Presentation on theme: "Rocket/Aircraft Propulsion* *Black Hole Performance May Vary."— Presentation transcript:
Rocket/Aircraft Propulsion* *Black Hole Performance May Vary
Rocket engines are, on the one hand, so simple that you can build and fly your own model rockets very easily. On the other hand, rocket engines (and their fuel systems) are so complicated that only three countries have actually ever put people in orbit. We will look at rocket engines to understand how they work, as well as to understand some of the complexity surrounding them.
A New Approach When most people think about motors or engines, they think about rotation. An electric motor produces rotational energy to drive a fan or spin a disk. A steam engine is used to do the same thing, as is a steam turbine and most gas turbines.electric motorsteam enginegas turbines Rocket engines are reaction engines. The basic principle driving a rocket engine is Newton's Third Law. A rocket engine is throwing mass in one direction and benefiting from the reaction that occurs in the other direction as a result.
How it Works If you have ever seen a big fire hose spraying water, you may have noticed that it takes a lot of strength to hold the hose (sometimes you will see two or three firefighters holding the hose). The hose is acting like a rocket engine. The hose is throwing water in one direction, and the firefighters are using their strength and weight to counteract the reaction. If they were to let go of the hose, it would thrash around with tremendous force. If the firefighters were all standing on skateboards, the hose would propel them backward at great speed! (Sounds fun!)fire hosefirefighters When you blow up a balloon and let it go so that it flies all over the room before running out of air, you have created a rocket engine.
So, How does it Work? If you throw a baseball, your body will react by moving in the opposite direction of the ball. The thing that controls the speed at which your body moves away is the mass of the baseball that you throw and the amount of acceleration that you apply to it.baseball Mass multiplied by acceleration is force (f = m * a). Whatever force you apply to the baseball will be equalized by an identical reaction force applied to your body (m * a = m * a). So let's say that the baseball weighs 1 pound, and your body plus the space suit weighs 100 pounds. You throw the baseball away at a speed of 32 feet per second (21 mph). That is to say, you accelerate the 1-pound baseball with your arm so that it obtains a velocity of 21 mph. Your body reacts, but it weighs 100 times more than the baseball. Therefore, it moves away at one-hundredth the velocity of the baseball, or 0.32 feet per second (0.21 mph).
How is thrust generated? Thrust is a mechanical force which is generated through the reaction of accelerating a mass of gas, as explained by Newton's third law of motion. A gas or working fluid is accelerated to the rear and the engine and aircraft are accelerated in the opposite direction. To accelerate the gas, we need some kind of propulsion system. If you want to generate more thrust from your baseball, you have two options: increase the mass or increase the acceleration. You can throw a heavier baseball or throw a number of baseballs one after another (increasing the mass), or you can throw the baseball faster (increasing the acceleration on it). But that is all that you can do. A rocket engine is generally throwing mass in the form of a high-pressure gas. The engine throws the mass of gas out in one direction in order to get a reaction in the opposite direction. The fact that the fuel turns from a solid or liquid into a gas when it burns does not change its mass. If you burn a pound of rocket fuel, a pound of exhaust comes out the nozzle in the form of a high-temperature, high-velocity gas. The form changes, but the mass does not. The burning process accelerates the mass.
Newton's Second Law - Variation
General Thrust Equation
Cont. If we keep the mass constant and just change the velocity with time we obtain the simple force equation - force equals mass time acceleration a F = m * a If we are dealing with a solid, keeping track of the mass is relatively easy; the molecules of a solid are closely bound to each other and a solid retains its shape. But if we are dealing with a fluid (liquid or gas) and particularly if we are dealing with a moving fluid, keeping track of the mass gets tricky. For a moving fluid, the important parameter is the mass flow rate. Mass flow rate is the amount of mass moving through a given plane over some amount of time. Its dimensions are mass/time (kg/sec, slug/sec,...) and it is equal to the density r times the velocity V times the area A. Aerodynamicists denote this parameter as m dot (m with a little dot over the top). m dot = r * V * A
Cont. So "m dot" is not simply the mass of the fluid, but is the mass flow rate, the mass per unit time. Since the mass flow rate already contains the time dependence (mass/time), we can express the change in momentum across the propulsion device as the change in the mass flow rate times the velocity. We will denote the exit of the device as station "e" and the free stream as station "0". Then F = (m dot * V)e - (m dot * V)0 A units check shows that on the right hand side of the equation: mass/time * length/time = mass * length / time^2
Alternate Formula This is the dimension of a force. There is an additional effect which we must account for if the exit pressure p is different from the free stream pressure. The fluid pressure is related to the momentum of the gas molecules and acts perpendicular to any boundary which we impose. If there is a net change of pressure in the flow there is an additional change in momentum. Recall that pressure is F/A so the force due to pressure is pA Across the exit area we may encounter an additional force term equal to the exit area Ae times the exit pressure minus the free stream pressure. The general thrust equation is then given by: F = (m dot * V)e - (m dot * V)0 + (pe - p0) * Ae
