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1. Some Introductory Concepts

1.1 Definition of Compressibility

constant all substances are compressible (especially gases)

if constant incompressible flow (Bernoullis eq. holds)

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if pressure is increased volume will change (decrease)

compressibility

dp

d

1

(1.1)

( specific volume)

usually temperature (T) increases, too

assume isothermal process: (T = constant)

TT

p

1

(1.2)

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assume isentropic process (adiabatic + reversible):

ss

p

1

(1.3)

liquids have low compressibility:

N

m105

210

T at 1 atm for water

gases have higher compressibility

N

m10

25

T at 1 atm for air

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use

dpd11

(1.4)

dpd (1.5)

for low velocities d)3.0M( small 0

constantflowibleincompress

for higher velocities d)3.0M( is larger

compressible )ct(

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1.2

Flow Regimes

(Incompressible if M < 0.3)

a

u

M

soundofspeed

magnitudevelocity

NumberMach

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Flow Regimes

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Flow Regimes

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Flow Regimes

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Why are reentry vehicles blunt?

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Knudsen number:

nk (1.6)

mean free path of a gas molecule

characteristic body dimension

if 01.0kn continuous flow

1kn rarefied gas (e.g. at 300,000 ft: 1ft1ft, )

10kn free molecular flow

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1.3

Brief Review of Thermodynamics

p RT (1.7)

R= specific gas constantk.kg

J287 =1716 ft lb/(slug R)

Or p (1.8)

specific volume =

this is accurate ( %1 error) for low pressures )atm1(

moderate temperatures )k273(

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For high pressures and cold temperatures molecules are more

closely packed together: Van der Wads equation

RT)bv(v

ap

2

(1.9)

a, b are constants depending on the type of gas

internal energy: )v,T(ee (1.10)

enthalpy: pve)p,T(hh (1.11)

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thermally perfect gas:

dTcdhdT,cde pv (1.12)

for vp c,c constant

calorically perfect gas

Tce v where:v

vT

ec

(1.13)

Tch p where:p

pT

hc

(1.14)

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v

pvp

c

cRcc (1.15)

n+2 =

n = 1.4 for diatomic gases

= 1.67 for triatomic gases

pp

c

c

c

R11

c

R1

p

v

pc (1.16)

vc (1.17)

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First Law of Thermodynamics:

system (fixed mass)

energyinchangeewq (1.18)

q heat added, w work done, e change in energy

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process: no heat added or taken away

process: no dissipative phenomena occur

process: adiabatic + reversible

for a reversible process

pdvw (1.19)

incremental change of volume due to boundary displacement

pdvq (1.20)

(alternative form of 1stlaw)

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Examples of Reversible processes

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Examples of irreversible processes

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Entropy and second law

Entropy is

ds (1.21)

s = entropy, revq amount of heat added reversibly

q = amount of heat

T = system temperature

irrevdsT

qds

(1.22)

0dsirrev (1.23)

(if 0dsirrev reversible process)

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2ndlaw:

Tqds (1.24)

for an adiabatic process:

0ds (1.25)

First law: you cannot

Second law: you always

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Calculation of entropy:

depdTdsTdsq rev

(using 1.20)

pddeTds (1.26)

use enthalpy peh

dppddedh

dpdhTds (1.27)

1.26 and 1.27 are alternative forms of 1stlaw

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for a thermally perfect gas: dTcdh p (1.12)

T

dp

T

dTcds p

RTp

1

2T

Tp12p

p

plnR

T

dTcss

p

dpR

T

dTcds

2

1

for a calorically perfect gas ( pc constant)

12 ss (1.28)

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similarly:

12 ss (1.29)

Isentropic relations

put 21 ss in (1.28)

1

2p

1

2

1

2

1

2p

T

Tln

R

c

p

pln

p

plnR

T

Tlnc0

1

2

p

p

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from (1.18)

1Rcp

1

2

p

p

(1.30)

similarly:

1

2v

1

2

1

2

1

2v

T

Tln

R

c

v

vlnlnR

T

Tlnc0

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)1(1

1

2

1

2R

c

1

2

1

2

T

T

T

Tv

(1.31)

1

2

(1.32)

in summary:

)1(

1

2

1

2p

p

T

T

1

2

(1.33)

(energy relation for an isentropic process) many processes can be

assumed isentropic, irreversible effects are usually constrained in

the boundary layer.

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Governing equations [review from 301]

incompressible flow: (= constant, T also constant)

Continuity: (1.34)

Momentum:2DV

p Dt V f (1.35)

Unknowns:

Equations:

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Energy:p v

DTC (k T)Dt

(1.36)

Bernoullis equation: p + = (1.37)

Assumptions: 1. Inviscid

2.

Steady

3.

Incomressible4.

5.

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Compressible flow: is variable

unknowns:

need energy equation

continuity: 0ddvt s

sV (1.38)

0t

p

V (1.39)

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momentum:

dvdp)d(dvt vsv

f

sVsVV (1.40)

xf

x

p

Dt

Du

(1.41)

v y y

w z z

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energy:

sV

VV d2

edv2

et

2

s

2

v

dv)(pdvq

vsv

VdsV

f (1.42)

)(.pq

Dt

2eD

2

VV

V

f (1.43)

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Definition of total (stagnation) conditions

Consider a fluid element: p, T, , M, V (static quantities); then

move adiabatically to 0V

(for incompressible flow: po= )

To (total temperature)

opo Tch (total enthalpycalorically perfect gas)

for adiabatic flow )0q( , negligible body forces:

the energy equation becomes

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VpDt

2

veD

2

(1.44)

using:

ppp VVV

(1.45)

Dt

Dp

Dt

DpDt/pDDt/Dp

Dt

)/p(D

2

definition ofDt

D

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using continuity: 0Dt

D

V

VVV

ppt

pp

Dt

Dp (1.46)

using (1.41), and adding (1.42) to (1.40):

VVVVV

pp

t

ppp

2

pe

Dt

D 2

t

p

Dt

2hD 2

V

(1.47)

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hpve

pe

for steady flow:

0Dt

2

hD2

V

2

h2

Vconstant along a streamline

constanth2

h o

2

V

; (stagnation enthalpy)

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etemperaturstagnationconstantTTch oopo

(for non-adiabatic process oo T,h change)

for isentropic (= adiabatic + reversible) process to 0V

oo,p,p (pressure and density)

for non-isentropic flow (e.g. across a shock wave) oo ,p change,

but Todoes not change, if adiabatic.

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Problem 7.8: In the reservoir of a supersonic wind tunnel, the

velocity is negligible, and the temperature is 1000K. The

temperature at the nozzle exit is 600K. Assuming adiabatic flow

through the nozzle, calculate the velocity at the exit.

Reservoir:

Exit:

Adiabatic flow:

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Problem 7.13: Bernoullis equation was derived in chapter 3 from

Newtons second law; it is fundamentally a statement that

force=mass x acceleration. However, the terms in Bernoulisequation have dimensions of energy per unit volume, which

prompt some argument that Bernoullis equation is an energy

equation for incompressible flow. If this is so, then it should de

derivable from the energy equation for compressible flow

discussed in the present chapter. Starting with Equation (7.53) forinviscid, adiabatic, compressible flow, make the appropriate

assumptions for an incompressible flow and see what you need to

do to obtain Bernoullis equation

Eq. 7.53:2Vh+ =constant

2

Rewrite enthalpy as:

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Hence eq. 7.53 can be written as2V

+ =constant2

For incompressible flow: