quasi one dimensional flow

8/3/2019 Quasi One Dimensional Flow

1/17

Quasi One-Dimensional Flow

I. Flow Assumptions and Integral Balance Equations

Consider the flow situation shown below

1 2

dx

x

We expect that given the boundary geometry the flow will be three-dimensional. The

Quasi one-dimensional assumption is that given the cross sectional area as a function of

the axial coordinate, A A( x ), all flow variables will be functions of the axialcoordinate and not coordinates in other directions, i.e.,

u u( x )

P P( x )

T T( x )

( x )

(1.1)

This approximation leads to results that have inestimable importance in gas dynamics but

we must bear in mind that it is an approximation. The results we obtain here are similar to

true one-dimensional flow with friction or heat addition but in this case changes along theflow direction are driven by the duct area change.

In what follows we will use the control volume integral balance formulation to derivedifferential balance equations. For this flow we will consider an inviscid and

nonconducting fluid and neglect body forces, wall friction forces and any applied heat

flux through the duct surface. The control volume equations for this case are


2/17

Continuity:

CV CS

dV v ndS 0t

(1.2)Momentum:

CV CS CS

vdV v v n dS PndS 0t

(1.3)Energy:

CV CS CS

1 1( e v v )dV ( e v v )v ndS Pv ndS 0

t 2 2 (1.4)

II. Conservative Differential Balance EquationsThe volume shown in the above figure has infinitesimal length dx and we may write

2dV A x dx O dx (2.1)This allows us to approximate the volume integral in the integral balance equations. Themass balance may be approximated as

x dx x

CV CS

dV v ndS A( x )dx uA uA 0t t

(2.2)or

CV CS

dV v ndS A( x )dx uA dx 0 t t x

(2.3)This may be evaluated in the limit dx to obtain the conservative differential form of

continuity

A uA0

t x

(2.4)Following along in a similar manner the x-component of the integral balance ofmomentum may be approximated as


3/17

L

x

CV CS CS

2

x

dS

udV u v n dS Pn dS uA dxt t

u A AP dx Pn dSx

0

(2.5)

The last term is the pressure integral along the lateral surface. The lateral surface element,

LdS may be related to the cross sectional surface element by using the figure belowdA

n

dSL

dA

From this figure we see that

x L n dS dA (2.6)

And the integral on the lateral surface becomes

L

x

dS

Pn dS PdA (2.7)The momentum balance in the x-direction then becomes

x

CV CS CS

2

udV u v n dS Pn dS uA dxt t

u A AP dx PdAx

0

(2.8)

This may be evaluated in the limit dx to obtain the x component of the conservative

differential momentum equation

2 AuA u A AP P 0t x

x

(2.9)or

2 PuA u A A 0 t x x

(2.10)


4/17

Finally, the energy eqation is approximated as

CV CS CS

2 2

1 1( e v v )dV ( e v v )v ndS Pv ndS

t 2 2

1 1( e u )A dx ( e u )uA PuA dx 0 t 2 x 2

(2.11)

Following the same steps as were used previously we obtain the conservative differentialenergy equation

2 21 1( e u )A ( e u )uA PuA 0 t 2 x 2

(2.12)

And noting that Ph e we arrive at2 21 1( e u )A ( h u )uA 0

t 2 x 2

(2.13)Summarizing, the conservative differential balance equations for quasi one-dimensional

flow are

2

2 2

A uA0

t x

PuA u A A 0

t x x

1 1( e u )A ( h u )uA 0

t 2 x 2

(2.14)

In what follows we will derive simpler forms of these equations and we note that theconservative form of these equations would be appropriate to use if we were considering

simulating flows with shocks as the conservative form is closely related to the control

volume form which is valid for flows containing shocks.


