Lecture-2. One-dimensional Compressible Fluid Flow in Variable Area

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1 Lecture-2 One-dimensional Compressible Fluid Flow in Variable Area

2 Summary of Results(Cont..) In isoenergetic-isentropic flow, an increase in velocity always corresponds to a Mach number increase and vice versa. Therefore, a converging passage always drive the Mach number towards the unity 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 = ( 1 ) da ) unless there is V Ma 2 1 A shock wave, which is specifically excluded in assumptions, the bracketed equation then implies that da = 0 when Ma = 1. Finally suppose that the compressible fluid is flowing in a passage that actually has an area minimum (called Throat). Is it always possible to conclude that the Mach number is unity at the throat? If da is zero, then two options are possible: Ma = 1 or dv = 0, simply knowing da = 0 does not allow us to determine which condition occurs.

4 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.

5 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 w shaft w shear = h 2 + V h 1 + V According to our assumptions, q and w shear are zero. But the shaft is not necessarily zero, energy equation becomes w shaft = h 2 + V h 1 + V When we use the definition of stagnation enthalpy, w shaft = h o2 If the fluid is an ideal gas, w shaft = C p ( T o2 h o1 T o1 ).(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.

6 Doing work on the fluid increases its stagnation temperature, even if there is no heat transfer. Now consider the changes of stagnation pressure in isentropic flow. By definition of stagnation state Ŝ = constant, (Ŝ 1 = Ŝ 2 ) Using the equation 1, we get Thus p o2 = ( T Cp o2 ) R p o1 T o1 = ( T o2 p o2 = (1 w shaft ) γ 1 p o1 C p T o1 A situation of considerable interest in compressible flow is the case with no shaft work. The following statement applies to this case: In the isentropic flow with no shaft work, all the stagnation properties are constant. T o1 ) γ γ γ 1 Stagnation properties are constant in isentropic flow only if there is no shaft work. This special case (w shaft = 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 to isentropic flow with no work. Equations relating property and velocity changes in isoenergetic-isentropic flow of an ideal gas can easily be developed from stagnation property definitions and constancy of stagnation properties. take, e.g. the calculation of change in temperature between two points in an isoenergetic-isentropic flow. Because To is constant, we can write T o1 = T 1 + V 1 2 = T 2C 2 + V 2 1 p 2C p = T o2

7 Using Mach number instead of velocity is sometimes convenient, above equation can be written in the following form T 2 = T 2 o2 1 + (γ 1) 2 M a1 T 2 1 T o1 1 + (γ 1) 2 M a2 A third approach involves the use of the table relating stagnation property ratioand Mach number. We calculate the temperature ratios by T 2 T T = M T 1 T 2 M o T 1 o Where T T o M is a numerical value read from the tabulated data at Mach Number M. In a fourth method, we make the calculations in two steps: T o2 = T o1 = T 1 T To M 1 and T 2 = T o2 T T o M 2

8 Any of the above equation could be used to relate the temperature and velocities ( or Mach number) at point 1 and 2. which equation is most convenient depends on the exact information available. Similarly pressure and velocity ( of Mach number) in iso-energetic and isentropic flow can be related by any of the following equations: p V 1 2 2T 1 C p γ γ 1 = p V 2 2 2T 2 C p γ γ 1 p 2 = 1 + p (γ 1) 2 M a1 2 γ γ 1 2 (γ 1) 2 M a2 γ γ 1 p 2 p 1 = p p M p 2 M o p 1 o p 2 = p o2 p p o M 2 where p o2 = p o1 = p 1 p p o M 1

Area-Mach Number Relation To 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.

9 Area-Mach Number Relation To 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) ρ 1 V 1 A 1 = ρ 2 V 2 A 2 For an ideal gas ρ = p RT and V = M ac = M a γrt Substituting the above equation into continuity equation gives γ R ρ 1 A 1 M a1 T 1 = γ R ρ 2 A 2 M a2 T 2 Simplifying the above equation gives the following A 2 A 1 = M a1 M a2 p 2 p 1 1 T2 T For isoenergetic-isentropic flow we can replace the temperature and pressure ratio A 2 = M a1 1 + γ 1 /2 M a2 A 1 M 2 a2 1 + γ 1 /2 M a1 If any of three quantities A 1, M a1, A 2, M a2 are specified, the fourth can be calculated. Usually A 1 and M a1 are known and M a2 or A 2 is to be found. If M a2 is given and A 2 is to be calculate then the calculations are simple, however, if A 2 is given and M a2 must be calculated, that is not so easy. 2 (γ+1) 2(γ 1)

Set M a1 = 1, A 1 = A* and drop the subscript 2 to get the following A A = 1 2 M a γ + 1 1 + γ 1 2 M a 2 γ+1 2(γ 1) (you may find tabulated values of

10 Set M a1 = 1, A 1 = A* and drop the subscript 2 to get the following A A = 1 2 M a γ γ 1 2 M a 2 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.

