ch.12: compressible flow

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

Compressible flow is a fun subject.

John Anderson

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, …)

Affect of Area Change

Assumptions = ?

ASSUMPTIONSALWAYS ~ • Steady Flow • Ideal Gas• Ignore Body Forces• only pressure work

(no shaft, shear or other work)• Constant specific heats• “One – Dimensional”

MOSTLY ~• ISENTROPIC

V(s) V(x,y)

“quasi-one-dimensional”

“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)


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)


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


Eqs. of

State

(steady)

Want to find qualitative relationships for: dT, dV, dA, d ISENTROPIC

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 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

Affect of Change in Velocity on Pressure

Isentropic, steady, no body forces, quasi-one-dimensional

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


dp/+d{Vx2/2}=0

Velocity Increases, then Pressure DecreasesVelocity Decreases, then Pressure IncreasesNOT DEPENDENT ON MACH NUMBER


Affect of Change in Area on Pressure and Velocity


EQ. 11.19b EQ.12.1a

isentropic, steady

{d(AV) + dA(V) +dV(A)}/{AV} = 0

steady

isentropic, steady

EQ. 12.5

EQ.12.6isentropic, steady

Affect of Change in Area on Velocity


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


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 dp > 0 or dp < 0

dVx > 0 or dVx < 0


M >1 dp > 0 or dp < 0

dVx > 0 or dVx < 0

isentropic, steady,~1-D

M<1 dVx > 0 or dVx < 0

dp > 0 or dp < 0

isentropic, steady,~1-D

M>1 dVx > 0 or dVx < 0

dp > 0 or dp < 0


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)


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 Area on Density



-(dA/A)/(1-M2) = -dA/A - d/

d/ = (dA/A)[1/(1-M2) - 1]

d/ = (dA/A)[M2/(1-M2)]

d/ = (dA/A)[M2/(1-M2)]

d =?

d/ = (dA/A)[M2/(1-M2)]

d/ < 0 d/ > 0

d/ > 0d/ < 0

WHAT HAPPENS AT M = 1 ?

? dp/(V2) = (dA/A)/(1-M2); dV/V = -(dA/A)/(1-M2);

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)]

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 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?

What happens to dp and d across the throat in a supersonic diffuser 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?

Same shape, but in one caseaccelerating flow, and in the other decelerating 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

Sketch a Supersonic Wind Tunnel

SUPERSONIC SONIC

Sneeze Breath

(180 mph)

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)


Cons. of massCons. of momentum

Cons. of energy

2nd Law of Thermodynamics.

Ideal gas



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)

Constant Area - Isentropic

Example ~ using 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

CAN NOT GET AREA INFOFROM STAGNATION REFERENCE

Calculating Area

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

Want to Relate Area to Mach Number and Critical Area

EQ.11.17 c = [kRT]1/2

EQ. 12.7b

EQ. 11.21b

EQ. 12.7c

EQ. 12.7b

AxAy = Ax+y

A = [1 + (k-1)M/2]1/(k-1)

1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)= (k+1)/(2(k-1))

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.

Sneeze Breath

(180 mph)

Isentropic Flow In A Converging Nozzle

What happens to p(x) as lower pb?

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

M < 1

throat

dp/ = - d{Vx2/2}

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])

dm/dt = ? = eVeAe

Ve* = c* = {2kRTo/(k+1)}1/2

* = e* = o [2/(k+1)]1/(k-1)

o = po/RTo

For air in kg/s:dm/dtchoked = 0.04 Aepo/To

1/2

For air in lbs/s:dm/dtchoked = 76.6 Aepo/To

1/2

GIVE ME TWO REASONS WHY WE CAN NOT INCREASE THE

MASS FLOW RATE ABOVE:

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

Schematic Ts diagram for choked flow through a

converging nozzle.

Affect of Area Change in a Converging-Diverging Nozzle

CONVERGINGCONVERGING-DIVERGING

I

S

E

N

T

R

O

P

I

C

NONISENTROPIC

Isentropic supersonic nozzle flow Isentropic subsonic nozzle flow

Infinite number of solutionsSingle solutions

??? – IF FRICTIONLESS WOULD WE FIND COMPLETE PRESSURE RECOVERY - ???

!!! – NO – ASSUMING FRICTIONLESS COMPRESSIBILITY AFFECT - !!!

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.

When M >1 and isentropic thenflow is said to be at:

design conditions.

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.

Affect of Area Change

Example

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

Me = ?

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


po/pe = {1 + [(k-1)/2]Me2}k/(k-1)

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


Ve = ?

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


Ve = Mece = Me(RTe)1/2

To/Te = 1 + [(k-1)/2]Me2

Ve = Me(RTe)1/2

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


Ae = ?

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


Ae/A* = [1/Me][(1 + {(k–1)/2}Me2/{(k+1)/2}](k+1)/[2(k-1)]

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


dm/dt = ?

?

Find: Me = ?, Ve = ?, Ae = ?, dm/dt = ?


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=?

po/pe = (1 + ½ [k-1]Me2) k/(k-1)

(Eq. 11.20a)

Me = 5.21


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


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

the end – class 12

It required an unhesitating boldness to undertake such

a venture …. an almost exuberant enthusiasm…but

most of all a completely unprejudiced imagination in

departing so drastically from the known way.

ch.12: compressible flow

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