5/17

III. Other Forms of the Differential Balance EquationsWe begin by factoring out a term proportional to the continuity equation in the

momentum equation

0

u u Pu A uA A uA A

t x t x x

0 (3.1)

or

u u P A u A

t x x 0 (3.2)

The area may be divided out and the resulting simplified momentum equation is

u u Pu

t x x 0 (3.3)

A similar operation on the energy equation may be performed

2 2

2

0

1 1( e u ) A uA A ( e u )

2 t x t 2

1 PuAuA ( e u ) 0 x 2 x

(3.4)

with the result

2 21 1 PuA A ( e u ) uA ( e u ) 0

t 2 x 2 x (3.5)

Now we multiply the momentum equation, equation (3.2), by the velocity to obtain

u u P A u u uA t x x

0 (3.6)or

2 2u u PuA uA2 2 A u P

t x x x

0 (3.7)


6/17

The represents the balance of mechanical energy and may be subtracted from the totalenergy equation to obtain the thermal energy equation

e e uA A uA P 0

t x x (3.8)

or De uA

A P Dt x

0 (3.9)We make the local equilibrium assumption (this assumption must be made for us to use

equations of state derived by considering equilibrium processes) so that we may apply theresult from classical thermodynamics

2

P de Td d 0 (3.10)

or

2

De D P DT

Dt Dt Dt

(3.11)

Substitution of this into equation (3.9) leaves us with

2

D P D uA A T P

Dt Dt x

0 (3.12)

The total derivative of density is substituted into the second term

2

D P A uA AT uA P

Dt t x x

0 (3.13)

And the continuity equation is used to eliminate the time derivative of density to obtain.

2

D P uA uA AT uA P

Dt x x x

0 (3.14)

or

D P uA uA AT P

Dt x x

0 (3.15)

This finally simplifies to

Du

Dt t x

0 (3.16)


7/17

and this indicates that the entropy of a fluid packet remains invariant. We would expectthis as all diffusion terms have been neglected. Further reductions may be implemented

by eliminating density derivatives using the equation of state for density.

P , (3.17)Note that entropy is chosen as one of the independent variables in the state equation for

density as we know the entropy is invariant. Total increments in density may be

calculated

P

0

P , P , d dP

P

d

(3.18)

or

2

1

d a dP (3.19)

wherea is the sound speed defined as

2

P ,1

a P

(3.20)

We use equation (3.19) to replace derivatives of density with derivatives of pressure. Thearea may be moved outside of the time derivative in the continuity equation and

derivatives of density may be isolated

uA A uA 0

t x x

(3.21)The density derivatives are now replaced with pressure derivatives

2 2

A P uA P uA0

a t a x x (3.22)

or

2 P P a uA

u t x A x

0 (3.23)

Summarizing, the reduces differential balance equations for quasi one-dimensional flow

are


8/17

2 P P a uAu 0

t x A x

u u Pu

t x x

u 0t x

0

(3.24)

Note that these equations are identical to the one-dimensional Euler equation except for

the last term in the continuity equation involving the area.

IV. The Area-Velocity Relations for Quasi One-Dimensional Flowand the De Laval Nozzle

Consider the differential balance equations derived in the previous section for the case of

steady flow (note that the energy equation simply tells us that the entropy remainsinvariant for a packet of fluid and need not be included here).

2 P a uAu

x A x

0 (4.1)

u Pu

x x

0

(4.2)

We introduce the Mach number

u

Ma (4.3)

And solve equations (4.1) and (4.2) for the spatial gradients of pressure and velocity

22 P a A1 Mx A

x

(4.4)

2 u u AM 1 x A x

(4.5)In what follows we consider a rightward directed flow, u>0 , and explore conditionswhere the area is either increasing or decreasing. Inspection of equations (4.4) and (4.5)

reveal the following cases


9/17

Diffusers (increasing pressure and decreasing velocity) occur when

u0

M 1 x

A P0 0

x x

(4.6)

Or

u0

M 1 x

A P0 0

x x

(4.7)

Nozzles (decreasing pressure and increasing velocity) occur when

u0

M 1 x

A P0 0

x x

(4.8)

u

0M 1 x

A P0 0

x x

(4.9)

Note that the cases given by equations (4.7) and (4.8) are counterintuitive given a

perspective based upon low speed flows. These cases indicate that for supersonic flows,

in order to increase the fluid speed we must increase the duct area and in order todecrease the fluid speed we must decrease the duct area.

Next we have to consider the behavior of the Mach number for nozzles and diffusers weconsider the case given by equation (4.8). We know that the pressure is decreasing and

becase the process is isentropic

PC (4.10)


10/17

which indicates that the density and hence the sound speed will decrease. With thevelocity increasing and the soundspeed decreasing the Mach number is increasing and we

may add to this case this condition

u 0M 1 x

M0

xA P

0 0x x

(4.11)

We will find that for the other three cases the behavior of the Mach number is the same,

i.e. the Mach number follows the velocity.