11 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 problem of calculating the area required to obtain a given Mach number seems easy 1. If A/A* is known, there are two corresponding Mach numbers. Choice of proper Mach number for known area ratio requires more information than just the area ratio itself. Generally we must know the pressure ratio or other information besides area ratio to conclude the Mach number 2. Warning!! In a passage with the throat, the throat area (A t ) does not necessarily equal the critical area unless other evidence indicates it. 1 (if the initial Mach number is M a1 is less than 1 and desired final Mach number M a2 is greater than 1, a minimum area is exactly equal to A* must occur between the planes 1 and 2, otherwise flow cannot pass through M a =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.

12 Mass Flow Relations and Chocking Mathematics ( on white board)

13 Conclusion Suppose that p o and T o are fixed (isoenergetic-isentropic flow). A specified mass flow rate can be forced through a limiting area. No smaller area can pass that flow. Suppose that p o and T o are fixed (isoenergetic-isentropic flow). Any particular fixed geometry duct, with fixed minimum area, can pass only a certain maximum flow. Passage of this maximum mass flow occurs when the Mach number at the minimum area is 1. once the Mach number at the minimum area becomes 1, it is not possible to increase the mass flow rate. It is possible to increase the limiting mass flow in a fixed geometry duct or to force a specified mass flow through a smaller area by increasing p o T o. Doing work on a gas (compression) increases both p o and T o ; however, the ratio p o T o is usually increased by compression. Chocking The chocking phenomenon occurs in compressible duct flow when the local Mach number reaches 1 at the minimum area in the duct. When this occurs, the mass flow rate through the duct cannot be increased unless the ratio of stagnation pressure to square root of stagnation temperature is increased.

14 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.

15 Discussion

16 Discussion (cont..) Curve labeled as a corresponds to a closed valve. There is no flow and pressure equals the reservoir pressure everywhere. Case b corresponds to slight opening of control valve. The back pressure is less than the supply pressure and there is flow. Minimum pressure and maximum Mach number are at the nozzle 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 pressure with correspondingly higher mass flow and exit Mach number and lower exit pressure. Note that 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 flow rate can be calculated from the suitable equation. Case e corresponds to opening the control valve farther than in the case d. After trying this we find that no changes occur in the nozzle. In the case d, we reached the limit of nozzle s capability. The exit Mach number cannot exceed 1, the exit pressure cannot drop below the critical 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 no longer equal. The flow must adjust to lower back pressure after leaving the nozzle. The down stream flow is multidimensional, so the pressure curve shown as a wavy line downstream from the nozzle exit. There are two regimes in simple converging nozzle If p B /p o > p * /p o, the flow is unchocked If p B /p o =< p * /p o, the flow is chocked

17 Flow through Converging-diverging Nozzle The nozzle is convergent-divergent, so the flow can pass through M a =1, the flow also can be subsonic everywhere. If M a =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.

18 Flow through a Converging-Diverging Nozzle

19 Discussion Case a corresponds to complete closure of control valve. Curves and points labeled b represents a slightly open control valve. The gas accelerates in the convergent portion of the nozzle, and the pressure drops. The throat Mach number is less than 1. The pressure and Mach number distributions are roughly symmetrical about the throat. Exit and back pressure are equal. Curve c represents a slightly larger control valve opening. The pressure and Mach number situation is qualitatively similar to case b but the mass flow is larger. The maximum Mach number occurs at the throat. In case d the control valve has been opened just enough to bring the throat Mach number exactly to 1. The flow is still subsonic everywhere except exactly at the throat. The mass flow has reached a maximum and flow has just become chocked. The pressure rises downstream form the throat, so the back pressure at which convergingdiverging nozzle chokes is greater than p * /p o. The Pressure ratio at which this occur is called 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.

20 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.

21 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-tothroat area ratio. The exit pressure is determined only by the stagnation pressure (p O1 ) 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.

22 Summary of Discussion We can summarize this information about converging diverging nozzle flow as follow. 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 divergent portion. 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 acceleration terminates in a shock that stands in the divergent portion at a location determined by the exact value of the back pressure. The flow experiences subsonic deceleration from the shock to the exit. The flow is choked. Over expanded regime (cases g i). The flow accelerates throughout the nozzle. The throat flow is sonic, and the exit flow is supersonic. The pressure of the gas increases to the back pressure downstream from the nozzle exit. Under expanded regime (cases i j). This case is similar to the over expanded regime, except that the external pressure adjustments are expensive rather than compressive.

Isentropic Duct Flows

An Internet Book on Fluid Dynamics Isentropic Duct Flows In this section we examine the behavior of isentropic flows, continuing the development of the relations in section (Bob). First it is important