Lecture-2

One-dimensional Compressible Fluid

Flow in Variable Area

Summary of Results(Cont..)

• In isoenergetic-isentropic flow, an increase in velocity always corresponds to aMach number increase and vice versa.

• Therefore, a converging passage always drive the Mach number towards theunity and diverging passage always drive the Mach number away from the unity.

• Suppose the Mach number is unity at some point in the flow. Since infinite

acceleration (dv → ∞) is not possible (Note:𝑑𝑉

𝑉= (

1

𝑀𝑎2−1)𝑑𝐴

𝐴) unless there is

shock wave, which is specifically excluded in assumptions, the bracketedequation then implies that dA = 0 when Ma = 1.

• Finally suppose that the compressible fluid is flowing in a passage that actuallyhas an area minimum (called Throat). Is it always possible to conclude that theMach number is unity at the throat?

• If dA is zero, then two options are possible: Ma = 1 or dV = 0, simply knowingdA = 0 does not allow us to determine which condition occurs.

Conclusion

The response of the flow to specific type of area change is exactly opposite for

subsonic and supersonic flow (as described in the last lecture in tabulated form).

Sonic flow (M = 1) can occur only at minimum area. Minimum areas occur at the

inlet of simply diverging passage, the outlet of simply converging passage, and

the throat of a converging and diverging passage.

It is possible, but not necessary, that the Mach number at the throat of a

converging-diverging passage be equal to 1. if the Mach number at the throat is

not 1, the velocity must pass through a maximum or minimum. If the throat Mach

number is 1, the fluid may either accelerate or de-accelerate down stream from

the throat.

Steady Isentropic flow of an ideal gas (do by yourself)

For flow of a compressible fluid in a small stream tube, we can make the following assumptions:

1. The flow is steady

2. There are no body forces(gravity, electromagnetic and others)

3. There are no shear stresses

4. There is no heat transfer

5. The stream tube does not pass through the a shock waves.*

The last three assumptions taken together imply that the flow is adiabatic and reversible, therefore the entropy of the

fluid is constant

Ŝ = constant, (Ŝ1 = Ŝ2)

For steady flow the energy equation is

𝑞 − 𝑤𝑠ℎ𝑎𝑓𝑡 − 𝑤𝑠ℎ𝑒𝑎𝑟 = ℎ2 +𝑉22

2− ℎ1 +

𝑉12

2

According to our assumptions, q and wshear are zero. But the shaft is not necessarily zero, energy equation becomes

−𝑤𝑠ℎ𝑎𝑓𝑡 = ℎ2 +𝑉22

2− ℎ1 +

𝑉12

2

When we use the definition of stagnation enthalpy,

−𝑤𝑠ℎ𝑎𝑓𝑡 = ℎ𝑜2 − ℎ𝑜1

If the fluid is an ideal gas,

− 𝑤𝑠ℎ𝑎𝑓𝑡 = 𝐶𝑝( 𝑇𝑜2 − 𝑇𝑜1 ) ……………………………………….(1)

*Assuming that there are no shocks If the flow is subsonic every where along the stream tube is not necessary, because the shocks are possible only if the

inlet flow is supersonic.

Doing work on the fluid increases its stagnation temperature, even if there is no heat transfer. Nowconsider the changes of stagnation pressure in isentropic flow. By definition of stagnation state

Ŝ = constant, (Ŝ1 = Ŝ2)

Thus 𝑝𝑜2

𝑝𝑜1= (

𝑇𝑜2

𝑇𝑜1)𝐶𝑝

𝑅 = (𝑇𝑜2

𝑇𝑜1)𝛾

𝛾−1

Using the equation 1, we get

𝑝𝑜2

𝑝𝑜1= (1 −

𝑤𝑠ℎ𝑎𝑓𝑡

𝐶𝑝𝑇𝑜1)𝛾

𝛾−1

A situation of considerable interest in compressible flow is the case with no shaft work. The followingstatement applies to this case:

• In the isentropic flow with no shaft work, all the stagnation properties are constant.

Stagnation properties are constant in isentropic flow only if there is no shaft work. This special case (wshaft

= 0) is so important in compressible flow that some engineers and authors use the phrase “isentropic flow”to mean “isentropic flow with no work”. Here use the phrase isoenergetic-isentropic flow to refer toisentropic flow with no work.

