Steady quasi-one-dimensional isentropic flow of a perfect gas

The flow direction in most of the duct will remain approximately the same (surely, near the walls, the flow may be moving at a substantial angle to the axis, but we can deal with the resulting effect in detail later). So, we specialize the conservation equations to a case where the cross-sectional area changes with distance downstream, but the flow direction is approximately along the axis. This is "quasi-1-dimensional".

Likewise, the specific heat of the gas may change a little as the temperature changes, but we will neglect these changes in this analysis. Once the essential features of the flow behavior are understood from the following analysis, we can incorporate the effects of varying specific heat, changing gas composition etc. in more sophisticated calculations, late (in a subject called High Temperature Gas Dynamics).

Here we will assume that the range of variation of temperature is not so large as to cause substantial changes in specific heat of the gas, or of the gas composition.

Assumed: No body forces; no viscous stresses.

Start with the integral form of the equations:

.............................(1)

.....................(2)

...............(3)

.................................(4)

...............................(5)

..................................................(6)

Steady:

1-dimensional: u>> v,w

The integral over the control surface reduces to just (downstream minus upstream) because there is no flow through the sides.

Adiabatic: No heat transfer, so that q = 0.

Isentropic: Reversible + adiabatic.

No body forces: 

No viscous stresses: 

If the only work involved is the work of volume change against pressure,

Perfect gas: Gas constant R, and specific heats are constant over the range of properties involved in the problem.
 

The conservation equations reduce, under these conditions, to:

Conservation of mass:

    1-dimensional:

(Integrals over the sides = 0 since  for these sides.

Also, since it is 1-d, velocity and density are constant over the area A1 and the area A2.

So, 

Thus, the 1-d, steady continuity equation is:

.

Or,

Since  and  are both zero, the above means that 

, or, 

Momentum Equation for Steady Quasi-1-D Flow

Neglecting body forces and viscous forces,  is simply the force due to pressure:

Thus, momentum equation is:

or

.

In differential form, this can be simplified to

Energy Equation for Steady, Quasi-1-D, flow :

The work term can be split into two: the work of volume expansion against pressure, the and external work. The vector theorems can be used to write the pressure work term as a surface integral.

Thus

Introducing the enthalpy per unit mass,

For 1-D flow, the integrals are simplified and evaluated at the inflow and outflow boundaries as before.

Note that

, the mass flow rate. So, per unit mass flow rate, we get:

If there is no heat transfer (i.e., the flow is adiabatic), and there is no external work, we get:

, the stagnation enthalpy per unit mass.

NOTE:

1. In steady flow, if there is no heat transfer, and no external work, the stagnation enthalpy remains constant along a streamline.

If heat is added, the stagnation enthalpy goes up.

If work is taken out, the stagnation enthalpy goes down.

In differential terms, for the case of adiabatic flow with no external work,

 Speed of Sound

Momentum

, or, 

, or, 

i.e.,  . Thus,  . Thus,

For an isentropic process,

, or, 

So,  . Therefore,

, and 

Stagnation Properties

Momentum: 

, or 

Now,  . So, 

, or,

, or, 

Bernoulli Equation for Compressible Flow

Binomial Theorem:

If M is <<1, the higher-order terms are negligible. Thus,

, or,
 

, or,  . This is the familiar Bernoulli equation for low-Mach-number flow.

For compressible flows in general, just use the isentropic relation

Sonic Conditions

and so on.

For air,  ,

;

;

So, if you have 1/0.5283 atmospheres pressure in a container, and open a valve to let the flow out, the flow speed should reach Mach 1. 0.

Assumed model:

Steady, quasi-one-dimensional. There are gradual variations in the geometry, so that the flow near the nozzle walls is not strictly along the x-direction. However, the flow angularity is very small. The variation in properties can be calculated asuming that the properties are constant in each cross-section. The cross-section area, A, is a function of x alone. Thus, all properties are functions of x alone.

A= A(x); u = u(x); T=T(x), p= p(x) etc.

Continuity:

, the mass flow rate, is constant.

.......................................(1)

Momentum: (no friction; differential form of the Euler equation)

. Hence,  ...................................(2)

Using (2) in (1),  .

Isentropic process:  . Thus,

. Also,

Consider several cases.

Case 1: M<1

dA, dp have the same sign. Thus, as A increases, p increases.

dA , du have opposite signs. Thus as A increases, u decreases.

Diverging duct in subsonic flow: pressure increases, speed decreases.

Converging duct in subsonic flow: pressure decreases, speed increases.

Case 2: M>1

dA, dp have opposite signs. Thus as A increases, p decreases.

dA, du have the same sign. Thus as A increases, u increases.

Diverging duct in supersonic flow: pressure decreases, speed increases.

Case 3: M = 1

dA/dx is 0. Thus we have either a maximum or minimum of area. The maximum area case is not of much interest, since there is no way to reach Mach 1 at this point, with flow from either direction.

So the case of interest is where the area becomes a minimum: a "throat".

From mass conservation,  where the * denotes conditions at Mach 1

So, 

Substitute into A/A*:

. Thus, for a given isentropic flow, i.e., a flow with mass flow rate, stagnation temperature and stagnation pressure all fixed, there are two solutions for a given value of A/A*: One solution is subssonic, the other is supesonic.

Mass Flow Rate Through a Nozzle

For given stagnation conditions  ;  are fixed.

For a given throat area, stagnation pressure and stagnation temperature, the maximum mass flow rate is the value where the Mach number at the throat reaches 1.0. This is called the "choked mass flow rate." To increase the mass flow rate, we have to increase the tagnation presure, decrease the stagnation temperature, or increase the throat area.

, or

. For M=1,  , 

Example:

1. Supersonic nozzle; reservoir at one end , throat in the middle, expanding section downstream of the throat.

(Note: What is a "reservoir"? It is a supply of stagnant fluid, so large enough that the conditions in the reservoir remain essentially unchanged for a long duration as the flow goes out through the nozzle, so that we can assume that its steady flow. Also, the area available to the flow inside the reservoir is so large that the flow velocity is essentially zero inside the reservoir, although there is flow going out through the nozzle.)

What is the ratio of A2 to A1 if M1 = 2, and M2 = 3?

2. Supersonic Wind Tunnel

Find the throat area needed to produce M=3 in a test section of cross-sectional area 0.2m2. Assume that 

3. Nozzle, with inlet area of 0.5 m2. Inlet Mach number is 0.2. How far can the area be reduced without changing the mass flow rate in the nozzle? (assuming that there is no friction).

The flow is choked when the area A reaches the sonic value A*.

4. Given the convergent-divergent nozzle shown, find the limits of possible M1 for this geometry.

The limits occur when the flow is choked, i.e., the Mach number is 1.0 at the throat.

. This gives the following results:

Maximum Mach number of 0.239, subsonic result.

Maximum Mach number of 2.44, supersonic result.

So, when the flow reaces sonic speed at the throat, one of two things can happen in the diverging area downstream of the throat:

1) go supersonic and accelerate.

2) go subsonic and decelerate.

The actual result depends on the conditions downstream. Before discussing this further, we need an understanding of the normal shock, discussed in the next section.