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ELEMENTS OF GAS DYNAMICS PDF

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Elements of Gas Dynamics. Home · Elements of Gas 41MB Size Report. DOWNLOAD PDF Elements of propulsion: gas turbines and rockets · Read more. PDF | Lecture notes on elementary Gas Dynamics writen by M.E.H. ρ(v·n)dA flows through a surface element dA, defined by its unit normal. This text covers introductory concepts from thermodynamics, one-dimensional gas dynamics and one-dimensional wave motion, waves in.


Elements Of Gas Dynamics Pdf

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H. W. Liepmann, A. Roshko - Elements of Gas Dynamics (, Dover Publications).pdf - Free ebook download as PDF File .pdf) or view presentation slides. Elements of Gasdynamics [H. W. Liepmann, A. Roshko] on sppn.info *FREE* shipping on qualifying offers. The increasing importance of concepts from. Elements of Gasdynamics, by H. W. LLEPMANN and A. ROSHKO. The term gas dynamics, which Liepmann and Roshko have chosen to combine into a single.

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Please review our Terms and Conditions of Use and check box below to share full-text version of article. Related Information. What are these conditions?

Summarizing the above: Since the losses affect the entropy changes through dsi , one generally uses either an h—s or T —s diagram. In the case of perfect gases, enthalpy is a function of temperature only and therefore the T —s and h—s diagrams are identical except for scale. Let us assume no heat transfer and no external work.

This is represented by the solid horizontal line in Figure 4. Two particular sections in the system have been indicated by 1 and 2. The actual process that takes place between these points is indicated on the T —s diagram. Notice that although the stagnation conditions do not actually exist in the system, they are also shown on the diagram for reference.

The distance between the static and stagnation points is indicative of the velocity that exists at that location since gravity has been neglected. What else do these points have in common? The same velocity? How about sonic velocity? Recall for gases that this is a function of temperature only. This means that points 1 and 3 would also have the same Mach number something that is not immediately obvious. One can now imagine that someplace on this diagram there is a horizontal line that represents the locus of points having a Mach number of unity.

Between this line and the stagnation line lie all points in the subsonic regime. Below this line lie all points in the supersonic regime. These conclusions are based on certain assumptions. What are they?

Sound waves fall into this latter category. Thus it will not be surprising to see Mach number become an important parameter.

Some of the most frequently used equations that were developed in this unit are summarized below. Compute and compare sonic velocity in air, hydrogen, water, and mercury. Assume normal room temperature and pressure. At what temperature and pressure would carbon monoxide, water vapor, and helium have the same speed of sound as standard air K and 1 atm?

Start with the relation for stagnation pressure that is valid for a perfect gas: See Problem 4. Note that these are gage pressures. Carbon dioxide with a temperature of K and a pressure of 1. Calculate the Mach number and stagnation pressure. Determine the velocity in the duct.

What temperature and pressure might you expect on the nose of the airplane? The stagnation pressure is 3. Determine the static conditions pressure, temperature, and velocity. A large chamber contains a perfect gas under conditions p1 , T1 , h1 , and so on. At one section, the total pressure is 50 psia, and at another section, it is At a point downstream, the Mach number is found to be unity. Pressure and temper- ature at section 1 are 7 atm and K. There is no heat transfer, work transfer, or losses as the gas passes through the device and expands to The duct has a cross-sectional area A and perimeter P.

The T —s diagram Figure CT4. State whether each of the following statements is true or false. Work Problem 4.

Chapter 5 Varying-Area Adiabatic Flow 5. Thus it is more than academic interest that leads to the separate study of each of the above-mentioned effects. In this manner it is possible to consider only the controlling factor and develop a simple solution that is within the realm of acceptable engineering accuracy.

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The chapter closes with a brief discussion of the various ways in which nozzle and diffuser performance can be represented. Compare the function of a nozzle and a diffuser. Under what conditions are these relations true? Explain how a converging nozzle performs with various receiver pressures. Do the same for the isentropic performance of a converging—diverging nozzle. To isolate the effects of area change, we make the following assumptions: In this manner we can distinguish the important differences between subsonic and supersonic behavior.