Equations relating property and velocity changes in isoenergetic-isentropic flow of an ideal gas can easilybe developed from stagnation property definitions and constancy of stagnation properties. take, e.g. thecalculation of change in temperature between two points in an isoenergetic-isentropic flow. Because To isconstant, we can write

• 𝑇𝑜1 = 𝑇1 +𝑉12

2𝐶𝑝= 𝑇2 +

𝑉12

2𝐶𝑝= 𝑇𝑜2

Using Mach number instead of velocity is sometimes convenient, above equation can be written in the

following form

𝑇2𝑇1=𝑇𝑜2 1 + (𝛾 − 1) 2 𝑀𝑎1

2

𝑇𝑜1 1 + (𝛾 − 1) 2 𝑀𝑎22

A third approach involves the use of the table relating stagnation property ratioand Mach number. We

calculate the temperature ratios by𝑇2𝑇1= 𝑇

𝑇𝑜𝑀2

𝑇

𝑇𝑜𝑀1

Where 𝑇

𝑇𝑜𝑀

is a numerical value read from the tabulated data at Mach Number M. In a fourth method, we make the

calculations in two steps:

𝑇𝑜2 = 𝑇𝑜1 =𝑇1

𝑇

𝑇𝑜𝑀1

and 𝑇2 = 𝑇𝑜2𝑇

𝑇𝑜𝑀2

Any of the above equation could be used to relate the temperature and velocities ( or Mach number) atpoint 1 and 2. which equation is most convenient depends on the exact information available. Similarlypressure and velocity ( of Mach number) in iso-energetic and isentropic flow can be related by any of thefollowing equations:

𝑝1 1 +𝑉12

2𝑇1𝐶𝑝

𝛾𝛾−1

= 𝑝2 1 +𝑉22

2𝑇2𝐶𝑝

𝛾𝛾−1

𝑝2𝑝1=1 + (𝛾 − 1) 2 𝑀𝑎1

2𝛾𝛾−1

1 + (𝛾 − 1) 2 𝑀𝑎22

𝛾𝛾−1

𝑝2𝑝1= 𝑝

𝑝𝑜𝑀2

𝑝

𝑝𝑜𝑀1

𝑝2 = 𝑝𝑜2𝑝

𝑝𝑜𝑀2 where

𝑝𝑜2 = 𝑝𝑜1 =𝑝1

𝑝𝑝𝑜𝑀1

Area-Mach Number RelationTo obtain an equation relating area and Mach number, use continuity equation for a control volume with inflow at plane

1 and outflow at plane 2. (following figure)

𝜌1𝑉1𝐴1 = 𝜌2𝑉2𝐴2

For an ideal gas 𝜌 =𝑝

𝑅𝑇and V = 𝑀𝑎𝑐 = 𝑀𝑎 𝛾𝑅𝑇

Substituting the above equation into continuity equation gives

𝛾

𝑅

𝜌1𝐴1𝑀𝑎1

𝑇1=

𝛾

𝑅

𝜌2𝐴2𝑀𝑎2

𝑇2

Simplifying the above equation gives the following

𝐴2𝐴1=𝑀𝑎1𝑀𝑎2

𝑝2𝑝1

−1𝑇2𝑇1

1 2

For isoenergetic-isentropic flow we can replace the temperature and pressure ratio

𝐴2𝐴1=𝑀𝑎1𝑀𝑎2

1 + 𝛾 − 1 /2 𝑀𝑎22

1 + 𝛾 − 1 /2 𝑀𝑎12

(𝛾+1) 2(𝛾−1)

If any of three quantities A1, Ma1, A2, Ma2 are specified, the fourth can be calculated. Usually A1 and Ma1 are known and

Ma2 or A2 is to be found. If Ma2 is given and A2 is to be calculate then the calculations are simple, however, if A2 is given

and Ma2 must be calculated, that is not so easy.

• Set Ma1 = 1, A1 = A* and drop the subscript 2 to get the following

𝐴

𝐴∗=

1

𝑀𝑎

2

𝛾 + 11 +

𝛾 − 1

2𝑀𝑎2

𝛾+1 2(𝛾−1)

(you may find tabulated values of A/A* in the various Fluid Mechanics Texts)

Plot of Area ratio function versus Mach number is given under and can be used to infer interesting information.

Conclusions

A/A* is always greater than unity except at Mach number unity where the A/A* is 1.

If the Mach number is given, there is single corresponding area ratio, so the problemof calculating the area required to obtain a given Mach number seems easy1.

If A/A* is known, there are two corresponding Mach numbers. Choice of properMach number for known area ratio requires more information than just the area ratioitself. Generally we must know the pressure ratio or other information besides arearatio to conclude the Mach number2.

