ch.12: compressible flow
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CH.12: COMPRESSIBLE FLOW
It required an unhesitating boldness to undertake sucha venture …. an almost exuberant enthusiasm…but most of all a completely unprejudiced imagination in departing so drastically from the known way. J. Van Lonkhuyzen, 1951, discussing designing Bell XS-1
Ch.12 - WHAT CAUSES FLUID PROPERTIES TO CHANGE IN A 1-D
COMPRESSIBLE FLOW FLOW?
(note if isentropic stagnation properties do not change)
Ch.12 - COMPRESSIBLE FLOW Flow can be affected by:
area change, shock, friction, heat transfer
Area1
Area2
Q
Q
friction
shock
shock
What can affect fluid properties? Changing area, normal shock, heating, cooling, friction.
- need 7 equations -
One-Dimensional Compressible Flow
dQ/dt
heat/cool
Rx
Surface forcefrom frictionand pressure
P1P2
(+ s1, h1,V1,…)(+ s2, h2, V2, …)
ASSUMPTIONSALWAYS ~ • Steady Flow • Ideal Gas• Ignore Body Forces• only pressure work
(no shaft, shear or other work)• Constant specific heats• “One – Dimensional”
MOSTLY ~• ISENTROPIC
“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 ofconservation laws that are applied to an approximate physical model
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)
One-Dimensional Compressible Flow
f(x1)f(x2)
+ s1, h1 +s, h2
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)
One-Dimensional Compressible Flow
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
Eqs. of
State
(steady)
Affect of Change in Velocity on Temperature
Isentropic, steady, cp&cv constant, no body forces, quasi-one-dimensional, ideal gas, only pressure work
For isentropic flow if the fluid accelerates what happens to the temperature?
if V2>V1
then h2<h1
if h2<h1
then T2<T1
Temperature Decreases !!!
Affect of Change in Velocity on Temperature
Velocity Increases, then Temperature DecreasesVelocity Decreases, then Temperature Increases
NOT DEPENDENT ON MACH NUMBER
Isentropic, steady, cp&cv constant, no body forces, quasi-one-dimensional, ideal gas, only pressure work
1-D, steady, isentropic flow with area change; no friction, no heat exchange or change in potential energy or entropy
Rx
Rx = pressure force along walls, no frictionp + ½ (dp/dx)dx
½ dA
FSx = (p + dp/2)dA + pA – (p+dp)(A + dA)
= pdA + dAdp/2 + pA – pA – pdA– dpA - dpdA
(dm/dt)(Vx+dVx) - (dm/dt)Vx == (Vx + d Vx) {VxA}- Vx{VxA} = Vx VxA + dVx VxA – VxVxA
-dpA = VxdVxA or dp/+d{Vx2/2}=0
Affect of Change in Velocity on Pressure
dp/+d{Vx2/2}=0
Velocity Increases, then Pressure DecreasesVelocity Decreases, then Pressure IncreasesNOT DEPENDENT ON MACH NUMBER
Isentropic, steady, no body forces, quasi-one-dimensional
Affect of Change in Area on Pressure and Velocity
Isentropic, steady, no body forces, quasi-one-dimensional
Affect of Change in Area on Velocity
Isentropic, steady, no body forces, quasi-one-dimensional
dV/V = - (dA/A)/[1-M2]M <1
Velocity and Area change oppositelyM > 1
Velocity and Area change the same
Affect of Change in Area on Pressure
Isentropic, steady, no body forces, quasi-one-dimensional
dp/(V2) = (dA/A)/[1-M2]M <1
Pressure and Area change the sameM > 1
Pressure and Area change oppositely
isentropic, steady, ~1-D
M<1
If M < 1 then [ 1 – M2] is +, then dA and dP are same sign;
and dA and dV are opposite signqualitatively like incompressible flow
Subsonic Nozzle Subsonic Diffuser(dp and dV are opposite sign)
isentropic, steady, ~1-D
M>1
If M > 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
Supersonic Nozzle Supersonic Diffuser(dp and dV are opposite sign)
If is constant then dA and dV must be opposite signs,
but for compressible flows all bets are off,e.g. for M>1 both dV and dA
can have the same sign
And this is the reason!!!!!!!
