, Solutions to problems for
Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2nd 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/m3
b. Calculations:
K J
h := 5000⋅ m ISA L1 := 6.5⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ m kg ⋅ K
Sea level: To := ( 15 + 273) ⋅ K = 288 K
5000 m: T5 := To − L1⋅ h = 255.5 K (Equ 1.6)
5.256
T5
P5 := Po ⋅ = 54000.3 Pa (Equ 1.16)
To
P5 kg
ρ5 := = 0.736 (Equ 1.23)
R1⋅ T5 3
m
Same results.
1
,1.2. Determine the pressure at 5,000 m and ISA-10 condition.
K J
h := 5000⋅ m ISA − 10 L1 := 6.5⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ m kg ⋅ K
Sea level: To := ( 15 + 273 − 10) ⋅ K = 278 K
5000 m: T5 := To − L1⋅ h = 245.5 K (Equ 1.6)
5.256
T5
P5 := Po ⋅ = 52714.2 Pa (Equ 1.16)
To
1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.
K J
h := 20000⋅ ft ISA + 15 L1 := 2⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ ft kg ⋅ K
Sea level: To := [ ( 15 + 273) + 15] ⋅ K = 303 K To = 545.4⋅ R
20000 ft: T20 := To − L1⋅ h = 263 K T20 = 473.4⋅ R (Equ 1.6)
5.256
T20 lbf
P20 := Po ⋅ = 48143.9 Pa P20 = 1005.5⋅ (Equ 1.16)
To ft
2
P20 kg slug
ρ20 := = 0.638 ρ20 = 0.001238⋅ (Equ 1.23)
R1⋅ T20 3 3
m ft
1.4. An aircraft is flying at an altitude at which its temperature is -4.5 oC. Calculate:
2
, a. Altitude in ISA condition
K
L1 := 6.5⋅ To := 15 °C Sea level: To = 288.15 K
1000⋅ m
ISA Talt := ( −4.5 + 273) ⋅ K Talt = 268.5 K TISA := Talt = 268.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h1 := = 3023 m
L1
b. Altitude in ISA+10 condition
ISA + 10 ∆T := 10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 258.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h2 := = 4562 m
L1
c. Altitude in ISA-10 condition
ISA + 10 ∆T := −10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 278.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h3 := = 1485 m
L1
3
Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2nd 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/m3
b. Calculations:
K J
h := 5000⋅ m ISA L1 := 6.5⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ m kg ⋅ K
Sea level: To := ( 15 + 273) ⋅ K = 288 K
5000 m: T5 := To − L1⋅ h = 255.5 K (Equ 1.6)
5.256
T5
P5 := Po ⋅ = 54000.3 Pa (Equ 1.16)
To
P5 kg
ρ5 := = 0.736 (Equ 1.23)
R1⋅ T5 3
m
Same results.
1
,1.2. Determine the pressure at 5,000 m and ISA-10 condition.
K J
h := 5000⋅ m ISA − 10 L1 := 6.5⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ m kg ⋅ K
Sea level: To := ( 15 + 273 − 10) ⋅ K = 278 K
5000 m: T5 := To − L1⋅ h = 245.5 K (Equ 1.6)
5.256
T5
P5 := Po ⋅ = 52714.2 Pa (Equ 1.16)
To
1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.
K J
h := 20000⋅ ft ISA + 15 L1 := 2⋅ R1 := 287⋅ Po := 101325⋅ Pa
1000⋅ ft kg ⋅ K
Sea level: To := [ ( 15 + 273) + 15] ⋅ K = 303 K To = 545.4⋅ R
20000 ft: T20 := To − L1⋅ h = 263 K T20 = 473.4⋅ R (Equ 1.6)
5.256
T20 lbf
P20 := Po ⋅ = 48143.9 Pa P20 = 1005.5⋅ (Equ 1.16)
To ft
2
P20 kg slug
ρ20 := = 0.638 ρ20 = 0.001238⋅ (Equ 1.23)
R1⋅ T20 3 3
m ft
1.4. An aircraft is flying at an altitude at which its temperature is -4.5 oC. Calculate:
2
, a. Altitude in ISA condition
K
L1 := 6.5⋅ To := 15 °C Sea level: To = 288.15 K
1000⋅ m
ISA Talt := ( −4.5 + 273) ⋅ K Talt = 268.5 K TISA := Talt = 268.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h1 := = 3023 m
L1
b. Altitude in ISA+10 condition
ISA + 10 ∆T := 10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 258.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h2 := = 4562 m
L1
c. Altitude in ISA-10 condition
ISA + 10 ∆T := −10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 278.5 K
(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h3 := = 1485 m
L1
3
