Straight and level flight PDF

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Notes for EG-194. Straight and level flight...

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3

AIRCRAFT PERFORMANCE IN STEADY FLIGHT

3.1 Straight and Level Flight at Constant Speed We have seen already that the forces acting must be in equilibrium because the motion is steady. Thus we have;

L=W T=D

( (

NB. We assume that the thrust acts in the flight direction. This will generally not be exactly the case but the small angles of misalignment due to incidence change and to initial offset do not make a significant difference. There are of course aircraft which use thrust vectoring to augment the lift for STOL (short take off and landing) and for enhanced manoeuvring in combat, or to provide complete support for VTOL, (vertical take off and landing). These are not considered here. From (,

1 2 W= ρ V S C L 2

(

which enables us to find one of the quantities V, CL, W, S, ρ, given values for the others. For example,

V=

√ √

W w = 1 1 ρ S CL ρC L 2 2

Where w is the wing loading and is equal to the weight, W, divided but the wing area, S. Note that slower flight can be obtained by,  

reducing the wing loading, (i.e. increasing wing area or reducing weight) increasing the lift coefficient, CL.

while, reducing density (i.e. increasing altitude), increases flight speeds. The other equation ( determines the performance in straight and level flight, i.e. how much thrust we need to balance the drag for any flight condition. We therefore need to consider the drag of the complete aircraft and how it varies with speed and altitude. Before going on to do this, we consider firstly the equivalent airspeed, a useful concept in performance calculation.

(

3.2 Equivalent Airspeed Density decreases with altitude and therefore, as we have just seen, with CL constant, the speed required to generate a given lift will increase with altitude. More general we can say that, aerodynamic force ∝ ρV 2 In view of this, we define the equivalent airspeed (EAS), VE, as the speed at standard sea level density which will give the same aerodynamic forces on the aircraft as a certain true airspeed at altitude for the same aerodynamic configuration, i.e. same lift coefficient etc. Thus,

1 1 2 2 L=W = ρV S C L = ρ0 V E S C L 2 2 or,

ρ V 2=ρ0 V E2 hence,

V E =V

√

ρ =V √ σ ρ0

Note that VE is always < V, unless flying at a density greater than the standard sea level value of 1.225 kg/m3. . Equivalent airspeed is a useful concept in performance calculations and in flight. In calculations:- because it is often possible to work only in terms of sea level conditions and then convert to the required altitude at the end. This will become more apparent as we proceed. In flight:- because EAS is close to the Indicated Airspeed or IAS as read by the standard pressure type airspeed indicator, the difference being only due to system errors. The reason for this close correspondence is that the ASI, reads the pressure difference from the pitotstatic tube i.e. ptotal - pstatic = ½ ρ V2. It has no means of determining the actual air density and therefore must be calibrated to read airspeed for one particular density. The standard sea level value is chosen so that, if perfect, the ASl will read EAS. This suits the pilot because it is the aerodynamic forces on the aircraft that he is most concerned with and a certain EAS implies the same forces regardless of altitude. Thus for example, the stall will always occur at the same EAS, even though at a high altitude the actual stalling speed will be considerably greater. An example of the difference between EAS and true airspeed is illustrated below; E.g. Typically, a high performance glider will have a maximum, never exceed speed in smooth air of 146 kt EAS. In mountain wave conditions the glider may well be flying at 20,000 ft. at which altitude the relative density is 0.53 if standard atmospheric conditions prevail. Thus, the true maximum airspeed will be 146/√0.53 = 200 kt, when 146 kt is indicated by the ASI at this height.

It might also be noted that the wind at high level under wave conditions could typically be as high as 80 kt. If the glider turned to fly downwind it would be covering the ground at 280 kt (320 mph or 144 m/s).

