Ch. 11: Introduction to

1
Ch. 11 Introduction to Compressible Flow

• Focus on 1-dimensional, compressible, inviscid
  flows

• Liquids, ? constant for us (1 increase in ?
  for every 1.6 km deep)

• Air, 1 change for every 26 m deep

• M 0.3 5 ??/? M 0.3 100 m/s or 230
  mph

• Significant density changes imply significant
  compression
• or expansion work on the gas, which can change
  T, e, s,

• Compressibility fluid acceleration because of
  friction, fluid deceleration in a converging
  duct, fluid temperature decrease
• with heating

• Ideal Gas p ?RT (simple, good approximations
  for our engineering
• applications, captures trends)

2

• Smits/A Physical Introduction To Fluid Mechanics

3
Density gradients will affect how light is
transmitted though medium (by affecting index of
refraction). By applying the Gladstone-Dale
formula it becomes evident that the shadowgraph
is sensitive to changes in the 2nd derivative of
the gas density. Strength of shock can be related
to width of dark band. - Methods of Experimental
Physics Vol 18, Martin
Deflection of light caused by shock compressed
gas ahead of a sphere flying at supersonic
speed.
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Thoughts on the increased complexity of
incompressible flow
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GOVERNING EQUATIONS FOR NEWTONIAN
FLUIDS INCOMPRESSIBLE
??/?t ?/?xk(?uk) 0 becomes
?uk/?xk 0 ??uj/?t ?uk?uk/?xk
-?p/?xj?/?xj(? ?uk/?xk)?/?xi?(?ui/?xj?uj/?xi)
?fj becomes ??uj/?t
?uk?uj/?xk -?p/?xj ?(?2ui/?xj?xj) ?fj
4 Equations continuity and three momentum 4
Unknowns p, u, v, w Know ?, ?, fj
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GOVERNING EQUATIONS FOR NEWTONIAN
FLUIDS COMPRESSIBLE
??/?t ?/?xk(?uk) 0 ??uj/?t ??uk/?xk
-?p/?xj ?/?xj(? ?uk/?xk) ?/?xk?(?ui/?xk
(?uj/?xi) ?fj p p(?,T) Thermal p ?RT e
e(?,T) Caloric e CvT ??e/?t ?uk?e/?xk
-p?uk/?xj ?/?xj(k ?T/?xj) ?(?uk/?xk)2
?(?ui/?xk ?uj/?xi)(?uk/?xk)
e is the internal energy
7 Equations continuity, momentum(3), energy,
thermal, state 7 Unknowns p, u, v, w, e, T,
? Know ?, fj, ?, k
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Thoughts on the speed of sound
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COMPRESSIBLE FLOW
front
c2 (?p/??)s
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COMPRESSIBLE FLOW
front
If fluid incompressible, gas would behave like
solid body and move everywhere at piston speed.
If pressure disturbance is small relative to p1
then front propagates at speed of sound. If
large shock waves occur where speed,
temperature, density and pressure change
significantly across shock. (Speed of shock is
between the speed of sound in the compressed and
undisturbed gas.)
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SPEED OF SOUND
Sound waves are pressure disturbances ltlt ambient
pressure. For loud noise p 1Pa whereas
ambient pressure is 105 Pa Speed of sound c2
(?p/??)s Assumptions ideal gas isentropic
p/?k const, or differentiating
dp/?k pk ?-k-1 d? 0 dp/(p/cont) (const
?k) k ?-k-1 d? 0 dp/p kd?/? 0
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SPEED OF SOUND
dp/p kd?/? 0 dp/d? kp/?
p ?RT for ideal gas then c2 kRT For
20oC and 1 atmosphere c 343 m/s 1126 ft/s
768 mph
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M 1
Dynamic Pressure ? Static Pressure
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M V/c M2 V2/c2 V2/kRT (ideal gas) p
?RT (ideal gas) M2 2(1/2V2/kRT) 2(1/2
V2/(kp/?)) M2 21/2 ?V2/(kp) 1/2
?V2/p M2 dynamic pressure/ static pressure
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Thoughts on the speed of sound
As related to the speed of the source
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• Regimes of flow
• Acoustics fluid velocities ltlt c, speed of
  sound
• fractional changes in p, T and ? are important.
• (2) Incompressible flow fluid velocities lt c,
  speed of sound
• fractional changes in ? are not significant
  fractional changes
• in p and T are very important
• (3) Compressible flow (gas dynamics) fluid
  velocities c,
• speed of sound fractional changes in p, T and ?
  are all
• important.

