Gas Dynamics and Jet Propulsion (For B.E. Mechanical Engineering Students) (As per Anna University and Leading Universities New Revised Syllabus) Prof. K. Pandian Dr. A.Anderson, M.E., Ph.D., Professor - Mechanical Dr. S.Ramachandran, M.E., Ph.D., Professor and Head Department of Mechanical Engineering Sathyabama University Jeppiaar Nagar, Chennai - 600 119 AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai - 600 004. Ph.: 466 1909, 94440 81904 Email: aishram006@gmail.com
First Edition: July 003 Second Edition: 16-11-013 All Rights Reserved by the Publisher This book or part thereof should not be reproduced in any form without the written permission of the publisher. Dedicated to Copies can be had from : Our Parents AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai - 600 004. Ph.: 466 1909, 9444 08 1904 Books available in all book stalls with attractive discounts Books will be door delivered after payment into AIR WALK PUBLICATIONS A/c No. 80160100001454 (IFSC: BKID0008016) Bank of India, Santhome branch, Mylapore, Chennai - 4 (or) S.Ramachandran, A/c.No.48894441 (IFSC:IDIB000S01), Indian Bank, Sathyabama University Branch, Chennai - 600119. Typeset by: aksharaa muthra aalayam, Chennai - 18. Ph.: 044-436 4303 Printed at: Abinayaram Printers, Chennai - 4. Ph.: 044-466 1909, 9444 08 1904
Contents 1 CONTENTS 1. FUNDAMENTAL OF GAS DYNAMICS 1.1 Energy Equation... 1. 1. Stagnation State and Stagnation Properties 1.6 1.3 Bulk Modulus of Elasticity... 1.9 1.4 Various Regions of Flow... 1.11 1.5 Acoustic Velocity or Sound Velocity... 1.14 1.6 Mach Angle and Mach Cone... 1.17 1.7 Reference Velocities... 1.0 Problems...1.7 to 1.49 1.8 Bernoulli s Equation... 1.49 1.9 Effect of Mach number on Compressibility.. 1.5 Problems and Answers...1.56 to 1.60 Questions and Problems...1.61 to 1.64. ISENTROPIC FLOW WITH VARIABLE AREA.1 Comparison between Isentropic and Adiabatic Processes....1. Mach Number Variation....3..1 Expansion in Nozzles....5.. Compression in Diffusers....6.3 Stagnation and Critical States....7.4 Area Ratio as a Function of Mach Number.10.5 Impulse Function....13
Gas Dynamics and Jet Propulsion.6 Mass Flow Rate....16 In terms of Pressure Ratio....17 In terms of Area Ratio....0 In terms of Mach Number....1 Numerical Value of Non-dimensional Maximum Mass Flow Parameter....3.7 Flow through Nozzles....3 Convergent Nozzles....4 Convergent - Divergent Nozzles....6 Over Expanding and Under Expanding.9 Nozzle Efficiency....9.8 Flow through Diffusers....9 Diffuser Efficiency....30 Problems and Answers...31 to.95 Questions and Problems...95 to.99 3. FLOW WITH NORMAL SHOCK WAVES 3.1 Development of a Normal Shock Wave... 3.1 3. Governing Equations... 3.3 Fanno Line... 3.5 Rayleigh Line... 3.7 3.3 Prandtl-Meyer Relation... 3.9 3.4 Mach Number Downstream of the Normal Shock Wave... 3.1 3.5 Static Pressure Ratio across the Shock... 3.14
Contents 3 3.6 Temperature Ratio across the Shock... 3.16 3.7 Density Ratio across the Shock... 3.17 3.8 Stagnation Pressure Ratio across the Shock 3.1 3.9 Change in Entropy across the Shock... 3.4 3.10 Impossibility of Rarefaction Shock Wave... 3.6 3.11 Strength of a Shock Wave... 3.7 3.1 Supersonic Wind Tunnels... 3.8 Problems and Answers...3.30 to 3.77 Questions and Problems...3.78 to 3.80 4. FLOW IN CONSTANT AREA DUCTS WITH FRICTION 4.1 Fanno Curves... 4.1 4. Fanno Flow Equations... 4.4 4.3 Solution of Fanno Flow Equations... 4.10 4.4 Variation of Flow Properties... 4.14 4.5 Variation of Mach Number with Duct Length... 4.19 4.6 Isothermal Flow in a Constant Area Duct with Friction... 4.0 Isothermal Flow Equations... 4. Variation of Flow Properties... 4.4 Problems and Answers...4.6 to 4.70 Questions and Problems...4.71 to 4.74
