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1 Tutorial Materials for ME 131B Fluid Mechanics (Compressible Flow & Turbomachinery) Calvin Lui Department of Mechanical Engineering Stanford University Stanford, CA March 1998

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3 Acknowledgments This work is specially dedicated to the graduating class of 1998 and my dear oce-mate, Zhongmin Xiong, here at Stanford University. They encouraged me to compile all these tutorial materials together into one single volume which serves as future references for the ME 131B class. I would like to express my gratitude to them for all their encouragement. Stanford, California March, 1998 Calvin Lui i

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5 Table of Contents Tutorial One: Thermodynamics Review 1 Solutions to Thermodynamics Review 4 Tutorial Two: Isentropic Flow I 10 Solutions to Isentropic Flow I 14 Tutorial Three: Isentropic Flow II 24 Solutions to Isentropic Flow II 28 Tutorial Four: Normal Shock 40 Solutions to Normal Shock 44 Tutorial Five: Fanno Flow 60 Solutions to Fanno Flow 66 Tutorial Six: Rayleigh Flow 86 Solutions to Rayleigh Flow 90 Tutorial Seven: Angular Momentum Principle 105 Solutions to Angular Momentum Principle 109 Tutorial Eight: Turbomachinery 123 Solutions to Turbomachinery 127 References 135 iii

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7 ME 131B Fluid Mechanics Tutorial One: Thermodynamics Review 1. What is a thermodynamic property? What is the dierence between an intensive and an extensive property? Give an example of both. 2. What is a simple compressible substance? What does the state principle for a simple compressible substance tell us? 3. What do the following laws of thermodynamics mean to you? Describe them in your own words. (a) Zeroth Law (b) First Law (c) Second Law 4. Apart from the above laws of thermodynamics, what other basic principle(s) do we usually apply in analyzing thermodynamic systems? 1

8 5. What are the dierent transfer modes for (a) Energy, (b) Entropy? 6. Write down the mathematical form of the First and Second Law for (a) a close system, (b) an open system. Describe the meaning of each term in the equation. 7. A patent application describes a closed system which at steady-state conditions receives a heat transfer of 500 W at a temperature of 400 K and develops a combined electrical and mechanical power output of 500 W. There are no other energy transfers. Is this claim thermodynamically feasible? 8. What is the ideal gas model? How about perfect gas model? Under what conditions will these models be appropriate in describing real-life phenomena? 9. What is the relation between the specic heats (C p ; C v ) for an ideal gas? 2

9 10. What is an adiabatic process? When will it be realized physically? 11. What is an isentropic process? What is its signicance in thermodynamic analysis? 12. What are some common causes for irreversibility in thermodynamic systems? 13. Write down the Gibbs equation. 14. Derive the P T relationships for a perfect gas undergoing an isentropic process. 15. Sketch the following curves on a T s diagram (a) constant pressure, (b) constant density. Based on the Gibbs equation, explain the dierence in the slope of the above two curves. 3

10 ME 131B Fluid Mechanics Solutions to Tutorial One: Thermodynamics Review 1. What is a thermodynamic property? What is the dierence between an intensive and an extensive property? Give an example of both. A thermodynamic property is a macroscopic characteristic which describes the state of a system. Extensive property depends on the size or extent of a system. Its value of an overall system is the sum of its individual parts, like entropy and internal energy. Intensive property is independent of the size of a system and is not additive, like temperature and pressure. 2. What is a simple compressible substance? What does the state principle for a simple compressible substance tell us? Simple { there is only one reversible work mode which can alter the energy of the system Compressible { work mode is associated with volume change R p dv The state principle states that two independent, intensive thermodynamic properties are sucient to fully determine the thermodynamic state of a simple compressible substance, like (T; v); (u; v). (Reminder: Pressure and temperature are not independent of each other in the two-phase region.) For a \non-simple" substance with n independent work modes, we need to know a total of n + 1 independent, intensive thermodynamic properties to completely specify its state. 3. What do the following laws of thermodynamics mean to you? Describe them in your own words. (a) Zeroth Law { Temperature Equality in temperature is a necessary and sucient condition for thermal equilibrium. (b) First Law { Energy Energy is conserved. (c) Second Law { Entropy 4

11 Entropy can only be produced but not destroyed. (Be careful that it does not mean that entropy of a system can never decrease. If we have enough heat transfer out of a system, it is possible to have a decrease in the entropy of the system.) It is a powerful tool for us to determine the possible direction of a thermodynamic process. 4. Apart from the above laws of thermodynamics, what other basic principle(s) do we usually apply in analyzing thermodynamic systems? Conservation of mass dm cv dt = X _m in X _mout Newton's law Z CV ~V ( dv ) + Z CS ~V ~ V d ~ A = X ~Fsurface + X ~ Fbody 5. What are the dierent transfer modes for (a) Energy, Mass transfer Heat transfer Work transfer (b) Entropy? Mass transfer Heat transfer There is no entropy transfer associated with work. This is a major dierence between the two energy transfer modes, work and heat. 6. Write down the mathematical form of the First and Second Law for (a) a close system, U = X Q in X Wout S = X Q in T + P s 5

12 (b) an open system. de dt = X _ Q in {z } heat trans. X _ W out {z } non-ow work + X _m ( enthalpy z} { kinetic z} { V 2 potential z} { gz ) {z } h mass trans. ds dt X _Q X in = + _ms {z T } {z } heat trans. mass trans. + _ P s {z} production Remarks: Equation can be expressed in an overall or rate form. Examples of non-ow work: { shaft work, { any R P dv type of work such as compression and expansion Enthalpy consists of internal energy and ow work, h = u + P= (Hence, do not double count the ow work in _ W out again!) Internal energy is a measure of microscopic molecular activities while kinetic and potential energies are measures of bulk uid motion. 7. A patent application describes a closed system which at steady-state conditions receives a heat transfer of 500 W at a temperature of 400 K and develops a combined electrical and mechanical power output of 500 W. There are no other energy transfers. Is this claim thermodynamically feasible? Since it is a closed system, there is no mass ow. Conservation of mass is trivial. First law (conservation of energy) is satised. Entropy production rate goes negative, hence, second law is violated. As a conclusion, the claim is not thermodynamically feasible. 8. What is the ideal gas model? How about perfect gas model? Under what conditions will these models be appropriate in describing real-life phenomena? Ideal gas model: { It satises the thermal equation of state: p = R T where R is the gas constant (dierent for dierent gases.) { u = u(t ) 6

