Airport Engineering Test 1

Concept:

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

Where,

Ta = monthly mean of the average daily temperature for the hottest month of the year

Tm = monthly mean of the maximum daily temperature for the same month

Calculation:

Ta =  30°C

Tm = 45°C

\(ART = 30 + \frac{{45 - 30}}{3}\)

∴ ART = 35 ° C

Concept:

The standard temperature at the airport site can be determined by reducing the standard mean sea level temperature of 15°C at the rate of 6.5°C per thousand meter rises in elevation.

Standard temperature = 15 - 0.0065 × elevation

Rise in temperature = (Reference Temp. - Standard Temp.)

Calculation:

Standard temperature =15 - 0.0065 × 2000 = 2°C.

Rise in temperature = 12 - 2 = 10°C

Concept:

The Turning Radius based on Lateral friction is given by 

\(R = \frac{{{V^2}}}{{125\;f}}\) m

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

Where, 

R is the Turning radius of taxiway

f is the coefficient of lateral friction

V is the design exit speed

Calculation:

Given,

V = 100 kmph ,R = 700 m

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

\(\Rightarrow f = \frac{{{{100}^2}}}{{125 \times 700}}\)

⇒ f = 0.114

∴ Coefficient of Lateral friction = 0.114

The assumed conditions at the airport are as follows:

(1) Airport altitude is at sea level. 

(2) Temperature at the airport is standard (15°C) 

(3) Runway is levelled in the longitudinal direction. 

(4) No wind is blowing on runway 

(5) Aircraft is loaded to its full loading capacity 

(6) There is no wind blowing enroute to the destination 

(7) Enroute temperature is standard.

Concept

Blast pads are specially designed shoulders provided at the toke off ends of the runway and along taxiway. They protect the shoulders from erosion due to high velocity of jet exhaust.

Cargo is used to indicate the freight, other than passengers, baggage and mail, which is carried by the transport aircraft.

Flight time is the total time from the moment an aircraft first moves under its own power for the purpose of taking off to the moment it comes to rest at the end of flight.

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:

Given

Runway length for takeoff = 2500 m

Runway length for landing = 2100 m

Elevation of Airport site = 300 m

Standard atmospheric temperature at given elevation (SAT) = 13.7 ° C

Airport reference temperature (ART) = 21.7 ° C

Note:

a) For landing elevation correction is necessary

b) For takeoff all corrections are necessary

i) For landing

Correction for elevation

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

∴ Correction for elevation \(= \frac{7}{{100}} \times 2100 \times \frac{{300}}{{300}} = \;147\;m\)

∴ Corrected length = 2100 + 147 = 2247 m

ii) For takeoff

Correction for elevation

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

∴ Correction for elevation \(= \frac{7}{{100}} \times 2500 \times \frac{{300}}{{300}} = \;175\;m\)

Corrected length = 2500 + 175 = 2675 m

Correction for temperature

Rise in temperature = ART - SAT  

Rise in temperature = 21.7 – 13.7 = 8° C

correction for temperature \(= 2675 \times \frac{1}{{100}} \times 8 = 214\;m\)

Thus corrected length = 2675 + 214 = 2889 m

As the airport is at MSL and the airport temperature is the same, the elevation and temperature correction is zero.

Gradient correction:

Effective rise or fall

\( = \frac{1}{{100}} \times 500 - \frac{{0.5}}{{100}} \times \left( {900 - 500} \right) + \frac{{0.5}}{{100}} \times \left( {1800 - 900} \right) + \frac{2}{{100}} \times \left( {2100 - 1800} \right)\)

= 13.5 m

Effective gradient \( = \frac{{ + 13.5}}{{2100}} \times 100 = 0.6428\% \)

∴ 0.6428% effective gradient = 20 × 0.6428% rise = 12.856% rise

∴ corrected length \( = \left( {1 + \frac{{12.856}}{{100}}} \right) \times 2100\) = 2370 m

Concept:

(i) Turning radius based on lateral friction

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

V = turn-off speed (kmph)

f = coefficient of lateral friction

(ii) Turning Radius based on Horonieff equation

\({\rm{R}} = \frac{{0.388{{\rm{w}}^2}}}{{\frac{{\rm{T}}}{2} - {\rm{s}}}}\)

W = wheelbase of aircraft (m)

T = width of taxiway pavement

S = distance between midway point of the main gear and the edge of the taxiway pavement (m)

iii) For Supersonic air crafts, the minimum radius of the taxiway is taken as 180 m

∴ Turning Radius will be maximum of above three

Calculation: 

(i) Turning radius based on lateral friction

\({\rm{R}} = \frac{{{{\rm{V}}^2}}}{{125{\rm{\;f}}}}\) =\(\frac{{{{70}^2}}}{{125 \times 0.14}}\)

= 280 m

(ii) Turning Radius based on Horonieff equation

\({\rm{R}} = \frac{{0.388{{\rm{w}}^2}}}{{\frac{{\rm{T}}}{2} - {\rm{s}}}}\)

T = 24 m

S = 9.5 m

W = 20 m

\({\rm{R}} = \frac{{0.388 \times {{20}^2}}}{{\frac{{24}}{2} - 9.5}}\)

R = 62.08 m

iii) Minimum Radius of taxiway is taken as 180 m 

So, the maximum of above three values is 280 m

∴ Taxiway Turning Radius (R) = 280 m

Concept:

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: is a graphical representation of the direction, intensity, and duration of wind is called wind rose.

The wind data should usually be collected for a period of at least 5 years and preferably of 10 years so as to obtain an average data with sufficient accuracy

Head Wind: The runway is oriented in the direction of the prevailing winds. It helps in the landing of vehicles by causing a breaking effect and during takeoff, it provided lift on the wings of the aircraft.

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