4 Gas Dynamics and Jet Propulsion 5. FLOW IN CONSTANT DUCTS WITH HEAT TRANSFER AND WITHOUT FRICTION (RAYLEIGH FLOW) 5.1 Rayleigh Line... 5.1 Slope of the Rayleigh Line... 5.4 Constant Entropy Lines... 5.4 Constant Enthalpy Lines... 5.6 5. General Equations in Rayleigh Flow Process... 5.8 5.3 Rayleigh Flow Relations... 5.10 5.4 Variation of Flow Properties... 5.14 5.5 Maximum Heat Transfer... 5.14 Problems and Answers...5.16 to 5.54 Questions and Problems...5.54 to 5.56 6. JET PROPULSION 6.1.1 Turbo Jet Engine... 6.1 6.1. Turbo Prop Engine... 6.4 6.1.3 Ram Jet Engine... 6.5 6.1.4 Pulse Jet or Flying Bomb... 6.8 6. Energy Relations and Efficiencies in a Turbo Jet Engine... 6.9 6.3 Thrust... 6.17 6.4 Propulsive, Thermal and Overall Efficiencies... 6.0 6.5 Specific Fuel Consumption... 6.3
Contents 5 6.5.1 Specific Thrust... 6.3 Specific Impulse... 6.4 6.6 Effect of Forward Speed... 6.4 6.7 Effect of Altitude... 6.4 6.8 Thrust Augmentation... 6.5 Problems and Answers...6.6 to 6.73 Questions and Problems...6.73 to 6.78 7. ROCKET PROPULSION 7.1 Comparison between Air Breathing Engines with Rocket Engines... 7.1 7. Classification of Rocket Engines... 7. 7.3.1 Solid Propellant Rockets... 7.3 7.3. Liquid Propellant Rockets... 7.6 7.3.3 Hybrid Propellant Rockets... 7.8 7.4 Liquid Propellants... 7.10 7.4.1 Mono Propellants... 7.10 7.4. Bipropellants... 7.11 7.4.3 Fuel... 7.1 7.4.4 Oxidizers... 7.1 7.4.5 Properties of Liquid Propellants... 7.14 7.5 Solid Propellants... 7.14 7.6 Restricted Burning... 7.15 7.7 Thrust and specific Impulse... 7.16 Specific Propellant Consumption... 7.19
6 Gas Dynamics and Jet Propulsion Total Impulse... 7.19 Weight Flow Co-efficient... 7.19 Thrust Co-efficient... 7.0 Impulse to Weight Ratio... 7.0 7.8 Propulsive, Thermal and Overall Efficiency 7.1 7.9 Applications of Rocket Engines... 7.3 7.10 RATO (or) JATO... 7.3 Problems and Answer...7.4 to 7.54 Questions and Problems...7.54 to 7.56 Marks Q &A University Solved Problems
Fundamentals of Gas Dynamics 1.1 Chapter - 1 Fundamentals of Gas Dynamics Gas dynamics deals with the study of compressible fluid flow when it is in motion. It analyses the high speed flows of gases and vapours with considering its compressibility. Applications The applications of Gas Dynamics are (i) used in steam and Gas turbines (ii) high speed aerodynamics (iii) Jet and Rocket propulsion (iv) high speed turbo compressors etc. The fluid dynamics of compressible flow problems which involves the relation between force, velocity, density and mass etc. Therefore, the following laws are frequently used for solving the Gas Dynamic problems. (i) (ii) Steady flow energy equation [derived from first law of Thermodynamics] Entropy relations [derived from second law of Thermodynamics] (iii) Continuity equation [derived from law of conservation of mass] (iv) Momentum equation [derived from Newton s second law of motion]
1. Gas Dynamics and Jet Propulsion Compressible Flow Incompressible Flow (i) The fluid velocities are appreciable compared with the velocity of sound. (ii) The fractional variation in density is significant i.e., the density is not constant. (iii) The fractional variations in temperature and pressure are all of significant magnitude (iv) Compressibility factor is greater than one. (i) The fluid velocities are small compared with the velocity of sound. (ii) The fractional variation in density are so small as to be negligible i.e., the density is constant. (iii) The variation in pressure and temperature may be very large (iv) Compressibility factor is one 1.1 ENERGY EQUATION The first law of thermodynamics states that when a system executes a cyclic process, the algebric sum of work transfers is proportional to the algebric sum of heat transfers. i.e., O dw Od Q O dw J O dq When heat and work terms are expressed in the same units