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SUB SONIC
SUPER
Propagation of Sound Waves from a Moving Source
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Some Assumptions
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It is assumed that the system is always in
equilibrium.
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It has been found by experiment that as long as
the temperatures and pressures are not too
extreme, the flow attains an instantaneous
equilibrium. This continues to hold even inside
shock waves. For all the flows examined here,
all systems will be assumed to be in equilibrium
at all times.
p1,?1, T1, s1, h1
p2,?2, T2, s2, h2
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It is assumed that all gases obey ideal gas law
p ? RT
Kelvin (or Rankine)
Not gauge pressure
R Ru/Mm 287.03 m2/(s2-K) (N-m)/(kg-K)
J/(kg-K) R 1716.4 ft2/(s2-R)
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Conservation of Energy
(note that u is now the internal energy)
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FIRST LAW OF THERMODYNAMICS
?Q ?W ?E ?(KE PE U)
?Q/m ?W/m ?E/m ?q ?w ?u
W - ?pdV
U, internal energy, is energy stored in
molecular bonding forces and random molecular
motion. (?KE and ?PE we will
ignore)
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Ideal gas is composed of point particles which
exhibit perfect elastic collisions. Thus internal
energy is a function of temperature only. U
f(T) Enthalpy, h, defined as h u pv
h f(T) since h(T) u(T) RT
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Specific Heat for Ideal Gas
dQ mCv,pdT
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Specific heat is defined as the amount of heat
required to raise the temperature of a unit mass
of substance by1oK. Different for constant
volume or pressure.
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Definition of heat capacity at constant volume
mCvdT dQ or CvdT dq
dq dw du
if Vol constant, w -pdv 0, then dq du,
Cv du/dT It can be
shown that du Cv dT even
if volume not held constant!!
- pg 41, Thermal-Fluid Engineering, Warhaft
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Definition of heat capacity at constant
pressure mCpdT dQ or Cp
dq/dT
dq dw dq pdv du h u pv dh du
pdv vdp
if pressure constant, dh du pdv dq Cp
dh/dT again can be shown to be true even if
pressure is not constant!!
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Cp dh/dT
Cv du/dT
h u pv u RT dh du RdT dh/dT du/dT
R Cp Cv R
IDEAL GAS
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cp/cv k
cp cv R
cp/cp cv/cp R/cp
1 1/k R/cp
(1 1/k)/R 1/cp
cp R/(1-1/k) kR/(k-1)
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cp/cv k
cp cv R
cp/cv cv/cv R/cv
k 1 R/cv
(k 1)/R 1/cv
cv R/(k 1)
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Another Assumption
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It is assumed that cp/cv is not a function of T
calorically perfect For a perfect gas cp/cv
1.4
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cp/cv k is not a function of temperature
k
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cp/cv k 1.4 for perfect gas
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The Second Law
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The second law of thermodynamics can be stated
in several ways, none of which is easy to
understand. Smits, A Physical Introduction to
Fluid Mechanics
