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JNTU B.Tech II INTRODUCTION TO SPACE TECHNOLOGY Previous Question Paper (August/September - 2022 - R13)

B.Tech | II | II-II | R13 | August/September - 2022

Answer PaperPaper Code: 114BJMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Explain Multi-stage Rocket system with neat schematic layout.

7 marks

Answer

A multi-stage rocket consists of several individual stages, each equipped with its own propulsion system and fuel supply, stacked vertically. As each stage exhausts its propellant, it is jettisoned to reduce the dead weight of the vehicle, allowing the remaining stages to accelerate more efficiently. This staging technique is essential for achieving orbital velocities, as it overcomes the limitations of the rocket equation by continuously improving the mass ratio.

The schematic layout typically follows a series configuration:

  1. First Stage (Booster): Provides high thrust to lift the vehicle off the launch pad and through the dense lower atmosphere.
  2. Interstage: The structural connection containing separation mechanisms (explosive bolts or pneumatic pushers) to discard the spent stage.
  3. Upper Stages: Designed for vacuum operation, these stages provide the high specific impulse needed for orbital insertion or deep space trajectory.
  4. Payload: Located at the tip, protected by a fairing until reaching space.
b

A rocket of total mass 100 tonnes carrying a spacecraft of 5 tonnes and engine develop a constant exhaust velocity of 3500 m/s. The structural mass is assumed to be 15% of the total mass. Calculate final velocity of rocket.

8 marks

Answer

To calculate the final velocity, we use the Tsiolkovsky rocket equation:

v = ve ln(m0 / mf)

Given data:

  • Total initial mass (m0) = 100 tonnes = 100,000 kg
  • Structural mass = 15% of total mass = 15,000 kg
  • Spacecraft mass = 5 tonnes = 5,000 kg
  • Exhaust velocity (ve) = 3500 m/s

The final mass (mf) includes the spacecraft and the structural mass remaining at burnout:

mf = 15,000 kg + 5,000 kg = 20,000 kg

Substituting the values into the equation:

v = 3500 × ln(100,000 / 20,000)
v = 3500 × ln(5)
v = 3500 × 1.6094
v ≈ 5632.9 m/s

The final velocity of the rocket is approximately 5632.9 m/s.

2

Obtain an expression for the minimum number of stages required in a multi-stage rocket assembly in order to carry maximum payload, for a given ideal velocity and structural efficiency. Further, discuss the method of optimizing the number of stages required.

15 marks

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315 marks
a

Derive an equation for Velocity burnout in Lifting body Re-entry.

8 marks

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b

Explain the process during flight in to orbit Vehicle must be rotated and lift acted down position.

7 marks

Answer

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415 marks
a

Using neat sketches, Derive an expression for Skip re-entry.

8 marks

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b

Derive an expression for Double-dip Re-entry.

7 marks

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515 marks
a

Provide the details of attitude determination during Hohmann transfer of a space vehicle to geosynchronous orbit.

8 marks

Answer

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b

Compare Hohmann transfer of orbit with Bi-elliptic transfer of orbit principles.

7 marks

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615 marks
a

Design a Hohmann transfer from a circular mars orbit of radius 8000 km to a circular mars orbit of a radius 15000 km (for mars μ = 42,828.3 km3/sec2) and calculate the period of transfer.

8 marks

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b

Explain how a simple plane change maneuver is different from Hohmann transfer maneuver. Mention the changes that are going to be associated with orbital parameters in a simple plane change of orbits.

7 marks

Answer

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715 marks
a

Explain about magnetic torques used in attitude control of spacecraft stabilization.

7 marks

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b

Explain how magnetic desaturation is useful in controlling attitude of a spacecraft.

8 marks

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815 marks
a

List out and describe in detail various test facilities used for space craft integration and testing.

8 marks

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b

Explain space mission types and objectives in mission diversity.

7 marks

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