jet engine propulsion systems review

8/22/2019 Jet Engine Propulsion Systems Review

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Jet Engine Propulsion

Systems


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Outlines of the Review:

Part I Jet Propulsion

Principles/Types/Parts/Functions

Part II Jet PropulsionAnalysis/Computations


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Part I:

Jet PropulsionPrinciples/Types/Parts/Functions


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ENGINE CLASSIFICATIONS

Heat Engines

Engines that convert heat energy to mechanical energy

TYPES OF COMBUSTION ENGINES (HEAT ENGINES)

1. External Combustion Engine

2. Internal Combustion Engine

TYPES OF INTERNAL COMBUSTION ENGINES

1. Reciprocating Engines (RE)

2. Jet Engines

Reaction Engines

Engines that generates thrust by its reaction to the flow in the opposite direction

of a mass of air


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Jet Propulsion Principle


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What is Thrust?

It is the force that pushes you deep into your seat as you

speed down the runway for take-off.

It is the force that propels an airplane forward through the

air.

THRUST


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How is THRUST created?

Balanced forces on an object prevent movement. If forces on anobject are the same in all directions the object will not move.


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Unbalanced forces cause an object to move.

During operation, a jet engine pushes gases out of the exhaust

nozzle. This makes an unbalanced force towards the front of the

engine. Unbalanced forces cause jet engines to make thrust.


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Newton's 2nd Law of Motion states "Force equals mass

accelerated (F = ma)."

When a jet engine accelerates air (mass), it makes thrust (force).

When a jet engine moves a small quantity of air, it makes a small

quantity of thrust.

When it moves a large quantity of air, it makes a large quantity of

thrust.


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Newton's 3rd Law of Motion states "Every action has an equal and

opposite reaction."

The ACTION is the jet exhaust going through the jet engine.

The REACTION is the jet engines then moves forward.


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Jet engine thrust can be increased in two ways:

increase the speed of exhaust gases.

increase the quantity of exhaust gases.

Increased engine thrust will make an aircraft fly faster

or with more weight.


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It has no static thrust. Must be carried aloft

by another high speed aircraft and

accelerated to operating speed.

Guided-missile systems, Space

vehicles use this type of jet


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Ram air pressure forces the shutters to open, fuel is

injected into the combustion chambers, and is

burned. Ignition is intermittent, timed to go on and off

as shutters open and close.


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Turbojet Engines Continued..

The turbojet engine is a reaction engine that uses only thrustdeveloped within the engine to produce propulsive force.Have no other features such as fan, propeller or free turbine.

Substantial increases in thrust can be obtained by employing

an afterburner Efficiencies of turbojets are attained at high altitude and

airspeed.

High thrust at low airspeed is not a turbojet characteristics.

To be at their best, turbojets need ram air pressure thatcomes only with high airspeed.

Need long runway for takeoff.


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Turboprop Engines Continued...

Deliver more thrust at low subsonic airspeeds.

Propulsive efficiency of a turboprop decreases as the

airspeed increases, while in a turbojet engine propulsive

efficiency increases as airspeed increases. Low of propulsive efficiency with speed that becomes a

limiting factor at airspeeds above Mach 0.6 for turboprop

aircraft.


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it provides power for a helicopter rotor or APU


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Air Inlet Duct


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Compressor Section


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Part II:

Jet PropulsionAnalysis/Computations

Useful Constants:


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Useful Constants:

Absolute Zero: 0R or - 460FAcceleration due to Gravity: g = 32.2 ft/s2 or 9.81m/s2

Gas Constant for Air: R = 53.342 ft-lbf/lbm-R

or 287.08 N-m/kgm-K

Mechanical Equivalent of Heat: J = 778.16 ft-lb/BTU Standard Sea Level Density: o = 0.002378 slugs/ft

3

or 1.225 kgm/m3

Standard Sea Level Pressure: Po = 14.7 psia, 29.92 inHg,

2116.8 lbf/ft2,

101325 Pa Standard Sea Level Temperature: To = 519R, 288.2K, 15C,59F


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The Brayton Cycle:Thermodynamic Analysis of the Operation of a

Gas Turbine Powerplant

Schematic of Simple Cycle:


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Schematic of Simple Cycle:

Air Standard Brayton Cycle


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Air Standard Brayton Cycle

1-2: Isentropic Compression2-3: Constant Presure addition of heat

3-4: isentropic expansion

4-1: Constant pressure rejection of heat

Analysis of the Brayton Cycle:


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Analysis of the Brayton Cycle:

QA = mcp (T3 T2)QR = mcp (T1 T4) = -mcp (T4 T1)