Warning!! In a passage with the throat, the throat area (At) does not necessarily equalthe critical area unless other evidence indicates it.

1(if the initial Mach number is Ma1 is less than 1 and desired final Mach number Ma2 is greater than 1, a minimum area is exactly equal to A* must occurbetween the planes 1 and 2, otherwise flow cannot pass through Ma=1)2(we can sometimes logically eliminate one of the possibilities. If we know that the flow is subsonic at a plane 1 and minimum area is exactly equal to A*does not occur between the planes 1 and 2, the flow at 2 cannot be supersonic.

Mass Flow Relations and Chocking

Mathematics ( on white board)

Conclusion

Suppose that po and To are fixed (isoenergetic-isentropic flow). A specified massflow rate can be forced through a limiting area. No smaller area can pass thatflow.

Suppose that po and To are fixed (isoenergetic-isentropic flow). Any particularfixed geometry duct, with fixed minimum area, can pass only a certain maximumflow. Passage of this maximum mass flow occurs when the Mach number at theminimum area is 1. once the Mach number at the minimum area becomes 1, it isnot possible to increase the mass flow rate.

It is possible to increase the limiting mass flow in a fixed geometry duct or toforce a specified mass flow through a smaller area by increasing 𝑝𝑜 𝑇𝑜. Doing

work on a gas (compression) increases both po and To; however, the ratio 𝑝𝑜 𝑇𝑜is usually increased by compression.

Chocking

“The chocking phenomenon occurs in compressible duct flow when the local Machnumber reaches 1 at the minimum area in the duct. When this occurs, the mass flowrate through the duct cannot be increased unless the ratio of stagnation pressure tosquare root of stagnation temperature is increased.

Flow through a converging Nozzle

The nozzle is only convergent , so the flow cannot pass through M = 1.

The flow at the nozzle inlet (in the large reservoir) is obviously subsonic (M= 0),

so the flow in the entire nozzle is subsonic , with the possible exception of the

nozzle exit.

The flow cannot be supersonic in the nozzle, so there can be no shocks and the

flow is isoenergetic-isentropic everywhere in the nozzle .the stagnation

properties are constant and equal to the gas properties in the reservoir.

The maximum possible Mach number in the nozzle is 1.0 .this value can occur

only at the nozzle exit (minimum area).

There is a maximum possible mass flow rate

that can occur only when the Mach number is 1

at the nozzle exit .

Discussion

Discussion (cont…..)

Curve labeled as “a” corresponds to a closed valve. There is no flow and pressure equals thereservoir pressure everywhere.

Case “b” corresponds to slight opening of control valve. The back pressure is less than thesupply pressure and there is flow. Minimum pressure and maximum Mach number are at thenozzle exit. We can calculate the mass flow rate by mass flow equation.

Case “c” is similar to case “b” except larger control valve opening permits lower back pressurewith correspondingly higher mass flow and exit Mach number and lower exit pressure. Notethat Mach number is still less than 1.

Case “d”, the control valve is sufficiently open to bring the exit Mach number to a value of 1.The exit pressure is now become critical pressure and also equals the back pressure. Mass flowrate can be calculated from the suitable equation.

Case “e” corresponds to opening the control valve farther than in the case “d”. After trying thiswe find that no changes occur in the nozzle. In the case “d”, we reached the limit of nozzle’scapability. The exit Mach number cannot exceed 1, the exit pressure cannot drop below thecritical pressure, which can be determined from critical property formulas.

The only difference between case “d” and case “e” is the back pressure and exit pressure are nolonger equal. The flow must adjust to lower back pressure after leaving the nozzle. The downstream flow is multidimensional, so the pressure curve shown as a wavy line downstream fromthe nozzle exit.

There are two regimes in simple converging nozzle

If pB/po > p*/po, the flow is unchocked

If pB/po =< p*/po, the flow is chocked

Flow through Converging-diverging Nozzle

The nozzle is convergent-divergent , so the flow can pass through Ma=1 , the flow

also can be subsonic everywhere.

If Ma=1 anywhere, it must be at the throat .

There may be supersonic flow in the divergent portion of the nozzle , so there may

be shocks in the flow . If there are shocks , the flow is not completely isentropic ,

although it is isoenergetic

If there are no shocks, the flow is isentropic. if there are shocks , the flow from the

reservoir up to the first shock is isentropic. Flow downstream from a shock also is

isentropic but with different values of entropy , stagnation pressure, and critical

area.