Inspection of equations (4.4) and (4.5) indicates thatM = 1 only ifA

0x

. This point inthe duct is known as the throat and will be denoted in the following discussion with asubscriptt. This result leads us to two conclusions

1. In a converging nozzle if A 0x

withM < 1 initially,Mcannot reach 12. In a converging nozzle if A 0

x

withM >1 initially,Mcannot reach 1The first conclusion indicates that if we wish to accelerate a fluid from subsonic to

supersonic velocity we need to use a duct that first converges to a throat with A 0x

and then diverges. This shape of duct is known as the De Laval nozzle and is

depicted below.

tM 1

tM 1

t

A0

x

M1


11/17

In practice the expansion of the gas can only go so far due to flow (boundary layer)separation and the attendant energy losses. The De Laval nozzle provides the means to

accelerate a subsonic flow to become a supersonic flow. The exit pressure of the nozzle is

how the flow characteristics are controlled. Consider the figure shown below. The two

exit pressures and correspond to the exit pressures for the subsonic exit

isentropic flow and the supersonic exit isentropic flow, respectively. In other words, if theexit pressure is set to

i _ subP i _supP

i _ subP the gas will reachM =1 at the throat and will then decelerate

until the exit is reached andM1 from the

throat to the exit. We will derive an analytic solution to each of these cases in the next

section.

i _supP

P

The exit pressures correspond to the most interesting case. The only

way that the flow physics can be resolved is if a shock exists between the throat and the

exit. The location of the shock is such that the isentropic flow after the shock will arrive

at the correct exit pressure.

i _ sup exit i _ sub P P P

If the exit pressure is less thani _ supP then the flow will be isentropic and supersonic after

the throat until the exit where an expansion must exist to bring the pressure to the low

exit pressure. Such a flow is termed under expanded. If the exit pressure is greater than

i _subP but less than rP (the reservoir pressure) then the flow will be isentropic but will

not reach Mach 1 at the throat. We will discuss the text figure 5.18 in class.

tx

i _ subP

i _ supP

shock 2P

shock 1P

M 1

x

tM 1

M 1

shock

rP


12/17

The other interesting feature of the De Laval nozzle is the chocked flow condition. Thiscondition may be observed in the previous figure where the exit pressure range

results inM =1 at the throat. The term chocked refers to the fact

that the throat conditions do not feel the effect of the pressure being lowered from

i _ sup exit i _ sub P P Pi _subP .

This choked condition is a manifestation of the fact that for supersonic flow

information cannot be propagated upstream and natures solution is to create a shockwave such that the pressure adjusts itself isentropically after the shock to meet the exit

pressure that is set. A more rigorous understanding of this phenomenon awaits the

development of the characteristic form of the differential balance equations.

V. Analytic Solution for a Calorically Perfect Gas in Steady-StateQuasi One-Dimensional Flow

We start with the continuity equation in steady form

uA0

x

(5.1)This equation tells us that

uA C (5.2)We fix the constant by setting it equal to the throat conditions

(5.3)* * *uA u A

where *A is the area of the throat and*

u a* as the throat is at Mach 1 thus

* * * *

o

*

o

A a a

A u u

(5.4)

Now since we have isotropic flow the density ratios are

1

12o 1

1 M2

(5.5)and (recall the starred state is at Mach 1)

1 1

1 1o

*

1 11

2 2

(5.6)


13/17

The energy equation may be used (in the text chapter 3 equation 3.37) to show that

2

*2

*2

1M

u 2M1a

1 M2

(5.7)

Equations (5.5), (5.6), and (5.7) may be substituted into equation (5.4) and after some

manipulation we arrive at the Area-Mach number relation

12

12

* 2

A 1 2 11 M

A M 1 2

(5.8)

This relation provides an interesting result in that

*AM f A (5.9)Suppose we know A A( x ). The analytic solution for isentropic flow in the De Laval

nozzle may be found by first solving the Area-Mach number relation for Mand then the

isentropic flow relations may be used to calculate all other variables. The Area-Mach

number relation provides two solutions forM, a subsonic and a supersonic solution. Thetable A.1 also includes the Area-Mach number relation as does the MATLAB file

isentrop.m. We will examine figures 5.13 and 5.14 in the text in class. Recall that the

isentropic flow relations are

2oT 1

1T 2

M (5.10)

12oP 11 M

P 2

(5.11)

1

12o 11 M2

(5.12)

This analytic solution and table A.1 (isentrop.m) will be the basis for problem solutions

for isentropic flow in nozzles. In the even a normal shock exists in the diverging sectionof the nozzle, this solution applies to either side of the shock and the normal shock

conditions (table A.2 and shock.m) apply through the shock. Two additional MATLAB


14/17

files are provided. The first, A_from_M.m calculates the area ratio given the Machnumber. The second, M_from_A.m calculates the Mach number given the area ratio.

