Midsummer 2016 EG2401 Solutions - Aerospace Part

Dr. M.C. Turner

April 3, 2017

Question 3

a) Equivalent airspeed is the velocity of the aircraft referenced to sea level.

ρoV 2EAS = ρV2

TAS

b) Something like:

AIRCRAFTDRAG

EQUIVALENT AIRSPEEDV V

S E,MD

TOTAL DRAG

NO−LIFTDRAG

LIFT DEPENDENT DRAG

c) Question makes use of following figure

mg

D T

L

V

Resolving horizontally

D =DL

L =CD

CLmg

Hence minimum drag whenCD/CL is minimum.

UsingCD =CD0+ εC2L we get

dCL

{

CD

CL

}

=d

CL

{

CD0

CL+ εCL

}

= −CD0C−2L + ε

Necessary condition for a minimum gives:

−CD0C−2L + ε = 0⇔CL =

√

CD0

ε

d) Similar to before, the required power is given by

Pr = DV =CD

CLmgV

1

In s/l flight, resolving vertically gives

V =

√

2mgρSCL

Hence

Pr =CD

CLmg

√

2mgρSCL

=

√

2ρS(mg)3/2 CD

C3/2L

Minimum power condition therefore occurs when

CD

C3/2L

is minimum

which implies

dCL

{

CD

C3/2L

}

=d

CL

{

CD0C−3/2L + εC1/2

L

}

= −32

CD0C−5/2L +

12

εC−1/2L

Necessary condition for a minimum gives:

−32

CD0C−5/2L +

12

εC−1/2L = 0⇔CL =

√

3CD0

ε

e) Resolving vertically gives

VEAS =

√

2mgρoSCL

Hence at minimum drag (after a little algebra) we get

VEAS,MD =

√

2mgρoS

(

εCD0

)14

Similarly at minimum power (after a little algebra) we get

VEAS,MP =

√

2mgρoS

(

ε3CD0

) 14

=

√

2mgρoS

(

εCD0

) 14(

13

) 14

= VEAS,MD

(

13

)14

≈ 0.75VEAS,MD

f) A pilot may choose to fly at the minimum drag condition if s/he is gliding and wants to save fuel - m/d condition

corresponds to the greatest range during gliding flight.

2

Question 4

a) Range - roughly speaking, the greatest distance an aircraft can fly on a tank of fuel.

Endurance - roughly speaking, the greatest time an aircraftcan stay aloft on a tank of fuel

b) A long range is desirable in long-distance passenger aircraft in order to minimise the number of stops an aircraft has

to make. The stops slow the journey time but are also expensive as take-off and landing can require large amounts

of fuel to be used. [Other answers acceptable too]

c) i) From question data we have

R = −∫ m1

m0

Vm f

dm

= −∫ m1

m0

ηVf TV

dm

= −∫ m1

m0

ηf D

(T = D)

Because, in s/l flight we have

D =CD

CLmg see above

if follows that

R = −∫ m1

m0

ηf

CL

CD

1mg

dm

= −ηf g

∫ m1

m0

CL

CD

1m

dm

Range maximum whenCL/CD is maximum (constant). Hence

R = −ηf g

(

CL

CD

)

max

∫ m1

m0

1m

dm

= −ηf g

(

CL

CD

)

max(lnm1− lnm0)

=ηf g

(

CL

CD

)

maxln

m0

m1

Now max range whenCL/CD is maximum or whenCD/CL is minimum. This occurs (see above) when

CL =

√

CD0

ε

This then implies (after a little algebra) that

CL

CD=

1

2√

CD0ε

Hence maximum range when

R =ηf g

1

2√

CD0εln

m0

m1

ii) As before

R = −ηf g

∫ m1

m0

CL

CD

1m

dm

But note that in s/l flight:

CD

CL=

CD0+ ε(

mg1/2ρV2

)2

mg1/2ρV2

3

Thus

R = −ηf g

∫ m1

m0

mg1/2ρV2

CD0+ ε(

mg1/2ρV2

)2

1m

dm

= −ηf

∫ m1

m0

11/2ρV2

CD0+ ε(

mg1/2ρV2

)2 dm

= −ηf

∫ m1

m0

1εg2

1/2CD0ρV 2

εg2 +m2dm

= −η

f εg2

∫ m1

m0

11/2CD0ρV 2

εg2 +m2dm

= −η

f εg2

√

εg2

1/2ρCD0V 2

(

tan−1 m1

√

εg2

1/2ρCD0V 2 − tan−1 m0

√

εg2

1/2ρCD0V 2

)

=η

f√

εCD0

√

2ρ

V

(

tan−1 m0

√

εg2

1/2ρCD0V 2 − tan−1 m1

√

εg2

1/2ρCD0V 2

)

d) Maximum range calculated as

RMAX =ηf g

1

2√

CD0εln

m0

m1

Using numerical data

RMAX =0.82

10−7×9.811

2√

0.014×0.05ln

70×103

58×103

HenceR = 2970600 m or≈ 2971 km

Similarly, at 100 ms−1 EAS we have

√

1/2CD0ρV2

εg2 =

√

0.014×0.5×1.225×1002

0.05×9.812 = 4.22

Thus we have

RMAX ,V EAS =0.82

10−7×√

0.014×0.05×

√

21.225

×100×(

tan−170000/4.22− tan−158000/4.22)

