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Horizontal Flight Performance Jet aircraft

Delft University of Technology

Dr. ir. Mark Voskuijl Daan Westerveld

Bryan - CC - BY 2.0

AE1110x - Introduction to Aeronautical Engineering

Horizontal flight performance of jet aircraft In the videos the horizontal flight performance of propeller aircraft is explained by means of the Spirit of St. Louis. Due to the different nature of the propulsion system, jet aircraft perform different from their propeller counterparts. Therefore, the horizontal flight performance of jet aircraft will be explained in this writing. Minimum airspeed The calculation of minimum airspeed of jet aircraft is the same as for propeller aircraft. It occurs at the maximum lift coefficient, and can be found from: s W2 1 Vmin = (1) S ρ CLmax Maximum airspeed The maximum airspeed of an aircraft occurs at maximum thrust (or maximum power). For a jet aircraft we speak in terms of thrust. In steady, horizontal flight for a normal jet aircraft (αt = 0) we have the following equations of motion: L=W

(2)

T =D

(3)

We want to find the maximum airspeed, and for that we need to find the CL for which we can get maximum airspeed. We know that the maximum airspeed occurs at the maximum thrust setting (T = Tmax ), which we assume as independent of airspeed. This can be done by starting from Equation 3 (the explanation of each subsequent step is written next to the formula): Tmax = D Tmax = D

L L

We multiply by 1

CD W CL CD0 + k1 CL + k2 CL2 = ·W CL

D CD = and L = W from Equation 2 L CL

Tmax = Tmax

From the lift-drag polar we know CD = CD0 + k1 CL + k2 CL2

This final equation can be rewritten as a parabolic relation as follows: CD0 + k1 CL + k2 CL2 ·W CL CD0 + k1 CL + k2 CL2 = CL

Tmax = Tmax W

Tmax · CL = CD0 + k1 CL + k2 CL2 W   Tmax 2 k2 CL + k1 − CL + CD 0 = 0 W

Divide by W Multiply with CL Group all terms with CL2 , CL and CD0

This quadratic equation can be solved for CL by means of the quadratic formula:
q
2 Tmax max − k1 − W ± k1 − TW − 4k2 CD0 CLopt = 2k2

(4)

Finally, the CL calculated with the equation above (take the smallest, positive value) can be inserted in Equation 1 to find the maximum airspeed. Horizontal flight performance of jet aircraft

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AE1110x - Introduction to Aeronautical Engineering

Maximum specific range In order to have maximum specific range, we want to use a minimum amount of fuel per unit of distance. This means that the specific range VF (speed over fuel flow) should be minimised. In this derivation we assume steady straight and symmetric flight, such that equations 2 and 3 hold. For jet aircraft the fuel flow can be expressed as follows, where cT is assumed a constant (in reality it can vary) relating the fuel flow F to the thrust T : F = cT · T

(5)

We know that T = D (we are cruising in a steady, straight, horizontal and symmetric flight), such that we can write F = cT · T = cT · D. Substituting this expression in the expression for specific range we find: V V /F = (6) cT · D If we want to maximise V /F , we need to maximise V /D (since cT = const). Furthermore, (V /D)max is the same as (D/V )min . This expression can be used in order to find the optimal aerodynamic condition as follows:     D L D 1 = We multiply with = 1 ·L· V min L V min L     D 1 CD CD D ·W · = and L = W = Since V min CL V min L CL     C 1 D D  = From Equation 1 ·W · q V min CL W 2 1 S ρ CL min     D W  = q We put everything under the square root 2 CL W V min 2 S ρ CD min     D CL → 2 V min CD max Now we know at what aerodynamic condition maximum specific range occurs, we need to find at what lift coefficient this is. For this we differentiate the expression just found with respect to CL , and put it equal to 0:   d CL =0 2 dCL CD   2 · 1 − C · 2C dCD CD L D dCL d CL = =0 Using the quotient rule 2 4 dCL CD CD 2 · 1 − C · 2C dCD CD L D dCL 4 CD

=0→

dCD 1 CD = dCL 2 CL

In order to rewrite the last expression into term of CL only, we need to use the lift-drag polar. First we will find the derivative of the lift-drag polar with respect to CL :
dCD d = CD0 + k1 CL + k2 CL2 = k1 + 2k2 CL dCL dCL

2

(7)

Horizontal flight performance of jet aircraft

AE1110x - Introduction to Aeronautical Engineering

If we fill that into the expression write:

dCD dCL

=

1 CD 2 CL ,

together with CD = CD0 + k1 CL + k2 CL2 we can

dCD 1 CD = dCL 2 CL 1 CD0 + k1 CL + k2 CL2 k1 + 2k2 CL = 2 CL 2CL (k1 + 2k2 CL ) = CD0 + k1 CL + k2 CL2 3k2 CL2

+ k1 CL − CD0 = 0

We filled in the equations Bring CL and 1/2 to the other side Bring everything to the left-hand side

Using the quadratic formula this brings us to an expression for the optimum CL for maximum specific range: q −k1 ± k12 + 12k2 CD0 (8) CL = 6k2 If the optimum flight speed is of interest the (positive) value of Equation 8 needs to be plugged into Equation 1. Maximum specific endurance In order to find an expression for the maximum specific endurance, we first need to state that the aircraft is performing steady, straight and symmetric flight such that equations 2 and 3 hold. Now we start looking at the physical condition for maximum specific endurance. If we want to stay in the air as long as possible, we should use as little fuel as possible per unit of time in order to make it last as long as possible. In other words: the fuel flow F should be minimised. We thus need to minimise Equation 5, repeated below: F = cT T F = cT D

From Equation 3

Because cT is constant, the only way to minimise the fuel flow F is to minimise the drag D: (F )min → (D)min   L (D)min = D L min   CD W (D)min = CL   min CL (D)min → CD max

We multiply with

 We cannot influence W and

CD CL

 min

L =1 L

We use L = W   CL can be expressed as CD max

This means that the optimum CL for maximum specific endurance occurs at:     CD0 + k1 CL + k2 CL2 d CD d = dCL CL dCL CL     d CD d CD 0 = + k1 + k2 CL dCL CL dCL CL   d CD CD = − 20 + k2 dCL CL C r L CD 0 CLopt = k2

(9)

If the optimum airspeed for this condition is of interest, Equation 9 needs to be put into Equation 1. Horizontal flight performance of jet aircraft

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