We start with the energy equation: Thus equation 1. For simplicity we shall assume that the pressure is always decreasing. Thus dp is negative. From equation 5. Now continue to assume that the pressure is decreasing.

Knowing the area vari- ation you can now consider equation 5. Fill in the following blanks with the words increasing or decreasing: Looking at equation 5. We summarize the above by saying that as the pressure decreases, the following variations occur: The appropriate shape of these curves can easily be visualized if one combines equations 5.

Consider the operation of devices such as nozzles and diffusers. From Figure 5. Figure 5. Notice that a device is called a nozzle or a diffuser because of what it does, not what it looks like. Further consideration of Figures 5. If one attached a converging section see Figure 5. On the other hand, if we made a converging—diverging device combination of Figure 5.

An example of this for the system shown in Figure 5. Recall from Chapter 4 that we also developed a general relationship between static and stagnation pressures for a perfect gas: Direct substitution of equations 5. At one section the Mach number is 1. Find the area ratio. We substitute directly into equation 5. We shall come back to this in Section 5. Before going any further with the working equations developed in Section 5. We now proceed to develop an extremely important relation.

We now apply equation 5. Then equation 5. Check Figure 5. We have previously developed a relation between the stagnation pressures which involves the same assumptions as equation 5. Learn equation 5. This leads to the natural use of the isentropic process as a standard for ideal performance with appropriate corrections made to account for losses when necessary. If we simplify equation 5. Our problem is to solve for the Mach number M2 at the other location.

Although this is not impossible, it is messy and a lot of work.

H. W. Liepmann, A. Roshko - Elements of Gas Dynamics (, Dover Publications).pdf

The problem posed earlier could then be solved as follows: A word of caution here! One value will be in the subsonic region and the other in the supersonic regime.

Note that the general problem with losses can also be solved by the same technique as long as information is available concerning the loss. All three of these represent equivalent ways of expressing the loss [through equations 4. These are obtained by taking the equations developed in Section 5. We proceed with equation 5. In this case we let 1 be the arbitrary point and its stagnation point is taken as 2.

The tabulation of equation 5. For example, assume that we are Given: Apply the previous techniques to equation 5. We shall see later that equation 5. To convince yourself that there is nothing magical about this table, you might want to check some of the numbers found in them opposite a particular Mach number.

Example 5. Find the possible values of M2. To determine conditions at section 2 in Figure E5. M2 and T2. Write an energy equation between 1 and 2. See the table in Appendix A for gas properties. Continue and compute p2 and T2. Two types of nozzles are considered: We start by examining the physical situation shown in Figure 5. Connected to the tank is a converging-only nozzle and it exhausts into an extremely large receiver where the pressure can be regulated.

We can neglect frictional effects, as they are very small in a converging section. Since the supply tank has a large cross section relative to the nozzle outlet area, the velocities in the tank may be neglected. There is no shaft work and we assume no heat transfer. We identify section 2 as the nozzle outlet.

The velocity will increase and the pressure will decrease as we progress through the nozzle until the pressure at the nozzle outlet equals that of the receiver. Recall Section 4. If the pressure in the receiver were lowered further, the air would expand to this lower pressure and the Mach number and velocity would increase.

Assume that the receiver pressure is lowered to If we now drop the receiver pressure below this critical pressure Why not? Assume that the nozzle outlet pressure could continue to drop along with the receiver. Under these conditions, the nozzle is said to be choked and the nozzle outlet pressure remains at the critical pressure. Expansion to the receiver pressure takes place outside the nozzle.

In reviewing this example you should realize that there is nothing magical about a receiver pressure of With sonic velocity at the exit, this ratio is 0. The analysis above assumes that conditions within the supply tank remain con- stant.

It is instructive to take an alternative view of this situation. You are asked in Problem 5. Converging—Diverging Nozzle Now let us examine a similar situation but with a converging—diverging nozzle some- times called a DeLaval nozzle , shown in Figures 5. We identify the throat or section of minimum area as 2 and the exit section as 3. The distinguishing phys- ical characteristic of this type of nozzle is the area ratio, meaning the ratio of the exit area to the throat area.

If the nozzle is to operate as desired, we know see Section 5. This mode is sometimes referred to as third critical. Recall that these two answers come from the solution of a quadratic equation. The device is nowhere near its design condition and is really operating as a venturi tube; that is, the converging section is operating as a nozzle and the diverging section is operating as a diffuser.

Once the receiver pressure has been lowered to Again, realize that it is not the pressure in the receiver by itself but rather the receiver pressure relative to the inlet pressure that determines the mode of operation.

At what pressure should the air in the chamber be for the nozzle to operate at its design condition third critical point? What will the outlet velocity be? With reference to Figure 5. Mach 1 in the throat 2.

Then properties in the throat can be found if desired. We shall learn later that under these conditions shocks will occur in either the diverging portion of the nozzle or after the exit. If the receiver pressure is below the third critical point, the nozzle operates internally as though it were at the design condition but expansion waves occur outside the nozzle. These operating modes will be discussed in detail as soon as the appropriate background has been developed.

Friction losses can then be taken into account by one of several methods. Direct in- formation on the entropy change could be given, although this is usually not available. Sometimes equivalent information is provided in the form of the stagnation pressure ratio. This is the case shown in Figure 5.

Also note that the ideal process is assumed to take place down to the actual available receiver pressure. The receiver pressure is 50 psia.

What is the actual nozzle outlet temperature? Nearly a dozen criteria have been suggested to indicate diffuser performance see p. The following discussion refers to the h—s diagram shown in Figure 5.

Most of the propulsion industry uses the total-pressure recovery factor as a mea- sure of diffuser performance. As we shall see in Chapter 12, for propulsion applications this is usually referred to the free-stream conditions rather than the diffuser inlet. The curves in Figure 5. However, they offer hints as to what magnitude of changes are to be expected in other cases.

There is almost always the need for interpolation. They do not necessarily have the required accuracy. Simply put, the computer can be programmed to do the hard and the easy numerical calculations. In this book we have deliberately avoided integrating any gas dynamics software some of which is commercially available into the text material, preferring to present computer work as an adjunct to individual calculations.

One reason is that we want you to spend your time learning about the wonderful world of gas dynamics and not on how to manage the programming.

Another reason is that both computers and packaged software evolve too quickly, and therefore the attention that must be paid just to use any particular software is soon wasted. Once you have mastered the basics, however, we feel that it is appropriate to discuss how things might be done with computers and this could include handheld programmable calculators.

In this book we discuss how the computer utility MAPLE can be of help in solving problems in gas dynamics. MAPLE is a powerful computer environment for doing symbolic, numerical, and graphical work. It is the product of Waterloo Maple, Inc. Other software packages are also popular in engineering schools. But we have chosen MAPLE because it can manipulate equations symbolically and because of its superior graphics.

The ex- perienced programmer can go much beyond these exercises. This section is optional because we want you to concentrate on the learning of gas dynamics and not spend extra time trying to demystify the computer approach. We focus on an example in Section 5. This is a problem where MAPLE can be useful because a built-in solver routine handles this type of problem easily.

If you are familiar with these, skip to the next paragraph. Statements terminated in a colon are also executed but no return is asked for. We are now looking for the ratio of static to stagnation temperature, which is given the symbol Z. This ratio comes from equation 5. A large number of useful equations were developed; however, most of these are of the type that need not be memorized. Equations 5. The working equations that apply to a perfect gas are summarized at the end of Section 5.

Equations used as a basis for the isentropic table are numbered 5. Those equations that are most frequently used are summarized below. You should be familiar with the conditions under which each may be used.

Go back and review the equations listed in previous summaries, particularly those in Chapter 4. Nozzle performance. Diffuser performance. Total-pressure recovery factor: The following information is common to each of parts a and b.

Determine the temperature at section 2. Two venturi meters are installed in a cm-diameter duct that is insulated Figure P5. Although each venturi is isentropic, the connecting duct has friction and hence losses exist between sections 2 and 3.

A smooth 3-in. Nitrogen is stored in a large chamber under conditions of K and 1. The gas leaves the chamber through a convergent-only nozzle whose outlet area is 30 cm2. The receiver pressure is psia. What are the actual outlet temperature, Mach number, and velocity? The air enters a converging— diverging nozzle which has an area ratio exit to throat of 3. Air enters a convergent—divergent nozzle at 20 bar abs.

At the end of the nozzle the pressure is 2. Assume a frictionless adiabatic process. The throat area is 20 cm2. It is fed by a large chamber of oxygen at Assuming the losses to be negligible, compute the velocity in the nozzle throat. A converging—diverging nozzle Figure P5. The exit area is such that the pressure at the nozzle exit exactly matches the receiver pressure. The nozzle is adiabatic and there are no losses. Figure P5. At one section the pressure is 2.

Indicate pertinent temperatures, pressures, etc. The exit Mach number is 2. Assume that a supersonic nozzle operating isentropically delivers air at an exit Mach number of 2. Use the following headings: List these parameters and show how they are related to one another. In the T —s diagram Figure CT5. Figure CT5. A supersonic nozzle is fed by a large chamber and produces Mach 3.

Sketch curves to no particular scale that show how properties vary through the nozzle as the Mach number increases from zero to 3. Sketch an h—s diagram to identify your state points. Work problem 5. Chapter 6 Standing Normal Shocks 6. Recall from Section 4. Due to the complex interactions involved, analysis of the changes within a shock wave is beyond the scope of this book. Thus we deal only with the properties that exist on each side of the discontinuity.

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The chapter closes with a discussion of shocks found in the diverging portion of supersonic nozzles. List the assumptions used to analyze a standing normal shock. Optional Given the working equations for a perfect gas, show that a unique relationship must exist between the Mach numbers before and after a standing normal shock. Optional Explain how a normal-shock table may be developed that gives property ratios across the shock in terms of only the Mach number before the shock.

Sketch a normal-shock process on a T —s diagram, indicating as many per- tinent features as possible, such as static and total pressures, static and total temperatures, and velocities. Indicate each of the preceding before and after the shock. Explain why an expansion shock cannot exist. Describe the second critical mode of nozzle operation. Given the area ratio of a converging—diverging nozzle, determine the operating pressure ratio that causes operation at the second critical point.

Demonstrate the ability to solve typical standing normal-shock problems by use of tables and equations. In this manner we deal only with the changes that occur across the shock. The area on both sides of the shock may be considered to be the same. There is negligible surface in contact with the wall, and thus frictional effects may be omitted. We begin by applying the basic concepts of continuity, energy, and momentum under the following assumptions: A typical problem would be: Continuity We start with the continuity equation developed in Section 6.

Thus equations 6. We proceed to combine the equations above and derive an expression for M2 in terms of the information given. First, we rewrite equation 6. To solve the nontrivial case, we square equation 6. Only if you have considerable moti- vation should you attempt to carry out the tedious algebra or to utilize a computer utility, see Section 6. To complete the picture, the total pressures pt1 and pt2 can be computed in the usual manner.

For this reason the shock process is usually shown by a dashed or wiggly line. Note that when points 1 and 2 are located on the T —s diagram, it can immediately be seen that an entropy change is involved in the shock process. This is discussed in greater detail in the next section. Example 6.

Determine the pressure ratio across the shock. We use equation 6. For example, equation 6. Such a table of normal-shock parameters is given in Appendix H. This table greatly aids problem solution, as the following example shows. If the conditions before the shock are: First we compute pt1 with the aid of the isentropic table. The fundamental relations of continuity 6. However, in the latter case the entropy change would be negative, which clearly violates the second law of thermodynamics for an adiabatic no-work system.

One can note from the table that as M1 increases, the pressure, temper- ature, and density ratios increase, indicating a stronger shock or compression. Thus as the strength of the shock increases, the losses also increase. Determine the density before and after the shock. The entering Mach number is 2.

Neglect all frictional losses. Its usefulness for solving certain types of problems will become apparent in Chapter 7. Remember that this type of nozzle is physically distinguished by its area ratio, the ratio of the exit area to the throat area.

It is also choked with Mach 1. This internal adjustment takes the form of a standing normal shock, which we now know involves an entropy change.