Fundamentals of Gas Dynamics 1.3 O dq O dw 0 The quantity dq and dw will follow the path function, but the quantity dq dw does not depend on the path of the process. Therefore, the change in quantity dq dw is a property called Energy (E). Thus de dq dw 1 de 1 dq 1 dw E E 1 Q W Q W E E 1... (1.1) In the above equation, the energy term E may includes kinetic energy, internal energy, gravitational potential energy, strain energy, magnetic energy, etc., By ignoring magnetic energy and strain energy the energy term E may be written as E U mg Z 1 mc... (1.) The differential form of Equation (1.) is de du mg dz 1 m c dc Integrating the above equation between the limits 1 and, then the equation becomes, E E 1 U U 1 mg Z Z 1 1 m C C1... (1.3)
1.4 Gas Dynamics and Jet Propulsion substituting (1.3) in equation (1.1) a general form of energy equation is obtained. i.e., Q W U U 1 mg Z Z 1 1 m C C 1... (1.4) 1.1.1. Energy equation for a flow process A change or a series of changes in an open system is known as a flow process examples are (i) (ii) Flow through nozzles, diffusers and ducts etc, expansion of steam and gas in turbines (iii) Compression of air and gases in turbo compressors etc. In such flow processes the work term W includes the flow work also i.e., W W s p v p 1 v 1 Shaft work Flow work... (1.5) substituting this in equation (1.4), we get Q W s P V P 1 V 1 U U 1 mg Z z 1 1 m C C1 but, we know that the enthalpy H U PV Q W s H H 1 mg Z mg Z 1 1 m C 1 m C 1 H 1 mg Z 1 1 mc 1 Q H mg Z 1 1 mc W s... (1.6) h 1 gz 1 C 1 q h gz C w s... (1.7) [per unit kg mass]
Fundamentals of Gas Dynamics 1.5 This is a steady flow energy equation which is generally used in flow problems of gases and vapours. 1.1. Adiabatic energy equation Compared to other quantities, the change in elevation g z z 1 is negligible in flow problems of gases and vapours. In a reversible adiabatic process the heat transfer q is negligibly small and can be ignored. Expansion of gases and vapours in nozzles and diffusers are examples of such process. For such processes equation (1.7) is reduced to h 1 C 1 h C... (1.8) [ w s 0 ] 1.1.3 Adiabatic energy transfer and energy transformation In an adiabatic energy transfer process, the shaft work will present (eg) expansion of gases in turbines and compression in compressors etc., In an adiabatic energy transformation process, the shaft work is zero. e.g. expansion of gases in nozzles and compression in diffusers etc., The adiabatic energy equation (1.7) is valid for processes involving both energy transfer and energy transformation. The energy equation for compressors and turbines is h 1 C 1 h C w s... (1.9) The energy equation for a nozzle and diffuser is
1.6 Gas Dynamics and Jet Propulsion h 1 C 1 h C... (1.10) 1.. STAGNATION STATE AND STAGNATION PROPERTIES The state of a fluid attained by isentropically decelerating it to zero velocity at zero elevation is referred to as the stagnation state. It is often used as a reference state. The properties of the fluid at stagnation state are the stagnation properties of the fluid. eg. stagnation temperature, stagnation pressure, stagnation enthalpy etc., 1..1 Stagnation enthalpy [ h 0 ] Stagnation enthalpy of a gas or vapour is its enthalpy when it is adiabatically decelerated to zero velocity at zero elevation. As per the definition, At the initial state h 1 h : C 1 C and At the final state h h 0 : C 0 By substituting this in equation (1.10), we get h 0 h C... (1.11) where h 0 stagnation ethalpy and h static enthalpy In an adiabatic energy transformation process the stagnation enthalpy remains constant.