The maximum possible Mach number that can occur anywhere in the passage

corresponds to acceleration of the fluid in an isentropic process from the reservoir

to the nozzle exit . The maximum possible Mach number can occur only at the

exit and is determined by the ratio of exit area to throat area .

The maximum possible mass flow rate in the nozzle is determined by the gas

constant , the reservoir properties, and the minimum area , which occurs at the

throat.

Flow through a Converging-Diverging Nozzle

Discussion Case “a” corresponds to complete closure of control valve.

Curves and points labeled “b” represents a slightly open control valve. The gas acceleratesin the convergent portion of the nozzle, and the pressure drops. The throat Mach number isless than 1. The pressure and Mach number distributions are roughly symmetrical about thethroat. Exit and back pressure are equal.

Curve “c” represents a slightly larger control valve opening. The pressure and Machnumber situation is qualitatively similar to case “b” but the mass flow is larger. Themaximum Mach number occurs at the throat.

In case “d” the control valve has been opened just enough to bring the throat Mach numberexactly to 1. The flow is still subsonic everywhere except exactly at the throat. The massflow has reached a maximum and flow has just become chocked.

The pressure rises downstream form the throat, so the back pressure at which converging-diverging nozzle chokes is greater than p*/po. The Pressure ratio at which this occur iscalled first critical pressure ratio.

What happens when the valve is opened beyond the case “d”? Because the throat is chocked, the conditions upstream of throat cannot be affected. There is a response in flow downstream from the throat, because at case “d” the down stream flow is subsonic.

Discussion (Cont.)

When the valve is opened and back pressure is lowered, the fluid begins to accelerate as it

enters the divergent portion of the nozzle, i.e., the flow become supersonic. If the control

valve is opened only slightly past the case d point, the downstream resistance is too large

for complete acceleration in the divergent portion of the nozzle, so a shock occurs in the

divergent portion.

Curves and points labeled as e represents the flow for a slightly more open control valve.

The flow accelerate in the converging portion of the nozzle, reaches sonic speed at the

throat, and accelerates to supersonic speed downstream from the throat. The supersonic

acceleration terminates in the shock wave. Downstream from the shock, the flow

experiences subsonic de-accleration and exits the passage with Ma<1. The exit and back

pressure are equal. The exact position of of the shock depends on the back pressure, and for

a given back pressure, is fixed. The mass flow rate for case e, as well as for all lower

values of back pressure, is the same as for the case of d.

Opening the control valve and lowering the back pressure causes the shock to move

downstream (case f). Note, that once the shock passes a plane, the flow up to that plane is

no longer affected by lowering the back pressure. As the valve is opened further, the shock

is eventually drawn to nozzle exit. The flow accelerates isentropically all the way from

reservoir to the exit shock. Exactly at the exit, the pressure jumps to the back pressure as

the fluid exits through the shocks. The situation is shown as case g. Note that the exit

pressure is doubled at the case of g.

Discussion (Cont.)

Lowering the back pressure further causes the shock to move out of the nozzle and become

multidimensional (case h). The flow accelerates isentropically from the reservoir to the

exit. The gas exits the nozzle supersonically with Ma corresponding to the nozzle’s exit-to-

throat area ratio. The exit pressure is determined only by the stagnation pressure (pO1) and

the exit Mach number and is not equal to the back pressure. The gas adjusts to the back

pressure externally.

If we continue to lower the back pressure, it eventually reaches equality with the exit

pressure (case i). At this condition, there is no pressure adjustment in the exiting gas.

Lowering the pressure further (case j) requires external expansion pressure adjustments but

does not the affect the nozzle flow.

Summary of DiscussionWe can summarize this information about converging – diverging nozzle flow asfollow. There are four regimes of flow in a converging – diverging nozzle.

Venturi regime (cases a – d). The flow is subsonic and isentropic everywhere.The flow accelerates in the convergent portion and decelerates in the divergentportion. Maximum Mach number and minimum pressure occur at the throat.

Shock regime (cases d – g). The flow is subsonic in the convergent portion,sonic at the throat, and partly supersonic in the divergent portion. The accelerationterminates in a shock that stands in the divergent portion at a location determinedby the exact value of the back pressure. The flow experiences subsonicdeceleration from the shock to the exit. The flow is choked.

Over expanded regime (cases g – i). The flow accelerates throughout thenozzle. The throat flow is sonic, and the exit flow is supersonic. The pressure ofthe gas increases to the back pressure downstream from the nozzle exit.

Under expanded regime (cases i – j). This case is similar to the over expandedregime, except that the external pressure adjustments are expensive rather thancompressive.