3.3 Drag of the complete aircraft So far we have looked mainly at the drag of the wing, consisting of the profile drag (boundary layer drag) and the induced drag (drag due to lift). There are many other sources of drag associated with the aircraft as a whole, these include; - the profile drag contributions from the other major components (fuselage, tail unit etc.) - interference drag between components (wing I fuselage junctions, regions between an engine nacelle and the adjacent wing or fuselage, etc.) - drag due to intake or exhaust of air for internal services (air conditioning, avionics cooling etc.), and due to leakage at joints or ill fitting removable panels. - slipstream drag for propeller driven aircraft. - jet scrubbing drag for some jet aircraft configurations. - trim drag due the aerodynamic load on the tailplane, (extra induced drag). - drag due to excrescences, (radio aerials beacons etc.). - drag due to surface roughness, waviness etc. In addition to the items listed above which are associated with the "clean aircraft", there will be large additions to the profile drag for other flight configurations such as landing, and take-off, due to deployment of undercarriage, flaps, airbrakes etc. The use of flaps and airbrakes will also modify the distribution of lift across the span and will therefore change the induced drag. Estimation of the total drag of an aircraft is a complex process carried out by the aerodynamics group in the design office. The methods used are usually developed in-house, based on experience and on data collected from many sources. In particular the group will extrapolate from previous aircraft they have designed. Often wind tunnel tests will be needed to improve estimates, especially for cruise drag where small differences can have profound effects on the overall operating economics and capabilities of the aircraft. Data sheet methods are available, for example published by ESDU in the UK, (the International Data Sciences Unit) and by NASA in the USA. There are also computer programmes for drag estimation, particularly for light aircraft and home designers, but these are not entirely satisfactory and should not be used without a good understanding of the subject. McCormick, "Aerodynamics, Aeronautics and Flight Mechanics" has a good section on drag estimation (Chapter 4), containing comparisons between aircraft and drag breakdowns for two aircraft, the Piper Cherokee 180, and the Gates Learjet Model 25. These examples will be discussed in the lecture. The total drag of an aircraft for a particular configuration (e.g. the clean configuration) is usually represented by the drag polar, a plot of CL against CD, with CD as the abscissa i.e.

Plotted in this form, a line joining the origin to a point on the curve represents the vector of the total aerodynamic force coefficient, CR, with correct direction relative to horizontal flight. From the Examples Sheet 2 you should be familiar with this plotted the other way around, i.e. with CL as the abscissa, and also with the plot of CD against CL2 i.e.

The examples you have been given show a linear region, (i.e. C D ∝C 2L ), covering the normal working range of CL. Although this is an idealised situation it is close to the actual drag variations found for many aircraft. Thus, we will assume for all our performance calculations that the drag consists of; -

a part which is independent of incidence, CD0, the zero lift drag coefficient

-

and a part proportional to C2L , which is comprised mainly of the induced drag contribution but is also taken to include the small part of the profile drag that results from changes of incidence.

Thus we have a drag equation of the form;

C D =C D 0 + K C2L We now go on to use this to define the variation of drag with speed and altitude.

3.4 Variation of Drag with Speed and Altitude We assume 2

C D =C D 0 + K C L Thus

1 2 1 D=C D 0 ρ V 2 S+ K C L ρ V 2 S 2 2 Using a rearrangement of (

C L=

W 1 ρV2S 2

We get

( )

2 1 1 1 K W2 W D=C D 0 ρ V 2 S+K ρ V 2 S=C D 0 ρ V 2 S+ 1 1 2 2 2 ρV2S ρV 2 S 2 2

❑

Or

D= A V 2 + 1 Where A= C D 0 ρ S 2

and B =

B V2

K W2 . Both these quantities are functions of the density, ρ, and 1 ρS 2

consequently dependent on the altitude.

The diagram on the left above defines the drag variation with speed for a fixed value of ρ. In the diagram on the right the drag curves are shown for different density (altitude) values. All the curves have the same minimum drag value. If we work in terms Equivalent Airspeed, i.e. ρ V 2=ρ0 V E2 , then the drag equation becomes,

1 KW2 2 D=C D 0 ρ0 V E S+ 1 2 2 ρ0 V E S 2

B1 2 D= A 1 V E + 2 VE

i.e. where

2

1 A 1=C D 0 ρ0 S 2

and

B 1=

KW 1 ρ S 2 0

A1 and B1 are independent of density and therefore altitude. Therefore a single curve now represents the drag as shown below.

