Question 1 |

For a 2^{\circ} curve on a high speed Broad Gauge (BG) rail section, the maximum sanctioned speed is 100 km/h and the equilibrium speed is 80 km/h. Consider dynamic gauge of BG rail as 1750 mm. The degree of curve is defined as the angle subtended at its center by a 30.5 m arc. The cant deficiency for the curve (in mm,round off to integer) is ________________

45 | |

65 | |

57 | |

28 |

Question 1 Explanation:

\begin{aligned} \text { Length of curve }&=\text { Radius } \times \text { Degree of curve }\\ \frac{30.5 \times 180}{2^{\circ} \times \pi}&=R\\ R &=873.76 \mathrm{~m} \\ C_{d} &=C_{t h}-C_{a c t} \\ C_{\mathrm{act}} &=\frac{G V^{2}}{127 R}=\frac{1750 \times 100^{2}}{127 \times 873.76}=157.70 \mathrm{~mm} \\ C_{\mathrm{ev}} &=\frac{G V^{2}}{127 R}=\frac{1750 \times 80^{2}}{127 \times 873.76}=100.930 \mathrm{~mm} \\ C_{\mathrm{def}} &=C_{\mathrm{act}}-C_{t h}=157.70-100.930=56.77 \mathrm{~mm} \\ &=57 \mathrm{~mm} \end{aligned}

Question 2 |

The longitudinal section of a runway provides the following data:

\begin{array}{|c|c|} \hline \text { End-to-end runway (m) } & \text { Gradient (\%) } \\ \hline 0 \text { to 300 } & +1.2 \\ \hline 300 \text { to } 600 & -0.7 \\ \hline 600 \text { to } 1100 & +0.6 \\ \hline 1100 \text { to } 1400 & -0.8 \\ \hline 1400 \text { to } 1700 & -1.0 \\ \hline \end{array}

The effective gradient of the runway (in %, round off to two decimal places) is _____________

\begin{array}{|c|c|} \hline \text { End-to-end runway (m) } & \text { Gradient (\%) } \\ \hline 0 \text { to 300 } & +1.2 \\ \hline 300 \text { to } 600 & -0.7 \\ \hline 600 \text { to } 1100 & +0.6 \\ \hline 1100 \text { to } 1400 & -0.8 \\ \hline 1400 \text { to } 1700 & -1.0 \\ \hline \end{array}

The effective gradient of the runway (in %, round off to two decimal places) is _____________

0.12 | |

0.32 | |

0.54 | |

0.69 |

Question 2 Explanation:

Assuming RL of start of runway as datum i.e., RL = 0 m)

\begin{aligned} \text { Effective gradient } &=\left[\frac{\text { Maximum difference in reduced level }}{\text { Total runway length }}\right] \\ &=\left[\frac{4.5-(-0.9)}{1700} \times 100\right] \% \\ &=0.3176 \% \simeq 0.32 \% \end{aligned}

\begin{aligned} \text { Effective gradient } &=\left[\frac{\text { Maximum difference in reduced level }}{\text { Total runway length }}\right] \\ &=\left[\frac{4.5-(-0.9)}{1700} \times 100\right] \% \\ &=0.3176 \% \simeq 0.32 \% \end{aligned}

Question 3 |

For the hottest month of the year at the proposed airport site, the monthly mean of the
average daily temperature is 39^{\circ}C. The monthly mean of the maximum daily temperature
is 48^{\circ}C for the same month of the year. From the given information, the calculated Airport
Reference Temperature (in ^{\circ}C), is

36 | |

39 | |

42 | |

48 |

Question 3 Explanation:

\begin{aligned} T_a&=39^{\circ}C\\ T_m&=48^{\circ}C\\ ATR&=T_a+\left ( \frac{T_m-T_a}{3} \right )\\ &=39+\left ( \frac{48-39}{3}\right )\\&=42^{\circ}C \end{aligned}

Question 4 |

The appropriate design length of a clearway is calculated on the basis of 'Normal Takeoff' condition. Which one of the following options correctly depicts the length of the
clearway?

A | |

B | |

C | |

D |

Question 4 Explanation:

For normal take off condition:

\begin{aligned} \text{Clearway} &\ngtr \frac{1}{2}(1.5 \; \text{of take off distance}\\ &-1.15 \text{of lift off diatance}) \\ &\ngtr \frac{1}{2}(1.15 \times 1625 -1.15 \times 875) \\ &\ngtr 431.25m \end{aligned}

So clearway is less then for 432 m.

\begin{aligned} \text{Clearway} &\ngtr \frac{1}{2}(1.5 \; \text{of take off distance}\\ &-1.15 \text{of lift off diatance}) \\ &\ngtr \frac{1}{2}(1.15 \times 1625 -1.15 \times 875) \\ &\ngtr 431.25m \end{aligned}

So clearway is less then for 432 m.

