High Bypass Gas Turbine Engine Coursework Example | Topics and Well Written Essays - 2000 words

High Bypass Gas Turbine Engine - Coursework ExampleApplication of atomic number 7s First law related to thrust If thrust and Drag are equal, the jobcraft maintains a regular speed. If thrust is increased, the speed of aircraft increases. Since drag is proportional to speed, drag likewise increases till it equals thrust. When drag again equals thrust, the aircraft travels at constant higher speed. Application of Newtons Second Law related to mass extend and devolve f number Force Mass * Acceleration F ma F=kma When SI trunk is used,the grassroots unit of force is the Newton, which is the force that will accelerate unit mass of 1 kg at a rate of 1 metre per second per second. Under these conditions, the constant k is unity. Therefore, F=ma F=ma=m di/dt=(m/ft) dV=d (mV)/dt =mass move rate times smorgasbord in velocity =(mv)dot Where m dot=Mass emanate rate is the amount of mass touching finished a given plane over a given period of time. Mass flow rate=r * V * A where r is the density and V is the velocity of the fluid strait through area A. This is denoted as m dot (m with a little dot over the top) m dot= r * V * A If we denote red ink of the turbojet by e and leave office stream by 0, then we get, F= (m dot*V)e-(m dot *V)0 Thus by maintaining the throttle velocity at much greater values than the velocity at intake, high thrust kindle be baffled in turbojet engines (High Exit Velocity). Application of Newtons Third Law related to thrust garget is the reaction force developed in the forward direction by accelerating a mass of fluid or gas backwards to the rear of the engine. The turboprop propulsion system consists of a core engine and a propeller. The general principles in Application of Newtons First Law and Third Law in Turboprop engines are the same as given in turbofan engine. Application of Newtons Second Law related to mass flow and exit velocity in Turbo Prop The general thrust equation is F= (m dot*V)e-(m dot *V)0 This means tha t if the exit velocity is maintained at a higher value than free stream velocity, and simultaneously, the engine flow rate (m dot) is kept as high as possible, the high engine flow will produce a high thrust in a turboprop engine. Even though a king-size amount of air is ingested, the change in velocity is very minimal between the intake and the exit so that the exit velocity is at a low value (Low exit velocity). Due to the vainglorious value of m dot, a high thrust is developed. Total Thrust= Thrust of Propeller Thrust of Core If we denote the free stream conditions by 0, the propeller exit conditions by 1, core exit conditions by e and core entrance conditions by c, then from the basic thrust equation we get F=(m dot)0 * V1 (m dot)0 * V0 + (m dot)e * Ve (m dot)c * V1 In Turboprop engine, the mass flow rate through the propeller is much greater than that of core engine(High mass flow ). The mass flow rate unveiling the core is nigh equal to the mass flow rate exiting the cor e. The exit velocity from the core is almost the same as inlet velocity into the core(Low exit velocity). Hence the thrust equation can be rounded off to get Thrust F=(m dot)0 * (V1-V0) + (m dot)e * (Ve-V1) High Bypass Gas Turbine Engine The 5 basic modules- Along with a Detailed Description of operation of each. Inlet Components and Purpose The intake also called the inlet serves three purposes, namely (1) recovering as much of the total pressure of the free air stream required for combustion, from free-stream conditions to the conditions and deliver this pressure to the entrance of fan or compressor.