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)]
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)]
dp/ = - d{Vx2/2}
d/ = (dA/A)[M2/(1-M2)]
Vx continues to increase so p continues to decrease
(dA/A)[M2/(1-M2)] is always negative so continues to decrease
What happens to dp and d across the throat in a supersonic nozzle with steady, isentropic flow?
dp/ = - d{Vx2/2}
d/ = (dA/A)[M2/(1-M2)]
Vx continues to decrease so p continues to increase
(dA/A)[M2/(1-M2)] is always positive so continues to increase
What happens to dp and d across the throat in a supersonic diffuser with steady, isentropic flow?
De Laval designed a turbine (1888)whose wheel was turned by jets
of steam flowing through a contraction - expansion section.
Goddard realizes (1917) that De Laval nozzle could be used
for rocket
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)
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)
One-Dimensional Compressible Flow
Cons. of massCons. of momentum
Cons. of energy
2nd Law of Thermodynamics.
Ideal gas
Ideal gas & constant cv, cp
Ideal gas & constant cv, cp
Have found qualitative relationships for dT, dV, dA, dNow want to find quantitative relationships.
IDEAL GAS, ISENTROPIC, QUASI-1-D, FBX=0, STEADY, only PRESSURE WORK
Coupled, nonlinear equations hard to solve.
Much easier to express flow variables in terms of stagnation quantities and local Mach number.
isentropic, steady, ideal gas, constant cp, cv
EQ. 11.19b
+
P / k = constant
EQ. 11.12c
Eqs. 11.20a,b,c = Eqs. 12.7a,b,c
(po, To refer to stagnation properties)
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
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]/2}M2
2)
T01 = T02; T1 = T01/(1 + {[k-1]/2}M12)
p2 = p02/(1 + {[k-1]/2}M22)k/(k-1)
p01 = p02; p1 = p01/(1 + {[k-1]/2}M22)k/(k-1)
T2 = T02/(1 + {[k-1]/2}M22)
T01 = T02
p2 = p02/(1 + {[k-1]/2}M22)
p01 = p02
p01 = p02 = p1(1 + ((k-1)/2)Ma12)k/(k-1) = 250[1 + 0.2(0.5)2]3.5
~ 297 kPap2 = p02/(1 + ((k-1)/2)Ma2
1)k/(k-1) = 297/[1 + 0.2(2.6)2]3.5 ~14.9 kPaT01 = T02
= T1(1 + {[k-1]/2}M12) = 573[1 +0.2((0.5)2]
~ 602oKT2
= T02/(1 + {[k-1]/2}M22) = 602/[1 +0.2((2.6)2]
~ 256oK
Then find:p02, p2, T2
Know:Ma1=0.5, p1=250 kPa, T1=300oC, Ma2=2.6
Missing relation for Area since stagnation state does not provide area information*.
So to get Area informationuse critical conditions as reference.
*mathematically the stagnation area is infinity
If M = 1 the critical state; p*, T*, *….
EQ. 11.17 c = [kRT]1/2 = [kRT*]1/2
isentropic, steady, ideal gas
Local conditions related to stagnation
Critical conditions related to stagnation
isentropic, ideal gas, constant specific heats
EQs. 12.7a,b,c,d
Provide property relations in terms of local Mach numbers, critical conditions,and stagnation conditions.
NOT COUPLED LIKEEqs. 12.2, so easier to use.
Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Area
Rati
o A/A
*
* *
Note two different M #’s for same A/A*
SUPERSONIC TUNNELS
Not built this way because of separation
A*
Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Are
a R
ati
o A
/A*
accelerating
• 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.
As lower pb by vacuum pump, how does p(x)/po change?
IDEAL GAS, ISENTROPIC, QUASI-1-D,
FBX=0, STEADY, only PRESSURE WORK,
Ignore gravity effects, cp & cv are constant
i: valve closed, no flow. Stagnation conditions everywhere e.g. p = p0 everywhere. ii: = pe= pb < p0
iii: = pe= pb << p0iv: = pe = p* = pb <<< p0
choked flowv: = pe = p* = pb <<< p0
Shock After Throat Not Isentropic
Isentropic Flow In A Converging Nozzle
What is mass flow when choked?
