CH.12: COMPRESSIBLE FLOW

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CH.12 COMPRESSIBLE FLOW
It required an unhesitating boldness to undertake
such a venture . an almost exuberant
enthusiasmbut most of all a completely
unprejudiced imagination in departing so
drastically from the known way. J. Van
Lonkhuyzen, 1951, discussing designing Bell XS-1
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Compressible flow is a fun subject. John
Anderson
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Ch.12 - WHAT CAUSES FLUID PROPERTIES TO CHANGE
IN A 1-D COMPRESSIBLE FLOW FLOW?
(note if isentropic stagnation properties do not
change)
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Ch.12 - COMPRESSIBLE FLOW Flow can be affected
by area change, shock, friction, heat transfer
shock
Q
friction
Area2
Q
shock
Area1
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One-Dimensional Compressible Flow
Rx
P1
P2
dQ/dt
Surface force from friction and pressure
heat/cool
( s1, h1,V1,)
( s2, h2, V2, )
What can affect fluid properties? Changing area,
normal shock, heating, cooling, friction. - need
7 equations -
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Affect of Area Change
Assumptions ?
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ASSUMPTIONS

• ALWAYS
• Steady Flow
• Ideal Gas
• Ignore Body Forces
• only pressure work
• (no shaft, shear or other work)
• Constant specific heats
• One Dimensional
• MOSTLY
• ISENTROPIC