Fundamentals of Gas Dynamics 1.7 1.. Stagnation temperature (or) Total temperature [ T 0 ] Stagnation temperature of a gas or vapour is defined as the temperature when it is adiabatically decelerated to zero velocity at zero elevation. as For a perfect gas, the equation (1.11) can be written C p T 0 C p T C equation divide by C p throughout the T 0 T C C p...(1.1) where, T 0 Stagnation temperature T Static temperature C C p Velocity temperature From equation (1.1) T 0 T 1 C C p T C 1 1 RT 1 1 C a C p a RT M C a 1 R T 0 T 1 1 M... (1.13)
1.8 Gas Dynamics and Jet Propulsion T 0 1 1 M T 1..3 Stagnation pressure [ P 0 ] (or) Total pressure Stagnation pressure is the pressure of the gas when it is adiabatically decelerated to zero velocity at zero elevation. The adiabatic relation for a perfect gas is T 0 T 1 P 0 P P 0 P T 0 1 T 1 1 M 1... (1.14) 1..4 Stagnation velocity of sound [ a 0 ] We know that the acoustic velocity of sound a RT. For a given value of stagnation temperature, the stagnation velocity of sound a 0 RT 1..5 Stagnation density 0 For the given values of stagnation pressure and temperature, the stagnation density is given by 0 P 0 RT 0 or from adiabatic relation T 0 T P 0 P 1 0 1
Fundamentals of Gas Dynamics 1.9 0 T 0 T 1 1 1 1 1.3 Bulk modulus of elasticity k by M 1... (1.15) The bulk modulus of elasticity of a fluid is defined K increase in pressure relative change in volume dp dv v... (1.16) (minus sign is because, increase in pressure takes place by decreasing the volume) But, v 1... (a) by differentiating the equation a dv 1 d substituting this in equation (1.16) we get, K v dp d K a dp d dp d dp d a 1.3.1 Adiabatic bulk modulus K In a reversible adiabatic process for an ideal gas pv constant, by differentiating this,
1.10 Gas Dynamics and Jet Propulsion p v 1 dv v dp 0 p v v dv v dp 0 p v dv v v dp v dp dv p... (1.17) but we know that from equation (1.16) vdp dv K, therefore equation 1.17 becomes K p where K adiabatic bulk modulus. 1.3. Reynold s number R e It is a non-dimensional number and is defined as R e inertia force viscous force AC C l... (1.18) where, c fluid velocity dynamic viscosity l characteristic length 1.3.3 Mach number M It is a non-dimensional number and is defined as M inertia force elastic force
Fundamentals of Gas Dynamics 1.11 where K bulk modulus a M C a C a AC KA C K... (1.19) M C a... (1.0) This gives the another definition of mach number and is the ratio of fluid velocity (c) to the local velocity of sound (a). Mach number is an important parameter in the analysis of compressible fluid flows. 1.4 VARIOUS REGIONS OF FLOW The adiabatic energy equation for a perfect gas is derived in terms of fluid velocity (c) and sound velocity (a). Then it is plotted graphically in the x axis C and y axis a respectively. From adiabatic energy equation We know that, h 0 h C constant h C p T 1 R T a 1 a RT a RT
1.1 Gas Dynamics and Jet Propulsion a=a O I II M < 1 III a M=1 Sonic M > 1 IV V o By substituting this in equation (1.11), we get h 0 C=C max C Fig 1.1 Various regions of flow a 1 C constant... (1.1) At T 0 h 0 a 0 and C C max Therefore equation (1.1) becomes h 0 C max... (1.) At C 0 ; a a 0 Therefore, from equation (1.1) a 0 h 0 1 constant... (1.3) h 0 a 1 C C max a 0 1 constant... (1.4)
Fundamentals of Gas Dynamics 1.13 Equation (1.4) is an another form of adiabatic energy equation. By substituting different values of (c) and (a) in the above equation and by plotting the values, a steady flow ellipse is obtained. It is shown in Fig.1.1 and there are five different regions on the ellipse. (i) Incompressible flow region The region of flow close to the axis a is an incompressible flow region. The fluid velocity is much smaller than the sound velocity which is shown in fig. 1.1. Therefore the Mach number M 1 and is very close to zero. (ii) Subsonic flow region The region on the right of the incompressible region and upto a mach number is less than unity. (iii) Transonic flow region When the Mach number of flow is unity, the flow is sonic flow. A small region slightly less than unity and just above the sonic point is referred as transonic flow region. The mach number in this region is in between 0.8 to 1.. (iv) Supersonic flow region The region is on the right side of the transonic flow region. The mach number in this region is always above unity and up to 5. (v) Hypersonic flow region In this region, the flow velocity is very high compared to the sound velocity and hence the mach