Minimum Drag:- is given by

D MIN =

( DL )

MIN

× L=

( ) CD CL

×W MIN

Thus to find minimum drag we want minimum CD/CL

C D C D 0 + K C 2L = CL CL If this equation is differentiated with respect to CL we get

( )

d

CD CL

d CL

=

−C D 0 C2L

+K

For a minimum of CD/CL the right hand side of the above equation must be equal to 0. Thus the minimum drag occurs when,

−C D 0 C As CD = CD0 + K CL2 then

2 L

2

+K =0 which meansC D 0=K C L

C DMIN =C D 0 +C D 0=2 C D 0=2 K C2L And

( ) ( CD CL

C D 0 +K C L2 = CL MIN

)

MIN

C D 0 +C D 0

(√ )

=

C D0 K

=2 √ C D 0 K

Minimum Drag Speed To calculate the minimum drag speed, the coefficient of lift which corresponds to the minimum drag ( C L =√ C D 0 / K ) is substituted into the equation which relates velocity to weight (. This gives

V MD =

√

W √K 1 ρS √ C D 0 2

=

( )( ) 2W ρS

1 2

K CD 0

1 4

3.5 Variation of the Power Required with Speed and Altitude The power required for steady level flight, ( Pr) = T V = D V

(

2

Thus Pr=V A V +

)

B 3 B =A V + 2 V V

(A & B as defined previously)

The diagram above on the left defines the variation of power required with speed for a fixed value of ρ. The second diagram shows the variation with ρ i.e. altitude. In terms of the equivalent airspeed, VE, and the previously defined coefficient A1 and B1 the power is given by the equation,

Pr=

VE

√σ

(

2

A1 V E+

B1 V

2 E

) ( =

B1 1 3 A 1 V E+ VE √σ

)

A single power curve is only obtained by plotting Pr √σ against VE as shown on the left above. If Pr is plotted against VE a series of curves are produced for each altitude (density ratio) as shown on the right. Minimum Power Required Starting with power equals drag times velocity we can obtain a relationship between power, coefficient of drag and coefficient of power,

Pr=D V =

√

C W D WV= DW C 1 L L ρS C L 2

Therefore power is proportional to C D / C L3/ 2 . For minimum power its rate of change with the lift coefficient must be zero, i.e.

( ) (

2

C C KC d 3/D 02 + 3 /2L d 3/D2 CL CL CL d Pr =k =k 0= d CL dCL d CL

)

Where k is a non-zero coefficient. Performing the differentiation one gets

0=

−3 C D 0 1 K which means + 2 C 5L /2 2 C 1L /2

3 C D 0=K C2L

therefore at minimum power, CD =CD0 + 3CD0 = 4CD0

√

and

C L=

3 CD0 K

Minimum Power Speed Substituting the lift coefficient corresponding to the minimum power into equation ( one gets

V mp=

√

√ ( ) W

= W 3 C D 0 1 ρS 1 ρS 2 K 2

1 /2

( ) K 3 C D0

1/4

Compare minimum power speed and minimum drag speed

V mp = V md

1 /2

( )( W 1 ρS 2

K 3C D 0

1 /4

)

( 2ρSW ) ( CK ) 1 2

1 4

1 = 1 /4 =0.76 3

D0

Minimum power occurs at a lower speed than minimum drag, requiring a higher lift coefficient, i.e. C L =√ 3 C D 0 / K compared to C L =√ C D 0 / K for minimum drag

3.6 Climbing Flight Consider the aircraft in a steady climb at constant speed along a straight path at angle θ to the horizontal. The forces acting will be in equilibrium and are shown in the diagram below.

Resolving along the flight path;

T- D = W sinθ Resolving perpendicular to flight path; L= W cosθ From (, the climb angle is given by;

sin θ=

T −D W

( ( (

If v is the rate of climb, and V is the speed along the flight path, then,

v =V sin θ=

V (T −D ) W

(

This can be re-written as, W v=T V - D V In this form, equation ( illustrates that the difference between the thrust power available, T V, and the power required to overcome the drag, D V, i.e. the excess power available for climb, equates to the rate of increase in potential energy, W v. Note that the maximum values of the rate of climb v and the angle of climb θ, do not occur together. From (, maximum rate of climb requires maximum excess power, ( TV - DV), and from (, maximum angle of climb requires maximum excess thrust, ( T- D). In general, climb rate is determined by the excess power available, while angle of climb is determined by the excess thrust. To illustrate this we will consider the climb performance of both jet powered and propeller driven aircraft, using simplifying assumptions about engine performance, i.e. for the jet aircraft we assume that thrust remains constant over the speed range and for the propeller driven

aircraft we assume that power output from the propellers remains constant with speed. The various cases of the excess thrust and power available for climb are shown in the figure below.

