Determine the density altitude for these conditions: Altimeter setting 29.25, Runway temperature +81°, Airport elevation 5,250 ft MSL.

• A. 7,500 feet
• B. 8,500 feet
• C. 9,000 feet
• D. 6,500 feet

Answer: (B)

Density altitude shows how the air would behave if it were in standard atmosphere, combining altitude and how warm or cold the air is. Warmer air makes air density lower, so the density altitude rises above the actual airport elevation.

First, find the pressure altitude using the observed altimeter setting. The formula uses a standard reference of 29.92 inHg. So, PA = field elevation + (29.92 − altimeter setting) × 1000. With a field elevation of 5,250 ft and a setting of 29.25, PA ≈ 5,250 + (0.67 × 1000) ≈ 5,920 ft.

Next, apply the temperature correction. A common quick method multiplies the difference between the actual temperature and standard temperature by about 120 ft per degree Fahrenheit. Treat the actual temperature as the runway temperature, 81°F, and use standard sea-level temperature, 59°F, for the difference. ΔT ≈ 81 − 59 = 22°F. So the correction is about 120 × 22 ≈ 2,640 ft.

Add that correction to the pressure altitude: 5,920 + 2,640 ≈ 8,560 ft, which rounds to about 8,500 ft. The warm temperature relative to standard air raises density altitude, and the lower-than-standard altimeter setting further increases the pressure altitude, combining to place density altitude around 8,500 ft.