1.14 Gas Dynamics and Jet Propulsion number is very high i.e., above 5. The region close to the c axis is called hypersonic region. 1.5 DERIVATION OF ACOUSTIC VELOCITY (or) SOUND VELOCITY It is the velocity of sound in a fluid medium or the speed with which a small disturbance is transmitted through the fluid. Consider a stationary fluid in an insulated cylinder fitted with a frictionless piston. The piston and gas in the tube are at rest originally at a pressure P. Let the parameters across the wave front (is a plane across which pressure and density changes suddenly and there will be a discontinuity in pressure, temperature and density) be as shown in Fig.1.. If a small impulse is given to the piston the gas immediately adjacent to the piston will experience a slight rise in pressure dp or in other words it will be compressed. The change in density d takes place because the gas is compressible and therefore, there is a lapse of time between the motion of the piston and the time this (motion of piston) is observed at the far end of the tube. Thus it will take certain time to reach far end of the tube or in otherwords there is a finite velocity of propagation which is acoustic velocity. It is shown in Fig.1.3. In this case the stagnant gas at pressure p on the right side moving with a velocity a towards left and thus its pressure is raised to p dp and its velocity lowered to a dc. This is because of the velocity of the piston dc acts opposite to the movement of gas a.
Fundamentals of Gas Dynamics 1.15 Constant area duct A=constant P + dp + d h + dh dc P a h C = O P + dp + d h + dh a P h Wave front Wave front Velocity dc Velocity a - dc a Distance Distance Pressure P + dp Wave p Pressure P + dp p Distance Distance Fig 1. Fig 1.3 Before deriving the equation, the following assumptions are made. 1. The fluid velocity is assumed to be acoustic velocity. There is no heat transfer in the pipe and the flow is through a constant area pipe. 3. The changes across an infinitesimal pressure wave can be assumed as reversible adiabatic (or) isentropic. Applying momentum equation between the two sides of the wave;
1.16 Gas Dynamics and Jet Propulsion PA p dp A m [ a dc a ] Pressure force Impulse force... (1.5) [ m A a ] A [ p p dp ] A a [ a dc a ]... A constant dp a dc dp a dc... (1.6) From continuity equation for the two sides of the wave m A a d A a dc a a a d dc d dc... (1.7) The product of d dc is very small. it is ignored. The equation (1.7) becomes adp dc Substituting this in equation (1.6), we get dp a d a dp d s cisentropic [or]... (1.8) For isentropic flow, pv C or p constant (or) p constant Differentiating the above equation p 1 d e dp 0
Fundamentals of Gas Dynamics 1.17 p 1 d dp 0 dp d p dp d RT Substituting this in equation (1.8) dp p d p R T p RT a dp d RT... (1.9) This is an important equation for solving gas dynamic problems. 1.6 MACH ANGLE AND MACH CONE When a body moves through a fluid or when fluid flows past a body or with in the walls of a duct, each element of solid surface tends to divert the fluid from its direction of flow. For example, in case of projectile moving through air, each element of the projectile s surface area pushes the neighboring air out of the way, and this local disturbance creates a pressure pulse which propagates in to the exterior air. The pressure field created by the most elementary type of moving disturbance is called Point source of disturbance. Point source may be imagined to emit infinitesimal pressure wave which spreads spherically from the point of emission with the speed of sound relative to the fluid. Fig. 1.4 (a, b, c and d) show the movement of a source of disturbance O at a velocity c in a fluid from
1.18 Gas Dynamics and Jet Propulsion Sound waves a O a 3a Fig 1.4 (a) Incompressible flow (C/a ~ o; M ~ o) Point source C t a a 3a 0 1 3 Sound waves Wave front Fig 1.4 (b) Subsonic flow (C < a) right to left. The disturbance travels distances of a, a and 3a meters when time is 1, and 3 unit times respectively. In an incompressible flow Fig. 1.4 (a) the velocity of source of disturbance C is negligibly small compared to the velocity of sound a. The sound waves generated which travel at a velocity a in all directions. The wave propagation will be a set of concentric circles as shown in Fig.