Cont. Normally, the magnitude of the pressure-area term is small relative to the m dot-V terms. Looking at the thrust equation very carefully, we see that there are two possible ways to produce high thrust. One way is to make the engine flow rate (m dot) as high as possible. As long as the exit velocity is greater than the free stream, entrance velocity, a high engine flow will produce high thrust. This is the design theory behind propeller aircraft and high-bypass turbofan engines. A large amount of air is processed each second, but the velocity is not changed very much. The other way to produce high thrust is to make the exit velocity very much greater than the incoming velocity. This is the design theory behind pure turbojets and turbojets with afterburners, and rockets. A moderate amount of flow is accelerated to a high velocity in these engines. If the exit velocity becomes very high, there are other physical processes which become important and affect the efficiency of the engine. These effects are described in detail on other pages at this site. There is a simplified version of the general thrust equation that can be used for gas turbine engines. The nozzle of a turbine engine is usually designed to make the exit pressure equal to free stream. In that case, the pressure-area term in the general equation is equal to zero. The thrust is then equal to the exit mass flow rate times the exit velocity minus the free stream mass flow rate times the free stream velocity. F = (m dot * V)e - (m dot * V)0
Specific Thrust Since the exit mass flow rate is nearly equal to the free stream mass flow rate, and the free stream is all air, we can call the mass flow rate through the engine the engine airflow rate. F = (m dot)eng * (Ve - V0) Now using a little algebra, we can define a new variable called the specific thrust Fs which depends only on the velocity difference produced by the engine: F / mdot 0 = Fs = (1 + f) * Ve - V0 We can further simplify by absorbing the engine airflow dependence into a more useful parameter called the specific thrust. Fs Specific thrust only depends on the velocity change across the engine. Fs = F /(m dot)eng = (Ve - V0)
F / mdot 0 = Fs = (1 + f) * Ve - V0
Rocket There is a different simplified version of the general thrust equation that can be used for rocket engines. Since a rocket carries its own oxygen on board, the free stream mass flow rate is zero and the second term of the general equation drops out. F = (m dot * V)e + (pe - p0) * Ae
The figure shows the situation at time t. The rocket and fuel have a total mass M and the combination is moving with velocity v as seen from a particular frame of reference. At a time t later the configuration has changed to that shown in Figure 1.2(b). A mass M has been ejected from the rocket and is moving with velocity u as seen by the observer. The rocket is reduced to mass M-deltaM and the velocity v of the rocket is changed to v+deltav. Because there are no external forces on the system, dP/dt=0. We can write, for the time interval t where P 2 is the final system momentum, Figure 1.2(b), and P 1 is the initial system momentum, Figure 1.2(a). We write
If we let t approach zero, v/t approaches dv/dt, the acceleration of the body. The quantity M is the mass ejected in t; this leads to a decrease in the mass M of the original body. Since dM/dt, the change in mass of the body with time, is negative in this case, in the limit the quantity M/t is replaced by -dM/dt. The quantity u-(v+v) is V rel, the relative velocity of the ejected mass with respect to the rocket. With these changes, the equation can be written as The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it. The rocket designer can make the thrust as large as possible by designing the rocket to eject mass as rapidly as possible (dM/dt large) and with the highest possible relative speed (V rel large). In rocketry, the basic thrust equation is written as
where q is the rate of the ejected mass flow, V e is the exhaust gas ejection speed, P e is the pressure of the exhaust gases at the nozzle exit, P a is the pressure of the ambient atmosphere, and A e is the area of the nozzle exit. The product qV e, which we derived above (V rel × dM/dt), is called the momentum, or velocity, thrust. The product (P e -P a )A e, called the pressure thrust, is the result of unbalanced pressure forces at the nozzle exit. Minimum thrust occurs when P e =P a.
Inherent Inefficiency of Fuel Rockets If you have ever seen the Space Shuttle launch, you know that there are three parts: The Orbiter The big external tank The two solid rocket boosters (SRBs) The Orbiter weighs 165,000 pounds empty. The external tank weighs 78,100 pounds empty. The two solid rocket boosters weigh 185,000 pounds empty each. But then you have to load in the fuel. Each SRB holds 1.1 million pounds of fuel. The external tank holds 143,000 gallons of liquid oxygen (1,359,000 pounds) and 383,000 gallons of liquid hydrogen (226,000 pounds). The whole vehicle -- shuttle, external tank, solid rocket booster casings and all the fuel -- has a total weight of 4.4 million pounds at launch. 4.4 million pounds to get 165,000 pounds in orbit is a pretty big difference! How this is possible will be explained in the future. Natural Log is key
Fuel Methods - Solid
Applications of Rocket Propulsion Space Launch Vehicles- each has a specific space flight objective, has between 2 and 5 stages Each stage is a complete vehicle, with its own propellant, rocket propulsion systems and its own control system Military (command and control satellites) Non-military Government (weather observation satellites, GPS) Space Exploration (space environment, planetary missions) Commercial (communication satellites)
Other Applications Military Missiles: Strategic Missiles- long range ballistic missiles which are aimed at military targets within an enemy country Tactical Missiles- intended to support or defend military ground forces, aircraft, or navy ships Other applications: Engines for Research airplanes, assist-take-off rockets for airplanes, and "Fourth of July" rockets