The shock will locate itself in a position such that the pressure changes that occur ahead of the shock, across the shock, and downstream of the shock will produce a pressure that exactly matches the outlet pressure.

In other words, the operating pressure ratio determines the location and strength of the shock. An example of this mode of operation is shown in Figure 6. As the pressure ratio is lowered further, the shock continues to move toward the exit.

When the shock is located at the exit plane, this condition is referred to as the second critical point. These effects sometimes cause what are known as lambda shocks.

If the operating pressure ratio is between the second and third critical points, a compression takes place outside the nozzle. This is called overexpansion i. If the receiver pressure is below the third critical point, an expansion takes place outside the nozzle. This condition is called underexpansion. We investigate these conditions in Chapters 7 and 8 after the appropriate background has been covered. The nozzle has an area ratio of 2. Thus the inlet conditions are essentially stagnation.

What receiver pressure do we need to operate at the second critical point? Figure 6. Suppose that we are given an operating pressure ratio of 0. The logical question to ask is: Where is the shock? This situation is shown in Figure 6.

We must take advantage of the only two available pieces of information and from these construct a solution. We know that Figure 6. For physical picture see Figure 6. More accurate answers could be obtained by interpolating within the tables.

At off-design condi- tions, the exit Mach number is observed to be 0. What operating pressure ratio would cause this situation? Using the section numbering system of Figure 6. A normal shock occurs at a section where the area is 1. Again, we use the numbering system shown in Figure 6. To operate economically, the nozzle—test-section combination must be followed by a diffusing section which also must be converging—diverging.

Starting up such a wind tunnel is another example of nozzle operation at pressure ratios above the second critical point.

A brief analysis of the situation follows. As power is increased further, a normal shock is formed just downstream of the throat, and if the tunnel pressure is decreased continuously, the shock will move down the diverging portion of the nozzle and pass rapidly through the test section and into the diffuser. We return to Figure 6. This is called the most unfavorable condition because the shock occurs at the highest possible Mach number and thus the losses are greatest.

We might also point out that the diffuser throat section 5 must be sized for this condition. Let us see how this is done. Keep in mind that this represents a minimum area for the diffuser throat. If it is made any smaller than this, the tunnel could never be started i. Thus, to operate as a diffuser, there must be a shock at this point, as shown in Figure 6. We have also shown the pressure variation through the tunnel for this running condition.

To keep the losses during running at a minimum, the shock in the diffuser should occur at the lowest possible Mach number, which means a small throat.

However, we have seen that it is necessary to have a large diffuser throat in order to start the tunnel. A solution to this dilemma would be to construct a diffuser with a variable- area throat. After startup, A5 could be decreased, with a corresponding decrease in shock strength and operating power. However, the power required for any installation must always be computed on the basis of the unfavorable startup condition. Equally important is the fact that it begins to focus our attention on some practical design applications.

Figures 6. On the other hand, as shown in Figure 6. For instance, we can easily calculate the left-hand side of equations 6. We can compute this from equation 6. A rather unique capability of MAPLE is its ability to solve equations symbolically in contrast to strictly numerically. This comes in handy when trying to reproduce proofs of somewhat complicated algebraic expressions. Above are the two roots of Y or M22 , because we are solving a quadratic. With some manipulation we can get the second or nontrivial root to look like equation 6.

It is easy to check it by substituting in some numbers and comparing results with the normal-shock table. The type of calculation shown above can be integrated into more sophisticated programs to handle most gas dynamic calculations. If con- ditions are known ahead of a shock, a precise set of conditions must exist after the shock. What has been lost? As in Chapter 5, most of the equations in this chapter need not be memorized. These are equations 6. Essentially, these say that the end points of a shock have three things in common: The same stagnation enthalpy 3.Mises von R.

All of these static and stagnation pressures and temperatures are shown in the T —s diagram of Figure E7. The stagnation pressure at the inlet is 8. This situation is shown in Figure 6. An isentropic process? State whether each of the following statements is true or false. We now ask what form the energy equation takes when applied to a control volume.

Mach number, wave motion, and sonic speed[ edit ] The Mach number M is defined as the ratio of the speed of an object or of a flow to the speed of sound.

Give examples of each.