Note:  

Max. Excess Power gives Max. Rate of Climb Max. Excess Thrust gives Max. Angle of Climb

i.e. from the diagrams: Max rate of climb (RoC) Max angle of climb (θ)

Propeller Aircraft Occurs at minimum power speed. i.e min D V Occurs at speed < min drag speed

Jet Aircraft Occurs at speed > min power speed Occurs at min drag speed

3.7 Excess Power and Thrust Available for Climb The diagrams on the previous page shows the excess power available for climb for a certain altitude and for typical turbojet and turboprop aircraft, assuming respectively T & P constant with speed. They have been drawn to show approximately the same level flight speed for both types of aircraft. Notice however that the propeller driven aircraft has a higher maximum excess power and hence will have a higher maximum rate of climb but that it occurs at a much lower speed than the maximum rate of climb of the turbojet. Calculation of Max. Rate of Climb and Max. Angle of Climb

We will do this under the assumption that the climb angles involved in these maximum cases are still shallow enough to assume that L =W. As we saw when considering gliding flight, this assumption can be used up to around 12° inclination. We consider four cases; a) Max. ROC for a turbojet aircraft - (assuming thrust is constant with speed). We have

v=

(

❑ V (T −D) VT V 1 KW2 C D 0 ρV 2 S + = − 1 2 W W W ρV 2 S 2

)

Assuming a drag equation of the form C D =C D 0 + K C2L . To determine the maximum value of v as velocity changes it is necessary to differentiate the above equation with respect to the velocity, V, and find we the resultant formula equals 0.

1 3 ρ V 2 S CD0 dv T 2 = − + dV W W

Multiplying through by

KW2 =0 1 2 W ( ρV S) 2

1 ρV 2 S W 2 3 CD0

This is a quadratic equation in

(

produces the following equation:

) (

)

2 1 1 2 ρ V 2 S −T ρ V 2 S − K W =0 2 2

1 ρV 2 S which has a single positive solution 2 T + √T +4 (3 C D 0 K W ) 1 2 ρV S= 6 CD0 2 2

2

This solution gives the speed for maximum rate of climb RoC, which is then used to find CL and subsequently D by substituting the speed into the following equations.

C L=

W 1 ρV2S 2

1 and D= ρV 2 S (C D 0+K C2L) 2

b) Maximum angle of climb for turbojet, (assuming thrust is constant with speed). From previous considerations we know that this occurs at minimum drag speed, thus

√

2 C D 0 =K C L , and C L =

therefore the speed for maximum climb angle is,

C D0 K

V=

We can now find D, and hence the maximum climb angle from, The corresponding rate of climb is,

W

√

√

CD0 1 S ρ K 2

1 D= ρ V 2 S 2 C D 0 , 2 T −D sin θ= W v =V sin θ

c) Max. rate of climb for a propeller driven aircraft, (assuming power is constant with speed). From previous considerations we know that this occurs at minimum power speed, for which,

3 C D 0=K C L2 ,and C L=

√

3 CD0 K

Therefore the speed for maximum rate of climb is,

V=

√

W

√

3C D 0 1 ρ S 2 K

We can now find drag D as in (b) and hence rate of climb v =

V (T −D) W

Alternatively, the same method as used in case (a) can be employed for the turbojet,

v=

(

2 ❑ V (T −D) VT V 1 KW = − C D 0 ρV 2 S + W W W 1 2 ρV 2 S 2

)

where, in this case TV = constant = P. Thus for vmax it is necessary to take the derivative of the right hand side of the above equation with respect to V and set the result to zero. This produces

1 −3 ρ V 2 S C D 0 2 + W

K W2 =0 1 W ( ρ V 2 S) 2

thus,

(

)

2 1 K W2 2 ρV S = 2 3 CD0

this gives the speed for maximum rate of climb ROC, and hence CL.

1 ρ V 2 S (4 C D 0) and substituting into the equation ( gives 2 V (T −D ) P−DV = v= W W

Then, D=

P −D v The corresponding angle of climb is, ¿ T −D V ¿ = V W W

[ ]

.

d) Maximum angle of climb for propeller driven aircraft, (assuming power is constant with speed). We have,

sin θ=

(

1 1 KW2 ❑ P T −D 2 − = C D 0 ρ V S+ 1 VW W 2 W 2 ρV S 2

)

For sin to be a maximum its differential with respect to velocity should be zero. This leads to a quadratic expression in V. We won’t pursue this approach as it is probably easier (using a spread sheet for example) to calculate D and thus sin for a range of speeds and then to plot the result in order to obtain the maximum value of sin. . Notes 1. If T (for a jet aircra...