DEFINITION ?S ?rev ?Q/T or dS (?Q/T)rev
dq du pdv Tds du pdv dh vdp
always true!
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DEFINITION ?S ?rev ?Q/T or dS (?Q/T)rev
Change in entropy intimately connected with the
concept of reversibility for a reversible,
adiabatic process entropy remains constant. For
any other process the entropy increases.
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What we can do with
Tds du pdv dh vdp ideal gas and constant
specific heats
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Cv du/dT Ideal Gas Cp dh/dT p ?RT
(1/v)RT
Tds du pdv dh vdp ds du/T ?RTdv/T
ds CvdT/T (R/v)dv
s2 s1 Cvln(T2/T1) Rln(v2/v1)
s2 s1 Cvln(T2/T1) - Rln(?2/?1)
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s2 s1 Cvln(T2/T1) - Rln(?2/?1)
If isentropic s2 s1 0 ln(T2/T1)Cv
ln(?2/?1)R
Cp Cv R R/Cv k 1
?2/?1 (T2/T1)Cv/R (T2/T1)1/(k-1)
assumptions ISENROPIC IDEAL GAS constant cp,
cv
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Cv du/dT Ideal Gas Cp dh/dT p ?RT
(1/v)RT
Tds du pdv dh vdp ds dh/T - vdp
ds CpdT/T - (RT/pT)dp
s2 s1 Cpln(T2/T1) - Rln(p2/p1)
s2 s1 Cpln(T2/T1) - Rln(p2/p1)
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s2 s1 Cpln(T2/T1) - Rln(p2/p1) If isentropic
s2 s1 0 ln(T2/T1)Cp ln(p2/p1)R
Cp Cv R R/Cp 1- 1/k
p2/p1 (T2/T1)Cp/R (T2/T1)k/(k-1)
assumptions ISENROPIC IDEAL GAS constant cp,
cv
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Stagnation Reference (V0)
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BE 1-D, energy equation for adiabatic and no
shaft or viscous work. (p2/?2) u2 ½ V22
gz2 (p1/?1) u1 ½ V12 gz1 Definition h
u pv u p/? assume z2 z1 h2 ½
V22 h1 ½ V12 ho 0 Cp dh/dT
(ideal gas) ho h1 cp (To T) ½ V12 T0
½ V12/cp T T (1 V2/2cpT)
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T0 ½ V12/cp T T (1 V2/2cpT)
cp kR/(k-1)
T0 T (1 V2/2(kR/(k-1)T)
T0 T (1 (k-1)V2/2kRT)
c2 kRT
T0 T (1 (k-1)V2/2c2)
V2/ c2
T0 T (1 (k-1)/2 M2)
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To/T 1 (k-1)/2 M2
STEADY, 1-D, ENERGY EQUATION FOR ADIABATIC FLOW
OF A PERFECT GAS
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?/?o (T/To)1/(k-1) To/T 1 (k-1)/2 M2 ?/?o
(1 (k-1)/2 M2 )1/(k-1) Ideal gas and
isentropic and constant cp, cv (isentropic
adiabatic reversible)
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p/p0 (T/To)k/(k-1) To/T 1 (k-1)/2 M2 p/p0
(1 (k-1)/2 M2)k/(k-1) Ideal gas and
isentropic and constant cp, cv (isentropic
adiabatic reversible)
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QUIZ

• When a fixed mass of air
• is heated from 20oC to 100oC
• What is the change in enthalpy?
• For a constant volume process,
• what is the change in entropy?
• For a constant pressure process,
• what is the change in entropy?
• For an isentropic process
• what are the changes in p and ??
• Compare speed of sound
• for isentropic and isothermal conditions.

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• h2 h1 Cp(T2- T1)
• s2 s1 Cvln(T2/T1)
• s2 s1 Cpln(T2/T1)
• ?100/ ?20 (T100/T20)2.5
• 2.5 1/(k-1) k 1.4 for ideal gas
• p100 / p20 (T100/T20)3.5
• 3.5 k/(k-1) k 1.4 for ideal gas

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(e) c2 dp/d? But c2 (?p/??)T does
not equal c2 (?p/??S) If isentropic p/?k
constant (ideal gas) Then c (?p/??)S1/2
(kRT)1/2 (1.4 287.03 (20 273.15))1/2
343.2 m/s If isothermal p ?RT (ideal
gas) Then c (?p/??)T1/2 (RT)1/2
(287.03 X (20 273.15)1/2 290.07 m/s 18
too low