13 { It is appropriate for high temperature and low pressure condition (negligible interaction between participating gas molecules.) Perfect gas model: { It is an ideal gas with constant specic heats C p ; C v. { It is an appropriate model if the temperature variations between states are not too large (together with the conditions for ideal gas behavior.) 9. What is the relation between the specic heats (C p ; C v ) for an ideal gas? According to denition C p and C v For ideal gases, they are reduced to ordinary derivatives C p = dh dt and C v = du dt Again, from denition h = u + P = u + R T (P = R T for an ideal gas) Take derivative with respect to temperature of above equation, we have C p = C v + R Recall the denition of specic heat ratio C p C v = k Solve the above two equations for C p and C v, we have C p = k k 1 R and C v = k 1 1 R 10. What is an adiabatic process? When will it be realized physically? Adiabatic process is a thermodynamic process in which no heat transfer occurs. It is a good model if { System has good insulation. 7

14 { Thermodynamic process proceeds at a much faster rate than heat transfer does. For example, ow in a nozzle, valve. 11. What is an isentropic process? What is its signicance in thermodynamic analysis? Isentropic process is reversible and adiabatic. It serves as a limit for real adiabatic process. 12. What are some common causes for irreversibility in thermodynamic systems? Dissipation like friction, viscous eects Mixing Spontaneous chemical reaction Unrestrained expansion 13. Write down the Gibbs equation. T ds = du P 2 d or T ds = dh dp 14. Derive the P T relationships for a perfect gas undergoing an isentropic process. T 1 k = constant T P 1 k k = constant P k = constant 15. Sketch the following curves on a T s diagram (a) constant pressure, T decreasing pressure s 8

15 (b) constant density. T decreasing density s Based on the Gibbs equation, explain the dierence in the slope of the above two curves. The slope of any curve on the T From the Gibbs equation, s plane is characterized by the derivative dt=ds. T ds = dh dp T ds = C p dt dp For a constant pressure process, dp = 0, we obtain dt ds = T C p Hence, the slope of a constant pressure curve on a T s diagram is equal to T=C p. Similarly, we can obtain the slope of a constant density curve to be equal to T=C v. Since C p = C v + R, the constant density curve has a steeper slope than the constant pressure curve at the same temperature. 9

16 ME 131B Fluid Mechanics Tutorial Two: Isentropic Flow I Choose the best answer in the following three questions: 1. Static property is dependent on/independent of the choice of reference frame. 2. Stagnation property is dependent on/independent of the choice of reference frame. 3. Stagnation property can/cannot be dened for a non-isentropic ow. 4. Under what conditions can we say that the stagnation enthalpy remains constant in a ow? How about stagnation pressure? 5. Is there any limitation on applying the following equations in a ow analysis? T 0 T = 1 + k 1 2 M 2 P 0 P = 1 + k 1! k k 1 M Given a thermodynamic state (T 1 ; P 1 ) and its speed in terms of Mach number, M 1. Can you locate its corresponding stagnation state on the T s diagram? T P 1 T 1 s 10

17 7. Consider a ow through a valve as follows: 1 2 Locate the static and stagnation states of both the upstream Station (1) and downstream Station (2) on the same T s diagram. 8. Recall that the incompressible Bernoulli's model gives us P 0 = P V 2 : But we obtain, from a compressible analysis, the following result: P 0 P = 1 + k 1! k k 1 M 2 : 2 These two equations give strikingly dierent expressions for the stagnation pressure. How do you reconcile the dierence between the two? What does this dierence depend on? 11

18 9. In a wind tunnel experiment, 1.0 kg/sec of air is accelerated through an adiabatic nozzle from an upstream section (P 1 = 2.0 bar, T 1 = 900 K, A 1 = 50 cm 2 ) to a Mach 1.2 ow in the downstream section.. m = 1.0 kg/s P 1 = 2.0 bar T 1 = 900 K A 1 = 50 cm Adiabatic Nozzle M 2 = 1.2 P 2 =? T 2 =? A 2 =? (a) Sketch the general shape of the nozzle section. (b) If the ow is further treated as isentropic, i. What is the downstream cross-sectional area (A 2 ), temperature (T 2 ) and pressure (P 2 )? ii. Sketch the variation of pressure, temperature, velocity and Mach number from Station (1) to Station (2). iii. Trace the process path from Station (1) to Station (2) on a T s diagram. 12

19 10. Air ows through a passage of unknown shape. The upstream state is characterized as follows: A 1 = 0.50 m 2 M 1 = 0.70 P 1 = 5.0 bar T 1 = 270 K 1 2 Passage with unknown shape A 2 = 1.0 m 2 M 2 =? P 2 =? T 2 =? Assume that the ow is isentropic. (a) Determine the possible downstream State (2) where A 2 = 1:0 m 2 (i.e. Find M 2 ; P 2 ; T 2.) (b) Sketch the shape of the associated ow passage found in Part (a). (c) Sketch the corresponding variation of density, velocity and Mach number from Station (1) to Station (2). 13

