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Runway Design – IES PYQ ​


11. At a certain station, the mean of the average temperature is 25°C and mean of the maximum daily temperature is 40°C. What is the airport reference temperature (ART) ?

[IES 2005]

  1. 45°C
  2. 35°C
  3. 30°C
  4. 46.6°C
Answer: Option C
Explanation:

Airport reference temperature = Ta +

Tm – Ta

3

= 25 +

40 – 25



3

⇒ Airport reference temperature = 30°C


12.  A runway is located 450 m above mean sea level. If the aeroplane reference field is 1800 m, what is the approximate corrected runway length for elevation ?

[IES 2006]

  1. 1889 m
  2. 1987 m
  3. 2013 m
  4. 1849 m
Answer: Option B
Explanation:

As per ICAO, for every 300 m above MSL, increase the basic runway length by 7%.
Hence for 450 m above MSL, percentage increase in runway length =

7%

300

x 450 = 10.5%

Increase in runway length =

10.5

100

x 1800 = 189 m

Thus, corrected runway length = 1800 + 189 = 1989 m
Here, the closest option is 1987 m.


13.  For design of a runway length, match List-I (Factor) with List-II (Correction) and select the correct answer using the codes given below the lists:

List-I

  1. Standard basic length
  2. Elevation
  3. Temperature
  4. Effective gradient

List-II

  1. 7% for every 303 metres
  2. 0.5%
  3. 20% for each 1% gradient
  4. Depends upon aircraft and obtained from ICAO Standard Tables

[IES 2006]

Codes: A B C D

  1.  2  3  4 1
  2.  4  3  2 1
  3.  2  1  4 3
  4.  4  1  2 3
Answer: Option D
Explanation:

  • Basic runway length depends on aircraft and are obtained from ICAO standard tables.
  • Elevation correction – For every 300 m above MSL, increase the runway length by 7%.
  • Temperature correction – For every 1°C above standard temperature, increase the runway length by 1% after applying elevation correction.
  • Gradient correction – For every 1% effective longitudinal gradient, increase the runway length by 20% after applying the above two corrections.


14.  The total correction percentage for altitude and temperature, in calculating the runway length from basic runway length, normally does not exceed _____ .

[IES 2006]

  1. 14
  2. 28
  3. 35
  4. 7
Answer: Option C
Explanation:
The total correction percentage in calculating the runway length after applying altitude and temperature corrections must not exceed 35%.

15.  As the elevation increases, the runway length has to be changed at what rate ?

[IES 2007]

  1. Increased @ 7% per 300 m rise in elevation above MSL
  2. Decreased @ 9% per 300 m rise in elevation above MSL
  3. Increased @ 15% per 300 m rise in elevation above MSL
  4. Decreased @ 5% per 300 m rise in elevation above MSL
Answer: Option A
Explanation:
As per ICAO, for every 300 m rise in elevation above MSL, increase the basic runway length by 7%.

16.  The orientation of runway is decided by which factors ?

  1. Maximum wind coverage and least cross winds
  2. Landing characteristics of aircrafts
  3. Scope for future expansion
  4. Obstruction-free approaches

Select the correct answer using the codes given below:

[IES 2008]

  1. 1, 2 and 3
  2. 1, 3 and 4
  3. 2, 3 and 4
  4. 1 and 2 only
Answer: Option B
Explanation:
The orientation of runway is independent of the landing characteristics of aircrafts.

17.  Assertion (A): Wind Rose Diagrams, showing wind direction, duration and intensity are an essential requirement for planning the best direction of main runway of the airport. 

Reason (R): Corrections for elevation, temperature  and gradient have to be applied to determine the length of the main runway of an airport, as per ICAO recommendations.

[IES 2008]

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not a correct explanation of A.
  3. A is true but R is false. 
  4. A is false but R is true.
Answer: Option C
Explanation:

  • The wind data (i.e., direction, duration and intensity) are graphically represented by a diagram called wind rose diagram. It helps in analysing the wind data and obtaining the most suitable orientation of the runway.
  • As per ICAO recommendations, the basic runway length is to be corrected for elevation and temperature. Correction due to gradient is applied under FAA recommendations.

18.  Wind-rose diagram is useful in deciding on the orientation of _____ .

[IES 2011]

  1. Hanger
  2. Apron
  3. Runway
  4. Taxiway
Answer: Option C
Explanation:
The wind data (i.e., direction, duration and intensity) are graphically represented by a diagram called wind rose diagram. It helps in analysing the wind data and obtaining the most suitable orientation of the runway.

19.  For safe landing and take-off, the following factors need to be carefully considered:

  1. Cross-wind
  2. Runway grade
  3. Runway width and side clearance
  4. Obstructions

[IES 2011]

  1. 1, 2, 3 and 4
  2. 1, 3 and 4 only
  3. 2, 3 and 4 only
  4. 1, 2 and 3 only
Answer: Option A
Explanation:
All the above mentioned factors are to be considered for safe landing and take-off.

20.  The runway length for an airport located at 460 m above MSL, corrected for elevation, is 3670 m. The monthly means of maximum and mean daily temperatures for the hottest month of the year are 27°C and 18°C respectively. What will be the final corrected length of the runway with correction incorporated also due to temperature effects ?

[IES 2015]

  1. 4000 m
  2. 3750 m
  3. 3400 m
  4. 4500 m
Answer: Option A
Explanation:
Runway length after applying elevation correction = 3670 m
Monthly mean of average daily temperature, Ta = 18°C
Monthly mean of maximum daily temperature, Tm = 27°C

Airport reference temperature = Ta +

Tm – Ta

3

= 18 +

27 – 18

3

= 21°C

Standard atmospheric temperature at 460 m above MSL = 15 – (0.0065 x elevation) = 15 – (0.0065 x 460) = 12.01°C ≈ 12°C
Rise in temperature = 21°C – 12°C = 9°C
As per ICAO, for every 1°C increase in temperature above the standard atmospheric temperature, increase the runway length by 1%. Hence, for 9°C rise in temperature, increase the runway length by 9%.

Runway length correction for temperature =

9

100

x 3670 = 330.3 m
Thus, final corrected runway length = 3670 + 330.3 = 4000.3 m

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