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quasi-one-dimensional
V(s)
V(x,y)
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quasi-one-dimensional
Flow properties are uniform across any given
cross section of area A(x), and that they
represent values that are some kind of mean of
the actual flow properties distributed over the
cross section. NOTE equations that we start
with are exact representation of conservation
laws that are applied to an approximate physical
model
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One-Dimensional Compressible Flow
f(x1)
f(x2)
s1, h1
s, h2
7 Variables T(x), p(x), ?(x), A(x), v(x), s(x),
h(x) 7 Equations mass, x-momentum, 1st and 2nd
Laws of Thermodynamics, Equations of State (3
relationships)
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One-Dimensional Compressible Flow
Variables T(x), p(x), ?(x), A(x), v(x), s(x),
h(x) Equations mass, momentum, 1st and 2nd Laws
of Thermodynamics, Equation of State (3
relationships)
Cons. of mass (steady / 1-D) Cons. of momentum (
no FB) Cons. of energy ( only pressure
work) 2nd Law of Thermodynamics . Ideal
gas Ideal gas constant cv, cp Ideal gas
constant cv, cp
(steady)
Eqs. of State
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Want to find qualitative relationships for dT,
dV, dA, d? ISENTROPIC
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Affect of Change in Velocity on Temperature
Isentropic, steady, cpcv constant, no body
forces, quasi-one-dimensional, ideal gas, only
pressure work
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For isentropic flow if the fluid accelerates
what happens to the temperature?
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For isentropic flow if the fluid accelerates
what happens to the temperature?
if V2gtV1 then h2lth1
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For isentropic flow if the fluid accelerates
what happens to the temperature?
if V2gtV1 then h2lth1
if h2lth1 then T2ltT1
Temperature Decreases !!!
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Affect of Change in Velocity on Temperature
Velocity Increases, then Temperature
Decreases Velocity Decreases, then Temperature
Increases NOT DEPENDENT ON MACH NUMBER
Isentropic, steady, cpcv constant, no body
forces, quasi-one-dimensional, ideal gas, only
pressure work
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Affect of Change in Velocity on Pressure
Isentropic, steady, no body forces,
quasi-one-dimensional
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Affect of Change in Velocity on Pressure
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(No Transcript)
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FSx (p dp/2)dA pA (pdp)(A dA) pdA
dAdp/2 pA pA pdA dpA - dpdA (dm/dt)(VxdVx
) - (dm/dt)Vx (Vx d Vx) ?VxA- Vx?VxA
Vx ?VxA dVx ?VxA Vx?VxA -dpA
?VxdVxA or dp/?dVx2/20
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Affect of Change in Velocity on Pressure
dp/?dVx2/20 Velocity Increases, then Pressure
Decreases Velocity Decreases, then Pressure
Increases NOT DEPENDENT ON MACH NUMBER
Isentropic, steady, no body forces,
quasi-one-dimensional
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Affect of Change in Area on Pressure and Velocity
Isentropic, steady, no body forces,
quasi-one-dimensional
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EQ. 11.19b
EQ.12.1a
d?(AV) dA(?V) dV(?A)/?AV 0
steady
isentropic, steady
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isentropic, steady
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EQ. 12.5
EQ.12.6
isentropic, steady
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Affect of Change in Area on Velocity
dV/V - (dA/A)/1-M2 M lt1 Velocity and Area
change oppositely M gt 1 Velocity and Area change
the same
Isentropic, steady, no body forces,
quasi-one-dimensional
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Affect of Change in Area on Pressure
dp/(?V2) (dA/A)/1-M2 M lt1 Pressure and Area
change the same M gt 1 Pressure and Area change
oppositely
Isentropic, steady, no body forces,
quasi-one-dimensional
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isentropic, steady, 1-D
Mlt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady, 1-D
Mlt1
dVx gt 0 or dVx lt 0
dp gt 0 or dp lt 0
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isentropic, steady, 1-D
M gt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady, 1-D
M gt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady,1-D
Mlt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady,1-D
Mlt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady,1-D
Mgt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady,1-D
Mgt1
dp gt 0 or dp lt 0
dVx gt 0 or dVx lt 0
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isentropic, steady, 1-D
Subsonic Nozzle
Subsonic Diffuser
(dp and dV are opposite sign)
Mlt1
If M lt 1 then 1 M2 is , then dA and dP
are same sign and dA and dV are opposite
sign qualitatively like incompressible flow
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isentropic, steady, 1-D
Supersonic Nozzle
Supersonic Diffuser
(dp and dV are opposite sign)
Mgt1
If M gt 1 then 1 M2 is -, then dA and dP
are opposite sign and dA and dV are the same
sign qualitatively not like incompressible flow
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And this is the reason!!!!!!!
If ? is constant then dA and dV must be opposite
signs, but for compressible flows all bets are
off, e.g. for Mgt1 both dV and dA can have the
same sign
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Affect of Change in Area on Density
Isentropic, steady, no body forces,
quasi-one-dimensional
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Affect of Change in Velocity on Pressure
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Affect of Change in Velocity on Pressure
-(dA/A)/(1-M2) -dA/A - d?/? d?/?
(dA/A)1/(1-M2) - 1 d?/? (dA/A)M2/(1-M2)
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d? ?
d?/? (dA/A)M2/(1-M2)
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d?/? (dA/A)M2/(1-M2)
d?/? gt 0
d?/? lt 0
d?/? gt 0
d?/? lt 0
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WHAT HAPPENS AT M 1 ?
?
dp/(?V2) (dA/A)/(1-M2) dV/V
-(dA/A)/(1-M2) d?/? (dA/A)M2/(1-M2)
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If M 1 have a problem, Eqs. blow up! Only if
dA ? 0 as M ? 1 can avoid singularity. Hence for
isentropic flows sonic conditions can only occur
where the area is constant.
dp/(?V2) (dA/A)/(1-M2) dV/V
-(dA/A)/(1-M2) d?/? (dA/A)M2/(1-M2)
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What happens to dp and d? across the throat in a
supersonic nozzle with steady, isentropic flow?
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What happens to dp and d? across the throat in a
supersonic nozzle with steady, isentropic flow?
dp/? - dVx2/2
Vx continues to increase so p continues to
decrease
d?/? (dA/A)M2/(1-M2)
(dA/A)M2/(1-M2) is always negative so ?
continues to decrease
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What happens to dp and d? across the throat in a
supersonic diffuser steady, isentropic flow?
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What happens to dp and d? across the throat in a
supersonic diffuser with steady, isentropic flow?
dp/? - dVx2/2
Vx continues to decrease so p continues to
increase
d?/? (dA/A)M2/(1-M2)
(dA/A)M2/(1-M2) is always positive so ?
continues to increase
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Same shape, but in one case accelerating flow,
and in the other decelerating flow
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Goddard realizes (1917) that De Laval nozzle
could be used for rocket
De Laval designed a turbine (1888) whose wheel
was turned by jets of steam flowing through a
contraction - expansion section.
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Sketch a Supersonic Wind Tunnel
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SONIC
SUPERSONIC
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Sneeze Breath
(180 mph)
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Affect of Area Change - Qualitative
7 Variables T(x), p(x), ?(x), A(x), v(x), s(x),
h(x) 7 Equations mass, x-momentum, 1st and 2nd
Laws of Thermodynamics, Equations of State (3
relationships)
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One-Dimensional Compressible Flow
Variables T(x), p(x), ?(x), A(x), v(x), s(x),
h(x) Equations mass, momentum, 1st and 2nd Laws
of Thermodynamics, Equation of State (3
relationships)
Cons. of mass Cons. of momentum Cons. of
energy 2nd Law of Thermodynamics . Ideal
gas Ideal gas constant cv, cp Ideal gas
constant cv, cp
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Have found qualitative relationships for dT, dV,
dA, d? Now want to find quantitative
relationships. IDEAL GAS, ISENTROPIC, QUASI-1-D,
FBX0, STEADY, only PRESSURE WORK
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Coupled, nonlinear equations hard to solve.
Much easier to express flow variables in terms
of stagnation quantities and local Mach number.
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Eqs. 11.20a,b,c Eqs. 12.7a,b,c
EQ. 11.19b