20 ME 131B Fluid Mechanics Solutions to Tutorial Two: Isentropic Flow I Choose the best answer in the following three questions: 1. Static property is independent of the choice of reference frame. Static property can be thought of being measured by someone who travels with the uid particles. Hence, it does not depend on the choice of reference frame. 2. Stagnation property is dependent on the choice of reference frame. Consider our everyday running or biking experiences. We feel a higher pressure on our face as we accelerate to faster speeds. It is because the air \appears" to travel faster with respect to us (a moving reference frame) as we accelerate. Accordingly, its stagnation pressure is higher with respect to a moving observer. Its value is given by the following isentropic relationship: P 0 P = 1 + k 1! k k 1 M 2 2 where P = 101 kpa and M is the Mach number of the observer. 3. Stagnation property can be dened for a non-isentropic ow. The stagnation process is a hypothetical process which is not necessarily found in the real physical ow. The stagnation state should be viewed as a reference thermodynamic state with which the ow is compared with. Based on this reason, the stagnation state or properties can be dened for any physical ow regardless it is isentropic or not. In the case of isentropic ow, the stagnation state is constant in the ow. Hence, it serves as a universal reference within the same ow. But the usefulness of the stagnation state as a reference will be highly degraded if the ow is non-isentropic. In this case, P 0 changes in the ow. (Depend on the importance of heat transfer in the ow, T 0 may also vary in the ow.) 14

21 4. Under what conditions can we say that the stagnation enthalpy remains constant in a ow? How about stagnation pressure? Stagnation enthalpy remains constant in a ow when { the ow is adiabatic, { the ow is not subject to any non-ow work. This can be inferred from the rst law of thermodynamics (conservation of energy). Stagnation pressure remains constant in a ow when { the ow is adiabatic, { the ow is not subject to any non-ow work, { the ow is reversible. This can be inferred from the rst and second law of thermodynamics. 5. Is there any limitation on applying the following equations in a ow analysis? T 0 T = 1 + k 1 2 M 2 P 0 P = 1 + k 1! k M 2 k 1 2 The only limitation is that the uid should behave as a perfect gas. 6. Given a thermodynamic state (T 1 ; P 1 ) and its speed in terms of Mach number, M 1. Can you locate its corresponding stagnation state on the T s diagram? By denition, the stagnation state should have the same specic entropy as the static state. Hence, they should both be on the same vertical line on the T s diagram. By denition, the stagnation temperature can never be lower than the static temperature. Hence, the stagnation state must be located somewhere above the static state. The dierence in temperature between the static and the stagnation states is a measure of the specic kinetic energy carried by the uid: h 0 = h V 2 T 0 = T T 0 T = 1 2 V 2 C p V 2 C p 15

22 The higher the uid velocity, the lower the static temperature compared with its stagnation temperature. This observation points to one important behavior of compressible ow: interchange between thermal and kinetic energy in an adiabatic ow. Based on the above conclusions, we can locate the stagnation state relative to its static state on the same T s diagram in the following gure: T P 0,1 T 0,1 P 1 T = V 2 2 C p T 1 s 1 = s 0,1 s From the T s diagram, we can conrm the fact that the stagnation pressure is always higher than the static pressure. 7. Consider a ow through a valve as follows: 1 2 Locate the static and stagnation states of both the upstream Station (1) and downstream Station (2) on the same T s diagram. Based on the procedure outline in the previous problem, we can locate the static and stagnation state of upstream Station (1) as our reference. Across the valve, { The ow is adiabatic (no heat transfer). Stagnation temperature remains constant (T 0;2 = T 0;1 ). { There are losses due to friction at the valve. Entropy is produced (s 2 > s 1 ). Both static and stagnation pressure drop across the valve (P 2 < P 1 ; P 0;2 < P 0;1 ). 16

23 { However, the non-trivial part is on the change in density. Does density increase or decrease across the valve? To reason it out, we need to invoke both conservation of mass and energy. { Assume the same cross-sectional area in both pipes, conservation of mass (COM) gives 1 V 1 = 2 V 2 { Conservation of energy (COE) gives T 1 + V 2 1 = T 2 + V C p 2 C p { Let us examine exhaustively two dierent alternatives: (a) 2 > 1 By COM, a density rise will result in a drop of ow velocity (V 2 < V 1 ). By COE, a drop of ow velocity will result in a rise in static temperature (T 2 > T 1 ). But a simultaneous rise in density and temperature cannot possibly produce a pressure drop across the valve. Hence, this case is not feasible. (b) 2 < 1 By COM, a density drop will result in a rise in ow velocity (V 2 > V 1 ). By COE, a rise in ow velocity will result in a drop of static temperature (T 2 < T 1 ). A simultaneous drop in density and temperature can produce a pressure drop across the valve. We can conclude that this is indeed the case. Based on the above conclusions, we can locate the static and stagnation state of downstream Station (2) with respect to those of upstream Station (1) on the same T s diagram as follows: T P 0,1 P 0,2 T 0,1 = T 0,2 P 1 P 2 T 1 T 2 s 1 s 2 s 17

24 8. Recall that the incompressible Bernoulli's model gives us P 0 = P V 2 : But we obtain, from a compressible analysis, the following result: P 0 P = 1 + k 1! k k 1 M 2 : 2 These two equations give strikingly dierent expressions for the stagnation pressure. How do you reconcile the dierence between the two? What does this dierence depend on? Assume a perfect gas model, we can write the incompressible model as: V 2 P 0 P = R T = k V 2 k R T = k M 2 Results of the incompressible model are compared with those of the compressible model in the following gure (for the k = 1:4 case): 2.0 Incompressible Compressible P 0 / P M We can observe that the two models give nearly identical results in the low Mach number range (M < 0:3). Dierence between the two models becomes apparent for M > 0:4 (compressibility is no longer negligible). The value of P 0 =P predicted by the two dierent models and the percentage dierence between them are tabulated in the following table for further reference: 18

