View the step-by-step solution to:

MAB120 - MAN120 - MAB125 semester one 2010 Problem Solving Assignment see Blackboard site for additional details, including submission instructions...

Rocket launching: In this problem we are looking at some of the aspects that need to be considered
in designing and describing the launching a single-stage rocket with a payload. In what follows we will be ignoring any e ects of cross-winds, rotation of Earth or air resistance. We begin by assuming that the rocket is launched from Earth's surface in such a way that it follows a simple vertical trajectory. As a single-stage rocket, immediately after launch there will be a period of rapid acceleration as the fuel consumed provides thrust for the upward motion. One of the complex
issues to be considered is that as the fuel is used, the mass of the rocket (i.e. fuel + engine + payload) decreases, and hence the acceleration will increase for the same amount of thrust. The period of time over which the rocket engine is ring is known as the `boost phase' of the trajectory, hence when the fuel runs out, the `boost phase' is at an end.Representing the rocket's total mass with the un-known function m(t), if we assume that one of thedesign parameters for the rocket engine is to set aconstant fuel burn rate, then we will have the relationship dm/dt = -b where b is the fuel burn rate constant measured in kg/s. The initial total mass of the rocket can be represented by the constant m0,and we can assume that the fuel is a signi cant frac-tion of this mass. In this case, the quantity of fuel can again be a design parameter, so we will take the mass of fuel to be X m0, where 0 < X < 1. If we further assume that the `boost phase' is short enough, and we assign the function h(t) to be the rocket's height above ground-level, then we can de-scribe the acceleration experienced by the rocket during this period as d2h/dt2 =[kb/(m0-bt)]-g where g is the acceleration due to gravity, and kb is the thrust provided by the rocket engine, k being an experimentally determined constant. If we launch the rocket from a stationary position, we can jus-ti ably include the initial conditions h(0) = 0 and dh=dt = 0 when t = 0.

(a) Show that the expression for the height h(t) of the rocket when it just runs out of fuel is given
by (km0/b) [(1-X)ln(1-X)+X[1-(gXm0/2kb)]

(b) What is the velocity of the rocket at the end of the 'boost phase'?

