4261 Cycle Analysis Overview

7/27/2019 4261 Cycle Analysis Overview

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MAE 4261: AIR-BREATHING ENGINES

Thermodynamics Review and Cycle Analysis Overview

September 1, 2009

Mechanical and Aerospace Engineering DepartmentFlorida Institute of Technology

D. R. Kirk


2/18


3/18

3

HEAT ENGINE: PROPULSION CHAIN

ChemicalEnergy

Heat

(ThermalEnergy)

MechanicalPower

Mech.

Power toGasFlow

ThrustPower

The overall efficiency for the propulsion chain is given by:

Combustion Thermal Propulsive

Thrust=F

speedFlighto

u

fueljetforJ/kg710x4.3=reaction)of(heatcombustionofHeatcomb

h

rateflowFuelf

m

comb

0=overall

hf

m

FU

usageenergychemicalofRate

powerThrust

forpayWhat we

getWhat we

propulsivemechthermalcombustionoverall

Mechanical


4/18

4

CONCEPTS / TOOLS FOR ENGINE IDEAL CYCLE ANALYSIS

Ideal gas equation of state, p = rRT

One-dimensional gas dynamics

Concepts of stagnation and static quantities (temperature, pressure, etc.)

Relations between Mach number and thermodynamic properties

Thermodynamics of propulsion cycle

Make use of 1st and 2nd Laws of Thermodynamics

Behavior of useful quantities: energy, entropy, enthalpy

Relations between thermodynamic properties in a reversible (lossless) process

Isentropic = reversible + adiabatic

Properties of cycles (it is cyclic)

Air starts at atmospheric pressure and temperature and ends up at atmospheric pressureand temperature

Definition of Open vs. Closed Cycles


5/18

5

STAGNATION QUANTITIES DEFINED

Quantities used in describing engine performance are the stagnation pressure,

enthalpy and temperature

Stagnation enthalpy, ht , enthalpy state if stream is decelerated adiabatically to zerovelocity

2

2

11or

2

2

2

11

2

1

)2(

21

2

2

2

2

MT

tT

a

u

T

tT

RTa

R

pc

Tpc

u

T

tT

pc

uT

tT

Tpch

uh

th

Ideal gas

Stagnation temperature

Speed of sound

Total to static temperature ratio

in terms of Mach number


6/18

6

FOR REVERSIBLE + ADIABATIC = ISENTROPIC PROCESS

flowspeedlowforEquation"Bernouli"

2

2

1

gettotheorembinomialtheusingexpand,12For

1

2

2

11

velocity)zeroally toisentropicddecelerateisstreamifpressureis(

pressurestagnationthedefines1

constant)1/(

findweusing

constant

upt

p

M

Mp

tp

tp

T

T

p

tp

T

pRTp

P

t

r

r

r


7/187

RECAP ON THERMODYNAMICS: 1st LAW

First law (conservation of energy) for a system: chunk of matter

of fixed identity

E0 = Q - W

Change in overall energy (E0 ) = Heat in - Work done

E0 = Thermal energy + kinetic energy ...

Neglecting changes in kinetic and potential energy

E = Q - W ; (Change in thermal energy)

On a per unit mass basis, the statement of the first law is thus:

e = q - w


8/188

RECAP ON THERMODYNAMICS: 2nd LAW

The second law defines entropy,s,by:

ds dqreversible

T

Where dqreversible is the increment of heat received in a reversible

process between two states

The second law also says that for any process the sum of the

entropy changes for the system plus the surroundings is equal

to, or greater than, zero

ssystem ssurroundings 0Equality only exists in a reversible (ideal) process


9/189

REPRESENTING ENGINE PROCESS

IN THERMODYNAMIC COORDINATES

First Law:E = Q - W, where E is the total energy of the parcel of air.

For a cyclic process Eis zero (comes back to the same state)

Therefore: Q (Net heat in) = W(Net work done)

Want a diagram which represents the heat input or output.

A way to do this is provided by the Second Law

Tdsreversible

dq

where ds is the change in entropy of a unit mass of the parcel and

dq is the heat input per unit mass

Thus, one variable should be the entropy ,s


10/1810

STEADY FLOW ENERGY EQUATION (I)

Shaft work

Heat input

Mass flow

Device

1 2

ht 2- ht 1= q - w shaft

qis heat input/unit mass

wshaft is the shaft work / unit mass

For any device in steady flow

previouslydefinedenthalpystagnationtheis2/quantityThe

doneshaft workofRate-inheatofRate=

2

shaft12

uhh

WQhhm

t

tt

Per unit mass flow rate:


11/1811

STEADY FLOW ENERGY EQUATION (II)

The form of the steady flow energy equation shows that enthalpy, h:

h = e + pv = e + p/r

Natural variable to use in fluid flow-energy transfer processes.

For an ideal gas with constant specific heat, dh = cpdT.

Changes in enthalpy are equivalent to changes in temperature.

To summarize, the useful natural variables in representing gas turbine engine

processes are h,s (or T, s).

Represent thermodynamic cycle (Brayton) for gas turbine engine on a T,s diagram


12/1812

THERMODYANMICS: BRAYTON CYCLE MODEL


13/1813

GAS TURBINE ENGINE COMPONENTS

Inlet: Slows, or diffuses, the flow to the compressor

Fan/Compressor: (generally two, or three, compressors in series) does work onthe air and raises its stagnation pressure and temperature

Combustor: Heat is added to the air at roughly constant pressure

Turbine: (generally two or three turbines in series) extracts work from the air to

drive the compressor or for power (turboshaft and industrial gas turbines)

Afterburner: (on military engines) adds heat at constant pressure

Nozzle: Raises the velocity of the exiting mass flow

Exhaust gases reject heat to the atmosphere at constant pressure


14/1814

THERMODYNAMIC CHARACTERISTICS

OF COMPONENTS (IDEAL COMPONENTS)

0=sexchange,heatno,shaft workNo:nozzleExhaust

losslessadiabatic,0s

,turbinefromoutputwork0,>shaft

wwh:Turbine

input)(heatqh:rafterburneandCombustor

losslessadiabatic,0s

,compressorinput towork0,wwh:Compressor

0h:Inlet

shaftt

int

shaftshaftt

t


15/1815

THERMODYNAMIC MODEL OF GAS TURBINE ENGINE CYCLE

[Cravalho and Smith]


16/1816

SCHEMATIC OF CONDITIONS THROUGH A GAS TURBINE

[Rolls-Royce]


17/1817

NOMINAL PRESSURES AND TEMPERATURES FOR A

PW4000 TURBOFAN [Pratt&Whitney]


18/1818

REVIEW OF STATION LOCATIONS

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4261 Cycle Analysis Overview