25 Incompressible Compressible % Dierence M 1 = 0: M 1 = 0: M 1 = 0: M 1 = 1: M 1 = 2: M 1 = 3: To nd out the dierence between the two models, we can invoke the binomial expansion technique as follows: P 0 P = 1 + k 1! k k 1 M 2 2 = 1 + k 2 M 2 + k 8 M 4 + k (2 k) M 6 + : : : 48 {z } correction terms The leading correction term of the compressible model to its incompressible counterpart scales with the fourth power of the ow Mach number. This explains why the dierence between the incompressible and compressible models increases so dramatically with compressibility (Mach number). 9. In a wind tunnel experiment, 1.0 kg/sec of air is accelerated through an adiabatic nozzle from an upstream section (P 1 = 2.0 bar, T 1 = 900 K, A 1 = 50 cm 2 ) to a Mach 1.2 ow in the downstream section.. m = 1.0 kg/s P 1 = 2.0 bar T 1 = 900 K A 1 = 50 cm Adiabatic Nozzle M 2 = 1.2 P 2 =? T 2 =? A 2 =? (a) Sketch the general shape of the nozzle section. For air, R = 286:9 J / kg K. 19

26 Given P 1 and T 1, the thermodynamic state at Station (1) is fully specied. We can use the thermal equation of state P = R T to compute the ow density, which gives Assume one-dimensional ow, 1 = 0:775 kg/m 3 _m = 1 A 1 V 1 We obtain V 1 = 258 m/sec The speed of sound at Station (1) is given by q c 1 = k R T 1 = 601 m/sec This gives a Mach number of M 1 = 0:429 at Station (1). Going from subsonic ow (M 1 = 0:429) to supersonic ow (M 2 = 1:2), we need to pass through a converging-diverging nozzle. (b) If the ow is further treated as isentropic, i. What is the downstream cross-sectional area (A 2 ), temperature (T 2 ) and pressure (P 2 )? For an isentropic ow, the following quantities are constant in the entire ow: A ; P 0 ; T 0 ; 0 They are invariant in the ow. Cross-sectional area (A 2 ), static temperature (T 2 ) and pressure (P 2 ) can all be computed in a similar manner as follows: A 2 A 1 = A 2=A 2 A 1 =A 1 = 1: :50072 = 0:6866 ) A 2 = 34:3 cm 2 T 2 = T 2=T 0;2 T 1 T 1 =T 0;1 = 0: :96434 = 0:8051 ) T 2 = 725 K P 2 = P 2=P 0;2 P 1 P 1 =P 0;1 = 0: :88065 = 0:4683 ) P 2 = 93:7 kpa ii. Sketch the variation of pressure, temperature, velocity and Mach number from Station (1) to Station (2). As the ow passes through the converging-diverging nozzle, velocity rises (nozzle is a ow accelerator), pressure drops (as a result of velocity rise, from momentum equation), temperature drops (as a result of velocity rise, from COE), Mach number rises (as a result of velocity rise and temperature drop). 20

27 iii. Trace the process path from Station (1) to Station (2) on a T s diagram. T P 0,1 = P 0,2 T 0,1 = T 0,2 P 1 T 1 P 2 T 2 s 1 = s 2 s 10. Air ows through a passage of unknown shape. The upstream state is characterized as follows: A 1 = 0.50 m 2 M 1 = 0.70 P 1 = 5.0 bar T 1 = 270 K 1 2 Passage with unknown shape A 2 = 1.0 m 2 M 2 =? P 2 =? T 2 =? Assume that the ow is isentropic. (a) Determine the possible downstream State (2) where A 2 = 1:0 m 2 (i.e. Find M 2 ; P 2 ; T 2.) This problem is very similar to the last one except that the Mach number at the downstream Station (2) is unknown. The only information we have about Station (2) is its cross-sectional area. Before we tackle this problem, let us examine the variation of A=A with Mach number (refer to Section 13-3 in Fox & McDonald). A A = 1 M 1 + k 1 2 M k 1 2! k ( k 1 ) 21

28 The above equation can be described graphically in the following gure: A / A * M From the above gure, we observe that there are two possible solutions to this question, namely a subsonic solution and a supersonic solution. Both solutions have the same A=A ratio: A 2 A 2 = A 2 A 1 A 1 A 1 A 1 A 2 = 1:0 0:50 (1:09437) (1) = 2:18874 From the isentropic ow table, { subsonic solution is M 2 = 0:277, { supersonic solution is M 2 = 2:298. To nd the pressure at Station (2), we can use the following procedure: P 2 P 1 = P 2 P 0;2 P 0;2 P 0;1 P 0;1 P 1 = P 2 P 0;2 (1) 1 0:72093 = 1:3871 P 2 P 0;2 Similarly, the temperature at Station (2) can be found in a similar manner: T 2 T 1 = T 2 T 0;2 T 0;2 T 0;1 T 0;1 T 1 = T 2 T 0;2 (1) For the subsonic solution (M 2 = 0:277), 1 0:91075 P 2 P 0;2 = 0:9481 ) P 2 = 6:58 bar T 2 T 0;2 = 0:9849 ) T 2 = 292 K For the supersonic solution (M 2 = 2:298), P 2 P 0;2 = 0:08025 ) P 2 = 55:7 kpa T 2 T 0;2 = 0:4864 ) T 2 = 144 K = 1:0980 T 2 T 0;2 22

29 (b) Sketch the shape of the associated ow passage found in Part (a). For the subsonic solution, the ow passage will be a diverging one. It serves as a diuser. For the supersonic solution, the ow needs to go from subsonic to supersonic. Hence, a converging-diverging passage will be necessary. It serves as a nozzle. (c) Sketch the corresponding variation of density, velocity and Mach number from Station (1) to Station (2). For the subsonic solution, { velocity drops (ow through a diuser), { pressure rises (as a result of velocity drop, from momentum equation), { temperature rises (as a result of velocity drop, from COE), { density rises (as a result of pressure rise and isentropic ow), { Mach number drops (as a result of velocity drop and temperature rise). For the supersonic solution, { velocity rises (ow through a nozzle), { pressure drops (as a result of velocity rise, from momentum equation), { temperature drops (as a result of velocity rise, from COE), { density drops (as a result of pressure drop and isentropic ow), { Mach number rises (as a result of velocity rise and temperature drop). 23