Fundamentals of Gas Dynamics 1.19 In a subsonic flow, the point source travelling with a velocity C a shown in Fig. 1.4 (b). At the reference time, the point of disturbance is assumed to be at O. At unit time later, the point source will have moved to 1 and the distance ct. At unit time units later, the source will have moved to ct and so on. It is observed that the wave fronts move ahead of the point source and the intensity is not symmetrical. The practical use of this is the case of automobiles, which move with C a. The horn is heard before the vehicle reaches a person standing on the road. In a sonic flow [ m 1 ], the point source travels with the same velocity as that of the wave. The wave fronts are always coincides with the point source and cannot move ahead of it. We won t hear any sound at the upstream side is called Zone of silence and the downstream is zone of action. In a supersonic flow, all the pressure disturbances are included in a cone which has the point source at its apex and the effect of the disturbance is not felt upstream of the source of disturbance. i.e., the point source is always ahead of the wave fronts. The cone with in which the disturbances are confined is called Mach Cone and the half angle of this cone is known as Mach angle. The space (or) zone outside the Mach Cone is called as Zone of silence i.e., C=a 0 Sound Waves 3a a 1 3 a Zone of action Fig. 1.4 (c.) Sonic flow (M = 1) c = a
1.0 Gas Dynamics and Jet Propulsion Mach cone Wave front Zone of silence O C>a Point source a 1 a 3 3a Zone of action Fig. 1.4 (d) Supersonic flow (C > a) (M > 1) there is no effect of disturbance in this region. While the region inside the Mach Cone is called Zone of action. In this region the fluid properties are affected by the disturbance. O Ct Mach angle at From the Fig., sin at Ct a C 1 C/a 1 M Mach angle sin 1 1 M... (1.30) 1.7 REFERENCE VELOCITIES In compressible fluid flow analysis, it is often convenient to express the fluid velocity in non-dimensional forms. The various reference velocities used are (i) Local velocity of sound, a
Fundamentals of Gas Dynamics 1.1 (ii) Stagnation velocity of sound, a 0 (iii) (iv) Maximum velocity of fluid, C max Critical velocity of fluid/sound, C a (i) Local velocity of sound a The local velocity of sound a RT dp d s Constant (c) Stagnation velocity of sound, a 0 It is a sound velocity at the stagnation conditions, and its value is constant. In an adiabatic flow for a given stagnation temperature. i.e., a 0 RT 0 (iii) Maximum velocity of fluid, C max From adiabatic energy equation h 0 h C. It has two components one is static enthalpy h and the another is kinetic energy C, when the static enthalpy is zero (or) when the entire energy is made up of kinetic energy only the above equation becomes h 0 and C C max h 0 C max C max h 0
1. Gas Dynamics and Jet Propulsion C p T 0 1 RT 0 C max 1 a 0 C max a 0 1 (iv) Critical velocity of fluid/sound, C a... (1.31) It is the velocity of a fluid at which the Mach number is unity. i.e., M critical C a 1 C a RT where T critical temperature At the critical state, the adiabatic energy equation becomes h 0 h C C p T 0 C p T C Divide throughout by C p T 0 T C C p C C p T 0 T... (1.3) We know that,