W = QA QR = mcp (T3 T2) mcp (T4 T1)

e = W = mcp

(T3

T2

) mcp (T4

T1

)

QA mcp (T3 T2)

e = 1 T4 T1T3 T2

e = 1 1 = 1 - 1rk

k-1 rpk-1

k


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where: rk = V1 , the compression ratio

V2

rp = p2 , the pressure ratio

p1


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Process 1-2:

T2 = V1k-1 = P2

k-1/k

T1 V2 P1

T2 = T1 rkk-1

rkk-1 = rp

k-1/k


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Process 3-4:

T3 = P3k-1/k = P2

k-1/k

T4 P4 P1

T3 = T4 rkk-1

e = 1 - 1 = 1 - 1

rkk-1 rp

k-1/k


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Total Compressor Work, Wc

= QH

Wc = -mcp(T2 T1)

Total Turbine Work, Wt = QH

Wt = -mcp (T4 T3)Wt = mcp (T3 T4)

Net Work, W or WB = Wt Wc

Pm = W / VD - Mean Effective Pressure

Example Problem


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Example Problem

1. The intake of the compressor of an air standardBrayton cycle is 40,000 cfm at 15 psia and 90 F.

The compression ratio, rk = 5 and the temperature

at the turbine inlet is 1440 F. The exit pressure ofthe turbine is 15 psia. Determine the net work,

thermal efficiency, and the mean effective pressure.

Solution


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Solution

Given:

V1 = 40,000 cfm

p1 = 15 psia

T1 = 550 R

T3 = 1900 R

p4 = 15 psiark = V1/V2 = 5


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m = p1V1 = (15)(144)(40,000)RT1 (53.34)(550)

Point 1:

v1 = V1 = 40,000 = 13.58 ft3/lb

m 2945Point 2:

v2 = v1/rk = 13.58/5 = 2.72 ft3/lb

p2 = p1rkk-1

= (15)(5)1.4

= 142.8 psiaT2 = T1 rk

k-1 = (550)(5)1.4 1 = 1047 R


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Point 3:v3 = v2 T3 = (2.72) 1900 = 4.94 ft

3/lbT2 1047

Point 4:

v4 = v3 P3 1/k = (4.94) 142.8 1/1.4 = 24.7 ft3/lbP4 15

T4 = T3 v3k-1 = (1900) 4.94 1.4 -1 = 998 R

v4 24.7


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Compressor Work, Wc = -cp (T2 T1)

= -(0.24)(1047-550)

= -119.3 BTU/lb

Turbine Work, Wt = cp (T3 T4)

= (0.24)(1900-998)

= 216.5 BTU/lb

Net Work, WB = Wt Wc = 216.5 119.3

= 97.2 BTU/lb

= (97.2)(2945) = 6751 hp42.4


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Heat added, QA = cp (T3 - T2)= (0.24)(1900 1047) = 204.7 BTU/lb

e = W = 97.2 = 0.4748 or 47.48 %QA 204.7

pm = W = W = (97.2)(778)

VD V4 V2 (24.7 2.72)(144)= 23.89 psi

Thrust Horsepower


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Thrust Horsepower

THP = THRUST x MPH375

Where: THP = The approximate thrusthorsepower produced by a jet

engine travelling at a givenairspeed

THRUST = The pounds of thrust beingdeveloped by the jet enginein flight

MPH = Airspeed in miles per hour

Newtons Second Law


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Newton s Second Law

F = MAWhere: F = force in pounds or newton

M = Mass in pound or kilograms

A = acceleration in ft/s2 or m/s2

Note: M = W/g

A = (V2 V1)

From Newtons Second Law, we can derive

the another useful formula for Force which


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the another useful formula for Force, which

is also a basic formula for jet engine thrust:F = w x (V2 V1)

g

Where: F = Force in pounds

w = Flow rate in lbs/sec of air, gas or a

liquid, such as fuel

V1 = Initial velocity of a mass of air, gas, or a liquid in

ft/secV2 = Final velocity of a mass of air, gas, or a liquid in ft/sec

g = gravitational acceleration

Note: M = w/g; where M = mass

Net Thrust


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Net Thrust

Fn = wa

+ wf

x Vj

- wa

x Va

+ wf

x Vf

g g g

Where: Fn = Net Thrust in lbs

wa = Airflow through the engine in

lbs/sec

wf= Fuel flow in lbs/sec

g = acceleration due to gravity

Vj = exhaust gas velocity in ft/sec

Va = incoming air velocity in ft/sec

Vf= incoming fuel velocity in ft/sec

By considering the Vfto be zero,

th f


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therefore....

