The length of runway under standard conditions is 2000 m. The elevation of airport site is 300 m. Its reference temperature is 33.05°C. If the runway is to be constructed with an effective gradient of 0.25 percent, the corrected runway length will be A. 2500 m B. 2600 m C. 2700 m D. 2800 m

The correct answer is $\boxed{\text{B) 2600 m}}$.

The corrected runway length is calculated using the following formula:

$$L_c = L_s \cdot \left(1 + \frac{0.0025 \cdot h}{100} + \frac{0.006 \cdot \Delta T}{100}\right)$$

where:

• $L_c$ is the corrected runway length
• $L_s$ is the length of runway under standard conditions
• $h$ is the elevation of airport site
• $\Delta T$ is the difference between the reference temperature and the actual temperature

In this case, we have:

• $L_s = 2000$ m
• $h = 300$ m
• $\Delta T = 33.05°C – 15°C = 18.05°C$

Substituting these values into the formula, we get:

$$L_c = 2000 \cdot \left(1 + \frac{0.0025 \cdot 300}{100} + \frac{0.006 \cdot 18.05}{100}\right) = 2600\text{ m}$$

Therefore, the corrected runway length is 2600 m.

The first term in the formula accounts for the effect of elevation. The higher the elevation, the thinner the air and the longer the runway needs to be. The second term accounts for the effect of temperature. The warmer the air, the thinner it is and the longer the runway needs to be.