Fundamentals of Gas Dynamics 1.3 T 0 1 1 M T T 0 1 1 1 1 T... M 1 T T 0 By sustituting this in equation 1.3 1 C 1 R T 0 T 0 1 RT 0 1 1 1 a 0 1 1 1... RT0 a 0 C C 1 a 0 1 1 a 0 1 a 0 1... (1.33) Divide equation (1.31) by (1.33), we get C max a 0 a 0 C 1 1
1.4 Gas Dynamics and Jet Propulsion 1 C max C 1... (1.34) We know that h 0 C max Substituting equation (1.34), we get h 0 C 1 1... (1.35) Therefore equation (1.4) becomes h 0 a 1 C a 0 1 C max a 1 1... (1.36) 1.7.1 Non-dimensional Mach Number M This is an another type of mach number and is defined as the ratio between the local velocity of fluid to the critical velocity of sound. i.e. M C C C a M C Multiply both sides by a C M C a a C a M a It is more convenient to use M because instead of M
Fundamentals of Gas Dynamics 1.5 But (i) At high fluid velocities M approaches infinity. M max C max C 1. Therefore, for doing 1 calculations it is very difficult if M. (ii) M C Since a is a. constant for any process. Therefore M is proportional to the fluid velocity only. But M C a where M is not proportional to the fluid velocity alone. 1.7. Relationship between M and M From equation (1.36) a 1 C a 1 1 a C 1 1 a 1 1 a a C 1 1 a a a C C M 1 1 M M M 1 1 M M 1 1
1.6 Gas Dynamics and Jet Propulsion M M 1 M 1 Comparison of M and M M 0 M 0 M 1 M 1 M 1 M 1 M 1 M 1 M M 1 M 1... (1.37) M M max 1 1 C max C 1.7.3 Crocco Number [ C r ] Crocco number is a non-dimensional fluid velocity which is defined as the ratio of fluid velocity to its maximum fluid velocity. C r C fluid velocity C max Max.fluid velocity Multiply both Nr. and Dr. by C C r C C C C max M 1 1 C 1 r M 1... (1.38)
Fundamentals of Gas Dynamics 1.7 By substituting equation (1.37) in (1.38) C r M 1 1 M 1 1 C r Cr M 1 M 1 C r M 1 Cr M 1 C r M 1 [ 1 Cr ] We know that, T 0 T M C r 1 C r 1... (1.39) 1 1 1 1 M C r 1 C 1 T 0 T 1 C r Cr 1 1 C 1 C r T 0 T 1 1 C r... (1.40) Problem 1.1 An air jet at 400 K has sonic velocity Determine (i) Velocity of sound at 400 K (ii) Velocity of sound at stagnation conditions (iii) Maximum velocity of jet.
1.8 Gas Dynamics and Jet Propulsion (iv) Stagnation enthalpy (v) Crocco number. [Oct 005, AU] Given Data: T 400 K, At sonic condition M 1 and C a, 1.4 (air) (i) Velocity of sound a RT (ii) T 0 T 1 1 M 1.4 87 400 400.899 m/sec 1 T 0.4 400 480 K. Velocity of sound at stagnation condition a 0 RT 0 1.4 87 480 439.1684 m/sec (iii) From adiabatic energy equation C max a 0 1 C max a 0 1 98 m/sec
Fundamentals of Gas Dynamics 1.9 (iv) Stagnation enthalpy h 0 C max 48.16 KJ/Kg (v) Crocco number C r C C max 400.899 98 0.40847 Result (i) Velocity of sound a 400.899 m/sec (ii) Velocity of sound at stagnation conditions a 0 439.16 m/sec (iii) Maximum velocity of jet C max 98 m/sec (iv) Stagnation enthalpy h 0 48.16 kj/kg (v) Crocco number Cr 0.40847 Problem 1.: The jet of gas at 593 K 1.3 and R 469 J/Kg K has a Mach number of 1.. Determine for static and stagnation conditions. (i) Velocity of sound (ii) Enthalpy (iii) What is the maximum attainable velocity of this jet? [Oct 005, AU] Given Data: T 593 K, 1.3, R 469 J/Kg K, M 1. C p R 1.3 469 03.3333 J/Kg K 1 0.3 T 0 1 1 M T 1 0.3 1. 593 71.088 K
1.30 Gas Dynamics and Jet Propulsion Velocity of sound a RT 1.3 469 593 601.9 m/sec a 0 RT 0 1.3 469 71.088 663.059 m/sec Enthalpy h C p T 105.17366 kj/kg Result Maximum attainable velocity C max h 0 (i) Velocity of sound (m/sec) h 0 C p T 0 1465.4911 kj/kg 1465491.1 171.0111 m/sec Static Condition Stagnation condition 601.9 663.059 (ii) Enthalpy (kj/kg) 105.17366 1465.4911 (iii) Maximum attainable 171.0111 velocity of jet C max m/sec
