Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2

1 Solutions to problems for

Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2

nd ed.Ch. 1 The software package Mathcad is used to solve problems.

1.1. Determine the temperature, pressure and air density at 5,000 m and ISA condition.

There are two methods:

a. Using appendix:

From Appendix A:

-Temperature: 255.69 K

-Pressure: 54,048 Pa

-Air density: 0.7364 kg/m

3

b. Calculations:

Same results.

Sea level:

5000 m: (Equ 1.6)

(Equ 1.16) (Equ 1.23)

h5000m⋅:= ISA L16.5

K 1000m⋅

⋅:=R1287

J kgK⋅

⋅:= P

o

1013 25Pa⋅:=

T o

152 73+( ) K⋅288K=:=

T 5 T o

L1h⋅− 255.5K=:=

P 5 P o T 5 T o       5.256

⋅ 54000.3Pa=:=

ρ 5 P 5 R1T 5 ⋅ 0.736 kg m 3

=:=

2

1.2. Determine the pressure at 5,000 m and ISA-10 condition.

1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.

1.4. An aircraft is flying at an altitude at which its temperature is -4.5

o

C. Calculate:

Sea level:

5000 m: (Equ 1.6)

(Equ 1.16)

Sea level:

20000 ft: (Equ 1.6)

(Equ 1.16) (Equ 1.23)

h5000m⋅:= ISA10− L16.5

K 1000m⋅

⋅:=R1287

J kgK⋅

⋅:= P

o

1 01325Pa⋅:=

T o

1 5 273+ 10−( ) K⋅278K=:=

T 5 T o

L 1h⋅−245.5K=:=

P 5 P o T 5 T o       5.256

⋅ 52714.2Pa=:=

h20000ft⋅:= ISA15+ L12

K 1000ft⋅

⋅:=R1287

J kgK⋅

⋅:= P

o

1 01325Pa⋅:=

T o

1 5 273+( ) 15+[ ] K⋅303K=:=T

o

5 45.4R⋅=

T 20 T o

L 1h⋅−263K=:= T

20

4 73.4R⋅=

P 20 P o T 2

T o       5.256

⋅ 48143.9Pa=:=P

20

1 005.5

lbf ft 2 ⋅= ρ 20 P 20 R1T 20 ⋅ 0.638 kg m 3

=:=ρ

20

0 .001238

slug ft 3 ⋅=

3

• Altitude in ISA condition

• Altitude in ISA+10 condition

• Altitude in ISA-10 condition

Sea level:

(Equ 1.6)

(Equ 1.6)

(Equ 1.6) L16.5 K 1000m⋅

⋅:= T

o

1 5°C:=T

o

2 88.15K=

ISA T alt

4 .5− 273+( ) K ⋅:= T

alt

2 68.5K=T

ISA T a

268.5K=:=

T ISA T o L 1h⋅−h1 T o T ISA

−( )

L1

3023m=:=

ISA10+ ∆T1:= T

ISA

4.5− ∆ T− 273+( ) K⋅:=T

ISA

2 58.5K=

T ISA T o L 1h⋅−h2 T o T ISA

−( )

L1

4562m=:=

ISA10+ ∆T 10−:= T

ISA

4.5− ∆ T− 273+( ) K⋅:=T

ISA

2 78.5K=

T ISA T o L 1h⋅−h3 T o T ISA

−( )

L1

1485m=:=

4

1.5. Determine relative density (σ) at ISA-20 condition and 80,000 ft altitude.

1.6. Determine the temperature at 70,000 ft and ISA condition.

Sea level:

Second layer:

(Equ 1.16) (Equ 1.23) (Equ 1.25)

Sea level:

Second layer:

h80000ft⋅:= ISA20− R1287

J kgK⋅

⋅:= P

o

1 01325Pa⋅:= ρ

o

0 .00237

slug ft 3

⋅:=

T o

1 5 273+( ) 20−[ ] K⋅268K=:=T

o

4 82.4R⋅=T

o

5 .15−°C=

T 80

5 6− 273+ 20−( ) K⋅:=T

80

3 54.6R⋅=T

80

1 97K=

P 80 P o T 8

T o       5.256

⋅ 20098.1Pa=:=P

80

4 19.8

lbf ft 2 ⋅= ρ 80 P 80 R1T 80 ⋅ 0.355 kg m 3

=:=ρ

80

0 .00069

slug ft 3 ⋅= σ ρ 80 ρ o

0.291=:=

h70000ft⋅:= ISA

T o

1 5 273+( ) K⋅288K=:=T

o

5 18.4R⋅=

T 70

5 6− 273+( ) K ⋅:=T

70

3 90.6R⋅=T

70

2 17K=