(c) Astronauts aboard the 'vomit comet' or its current equivalent, experience a rapid loss of weight as the acceleration due to gravity is counteracted by the acceleration provided by the aircraft. The rate of weight-loss experienced is equivalent to a rate of decrease in the acceleration due to gravity of the order of -10/15ms^-2/s = -2/3ms^-3. For our rocket and its payload, the accleration due to gravity decreases as the rocket gets further from the launch site - still assming a vertical trajectory. If we take the radius of Earth to be R = 6.38x10^6m, then the acceleration due to gravity goes like ab = MG/(h(t)+R)^2, where M and G are constants and as before h(t) is the rocket's height above Earth's surfacae as a function of time. For an object of constant mass, but changing height, the rate of weight-loss experienced would be given by d(a)/dt.
Using the information gained previously for the height and velocity of the rocket in the 'boost phase', find an expression for the rate of weight-loss experienced by the rocketand it's payload as it continues to climb just at the end of the boost phase. You should use the operating parameters of the Shahab 3 missle employed by Iran, which has an initial mass of 16000kg of which 90% can be considered as fuel. Furthermore the Shahab 3 has a boost phase of 95s by which time it reaches a height of 75km. How does this rate of weight-loss compared with that experienced in the 'vomit comet'?
MAB120 - MAN120 - MAB125 semester one 2010 see Blackboard site for additional details, including submission instructions Problem Solving Assignment 1. Rocket launching: In this problem we are look- ing at some of the aspects that need to be con- sidered in designing and describing the launching a single-stage rocket with a payload. In what fol- lows we will be ignoring any effects of cross-winds, rotation of Earth or air resistance. We begin by assuming that the rocket is launched from Earth’s surface in such a way that it follows a simple vertical trajectory. As a single-stage rocket, immediately after launch there will be a period of rapid acceleration as the fuel consumed provides thrust for the upward motion. One of the com- plex issues to be considered is that as the fuel is used, the mass of the rocket ( i.e. fuel + engine + payload) decreases, and hence the acceleration will increase for the same amount of thrust. The period of time over which the rocket engine is fir- ing is known as the ‘boost phase’ of the trajectory, hence when the fuel runs out, the ‘boost phase’ is at an end. Representing the rocket’s total mass with the un- known function m ( t ), if we assume that one of the design parameters for the rocket engine is to set a constant fuel burn rate, then we will have the relationship dm/dt = - b where b is the fuel burn rate constant measured in kg s - 1 . The initial to- tal mass of the rocket can be represented by the constant m 0 , and we can assume that the fuel is a significant fraction of this mass. In this case, the quantity of fuel can again be a design parameter, so we will take the mass of fuel to be X m 0 , where 0 < X < 1. If we further assume that the ‘boost phase’ is short enough, and we assign the function h ( t ) to be the rocket’s height above ground-level, then we can describe the acceleration experienced by the rocket during this period as d 2 h dt 2 = κb m 0 - bt - g , where g is the acceleration due to gravity, and κb is the thrust provided by the rocket engine, κ being an experimentally determined constant. If we launch the rocket from a stationary posi- tion, we can justifiably include the initial condi- tions h (0) = 0 and dh/dt = 0 when t = 0. (a) Show that the expression for the height h ( t ) of the rocket when it just runs out of fuel is given by κ m 0 b ± (1 - X ) ln(1 - X ) + X ± 1 - g 2 κb X m 0 ²² . (b) What is the velocity of the rocket at the end of the ‘boost phase’? (c) Astronauts aboard the ‘vomit comet’ or its current equivalent 1 , experience a rapid loss of weight as the acceleration due to gravity is counteracted by the acceleration provided by the aircraft. The rate of weight-loss ex- perienced is equivalent to a rate of decrease in the acceleration due to gravity of the order of - 10 / 15 ms - 2 / s = - 2 / 3 ms - 3 . For our rocket and its payload, the acceler- ation due to gravity decreases as the rocket gets further from the launch site - still as- suming a vertical trajectory. If we take the radius of Earth to be R 0 = 6 . 38 × 10 6 m, then the acceleration due to gravity goes like a g = M G ( h ( t ) + R 0 ) 2 , where M and G are constants 2 and as before, h ( t ) is the rocket’s height above Earth’s sur- face as a function of time. For an object of constant mass, but changing height, the rate of weight-loss experienced would be given by d ( a g ) /dt . Using the information gained previously for the height and velocity of the rocket in the ‘boost phase’, find an expression for the rate of weight-loss experienced by the rocket and it’s payload as it continues to climb just at the end of the boost phase. You should use the operating parameters of the Shahab 3 missile employed by Iran (see notes 3 4 ), which has an initial mass of 16000 kg of which 90% can be considered as fuel. Furthermore the Shahab 3 has a boost phase of 95 s by which time it reaches a height of 75 km. How does this rate of weight-loss compare with that ex- perienced in the ‘vomit comet’? 1 Wikipedia article at http://en.wikipedia.org/wiki/Vomit Comet 2 The constants are the mass of Earth (M = 5 . 9742 × 10 24 kg) and the universal gravitational constant (G = 6 . 673 × 10 - 11 m 3 kg - 1 s - 2 ) 3 http://www.princeton.edu/sgs/publications/sgs/pdf/12 1- 2 wilkening.pdf 4 http://www.armscontrol.org/print/2617 version 02 - updated April 29, 2010 1
Background image of page 1
MAB120 - MAN120 - MAB125 semester one 2010 see Blackboard site for additional details, including submission instructions Problem Solving Assignment 2. Work and Heat Engines Thermodynamics al- lows us to study ideal models for a heat engine that converts heat energy to mechanical energy. For an idealised engine, we can assume that gases obey ideal gas laws such as PV = nRT , or for adiabatic expansions, PV γ = constant ( γ 1 . 4 for air). Note that P = pressure, V = volume and T = temperature of the gas, and n and R can be taken to be constants. The term adiabatic simply implies that in the process of compressing (or expanding) a gas by changing its pressure and volume, the temperature of the gas also changes. The work done by the system in one complete cycle in an ideal heat engine is given by the area enclosed in the cycle when illustrated on a P - V (pressure versus volume) diagram. By definition, as the gas changes from volume V a to V b the work done by the gas is Work = Z V b V a P dV . An example of a heat engine is the Rankine cycle, which is an idealised version of a steam engine. Details of such a cycle are shown in the diagram. V P 0 . 05 0 . 15 0 . 25 10 20 c = (0 . 01 , 20 . 0) d = (0 . 03 , 20 . 0) b = (0 . 01 , 1 . 03) a = (0 . 25 , 1 . 03) In this cycle, liquid water at low temperature and pressure is heated at constant volume within a boiler. This occurs along path bc . The water is converted to steam and expands along path cd . The expansion continues adiabatically along path da . The water is then cooled and condensed back to liquid along path ab . There is no work done along path bc because the volume remains con- stant. (a) Using the ideal gas law PV = nRT , show that the appropriate expression for work along an isothermal (constant temperature) path from α to β is Work = nRT ln( V β /V α ) . (b) Show that the appropriate expression for work along an adiabatic path from α to β is Work = ( P β V β - P α V α ) / (1 - γ ) . (c) Determine the total work done in single cycle of the Rankine cycle, using the information provided in the figure above. version 02 - updated April 29, 2010 2
Background image of page 2

Recently Asked Questions

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

-

Educational Resources
  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question