A_from_M.m

function arear = A_from_M(mach)

gamma = 1.4;

term = 1 + .5*(gamma-1)*mach^2;

arear = 2/( gamma+1)*term;arear = arear^((gamma+1)/(gamma-1));arear = arear/( mach^2);arear = sqrt(arear);

return

M_from_A.m

function mach = M_from_A(arear, mguess)

% secant method to get mach # given area ratio

mo = mguess;dm = 1/1000.;delm = 1.0;

while abs(delm) > .0001,mr = mo+dm; ml = mo-dm;dadm = ( m_a(mr) - m_a(ml) )/(2*dm);delm = -(m_a(mo)-arear)/dadm;mo = mo + delm;

end

mach = mo;


15/17

VI. The Characteristic form of the Differential Balance Equations,Boundary Conditions and Choking

The differential balance equations we start with are

2 P P a uA

u t x A x

0 (6.1)

u u Pu

t x x 0 (6.2)

ut x

0 (6.3)

We seek to form linear combinations of the continuity and momentum equations

,equations (6.1) and (6.2), respectively, that result in the characteristic form of the

equations. First we add the continuity equation to the momentum equation multiplied bythe sound speed. The result is the first characteristic form

2u u P P ua a u a u a t x t x A

A

x

(6.4)

Next we subtract the continuity equation from the momentum equation multiplied by the

sound speed. The result is the second characteristic form

2u u P P ua a u a u a t x t x A

A

x

(6.5)The energy equation, equation (6.3), is already in characteristic form. In order to interpretthese equations we must introduce the concept of a derivative along a characteristic

curve. Consider the total derivative of a function ofx andt

f f df dt dx

t x

(6.6)Suppose we constrainx andt to fall along the curve

dx adt

(6.7)The total derivative along this curve is then

f f df dt adt

t x

(6.8)or


16/17

f f

df a d t x

t (6.9)

Thusf f

at x

representsdf

dtalong the curve

dxa

dt

For our case the characteristic form of the equations is

u u 1 P P uu a u a t x a t x

a A

A x

(6.10)

u u 1 P P uu a u a t x a t x A

a A

x

(6.11)

u

t x

0

(6.12)

and this set of equations may be interpreted as

du 1 dp ua A dx along u a

dt a dt A x dt

(6.13) du 1 dp ua A dx

along u a dt a dt A x dt

(6.14)

dxu 0 along u t x dt (6.15)

There are two important concepts associated with this characteristic from of the

equations:

1. The characteristic curves represent the directions along which information isbeing propagated.

2. The characteristic equations tell us what information is propagated along thecharacteristic curves.

These concepts will be useful to determine what the appropriate boundary conditions are

and will help us explain how the phenomena known as choking occurs. We consider theDe Laval nozzle introduced earlier where the throat is sonic and the diverging section is

supersonic at the exit.


17/17

tM 1

M>1

t

A0

x

At the left boundary we have two characteristic curves propagating information from the

outside. We replace the characteristic equations associated with these curves with

boundary conditions, say pressure and temperature are specified. We then must solve theu-a characteristic equation to obtain the third variable at the boundary. At the supersonic

exit all information is propagating from the interior and we cannot specify a boundary

condition. We must solve all three characteristic equations to find the three flow variables

at the exit.

Consider next the case shown above where the throat is choked and supersonic flow

exists in the diverging section until a shock develops in order to allow the exit pressure to

match the specified exit pressure. The characteristics are all directed downstream thus noinformation is propagated upstream. This is a unique aspect of supersonic flow. The

throat will remain at Mach 1 as the direction of information propagation is downstream

and any change in the exit pressure will not be felt at the throat. Since the flow issubsonic at the exit, we may replace the upstream directed characteristic equation with a

boundary condition specifying the exit pressure.

dxu a

dt

dxu

dt

dxu a

dt

dx

u adt

dx u adt

dx

udt

M

quasi one dimensional flow

Documents