SME
Flight Mechanics, Stability & Control · Problem type
Density at an altitude (standard atmosphere)
Use this for “Find the standard-atmosphere temperature, pressure and density at 3000 m.”
Diagram
In the troposphere the air cools about 6.5 °C for each km up: −4.5 °C at 3 km.
This is an example. Type your own values to start a new problem.
h
Altitude
example
T
Temperature
calculated · K
p
Pressure
calculated
ρ
Air density
calculated
Formulas
T
=
288.15
−
6.5
×
h
268.7 = 288.15 − 6.5 × 3
p
=
101,325
×
(
288.15 − 6.5 ×
h
288.15
)
5.2559
70,110
=
101,325
×
(
288.15 − 6.5 × 3
288.15
)
5.2559
ρ
=
p
287 ×
T
0.9093
=
70,110
287 × 268.7
Formulas are worked in km, Pa. The step-by-step shows the conversions.
Assumptions
•
The standard atmosphere’s layers start at 0, 11, 20, 32, 47, 51 and 71 km, to 84.852 km; their lapses are −6.5, 0, +1, +2.8, 0, −2.8 and −2 K per km.
•
Their base temperatures are 288.15, 216.65, 216.65, 228.65, 270.65, 270.65 and 214.65 K.
•
Their base pressures are 101,325, 22,632, 5,474.9, 868.03, 110.91, 66.941 and 3.9567 Pa.
•
Air is a perfect gas with R = 287 J/(kg·K) and g ÷ R = 34.163 K/km; a layer’s pressure power is about 34.163 ÷ |L|: 5.2559, 34.163, 12.201 or 17.082.
Step-by-step
WE KNOW
h = 3,000 m
FIND
temperature (T), pressure (p), air density (ρ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to km, Pa).
h = 3,000 m = 3 km (1 km = 1000 m)
Step 2 · Find temperature (T)
Use T = 288.15 − 6.5 × h
In the troposphere start from 288.15 K at 0 km and take off 6.5 K for each km above it.
3 ≤ 11: troposphere
T
=
288.15
−
6.5
×
3
T
=
288.15
−
19.5
T
=
268.65
K
Step 3 · Find pressure (p)
Use p = 101,325 × ((288.15 − 6.5 × h) ÷ 288.15)5.2559
Where the temperature changes, the pressure is 101,325 Pa at 0 km times the temperature ratio to the power 5.2559.
p
=
101,325
×
(
288.15 − 6.5 × 3
288.15
)
5.2559
p
=
101,325
×
(
288.15 − 19.5
288.15
)
5.2559
p
=
101,325
×
(
268.65
288.15
)
5.2559
p
=
101,325
×
0.9323
5.2559
p
=
101,325
×
0.6918
p
=
70,108.4273
Pa
Step 4 · Find air density (ρ)
Use ρ = p ÷ (287 × T)
The perfect-gas law: pressure over R times the temperature.
ρ
=
70,108
287 × 268.65
ρ
=
70,108
77,102.55
ρ
=
0.9093
kg/
m
3
Step 5 · Convert the answers
Change each answer back to the unit you chose.
p = 70,110 Pa = 70.108 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
268.65 = 288.15 − 6.5 × 3 ✓
70,108
=
101,325
×
(
288.15 − 6.5 × 3
288.15
)
5.2559
✓
0.9093
=
70,108
287 × 268.65
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Aircraft performance
Main lesson
Stall speed
Use this for “A 10,000 N airplane has S = 16 m² and CLmax = 1.6. Find its stall speed at sea level, ρ = 1.225 kg/m³.”
Jet range (Breguet)
Use this for “A jet cruises at 230 m/s with L/D = 16 and cₜ = 0.6 per hour. It starts the cruise at 600,000 N and burns 120,000 N of fuel. Find its range.”
Rate of climb from excess thrust
Use this for “A 50,000 N airplane at 80 m/s has 6000 N of thrust and 3553 N of drag. Find its rate of climb and climb angle.”
Load factor and radius of a level turn
Use this for “An airplane flies a level turn at 80 m/s, banked 60°. Find the load factor, the turn radius and the turn rate.”