Question 5 |

A broad gauge railway line passes through a horizontal curved section (radius = 875 m) of length 200 m. The allowable speed on this portion is 100 km/h. For calculating the cant, consider the gauge as centre-to-centre distance between the rail heads, equal to 1750 mm. The maximum permissible cant (in mm, round off to 1 decimal place) with respect to the centre-to-centre distance between the rail heads is_____

157.5 | |

182.6 | |

126.4 | |

187.6 |

Question 5 Explanation:

For a railway track

\begin{array}{l}\mathrm{Allowable}\;\mathrm{cant}=\frac{\mathrm G.\mathrm V^2}{127\mathrm R}\;\;\mathrm{where}\;\mathrm V\;\mathrm{is}\;\mathrm{in}\;\mathrm{km}/\mathrm{hr}=100\\\mathrm{Allowable}\;\mathrm{cant}=\frac{1750\times100^2}{127\times895}=157.5\;\mathrm{mm}\end{array}

\begin{array}{l}\mathrm{Allowable}\;\mathrm{cant}=\frac{\mathrm G.\mathrm V^2}{127\mathrm R}\;\;\mathrm{where}\;\mathrm V\;\mathrm{is}\;\mathrm{in}\;\mathrm{km}/\mathrm{hr}=100\\\mathrm{Allowable}\;\mathrm{cant}=\frac{1750\times100^2}{127\times895}=157.5\;\mathrm{mm}\end{array}

Question 6 |

For a broad gauge railway track on a horizontal curve of radius R (in m), the equilibrium cant e required for a train moving at a speed of V (in km per hour) is

e = 1.676\frac{V^{2}}{R} | |

e = 1.315\frac{V^{2}}{R} | |

e = 0.80\frac{V^{2}}{R} | |

e = 0.60\frac{V^{2}}{R} |

Question 6 Explanation:

\theta(\text { in } \mathrm{cm})=\frac{G V^{2}}{127 R}=\frac{1.676 \mathrm{V}^{2}}{127 \mathrm{R}} \times 100=1.315 \frac{\mathrm{V}^{2}}{\mathrm{R}}

Question 7 |

A runway is being constructed in a new airport as per the International Civil Aviation Organization (ICAO) recommendations. The elevation and the airport reference temperature of this airport are 535 m above the mean sea level and 22.65^{\circ}C, respectively. Consider the effective gradient of runway as 1%. The length of runway required for a design-aircraft under the standard conditions is 2000 m. Within the framework of applying sequential corrections as per the ICAO recommendations, the length of runway corrected for the temperature is

2223 m | |

2250 m | |

2500 m | |

2750 m |

Question 7 Explanation:

Elevation =535 m

Airport Reference Temperature =22.65^{\circ} \mathrm{C}

Effective Gradient = 1%

Runway length =2000 m under standard conditions.

Runway length correction for T^{\circ} \mathrm{C} .

\Rightarrow Correction for elevation

=\frac{7}{100} \times 2000 \times \frac{535}{300}=249.67 \mathrm{m}

Corrected runway length =2249.67 \mathrm{m}

\Rightarrow Corrected standard T^{\circ} at 535 m elevation

=15^{\circ} \mathrm{C}-(0.0065 \times 535)=11.52^{\circ} \mathrm{C}

Correction for T^{\circ}

\begin{aligned} &=\frac{1}{100} \times 2249.67 \times \frac{\Delta T^{\circ} \mathrm{C}}{1^{\circ} \mathrm{C}} \\ \Delta T^{\circ} \mathrm{C} &=22.65-11.52=11.13^{\circ} \mathrm{C} \\ &=\frac{1}{100} \times 2249.67 \times \frac{11.13^{\circ} \mathrm{C}}{1^{\circ} \mathrm{C}} \\ &=250.39 \mathrm{m} \end{aligned}

Corrected Runway length

\begin{array}{l} =2249.67+250.39 \\ =2500.06 \mathrm{m} \simeq 2500 \mathrm{m} \end{array}

Airport Reference Temperature =22.65^{\circ} \mathrm{C}

Effective Gradient = 1%

Runway length =2000 m under standard conditions.

Runway length correction for T^{\circ} \mathrm{C} .

\Rightarrow Correction for elevation

=\frac{7}{100} \times 2000 \times \frac{535}{300}=249.67 \mathrm{m}

Corrected runway length =2249.67 \mathrm{m}

\Rightarrow Corrected standard T^{\circ} at 535 m elevation

=15^{\circ} \mathrm{C}-(0.0065 \times 535)=11.52^{\circ} \mathrm{C}

Correction for T^{\circ}

\begin{aligned} &=\frac{1}{100} \times 2249.67 \times \frac{\Delta T^{\circ} \mathrm{C}}{1^{\circ} \mathrm{C}} \\ \Delta T^{\circ} \mathrm{C} &=22.65-11.52=11.13^{\circ} \mathrm{C} \\ &=\frac{1}{100} \times 2249.67 \times \frac{11.13^{\circ} \mathrm{C}}{1^{\circ} \mathrm{C}} \\ &=250.39 \mathrm{m} \end{aligned}

Corrected Runway length

\begin{array}{l} =2249.67+250.39 \\ =2500.06 \mathrm{m} \simeq 2500 \mathrm{m} \end{array}

There are 7 questions to complete.

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