EQ. 11.12c
P / ?k constant
(po, To refer to stagnation properties)
isentropic, steady, ideal gas, constant cp, cv
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Constant Area - Isentropic
Example using stagnation properties
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Example Air flows isentropically in a duct. At
section 1 Ma1 0.5,
p1 250kPa, T1 300oC At
section2 Ma2 2.6 Then find T2,
p2 and po2
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Example Air flows isentropically in a duct. At
section 1 Ma1 0.5, p1 250kPa, T1 300oC
At section2 Ma2 2.6 Then find
T2, p2 and p02
Equations T2 T02/(1 k-1/2M22)
T01 T02 T1 T01/(1 k-1/2M12)
p2 p02/(1 k-1/2M22)k/(k-1)
p01 p02 p1 p01/(1 k-1/2M22)k/(k-1)
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Know Ma10.5, p1250 kPa, T1300oC, Ma22.6
T2 T02/(1 k-1/2M22) T01 T02 p2 p02/(1
k-1/2M22) p01 p02
Then find p02, p2, T2
p01 p02 p1(1 ((k-1)/2)Ma12)k/(k-1) 2501
0.2(0.5)23.5 297 kPa p2 p02/(1
((k-1)/2)Ma21)k/(k-1) 297/1 0.2(2.6)23.5
14.9 kPa T01 T02 T1(1 k-1/2M12)
5731 0.2((0.5)2 602oK T2 T02/(1
k-1/2M22) 602/1 0.2((2.6)2 256oK
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CAN NOT GET AREA INFO FROM STAGNATION REFERENCE
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Calculating Area
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Missing relation for Area since stagnation state
does not provide area information. So to get
Area information use critical conditions as
reference.
mathematically the stagnation area is infinity
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If M 1 the critical state p, T, ?.
Critical conditions related to stagnation
Local conditions related to stagnation
EQ. 11.17 c kRT1/2 kRT1/2
isentropic, steady, ideal gas
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Want to Relate Area to Mach Number and Critical
Area
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Want to Relate Area to Mach Number and Critical
Area
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EQ.11.17 c kRT1/2
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EQ. 12.7c
EQ. 11.21b
EQ. 12.7b
EQ. 12.7b
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AxAy Axy A 1 (k-1)M/21/(k-1) 1/(k-1) ½
2/2(k-1) (k-1)/2(k-1) (k1)/(2(k-1))
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EQs. 12.7a,b,c,d
Provide property relations in terms of local
Mach numbers, critical conditions, and
stagnation conditions. NOT COUPLED LIKE Eqs.
12.2, so easier to use.
isentropic, ideal gas, constant specific heats
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A
Note two different M s for same
A/A SUPERSONIC TUNNELS Not built this way
because of separation
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• For accelerating flows, favorable pressure
  gradient,
• the idealization of isentropic flow is generally
  a
• realistic model of the actual flow behavior.
• For decelerating flows (unfavorable pressure
• gradient) real fluid tend to exhibit
  nonisentropic
• behavior such as boundary layer separation, and
• formation of shock waves.