30 ME 131B Fluid Mechanics Tutorial Three: Isentropic Flow II 1. From an energy view point, (a) a nozzle is a device that converts into. (b) a diuser is a device that converts into. 2. For a steady, quasi-one-dimensional, adiabatic ow without wall friction, what do the following principles simplify to: (a) Conservation of Mass: (b) Momentum Equation: (c) Conservation of Energy: (d) Second Law of Thermodynamics: 3. We have discussed how to locate the stagnation state of a given ow state (T 1 ; P 1 ; M 1 ) last week. How about its sonic ( ) state? Can you locate it on the T s diagram? T P 1 T 1 s 24

31 4. Complete the following table with increases, decreases, remains constant for an isentropic ow: P T V c M P 0 0 T 0 A P T P T V c M P 0 0 T 0 A P T Subsonic Flow: Converging Channel Diverging Channel Supersonic Flow: Converging Channel Diverging Channel Trace each process path on a T s and a P diagram. 25

32 5. Choose the best answer in the following questions which concern the sonic state in an adiabatic, non-isentropic ow: (a) T decreases/increases/remains constant in the ow. (b) P decreases/increases/remains constant in the ow. (c) A decreases/increases/remains constant in the ow. 6. A large supply chamber containing air at 6.0 atm and 300 K is connected to a converging nozzle on the left side and a converging-diverging (C-D) nozzle on the right side. Both nozzles share the same minimum passage area of 100 cm 2. The C-D nozzle has an exit-to-throat area ratio of 1.2. P amb B P = 6.0 atm T = 300 K D C A Converging Nozzle C D Nozzle (a) Let us consider the converging nozzle on the left. i. Compare the pressure level at Point A, B, C and D. ii. If the ambient pressure is reduced to 5.0 atm, what is the mass ow rate in the nozzle? iii. How much do we need to lower the ambient pressure (relative to the chamber pressure) to reach the choking point of this converging nozzle? iv. What is the corresponding mass ow rate at the choking condition? (b) Let us consider the C-D nozzle on the right. i. If the ambient pressure is set at 5.0 atm, do you expect the mass ow rate in the C-D nozzle to be the same as that in the converging nozzle computed before? 26

33 ii. How much do we need to lower the ambient pressure for the nozzle to operate at its rst critical point? iii. What is the corresponding mass ow rate at the rst critical point? iv. At the design point (third critical), A. what is the ambient pressure? B. determine the density and velocity at the exit plane. (c) Look back to your calculations, i. How do you compare the ambient pressure which is required to choke the converging and C-D nozzle? Which one is higher? Can you explain it? ii. How do you compare the mass ow rate between the two nozzles: A. before choking? B. after choking? 7. You are asked to build a supersonic wind tunnel with operating Mach number of 2.0 in the test section. The plenum conditions are constantly kept at 300 K and 10.0 bars. Due to cost factor, air ow is delivered at a rate of 1 kg/sec. (a) If the ow is treated as isentropic, what is the downstream cross-sectional area? (b) If the entropy change between the plenum and the test section is 40 J/kg-K, what will be the cross-sectional area in the test section? Compare the result with Part (a) and label the two states on the same T s diagram. 27

34 ME 131B Fluid Mechanics Solutions to Tutorial Three: Isentropic Flow II 1. From an energy view point, (a) a nozzle is a device that converts static enthalpy into kinetic energy. (b) a diuser is a device that converts kinetic energy into static enthalpy. In the absence of heat transfer and non-ow work, the stagnation enthalpy of the ow is constant. 2. For a steady, quasi-one-dimensional, adiabatic ow without wall friction, what do the following principles simplify to: (a) Conservation of Mass: V A = constant Remarks: The mass ow rate (product of density, velocity and area) is required to be a constant to conserve mass. In subsonic ow, the change in density is not so drastic. Velocity varies in a way which is consistent with our everyday experience. But the situation is so much dierent in the supersonic regime. Density change becomes very appreciable. Take the case of a supersonic ow in a converging passage: the density increase outweighs the area decrease and forces velocity to go down in order to conserve mass ow. Similar behavior is found in a supersonic ow in a diverging passage but opposite eects are observed. In summary, the \strange" behavior of supersonic ow is caused by the appreciable density change. It seems counter-intuitive because the world we encounter with on a daily basis operates mostly in the incompressible regime. (b) Momentum Equation: dp = V dv Remarks: Pressure and velocity change in opposite direction to each other in both subsonic and supersonic regimes. (c) Conservation of Energy: h 0 = h + V = constant

35 Remarks: When the ow speeds up, the uid cools down and vice versa. This interchange between static enthalpy and kinetic energy is fundamental in understanding an adiabatic ow. (d) Second Law of Thermodynamics: s = constant 3. We have discussed how to locate the stagnation state of a given ow state (T 1 ; P 1 ; M 1 ) last week. How about its sonic ( ) state? Can you locate it on the T s diagram? By denition, the sonic ( ) state should have the same specic entropy as its static state. Hence, they should both be on the same vertical line on the T s diagram. By denition, the Mach number of the sonic state is unity. Hence, its location relative to that of its static state depends on the ow Mach number. If the ow is subsonic (M < 1), the sonic state will be below its static state. If the ow is supersonic (M > 1), the sonic state will be above its static state. Based on the above conclusions, we can locate the sonic state relative to its static state on a T s diagram in the following gures: Subsonic Case Supersonic Case T P 1 P * 1 T P * 1 T 1 P 1 T * 1 T * 1 T 1 s 1 s s 1 s 29

36 4. Complete the following table with increases, decreases, remains constant for an isentropic ow: Subsonic Flow: Converging Channel Diverging Channel P decreases increases decreases increases T decreases increases V increases decreases c decreases increases M increases decreases P 0 remains constant remains constant 0 remains constant remains constant T 0 remains constant remains constant A remains constant remains constant P remains constant remains constant remains constant remains constant T remains constant remains constant Remarks: Supersonic Flow: Converging Channel Diverging Channel P increases decreases increases decreases T increases decreases V decreases increases c increases decreases M decreases increases P 0 remains constant remains constant 0 remains constant remains constant T 0 remains constant remains constant A remains constant remains constant P remains constant remains constant remains constant remains constant T remains constant remains constant Both the stagnation state and the sonic state are constant in an isentropic ow. They serve as convenient reference states for the ow. 30

