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Gas Turbine Engine Basics - Coursework Example

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Gas Turbine Engine Basics

22-23 Application of Newton’s First Law related to thrust: # If thrust and Drag are equal, the aircraft maintains constant speed. # If thrust is increased, the speed of aircraft increases. Since drag is proportional to speed, drag also increases till it equals thrust. When drag again equals thrust, the aircraft travels at constant higher speed. Application of Newton’s Second Law related to mass flow and exit velocity: Force Mass * Acceleration F ma F=kma When SI system is used,the basic unit of force is the Newton, which is the force that will accelerate unit mass of 1 kilogram at a rate of 1 metre per second per second. Under these conditions, the constant k is unity. Therefore, F=ma F=ma=m dV/dt=(m/dt) dV=d (mV)/dt =mass flow rate times change in velocity =(mv)dot Where “m dot”=Mass flow rate is the amount of mass moving through 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 passing 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 exit of the turbojet by ‘e’ and free stream by ‘0’, then we get, F= (m dot*V)e-(m dot *V)0 Thus by maintaining the exit velocity at much greater values than the velocity at intake, high thrust can be produced in turbojet engines (High Exit Velocity). Application of Newton’s Third Law related to thrust: Thrust 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 Newton’s First Law and Third Law in Turboprop engines are the same as given in Turbojet engine. Application of Newton’s 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 that 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 large 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 large 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 entering the core is almost equal to the mass flow rate exiting the core. 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, (2) ...Show more


GAS TURBINE ENGINE BASICS Academia-Research Order #: 614838 Writer ID: 21516 CONTENTS Page No. TASK 1 3-8 Turbojet ……………………. 3-4 Turboprop ……………………. 5-6 Turbofan ……………………. 7-8 TASK 2 ………………………
Author : fayesporer
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