accelerating
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Sneeze Breath
(180 mph)
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Isentropic Flow In A Converging Nozzle
What happens to p(x) as lower pb?
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As lower pb by vacuum pump, how does p(x)/po
change?
IDEAL GAS, ISENTROPIC, QUASI-1-D, FBX0, STEADY,
only PRESSURE WORK, Ignore gravity effects, cp
cv are constant
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M lt 1
throat
dp/? - dVx2/2
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i valve closed, no flow. Stagnation conditions
everywhere e.g. p p0 everywhere. ii pe pb
lt p0 iii pe pb ltlt p0
iv pe p pb ltltlt p0 choked
flow v pe p pb ltltlt p0 Shock After
Throat Not Isentropic
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(No Transcript)
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Isentropic Flow In A Converging Nozzle
Given stagnation conditions and throat area
What is mass flow when choked?
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dm/dtchoked ? ?eVe Ae ?e Ve Ae
?o/? 1 (k-1)M2/2 1/(k-1) ?o/? 1
(k-1)/2 1/(k-1) (k1)/21/(k-1) ? ?e
?o 2/(k1)1/(k-1)
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dm/dtchoked ? ?eVe Ae ?e Ve Ae
Ve c kRT1/2 To/T 1 (k-1)M2/2 To/T
(k1)/2 T To2/(k1) Ve kR2To/(k1)1/2
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dm/dt ? ?eVeAe
Ve c 2kRTo/(k1)1/2 ? ?e ?o
2/(k1)1/(k-1)
2/(k1)1/2 2/(k1)1/(k-1) 2/(k
1)k-12/(2k-1) 2/(k1)k1/(2k-1)
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dm/dt ? ?eVeAe
Ve c 2kRTo/(k1)1/2 ? ?e ?o
2/(k1)1/(k-1) ?o po/RTo
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For air in kg/s dm/dtchoked 0.04
Aepo/To1/2 For air in lbs/s dm/dtchoked 76.6
Aepo/To1/2
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GIVE ME TWO REASONS WHY WE CAN NOT INCREASE THE
MASS FLOW RATE ABOVE
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GIVE ME TWO REASONS WHY WE CAN NOT INCREASE THE
MASS FLOW RATE ABOVE

• Downstream conditions-influences propagate at the
• speed of sound so cant move upstream past
  throat
• (2) Cant exceed sonic conditions at throat cause
  that
• would require the flow to become sonic upstream
  in
• the converging section