37 Trace each process path on a T s and a P diagram. Subsonic Case Supersonic Case T T Diverging T * 1 T 1 Converging T 1 Converging Diverging T * 1 s 1 s s 1 s P P ρ k = constant P P ρ k = constant P * 1 P 1 Diverging P 1 P * 1 ρ * 1 Converging ρ 1 ρ Diverging ρ 1 Converging ρ * 1 ρ 5. Choose the best answer in the following questions which concern the sonic state in an adiabatic, non-isentropic ow: (a) T remains constant in the ow. For any specic gas, the ratio of stagnation temperature to sonic temperature is a constant: T 0 T = 1 + k 1 2 Since the stagnation temperature of an adiabatic ow is constant, so is the sonic temperature. (b) P decreases in the ow. For any specic gas, the ratio of stagnation pressure to sonic pressure is a constant: P 0 P = 1 + k 1! k k

38 Since the stagnation pressure of an adiabatic, non-isentropic ow decreases in the ow direction, so does the sonic pressure. (c) A increases in the ow. To see this point clearly, we can evaluate the mass ow rate at the sonic point: _m = A c where c = p k R T : We know from previous results that T ; c remain constant but P decreases in the ow. This leads us to conclude that decreases in the ow also, from the ideal gas equation. Hence, A has to increase to conserve the same mass ow rate. 6. A large supply chamber containing air at 6.0 atm and 300 K is connected to a converging nozzle on the left side and a converging-diverging (C-D) nozzle on the right side. Both nozzles share the same minimum passage area of 100 cm 2. The C-D nozzle has an exit-to-throat area ratio of 1.2. P amb B P = 6.0 atm T = 300 K D C A Converging Nozzle C D Nozzle (a) Let us consider the converging nozzle on the left. i. Compare the pressure level at Point A, B, C and D. The main point of this part is to visualize the pressure variation and uid acceleration within the supply chamber. When the uid \senses" the pressure dierential between the inner chamber, P 0, and the surrounding ambient, P amb, it accelerates from negligible velocity at chamber pressure to some nite velocity closed to the nozzle inlet. Associate with this ow acceleration, there is a corresponding pressure drop. We can treat the ow going through an \imaginary" converging from the inner chamber to the nozzle inlet. passage 32

39 Between the inlet and exit, the ow continues to accelerate and pressure continues to drop. We can conclude that P A > P C > P D The cause of pressure dierence between Point B and C is apparent after we draw the streamlines around the inlet of the converging nozzle. Since the streamlines curve around the corner, there is a positive pressure gradient developed in the normal (to the streamline) direction. Hence, P C > P B The comparison between pressure level at Point B and D depends on the exact nozzle geometry and requires further quantitative analysis. ii. If the ambient pressure is reduced to 5.0 atm, what is the mass ow rate in the nozzle? In this type of problem, we always need to check if the converging nozzle is choked at P amb = 5:0 atm. For a converging nozzle, we learned that the ambient pressure has to be lower than 52.8 % of the chamber pressure before choking occurs. In this case, P amb P 0 = 5:0 6:0 = 0:833 > 0:528 Hence, the nozzle is not choked. Furthermore, we can conclude that the pressure at the exit plane is the same as the ambient value. For the given pressure ratio P exit P 0 = 0:833 We can nd out from the isentropic ow table that And the temperature ratio is M exit = 0:517 T exit T 0 = 0:94924 which gives an exit temperature of T exit = 284:8 K. Using the thermal equation of state for an ideal gas P = R T we obtain an exit density of exit = 6:201 kg / m 3. 33

40 The mass ow rate can then be computed by _m = exit V exit A exit = exit M exit qk R T exit A exit ) _m = 10:8 kg/sec iii. How much do we need to lower the ambient pressure (relative to the chamber pressure) to reach the choking point of this converging nozzle? For a converging nozzle, the ambient pressure has to be lower than 52.8% of the chamber pressure to choke the converging nozzle. This corresponds to an ambient pressure of P amb 3:17 atm If P amb is lower than 3.17 atm, the exit plane pressure will not be the same as the ambient value (pressure mismatch). P amb will keep staying at 3.17 atm. This is because no downstream pressure information can propagate upstream past the sonic point (exit plane). The ow within the nozzle becomes invariant once the sonic condition is attained at the exit. iv. What is the corresponding mass ow rate at the choking condition? When P amb = 3:17 atm, the Mach number at the exit plane just reaches unity. Pressure at the exit plane equals to the ambient pressure P amb = 3:17 atm From the isentropic ow table, we obtain T exit T 0 = 0:8333 ) T exit = 250 K Using the ideal gas equation, we obtain The mass ow rate is (b) Let us consider the C-D nozzle on the right. exit = 4:478 kg/m 3 _m = 14:2 kg/m 3 i. If the ambient pressure is set at 5.0 atm, do you expect the mass ow rate in the C-D nozzle to be the same as that in the converging nozzle computed before? 34

41 For this C-D nozzle case, we also need to check if the nozzle is choked at P amb = 5:0 atm. The main dierence between the C-D nozzle and the converging nozzle is that the choking pressure ratio is dependent on the exit-to-throat area ratio (not a universal constant anymore). With an area ratio of 1.20, we nd from the isentropic ow table that the subsonic solution gives a pressure ratio P = 0:78997 < 5:0 P 0 6:0 Hence, we conclude that { the ambient pressure is high enough that the ow is not choked { the ow remains subsonic within the C-D nozzle { exit ; M exit ; T exit are the same as those in the converging nozzle case Since the exit area is 1.20 times as large as that of the converging nozzle, we expect a 20 % increase in the mass ow rate. Hence, _m = 13:0 kg/m 3 ii. How much do we need to lower the ambient pressure for the nozzle to operate at its rst critical point? The rst critical point corresponds to an isentropic, subsonic solution with Mach 1.0 ow at the throat. We obtain from the isentropic ow table that P amb P 0 = 0:78997 ) P amb = 4:74 atm iii. What is the corresponding mass ow rate at the rst critical point? Once this converging-diverging nozzle is choked at its rst critical point, we know that Mach 1.0 is achieved at its minimum ow area, i.e. at the throat. Furthermore, P throat ; T throat ; M throat are the same as those of the converging nozzle choked case. Hence, we expect the same mass ow rate as that of the converging nozzle choked case _m = 14:2 kg/m 3 iv. At the design point (third critical), A. what is the ambient pressure? The third critical point corresponds to an isentropic, supersonic solution in the C-D nozzle. 35

