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Part II: Rocket Mechanics. Rocket Stability Stability is maintained through two major ideas First is the Center of Mass (CM) –This is the point where.

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Presentation on theme: "Part II: Rocket Mechanics. Rocket Stability Stability is maintained through two major ideas First is the Center of Mass (CM) –This is the point where."— Presentation transcript:

1 Part II: Rocket Mechanics

2 Rocket Stability Stability is maintained through two major ideas First is the Center of Mass (CM) –This is the point where the mass of the rocket is perfectly balanced –It should be about halfway up the rocket

3 CM The CM point is where the three axes of pitch, roll and yaw all intersect

4 Center of Pressure (CP) This is the second important concept It only exists when air is flowing over the moving rocket The rocket’s surface area helps to determine the total CP The CP ought to be between the CM and the engine end of the rocket The CP location can be controlled by using movable fins, a gimbaled (movable nozzle) and of course, the overall rocket design

5 Your Water Rocket In the diagram of a water rocket, can you determine where the CM and CP should be located at?

6 Rocket Mass Ideally – the mass of a rocket should be about 91 % fuel; 3 % rocket and 6% payload This is determined through the use of the mass fraction (MF) MF = m propellant m rocket

7 Rocket Propulsion Rockets use either liquid fuel or solid fuel Liquid fuel is a mix of liquid oxygen and liquid hydrogen These gases must be at -423 o F to become liquefied

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9 de Laval Rocket Nozzle

10 The linear velocity of the exiting exhaust gases can be calculated using the following equation:[[ where: Ve= Exhaust velocity at nozzle exit, m/sT= absolute temperature of inlet gas, KR= Universal gas law constant = 8314.5 J/(kmol·K)M= the gas molecular mass, kg/kmol (also known as the molecular weight)k= cp/cv = isentropic expansion factorcp= specific heat of the gas at constant pressurecv= specific heat of the gas at constant volumePe= absolute pressure of exhaust gas at nozzle exit, PaP= absolute pressure of inlet gas, Pa temperatureUniversal gas law constantmolecular massisentropic expansion factorspecific heatabsolute pressurePa Some typical values of the exhaust gas velocity Ve for rocket engines burning various propellants are: –1700 to 2900 m/s (3,800 to 6,500 mph) for liquid monopropellantsmonopropellants –2900 to 4500 m/s (6,500 to 10,100 mph) for liquid bipropellantsbipropellants –2100 to 3200 m/s (4,700 to 7,200 mph) for solid propellantssolid propellants As a note of interest, Ve is sometimes referred to as the ideal exhaust gas velocity because it based on the assumption that the exhaust gas behaves as an ideal gas.

11 Newton’s Laws and Rockets Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector. For every action there is an equal and opposite reaction.

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15 Other Concepts:

16 Escape Velocity Isaac Newton's analysis of escape velocity. Projectiles A and B fall back to earth. Projectile C achieves a circular orbit, D an elliptical one. Projectile E escapes.

17 Escape velocity is defined to be the minimum velocity an object must have in order to escape the gravitational field of the earth, that is, escape the earth without ever falling back. The object must have greater energy than its gravitational binding energy to escape the earth's gravitational field. So: 1/2 mv2 = GMm/R Where m is the mass of the object, M mass of the earth, G is the gravitational constant, R is the radius of the earth, and v is the escape velocity. It simplifies to: v = sqrt(2GM/R) - or - v = sqrt(2gR) Where g is acceleration of gravity on the earth's surface. The value evaluates to be approximately: 11 100 m/s 40 200 km/h 25 000 mi/h So, an object which has this velocity at the surface of the earth, will totally escape the earth's gravitational field (ignoring the losses due to the atmosphere.)

18 Now – you are becoming real rocket scientists!!!!!


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