravity acting on the mass inside the element

For the time being, let us ignore the frictional andgravitational forcesThen the only source of force on the fluid element is due topressure


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.14

Momentum Equation III

Let the dimensions of the fluid element be dx , dy and dzConsider the left and right faces, which are perpendicularto the x axisLet the pressure on the left face be pThe force on this face is p dy dzThis force in positive x directionPressure varies from one point to another


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.15

Momentum Equation IV

Let us denote the pressure per unit length in x direction asdp/dxTherefore, the pressure on the right face is p + dp/dx dxHence the force on the right face is [p + (dp/dx)dx ]dy dzThis force acts in the negative directionThe net force in the x direction is the sum of the two

F = p dy dz −(

p +dpdx

dx)

dy dz

F = −dpdx

(dx dy dz)


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.16

Momentum Equation V

This give the force on the fluid element due to pressureSince we took x-axis to be parallel to the streamline,pressure on the faces parallel to the streamline do notaffect the motion of the element along the streamlineThe mass of the fluid element is

m = ρ(dx dy dz)

Acceleration of the fluid element is defined as a = dV/dt

a =dVdt

=dVdx

dxdt

=dVdx

V


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.17

Momentum Equation VI

Using these in the Newton’s second law

F = ma

−dpdx

(dx dy dz) = ρ(dx dy dz)VdVdx

dp = − ρV dV

This is the Euler’s equation

This is a form of the momentum equation

Assumptions made: neglect friction and gravity, steady

Frictionless flows are called inviscid flow


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.18

Momentum Equation VIIConsider the x-component of momentum equation,without viscous and body forces

ρDuDt

= −∂p∂x

ρ∂u∂t

+ ρu∂u∂x

+ ρv∂u∂y

+ ρw∂u∂z

= −∂p∂x

For steady flow

u∂u∂x

+ v∂u∂y

+ w∂u∂z

= −1ρ

∂p∂x

Multiplying by dx

u∂u∂x

dx + v∂u∂y

dx + w∂u∂z

dx = −1ρ

∂p∂x

dx

From equation of streamline in a 3D space

udz − wdx = 0vdx − udy = 0


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.19

Momentum Equation VIII

Substituting into the x-momentum equation

u∂u∂x

dx + u∂u∂y

dy + u∂u∂z

dz = −1ρ

∂p∂x

dx

u(∂u∂x

dx +∂u∂y

dy +∂u∂z

dz)

= −1ρ

∂p∂x

dx

For a function u = u(x , y , z), the differential of u is

du =∂u∂x

dx +∂u∂y

dy +∂u∂z

dz

Therefore

u du = −1ρ

∂p∂x

dx

12

d(u2) = −1ρ

∂p∂x

dx


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.20

Momentum Equation IXSimilarly expression can be obtained for the y -componentand z-component of momentum equation along thestreamline

12

d(u2) = −1ρ

∂p∂x

dx

12

d(v2) = −1ρ

∂p∂y

dy

12

d(w2) = −1ρ

∂p∂z

dz

Adding, the above three

12

d(u2 + v2 + w2) = −1ρ

(∂p∂x

dx +∂p∂y

dy +∂p∂z

dz)

or dp = −ρVdV

This is called Euler’s equationApplies to an inviscid flow with no body forcesRelates the change in velocity along a streamline to thechange in pressure along the same streamline


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.21

Momentum Equation X

dp = −ρV dV

The equation relates a change in pressure to a change invelocityIt is a differential equation and describes the phenomenain the infinitesimal neighborhood of point P

Now consider two points, 1 and 2, far from each other, buton the same streamlineTo relate p and V at these two points, the differentialequation must be integratedThis integration depends on whether the flow iscompressible or incompressible


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.22

Momentum Equation XI

Let us consider the incompressible case

dp + ρV dV = 0

ρ is constant∫ p2

p1

dp + ρ

∫ V2

V1

VdV = 0

p2 − p1 + ρ

(V 2

22

−V 2

12

)= 0

p2 + ρV 2

22

= p1 + ρV 2

12

p + ρV 2

2= const along a streamline


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.23

Momentum Equation XII

This is called the Bernoulli’s equationBernoulli’s equation

holds only for inviscid incompressible flowrelate properties at two points along a streamline

When the flow is compressible, changes in density ρ mustbe taken into accountIf we assume that changes in density are related to thosein pressure by the isentropic relation

pργ

= constant c

dp = cγργ−1dρ

Then integration of Euler’s equation gives

γ

γ − 1pρ

+V 2

2= constant


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.24

Momentum Equation XIIIProblem: Consider a semi-circular cylinder of unit span.This cylinder is immersed in a flow of air, with its axisperpendicular to the flow. The rounded section of thecylinder is facing into the flow (front face). The radius ofthe semi-circular cross section is R = 0.5m. The velocityof the flow far ahead of the cylinder (freestream) isV∞ = 10m/s. Assume inviscid flow. The velocity of theflow along the surface of the rounded front face of thecylinder is a function of the location on the surface, i.e.V = V (θ) in (r , θ) coordinates. This variation is given by

V = 2V∞ sin θ

On the back face, the pressure, denoted by pB, isconstant. The back face pressure is given by

pB = p∞ − 0.7ρ∞V 2∞

where p∞ and V∞ are the pressure and density,respectively in the free stream. ρ∞=1.225 kg/m3.Calculate the aerodynamic force exerted by the surfacepressure distribution on the unit space cylinder.


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.25

Momentum Equation XIV

R = 0.5m, V∞ = 10m/sV = 2V∞ sin θpB = p∞ − 0.7ρ∞V 2

∞


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.26

Momentum Equation XV

ds = R dθ

The vertical projection of ds, denoted by dy , is

dy = ds cos θdy = R cos θ dθ

The horizontal force is

(p ds) cos θ = pR cos θ dθ

Integrating this would give the total horizontal force

Df =

∫ π/2

−π/2pR cos θ dθ

From Bernoulli’s equation

p = p∞ +12ρ(V 2∞ − V 2)


Engineering

Basic Aerodynamics

Continuity Equation

Momentum Equation

06.27

Momentum Equation XVI

Therefore

Df =

∫ π/2

−π/2

[p∞ +

12ρ(V 2∞ − V 2)]R cos θ dθ

But V = 2V∞ sin θ

Df =

∫ π/2

−π/2

[p∞ +

12ρ(

V 2∞ − 4V 2

∞ sin2 θ)]

R cos θ dθ

Df =

∫ π/2

−π/2

[p∞ +

12ρV 2∞

(1 − 4 sin2 θ

)]R cos θ dθ

Force on the base of the cylinder is

Db = (p∞ − 0.7ρV 2∞)2R

Total force is D = Df − Db

D = 1.067ρV 2∞R

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lecture 6