42 For an area ratio of 1.20, we obtain a supersonic solution from the isentropic ow table M exit = 1:534 This solution gives a pressure ratio of P amb P 0 = 0:25922 ) P amb = 1:55 atm B. determine the density and velocity at the exit plane. For the M exit = 1:534 solution, we obtain a temperature ratio of T exit T 0 = 0:67995 ) T exit = 204 K Using the ideal gas model, we obtain (c) Look back to your calculations, exit = 2:693 kg/m 3 and V exit = 439 m/sec i. How do you compare the ambient pressure which is required to choke the converging and C-D nozzle? Which one is higher? Can you explain it? For the converging nozzle, P choke = 3:17 atm. For the C-D nozzle, P choke = 4:74 atm. We conclude that the C-D nozzle is choked at a higher ambient pressure than the converging nozzle. This conclusion can be explained by the following pressure plot for C-D nozzle operation: P amb / P First Critical (C D nozzle) Choking Point (Converging nozzle) Third Critical (C D nozzle) x Due to the pressure recovery in the diverging section of a C-D nozzle (subsonic ow), the C-D nozzle is choked at a higher back-to-plenum pressure ratio. The exact value of the this pressure ratio depends only on the exit-to-throat area ratio. 36

43 In summary, there are three operating regimes: A. P amb =P 0 > 0:790, both nozzles are not choked. B. 0:790 > P amb =P 0 > 0:528, only the C-D nozzle is choked. C. P amb =P 0 < 0:528, both nozzles are choked. ii. How do you compare the mass ow rate between the two nozzles: A. before choking? Before any choking occurs, the C-D nozzle has a higher mass ow rate (20 % higher) than the converging nozzle simply because the exit area of the C-D nozzle is 20 % larger than that of the converging nozzle. As the ambient pressure is reduced, the C-D nozzle gets choked rst. Once it is choked, its mass ow rate is not aected by the ambient pressure anymore. Meanwhile the mass ow rate of the converging nozzle keeps increasing as the ambient pressure is reduced. B. after choking? The mass ow rate is the same in both nozzles after they are both choked. Graphically, the mass ow rate of the two nozzles can be compared as follows:. m 14.2 kg/sec C D Nozzle Converging Nozzle P amb / P 0 7. You are asked to build a supersonic wind tunnel with operating Mach number of 2.0 in the test section. The plenum conditions are constantly kept at 300 K and 10.0 bars. Due to cost factor, air ow is delivered at a rate of 1 kg/sec. (a) If the ow is treated as isentropic, what is the downstream cross-sectional area? To achieve a supersonic ow in the test section, we need a C-D nozzle connecting the plenum and the test section, with a Mach 1.0 ow (sonic state) right at the minimum throat area. 37

44 From the isentropic ow table, we obtain P P 0 = 0:52828 ) P = 5:283 bar T T 0 = 0:83333 ) T = 250 K The ideal gas equation further gives us the density at the throat: = 7:365 kg/m 3 The size of the throat can be found from the mass ow rate equation _m = V A throat = p k R T A throat ) A throat = 4:285 cm 2 To nd out the cross-sectional area at the test section, we need to relate the Mach 2.0 ow in the test section with the sonic state at the throat. From the isentropic ow table, M section = 2:0 gives A section A throat = 1:68750 ) A section = 7:230 cm 2 (b) If the entropy change between the plenum and the test section is 40 J/kg-K, what will be the cross-sectional area in the test section? Compare the result with Part (a) and label the two states on the same T s diagram. The ow remains to be adiabatic. Hence, T 0 ; T remains constant even in this non-isentropic ow. However, the entropy increase in the nozzle causes a drop in the stagnation pressure. (Take State 1 to be the plenum state and State 2 to be the test section state in the following analysis.)! s 2 s 1 = C p log T 0;2 T 0;1 P ) 0;2 s2 s 1 = exp P 0;1 R = 0:870 ) P 0;2 = 8:70 bar From the isentropic ow table, M 2 = 2:0 gives! R log P 0;2 P 0;1 P 2 P 0;2 = 0:12780 ) P 2 = 1:11 bar T 2 T 0;2 = 0:55556 ) T 2 = 167 K 38

45 The ideal gas equation further gives us the density in the test section: 2 = 2:32 kg/m 3 The cross-sectional area in the test section can be found from the mass ow rate equation _m = 2 V 2 A 2 = 2 M 2 qk R T 2 A 2 ) A 2 = 8:31 cm 2 Comparing this result with that of Part (a), we conclude that a larger test section area is necessary when irreversible eects are taken into account. It is also interesting to point out that the ratio between the area obtained in Part (a) and Part (b): A section;a A section;b = 7:230 cm2 8:310 cm 2 = 0:870 is the same ratio as the stagnation pressure loss. Hence, we conclude from this observation that P 0 A = constant for an adiabatic ow. When stagnation pressure drops, the sonic area increases. Both the isentropic and non-isentropic solutions are shown in the following T s diagram for reference: T P 2,a P 2,b T * = T* 2,a 2,b T = T 2,a 2,b Isentropic Solution Non Isentropic Solution s 2,a s 2,b s Comments: The ridiculously small test section area is not reasonable for conducting wind tunnel experiments. We can increase its size by { reducing the plenum pressure { increasing the plenum temperature { paying a higher cost to allow a higher mass ow rate 39