Fn = wa + wf x Vj - wa x Va + 0

g g

By Transposing:

Fn = wa (Vj - Va) wf(Vj )

g g


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Fn = Wa (Vj Va) + Wf(Vj) + Aj (Pj Pam)g g

Where: Aj = Area of Engine Jet Nozzle in sq ft.

Pj = Static Pressure at the jet nozzle discharge inlbs/sq ft.

Pam = static pressure of the ambient (outside) air at

the jet nozzle in lbs/sq ft.

Without Considering Fuel flow:


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g

Fn = Wa (Vj - Va) + Aj (Pj Pam)

g

Gross Thrust:


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Fg = Wa (Vj) + Aj (Pj Pam)g

Where: Fg = Gross Thrust in lbs

other units the same as before

Np

= 2Va

- propulsive efficiency

Vj + Va

Net Thrust for a single fluid:


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g Fn = W (Vj Va) + Aj (Pj Pa)

g

Where:

Fn = Net Thrust

W = Weight

Va = Jet Engine Velocity (Forward Velocity)

Vj = Jet Exit Velocity

Aj = Projected Exhaust Area

Pj = Pressure at Exhaust

Pa = Atmospheric Pressure

Example Problems:


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p

1. A turbojet engine is travelling at 500 mph.The jet velocity is 1900 ft/s. The pressure atthe jet exhaust is atmospheric. The air flow is300,000 lb/hr and the fuel flow is 4300 lb/hr.

a. Calculate the net thrust velocityneglecting the fuel flow.

b. Calculate the net thrust velocity

considering the fuel flow.c. What is the propulsive efficiency?

S l ti


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Solution:

Given:Va = 500 mph = 733.5 ft/s

Vj = 1900 ft/s

Pj = Pa

Wa = 300,000 lb/hr

Wf = 4300 lb/hr


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a. Fn = Wa (Vj

Va)g

= 300,000 lb/hr [1900 733.5] 1 hr

32.2 ft/s2 3600 s

Fn = 3018.89 lbs

b Fn Wa (Vj Va) + Wf Vj + Aj (Pj Pa)


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b. Fn = Wa (Vj Va) + Wf Vj + Aj (Pj Pa)

g g

= 300,000 1900 733.5 1 hr + 4300

32.2 3600 s (32.2)(3600)

Fn = 3089.37 lbs

c. Np = 2 Va = 2(733.5)

Vj + Va 1900 + 733.5

Np = 55.71%


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2. Turbojet Engine on a test bed run have the following data.:

Thrust = 1400lb

AirFlow = 25lb/s

Fuel Flow = 1260 lb/hr

Jet Exit Pressure = 25Hg absa. Calculate the Jet Velocity Neglecting the fuel flow.

b. Calculate the Jet Velocity considering the fuel flow.

C. What is the propulsive efficiency?

Problems:


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1. Given is a turbo jet engine with flight velocity of 300meters/sec and exhaust gas exit velocity of 700

meters/sec. Neglect fuel, Calculate the propulsive

efficiency. (Aero Board 1983)

2. An aircraft powered by a gas turbine has an engine air

flow rate of 96.60 lbs. per second and a flight speed of

700 kilometers per hour at sea level standard condition.

The exhaust gas leaves the engine at a velocity of 1600kilometers per hour and at atmospheric pressure.

Neglecting fuel, determine the net thrust (newton) of the

engine. ( Aero Board 1984)


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3. The Turbo jet engine of an airplane in flight at 600 ft persecond consumes 50 lbs of air per second and 120 lbs of

fuel per minute. The exhaust gas leaves the engine tailpipe

at a velocity of 1,800 feet per second at atmospheric

pressure. Calculate: (Aero Board 1985)

a. Engine fuel air ratio

b. Engine thrust (newtons)


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4. For a turbo jet with flight velocity of 310 kilometers per hourand exhaust gas jet velocity of 690 kilometers per hour at

atmospheric pressure, What is the propulsive efficiency

with the fuel considered negligible? (Aero Board 1986)

5. For a rocket engine to develop a thrust of 10,000 newtons,

the 19.2 lb propellant consumed by the rocket per second

must generate an exhaust gas jet velocity at atmospheric of

what? (Aero Board 1986)


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6. A missile using a rocket engine in flight at sea level burnsits propellant at the rate of 15 lbs/s. The exhaust gas exit

velocity is 3820 fps, pressure is equal to ambient, at this

instant the rocket develops 2500 lbs thrust horsepower.

Assume level flight. (Aero Board 1988)

a. Thrust

b. Flight Velocity

jet engine propulsion systems review

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