Given stagnation conditions and throat area
dm/dtchoked = ? = eVe Ae = e* Ve
* Ae
o/ = [1 + (k-1)M2/2] 1/(k-1)
o/* = [1 + (k-1)/2] 1/(k-1) = [(k+1)/2]1/(k-1)
* = e* = o [2/(k+1)]1/(k-1)
dm/dtchoked = ? = eVe Ae = e* Ve
* Ae
Ve* = c* = {kRT*}1/2
To/T = 1 + (k-1)M2/2To/T* = (k+1)/2T* = To2/(k+1)
Ve* = {kR2To/(k+1)}1/2
dm/dt = ? = eVeAe
Ve* = c* = {2kRTo/(k+1)}1/2
* = e* = o [2/(k+1)]1/(k-1)
[2/(k+1)]1/2 [2/(k+1)]1/(k-1)
= [2/(k + 1)][k-1+2]/(2[k-1)] = [2/(k+1)][k+1]/(2[k-1])
GIVE ME TWO REASONS WHY WE CAN NOT INCREASE THE MASS FLOW RATE ABOVE:
(1)Downstream conditions-influences propagate at the speed of sound so can’t move upstream past throat(2) Can’t exceed sonic conditions at throat cause that would require the flow to become sonic upstream in the converging section
Isentropic supersonic nozzle flow Isentropic subsonic nozzle flow
Infinite number of solutionsSingle solutions
i – if flow is slow enough, V<0.3M, 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; Mt=1
i, ii, iii (M<1) and iv (M>1) are all isentropic flows
But replace Ae by At.
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 < 1, one >1 for a given C-D nozzle which still supports isentropic flow when M=1 at throat.
* *
Flow can not contract isentropically to pb so contracts through a shock.
Flows are referred to as being overexpanded because pressure
pe < pb.
Lowering pb further will have noeffect upstream, where flow remainsisentropic. Flow will go through 3-Dirreversible expansion. Flow is called underexpanded, since additional expansion takesplace outside the nozzle.
Consider a rocket engine:
hydrogen
oxygen
3517 K
25 atm
= 1.22; Molecular Weight = is 16, R = 8314/16 = 519.6 J/kgCalorically perfect gas, isentropic flow
1.174 x 10-2 atm
0.4 m
Me = ?Ve = ?Ae = ?
dm/dt = ?
Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?
Given: To, p0, A*, pe
Ve = Mece = Me(RTe)1/2
To/Te = 1 + [(k-1)/2]Me2
Ve = Me(RTe)1/2
Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?
Given: To, p0, A*, pe
Ae/A* = [1/Me][(1 + {(k–1)/2}Me2/{(k+1)/2}](k+1)/[2(k-1)]
Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?
Given: To, p0, A*, pe
dm/dt = eAeVe
po = oRTo
o/e = [1 + (k-1)/(2Me2)]1/(k-1)
dm/dt = eAeVe
eAeVe = constant
hydrogen
oxygen
To =3517K po= 25 atmMW = 16 = 1.22
pe = 0.01174 atm
At= 0.4 m2
Me=?Ve =? Ae=?
dm/dt=?
With Numbers
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
Know: To; po; At; pe Me=?Ve =? Ae=?
dm/dt=?
dm/dt = * A* V*
*/ o = (2/[k+1])1/(k-1) (Eq. 11.20c)
= 0.622o = po/(RTo) = 1.382 kg/m3
* = 0.860 kg/m3
dm/dt = * A* V*
V* = c = (kRT*)T*/To = 2/(k+1)
(Eq. 11.21b)T* = 3168K
V* = (kRT*)1/2 = 1417 m/s
dm/dt = * A* V* = 487.4 kg/s
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(k–1)/2)](k+1)/[2(k-1)]
[(k+1)/2](k+1)/[2(k-1)]
Eq. 12.7d
Ae = 48.5 m2
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