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Schematic Ts diagram for choked flow through a
converging nozzle.
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Affect of Area Change in a Converging-Diverging
Nozzle
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CONVERGING
CONVERGING-DIVERGING
I S E N T R O P I C
NONISENTROPIC
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Isentropic subsonic nozzle flow
Isentropic supersonic nozzle flow
Infinite number of solutions
Single solutions
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??? IF FRICTIONLESS WOULD WE FIND COMPLETE
PRESSURE RECOVERY - ???
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!!! NO ASSUMING FRICTIONLESS COMPRESSIBILITY
AFFECT - !!!
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i if flow is slow enough, Vlt0.3M, then ?
incompressible so B. Eq. ? holds. ii still
subsonic but compressibility effects more
apparent, B.Eq. Not Good. iii highest pb where
flow is choked Mt1 i, ii, iii (Mlt1) and
iv (Mgt1) are all isentropic flows
But replace Ae by At.
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Note diverging section decelerates subsonic
flow, but accelerates supersonic flow. What is
does for sonic flow depends on downstream
pressure, pb. There are two Mach numbers, one lt
1, one gt1 for a given C-D nozzle which still
supports isentropic flow when M1 at throat.
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Flow can not contract isentropically to pb so
contracts through a shock. Flows are referred to
as being overexpanded because
pressure pe lt pb.
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When M gt1 and isentropic then flow is said to be
at design conditions.
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Lowering pb further will have no effect
upstream, where flow remains isentropic. Flow
will go through 3-D irreversible expansion. Flow
is called underexpanded, since
additional expansion takes place outside the
nozzle.
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Affect of Area Change
Example
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Consider a rocket engine
Me ? Ve ? Ae ?
hydrogen
3517 K
0.4 m
25 atm
1.174 x 10-2 atm
oxygen
dm/dt ?

• 1.22 Molecular Weight is 16, R 8314/16
  519.6 J/kg
• Calorically perfect gas, isentropic flow

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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
Me ?
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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
po/pe 1 (k-1)/2Me2k/(k-1)
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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
Ve ?
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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
Ve Mece Me(?RTe)1/2 To/Te 1
(k-1)/2Me2 Ve Me(?RTe)1/2
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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
Ae ?
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Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
Ae/A 1/Me(1 (k1)/2Me2/(k1)/2(k1)/
2(k-1)
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?
Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
dm/dt ?
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?eAeVe constant
Given To, p0, A, pe
Find Me ?, Ve ?, Ae ?, dm/dt ?
dm/dt ?eAeVe po ?oRTo ?o/?e 1
(k-1)/(2Me2)1/(k-1) dm/dt ?eAeVe
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With Numbers
hydrogen
pe 0.01174 atm
To 3517K po 25 atm MW 16 ? 1.22
Me? Ve ? Ae? dm/dt?
At 0.4 m2
oxygen
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Know To po At pe
Me? Ve ? Ae? dm/dt?
po/pe (1 ½ k-1Me2) k/(k-1) (Eq.
11.20a) Me 5.21
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Know To po At pe
Me? Ve ? Ae? dm/dt?
Ve Mec Me(kRTe)1/2 R 8314/16 519.6
J/(kg-K) To/Te 1 ½ (k-1)Me2 po/pe(k-1)/k
(Eq. 11.20b) Te 885.3 K Ve Me(kRTe)1/2
1417 m/s
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Know To po At pe
Me? Ve ? Ae? dm/dt?
dm/dt ? A V V c (kRT) T/To
2/(k1) (Eq. 11.21b) T 3168K V (kRT)1/2
1417 m/s dm/dt ? A V 487.4 kg/s
dm/dt ? A V ?/ ?o (2/k1)1/(k-1)
(Eq. 11.20c) 0.622 ?o po/(RTo) 1.382
kg/m3 ? 0.860 kg/m3
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Know To po At pe
Me? Ve ? Ae? dm/dt?
dm/dt ?e Ae Ve ?e pe/(RTe) 0.00258
kg/m3 dm/dt ?e Ae Ve Ae (dm/dt)/(?eVe)
487.4/(0.00258)(3909) 48.5 m2
Ae/A 1/Me x 1 Me2(k1)/2)(k1)/2(k-1)
(k1)/2(k1)/2(k-1) Eq. 12.7d Ae 48.5 m2
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the end class 12
It required an unhesitating boldness to undertake
such a venture . an almost exuberant
enthusiasmbut most of all a completely
unprejudiced imagination in departing so
drastically from the known way.