Design of Airports MCQ Quiz in मल्याळम - Objective Question with Answer for Design of Airports - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Correction for basic Runway length (l)

i) Elevation correction

For every 300-meter rise above MSL, the length will be increased by 7%

Corrected length \(\Rightarrow {{\rm{l}}_1} = {\rm{l}} + \left( {\frac{{\rm{x}}}{{300}} \times \frac{7}{{100}}} \right) \times {\rm{l}}\)

ii) Temperature correction

Airport reference Temperature (ART)

\({\rm{T}} = {{\rm{T}}_{\rm{a}}} + \frac{{{{\rm{T}}_{\rm{m}}} - {{\rm{T}}_{\rm{a}}}}}{3}\)

Tm → monthly mean of maximum daily temp of hottest month

Ta → monthly mean of average daily temp of hottest month

For energy 1°C rise of ART above (SATx i.e. Standard Airport Temperature) length will be increased by 1%

Corrected length \( \Rightarrow {{\rm{l}}_2} = {{\rm{l}}_1} + \left( {\frac{{{\rm{\Delta T}}}}{{100}}} \right){{\rm{l}}_1}\)

Note: As per ICAO, the cumulative connection for elevation and temperature together should not be ≯ 35%.

Cumulative (%) correction \(= \frac{{{{\rm{l}}_2} - {{\rm{l}}_1}}}{{\rm{l}}} \times 100 \not > 35\% \)

If exceeds modify, l2 = 1.35 × l

iii) Gradient correction

For every 1% effective gradient, runway length will be increased by 20%

\({\rm{Final\;length\;}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}_2} + \left( {20 \times \frac{{{{\rm{G}}_{{\rm{effective}}}}}}{{100}}{\rm{\% }}} \right){{\rm{l}}_2}\)

\({{\rm{G}}_{{\rm{effective}}}}\left( {\rm{\% }} \right) = \frac{{{\rm{R}}{{\rm{L}}_{{\rm{highest\;point}}}} - {\rm{R}}{{\rm{L}}_{{\rm{lower\;point}}}}}}{{{\rm{Runway\;unit\;}}\left( {{{\rm{l}}_2}} \right)}}\).

Head wind: The runway is oriented in the direction of the prevailing winds. It helps in the landing of vehicle by causing breaking effect and during takeoff, it provides grater lift on wings of the aircraft.

Cross Wind: Component of wind perpendicular to runway. It abrupts the landing and takeoff of aircraft.

Wind Cover: The percentage of time in a year during which the cross-wind component of wind is within permissible limit of 25 kmph.

Wind Rose: It is a graphical representation of direction, intensity & duration of wind is called wind rose. The name of diagram is wind rose because of its irregular shape resembling a rose.

The study of wind rose helps in determining the most suitable orientation of runway and useful device for estimating the runway capacity.

Wind rose diagram are of two types:

Explanation:

The airport reference temperature is calculated using the following formula:

\(ART= T_a + \frac{T_m - T_a}{3}\)

Here,

T_a is the monthly mean of the average daily temperature for the hottest month of the year 

T_m is the monthly mean of the maximum daily temperature for the same month of the same year

Given that

T_a = 24⁰ C and T_m = 30⁰ C

Substitute these values in the given formula:

\(ART= 24 + \frac{30 - 24}{3}\)

ART = 24 + 2 = 26⁰ C

Additional Information

Procedure to find T_a

Step 1: Select any month for a given year.

Step 2: Evaluate the average daily temperature variation for a day and this will be called the daily average temperature for that day. This procedure is repeated for all the other days of that month.

Step 3: Evaluate the month's mean temperature as;

\(\text{Mean monthly temp.} = \frac{\text{ summation of average daily temperature of all the days in the month }}{\text{total number of days in that month}}\)

Step 4: Repeat the above procedure (Step 2 and Step 3) for all other months of the year. The hottest month is that which has the maximum monthly mean temperature and that will be called T_a.

Concept:

Airport reference temperature \(={{T}_{a}}+\frac{1}{3}\left( {{T}_{m}}-{{T}_{a}} \right)\)

Where, T_a = Average temperature of the hottest month

T_m = Monthly mean of the maximum daily temperature of the same month

Calculation:

Given:

T_m = 48°C, T_a = 39°C

∴ Airport reference temperature \(=39+\frac{1}{3}\left( 48-39 \right)\)

= 42°C

Additional Information

Basic runway length is calculated with the following assumptions:

i) It is calculated at Mean sea level (MSL)

ii) It is calculated for standard temp 15°C at MSL.

iii) The gradient is assumed to be zero.

Due to the variance of these assumptions, some corrections are applied.

i) Elevation Correction: 7% increase in basic runway length for every 300 m rise above MSL

ii) Temperature correction: 1% increase for every 1°C rise in airport reference temperature with respect to standard temperature at elevation.

