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

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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 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 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 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 firing 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 rela- tionship dm/dt = - b where b is the fuel burn rate constant measured in kg s - 1 . The initial total mass of the rocket can be represented by the constant m 0 , and we can assume that the fuel is a significant 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 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 de- scribe 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 position, we can jus- tifiably 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 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 expe- rienced 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 acceleration due to gravity decreases as the rocket gets fur- ther from the launch site - still assuming a ver- tical 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 surface 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 pa- rameters 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 experienced in the ‘vomit comet’? 1 Wikipedia article at 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 1- 2 wilkening.pdf 4 version 01 - updated April 15, 2010 1
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MAB120 - MAN120 - MAB125 semester one 2010 see Blackboard site for additional details, including submission instructions Problem Solving Assignment 2. Work and Heat Engines Thermodynamics allows us to study ideal models for a heat engine that con- verts heat energy to mechanical energy. For an ide- alised engine, we can assume that gases obey ideal gas laws such as PV = nRT , or for adiabatic expan- sions, 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 con- stants. 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. De- tails 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 con- verted to steam and expands along path cd . The ex- pansion 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 constant. (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 pro- vided in the figure above. version 01 - updated April 15, 2010 2
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