Question is ⇒ 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, Options are ⇒ (A) 2500 m, (B) 2600 m, (C) 2700 m, (D) 2800 m, (E) , Leave your comments or Download question paper.

Question is ⇒ 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, Options are ⇒ (A) 2500 m, (B) 2600 m, (C) 2700 m, (D) 2800 m, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	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 [Wrong Answer]
B.	2600 m [Wrong Answer]
C.	2700 m [Correct Answer]
D.	2800 m [Wrong Answer]
View Answer 19 -7 Explanation:- Answer : C Discuss it below : Madan Bhusal Said on 2019-09-06 07:13:06 Correction of elevation = 20007/100300/300 = 140m. Length of runway after correction of elevation = 2000+140 = 2140m. Correction of temperature: Difference b/n airport reference temp. and Std. temp. @ Elevation = 33.05-(15-0.0065300) = 20 C Applying correction at the rate of 1% for every 1 C, Correction of temperature = (1/1002140)20 = 428 = 430 (Approx.) Corrected runway length = 2140+430 = 2570 m. Correction for gradient : Applying correction for effective gradient at the rate of 20% for each 1% effective gradient, correction for gradient = (20/1002570)0.25/1 = 128.5 = 130m (Approx.) Actual length of runway = 2570+130 = 2700m. !! OOPS Login [Click here] is required to post your answer/result* Help other students, write article, leave your comments

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