Where,

Standard temperature at elevation = Temperature at MSL – 0.0065 × Elevation

This increase is made on the already corrected runway length for elevation.

iii) Gradient correction: 20% increase for 1% of effective gradient.

• These corrections are always applied sequentially and on the already corrected runway length value.

Explanation:

The basic runway length is determined in three cases

1. Normal landing case - Aircraft should stop within 60% of landing distance or runway length
2. Normal take-off case - minimum clearway width should be 150m and it is also kept free from obstructions.
3. Engine failure case - In this case, basic runway length may consider either clearway or stop way or both.

Runway Capacity:

• The ability of a runway system to accommodate aircraft landing and take-off. It is usually expressed in operation per hour or operation per year.
• The location of exit taxiways depends on the mix of aircraft, the approach and touchdown speeds, the point of touchdown, the exit speed, the rate of deceleration, which in turn depends on runway capacity and the condition of the pavement surface that is, dry or wet, and the number of exits.

The following items are to be considered in the geometric design of the runway.

1. Runway length
2. Runway width
3. Width and length of safety area
4. Transverse gradient
5. Longitudinal and effective gradient
6. Rate of change of longitudinal gradient
7. Sight distance

Hanger

• It is the covered area for repair and servicing the aircraft.
• The hanger depends upon the size of the aircraft and turning radius.
• The number of hangers depends upon peak hour per volume of aircraft.

Explanation:

Correction to runway take-off length

(i) Correction due to elevation \(= \frac{7}{{100}} \times 2500 \times \frac{{600}}{{300}}\)

Corrected length = 2500 + 350 = 2850 m

(ii) Correct for temperature

Rise of temperature = 18°C – 8°C = 10°C

Correction \(= \frac{1}{{100}} \times 2850 \times 10^\circ C = 285\)

Corrected length = 2850 + 285 = 3135 m = 3.135 km

Check

Concept:

Correction for elevation

ICAO recommends that basic runway length should be increased at the rate of 7% per 300 m rise in elevation above mean sea level.

∴ Correction for elevation \( = \frac{7}{{100}} \times runway\;length \times \frac{{Airport\;elevation}}{{300}}\)

Correction for temperature

Rise in temperature = ART - SAT  

Where,

ART is Atmospheric reference temperature

SAT Standard atmospheric Temperature

As per ICAO, Basic runway length after correction for elevation should be further increased at the rate of 1% for every 1° C rise of airport reference temperature

correction for temperature \(= Correctedlength \times \frac{1}{{100}} \times Rise\;in\;temperature\)

Calculation

Correction for elevation

7% increase for 300 m rise

So for 600 m rise = 7 × 2 = 14% increase

∴ Runway length after elevation correction.

= 1800 + 1800 × 0.14

= 2052 m

Temperature Correction

Airport Reference temperature

\({\rm{ART}} = {\rm{Ta}} + \frac{{{\rm{Tm}} - {\rm{Ta}}}}{3}\)

\(= 15 + \frac{{30 - 15}}{3} = 20^\circ {\rm{c}}\)

Standard temperature = 15 – 0.0065 × 600

= 11.10c [6.5°c fore 1000m]

DT = 20 – 11.1 = 8.9°c

∴ Corrected = 1% for 1⁰c difference \(= \frac{{8.9}}{{100}} \times 2052 = 182.6{\rm{\;m}}\) 

Corrected value = 2052 + 182.6

= 2234.6 m = 2235 m

∴ % increase in length \(= \frac{{2235 - 1800}}{{1800}} \times 100\% \;\)

Maximum difference in elevation = (103.6 – 98.0) = 5.6 m

Total runway length = 1600 m

Effective gradient of runway \( = \frac{{5.6}}{{1600}} \times 100\) = 0.35% 

Concepts:

For taxiway design, the turning radius is given as:

\(R = \frac{V^2}{125 f}\)

Where,

 R is the radius of the curve in m,

V is the speed in km/hr and

f is the coefficient of friction between the tire and pavement surface

Further,

For airports serving large subsonic jet transports, the minimum value of the radius of curvature is 120 m and for supersonic transports, a minimum radius is 180 m.

Calculation:

Given: f = 0.15, V = 60 km/hr

The turning radius based on the above formula is calculated as:

\(R = \frac{60^2}{125 \times 0.15 }\)

R = 192 m

For subsonic transports, the minimum radius recommend is R_min = 180 m

Therefore, the final design radius will be the maximum of the above two values i.e The radius for the taxiway is 192 m.

Concept:

The radius of exit taxiway is given by,

\(R = \frac{{{V^2}}}{{125\;f}}\)

Where,

V = Design exit speed

f = Coefficient of friction

Calculation:

\(R = \frac{{{V^2}}}{{125\;f}}\)

\(\Rightarrow R = \frac{{{{90}^2}}}{{125 \times 0.13}}\)