46 ME 131B Fluid Mechanics Tutorial Four: Normal Shock 1. In normal shock analysis, what do the following principles simplify to: (a) Conservation of Mass: (b) Momentum Equation: (c) Conservation of Energy: (d) Second Law of Thermodynamics: 2. In what frame of reference are the shock jump relationships derived in? 3. How is the strength of a shock measured by? What does it depend on? 4. Derive the relationship between the entropy production within a shock and the corresponding loss in stagnation pressure across it. 5. Choose the best answer in the following question and explain it: The Mach number upstream of a shock can/cannot be less than unity. 40

47 6. Complete the following table with increases, decreases, remains constant for a ow across a normal shock: P T V c M P 0 0 T 0 A P T Across a Normal Shock Trace the process path across a shock on a T s diagram. (Indicate clearly its position relative to the stagnation state and the sonic state.) 7. Moving Shock Problem: A shock wave is propagating at Mach 2.0 into still air which is at atmospheric condition. What is the change of stagnation pressure observed by someone standing on the ground? How about the change in stagnation temperature? P 0 =? T 0 =? M = 2.0 Still air at atm. condition Moving shock 41

48 8. Shock Location Specication Problem: Air enters a C-D nozzle which has an exit-tothroat area ratio of 1.8 (A 5 =A 2 = 1:8). A normal shock occurs at a location where the cross-sectional area is 1.2 times that of the throat (A 3 =A 2 = 1:2). The schematic is shown below: (3) (4) (1) (2) Shock (5) (a) Sketch the pressure variation with streamwise location. (b) Locate the static and stagnation states of 1, 2, 3, 4 and 5 on a T s diagram. (c) What is the operating pressure ratio P 5 =P 0;1? 9. Back Pressure Specication Problem: Let us return to the C-D nozzle (with an exitto-throat area ratio of 1.2) we worked with last week. It connects again the same air supply chamber at 300 K and 6.0 bars to the ambient. P = 6.0 atm T = 300 K P amb C D Nozzle 42

49 (a) Last week, we have found that the rst critical point operates at P amb =P 0 = 0:78997 and the third critical point operates at P amb =P 0 = 0: What is the ambient-to-chamber pressure ratio at the second critical point? (b) We have learned last week how to compute the mass ow rate for both choked and unchoked cases. Let us focus on some o-design operations of this C-D nozzle this time. Describe qualitatively what happens inside the nozzle for the following ambient pressure values: i. P amb = 5:4 bars ii. P amb = 4:5 bars iii. P amb = 4:2 bars iv. P amb = 2:4 bars v. P amb = 0:6 bars (c) In the cases where standing normal shock occurs in the nozzle, compute its location in terms of area ratio with respect to the throat. (d) Look back to your calculations, how do you compare the shock location for different ambient pressure? Do you expect the shock to be closer to the throat or to the exit for a high ambient pressure? 43

50 ME 131B Fluid Mechanics Solutions to Tutorial Four: Normal Shock 1. In normal shock analysis, what do the following principles simplify to: (a) Conservation of Mass: For a steady, quasi-one-dimensional analysis, the mass conservation equation is simply V A = constant. Shock thickness is usually of the order of a few microns which is much shorter than most representative length scales in the ow. Hence, we can treat the ow cross-sectional area to be constant across the shock. This reduces the mass conservation equation to V = constant (b) Momentum Equation: Apply the results from mass conservation, we can directly integrate the momentum equation dp = V dv to obtain the following results: P + V 2 = constant (c) Conservation of Energy: The shock process is adiabatic. h 0 = h + V 2 2 = constant (d) Second Law of Thermodynamics: The shock process is irreversible. Entropy is produced. s > 0 2. In what frame of reference are the shock jump relationships derived in? The shock jump relationships are derived in the shock frame of reference. Recall that the ow is treated as steady in the derivation, which is made possible by a Galilean transformation from a stationary observer to one traveled with the shock. We need to be careful when we apply these jump relationships to problems in which the shock is propagating. A change of reference frame to that of the shock will be necessary. 44

51 Associated with this change of reference frame, the stagnation properties will be dierent but the static properties will remain the same. 3. How is the strength of a shock measured by? What does it depend on? The strength of a shock can be measured by the pressure jump across it: = P 2 P 1 P 1 where P 1 and P 2 are the static pressure upstream and downstream of the shock respectively. The larger the pressure jump, the stronger the shock is. The strength of a shock is dependent on the upstream Mach number: P 2 = 2 k M 2 1 ( k 1 ) P 1 k + 1 ) = 2 k ( M ) k + 1 The variation of shock strength with upstream Mach number is shown graphically in the following gure (for the k = 1:4 case): 10 8 (P2 - P1) / P Sound Wave Limit M 1 As shown in the gure, the strength of a shock increases with the upstream Mach number in a non-linear manner. Much higher pressure jump can be achieved at high Mach numbers. More quantitative results are tabulated in the following table for further reference: M

52 Remarks: We can treat sound wave (M = 1) as a shock wave with zero strength. 4. Derive the relationship between the entropy production within a shock and the corresponding loss in stagnation pressure across it. We can apply the Gibbs equation to evaluate the pressure drop between the upstream and downstream stagnation states. T 0 ds 0 = dh dp 0 The shock process is an adiabatic process, hence, the stagnation enthalpy (h 0 ) is constant across the shock ) d h 0 = 0 The Gibbs equation is then reduced to ds 0 = = 1 dp 0 0 T 0 R dp 0 P 0 Direct integration of the above equation between the upstream State (1) and the downstream State (2) gives! s 0;2 s 0;1 = R log P 0;2 P 0;1 Since a stagnation state has the same specic entropy as its static state, i.e. s 0;1 = s 1 ; s 0;2 = s 2 we can also write the previous results as follows:! s 2 s 1 = R log P 0;2 P 0;1 The above equation relates the stagnation pressure drop to the entropy increase across a shock. 5. Choose the best answer in the following question and explain it: The Mach number upstream of a shock cannot be less than unity. 46

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