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# EASA Part 66 Module 15.1 : Fundamental

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# EASA Part 66 Module 15.1 : Fundamental

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#### Description

Fundamental of jet engine

#### Transcript

1. 1. 15.1 PRINCIPLE http://part66.blogspot.com/
2. 2. ENERGY • Energy : Capacity to do work (Joule) • Potential Energy : Result from influence of gravity on mass due to its position PE = mass x gravity x height • Kinetic Energy : Result from motion of mass KE = 1/2mv2
3. 3. NEWTON LAW • 1ST Law : A body will continue its state of rest or uniform motion unless external force changing it. 2nd Law : Rate of change of momentom of a body is proportional to the applied force. 3rd Law : In every action there will be equal and opposite reaction
4. 4. BRAYTON CYCLE • Brayton cycle : – The process of converting heat to do the work in jet engine. – Air is compress,heat and accelerate due to air molecule expand to produce the reaction force
5. 5. RELATION • Application to Jet engine : u = 0m/s mair= 260kg/s v=500m/s Acc = (V–U) ÷ t Work = F x D Acc =(500 – 0) ÷ 1s Work = Energy Acc = 500 m/s2 KE = ½(mv2) KE = ½(260)(500)2 KE = 32.5MJ F = ma F =260 x 500 Power = Work/Time Thrust = 130kN Pow = 32.5/1s Pow = 32.5Mwatt
6. 6. TURBO JET
7. 7. TURBOFAN
8. 8. TURBOSHAFT
9. 9. TURBOPROP

#### Description

Fundamental of jet engine

#### Transcript

1. 1. 15.1 PRINCIPLE http://part66.blogspot.com/
2. 2. ENERGY • Energy : Capacity to do work (Joule) • Potential Energy : Result from influence of gravity on mass due to its position PE = mass x gravity x height • Kinetic Energy : Result from motion of mass KE = 1/2mv2
3. 3. NEWTON LAW • 1ST Law : A body will continue its state of rest or uniform motion unless external force changing it. 2nd Law : Rate of change of momentom of a body is proportional to the applied force. 3rd Law : In every action there will be equal and opposite reaction
4. 4. BRAYTON CYCLE • Brayton cycle : – The process of converting heat to do the work in jet engine. – Air is compress,heat and accelerate due to air molecule expand to produce the reaction force
5. 5. RELATION • Application to Jet engine : u = 0m/s mair= 260kg/s v=500m/s Acc = (V–U) ÷ t Work = F x D Acc =(500 – 0) ÷ 1s Work = Energy Acc = 500 m/s2 KE = ½(mv2) KE = ½(260)(500)2 KE = 32.5MJ F = ma F =260 x 500 Power = Work/Time Thrust = 130kN Pow = 32.5/1s Pow = 32.5Mwatt
6. 6. TURBO JET
7. 7. TURBOFAN
8. 8. TURBOSHAFT
9. 9. TURBOPROP

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