HenceRMAX ,VEAS = 1663547m or≈ 1664 km i.e. much lower

4

Question 5

a) Diagram should look something like:

mg

L

γ

γ

γ

V T

D

b) Resolving:

Lcosγ = mg

T = D+Lsinγ

Hence from the second equation:

sinγ =T −D

mg

=T −1/2ρSCDV 2

mg

=T −1/2ρS(CD0+ εC2

L)V2

mg

=1

mg

(

T −1/2ρSCD0V2−1/2ρSV2ε

(

mgcosγ1/2ρSV2

)2)

=1

mg

(

T −1/2ρSCD0V2−

(mg)2(cosγ)2

1/2ρSV2

)

Assumingγ is small

sinγ =1

mg

(

T −1/2ρSCD0V2−

(mg)2

1/2ρSV2

)

Inserting the question data then gives:

sinγ = 0.106⇒ γ = 6.1◦

This implies the climb rate is given as

Vc =V sinγ = 9.77m/s

c) Max climb angle is found again by resolving horizontally,which, as before gives:

sinγ =T −D

mg

=T

mg−

DLcosγ

≈T

mg−

CD

CL

5

Hence maximum climb angle occurs at the mininum drag conditions, which (see above) means that

CL =

√

CD0

ε

This implies thatCD

CL= 2√

CD0ε

and thus that

sinγ =T

mg−2√

CD0ε

Inserting the numerical data we thus get that

sinγ = 0.109⇒ γ = 6.3◦

To find the airspeed, we again resolve “vertically” to get

V =

√

2mgcosγρSCL

=

√

2mgcosγρS

(

εCD0

)12

Inserting the numerical data we then getV = 108.3m/s

d)

Vc = V sinγ

= V

(

Tmg

−CD

CL

)

(from earlier)

=

√

2mgcosγρSCL

(

Tmg

−CD

CL

)

≈

√

2mgρSCL

(

Tmg

−CD

CL

)

(since angle is small)

=

√

2mgρS

(

Tmg

C− 1

2L −CD0C

− 32

L − εC12L

)

Hence the minimum occurs when

ddCL

(

Tmg

C− 1

2L −CD0C

− 32

L − εC12L

)

= 0

Differentiating:

− f rac12C−3/2L

(

Tmg

)

+32

CD0C−5/2L −

12

εC−1/2L = 0

Multiplying through byC1/2L then gives

−12

C−1L

(

Tmg

)

+32

CD0C−2L −

12

ε = 0

−T

mgC−1

L +3CD0C−2L − ε = 0

Solving the quadratic equation forC−1L then gives:

C−1L =

T/mg+√

(T/mg)2+12CD0ε6CD0

Evaluating this using the question data then gives

C−1L = 4.12⇒CL = 0.24

Using this in the expression forV then gives

V ≈√

2mgρSCL = 161ms−1

6

Similarly using the obtainedCL in the expression for sinγ gives

sinγ =T

mg−

CD

CL

=T

mg−CD0C

−1L − εCL

which impliesγ ≈ 5.3◦

7

Question 6

a) Correctly banked turn: a turn in which the aircraft turns due to the centripetal acceleration generated by rolling the

aircraft. There should be no sideslip.

b) Consultα-CL diagram. Note straight-line equation isy = mx+ c i.e.

CL =CL0+CLαα

Thus atCL = 0 we have

CL0 = 4.5×2.5∗π

180= 0.196

NB - Angle should be measured should be radians!

c) Something like:

mg

α

φ

Aircraft nosepoints into

slide

L t

d) Resolving gives

Lt cosφ = mg

Lt sinφ = mα = mV 2/R

Dividing one by the other gives

tanφ =V 2

gR

Now tanφ =√

sec2 φ−1 and becauseN = secφ we have

tanφ =√

N2−1

Equating the two expressions for tanφ then gives

R =V 2

g√

N1−1

However, we also have that

Lt cosφ = mg

Lt =mg

cosφ= Nmg

and that

Lt =12

ρSCL,tV2

which implies

V =Nmg

1/2ρSCL,t

8

This then implies

R =2Nmg

ρCL,tSg√

N1−1

=2Nm

ρCL,tS√

N1−1

d)

CL,t =CL0+CLαα = 0.633

So

R =2×4×10,000

1.225×0.74×50×0.633×√

42−1≈ 720m

e) Minimum radius occurs atCL,max which means

Rmin =2Nm

ρSCL,max√

N2−1

Now CL,max = 0.0044×4.5= 1.18. Hence

Rmin =8000

1.225×0.74×1.18×√

42−1= 386m

9