Fundamentals of Gas Dynamics 1.31 Problem 1.3: (a) Determine the velocity of air 1.4 C p 1.005 kj/kgk corresponding to a velocity of temperature of 1C. (b) Determine the Mach number of an aircraft at which the velocity temperature of air at the entry of the engine equals the static temperature. Given Data: (a) 1.4, C p 1.005 KJ/Kg K T c C C p 1 73 74 K T c C C p 74 C 74 1005 74.1185 m/sec (b) T c T We know that, T 0 T T c T T 0 T 1 1 M 1 0. M M 1 0..36 Result (a) Velocity of air C 74.1185 m/sec (b) Mach number of an aircraft M.36
1.3 Gas Dynamics and Jet Propulsion Problem 1.4: Air flows from a reservoir at 550 KPa and 70C. Assuming isentropic flow. Calculate the velocity, temperature, pressure and density at a section where M 0.6. [Apr. 005 AU] Given Data: [In a reservoir the fluid is in a stagnation state i.e., the velocity of the fluid C 0] P 0 550 KPa T 0 70 73 343 K since the flow is isentropic. from isentropic table 1.4 and M 0.6 T T 0 0.933, P P 0 0.784 T 30.019 K and P 4.31 bar M C a C M RT We know that 0.6 1.4 87 30.019 15.151 m/sec P RT P RT 431. 0.87 30.019 Result 4.6948 kg/m3 (i) Velocity of flow C 15.151 m/sec (ii) Temperature at the section T 30.019 K (iii) Pressure at the section P 4.31 bar (iv) density at the section 4.6948 kg/m 3.
Fundamentals of Gas Dynamics 1.33 Problem 1.5: An air stream at 1 bar and 400 K flowing with a velocity of 400 m/sec is brought to rest isentropically. Determine the stagnation pressure and temperature. [MKU.Nov 95], [April 01 - AU] Given Data: P 1 bar, T 400 K, C 400 m/sec Result: (i) Mach number M C a We know that T 0 T 1 1 C RT 400 1.4 87 400 0.99775 M 1 0.4 From adiabatic relations, T 0 T 1 P 0 P P 0 T 0 T P 0 1.8879 bar. 479.6416 K. 1 P 476.6416 400 0.99775 400 1.4 0.4 1 (i) Stagnation Pressure P 0 1.8879 bar (ii) Stagnation temperature T 0 479.6416 K.
1.34 Gas Dynamics and Jet Propulsion Problem 1.6: Air enters a straight axisymmetric duct at 7C, 3.45 bar and 150 m/sec and leaves 4C,.058 bar and 60 m/sec. Under adiabatic flow conditions, for an inlet cross sectional area of 500 sq.cms, estimate the stagnation temperature, maximum velocity, mass flow rate and the exit area. [MKU. Nov.95], [Apr 010 - AU] Given Data: T 1 7 73 300 K P 1 3.45 bar, C 1 150 m/sec T 77 K P.058 bar, C 60 m/sec A 1 500 cm M 1 C 1 RT 0.43 T 0 1 1 M T 1 1 T 0 1 0.4 0.43 300 T 0 311. K We know that a 0 RT 0 1.4 87 311. 353.61 m/sec C max a 0 1 C max a 0 1
Fundamentals of Gas Dynamics 1.35 C max 790.693 m/sec Mass flow rate m 1 A 1 C 1 From continuity equation 3.45 105 87 300 500 10 4 150 30.056 kg/sec m 1 A 1 C 1 A C m P RT A C Result A m RT 30.056 87 77 P C.058 10 5 60 0.044649 m 446.4986 cm (i) Stagnation temperature T 0 311. K (ii) maximum velocity C max 790.693 m/sec (iii) Mass flow rate m 30.056 kg/sec (iv) Exit area of the duct A 466.4986 cm Problem 1.7: The flight speed is 800 km/hr. The stagnation conditions are 105 kpa and 35C. Find the static conditions and the flight mach number. [Apr.96 Bharathidasan] [Oct.00 AU] Given Data 800 1000 C 3600. m/sec, P 0 105 kpa
1.36 Gas Dynamics and Jet Propulsion T 0 35 73 308 K From adiabatic energy equation h 0 a 1 C a 0 1 a 0 RT 0 1.4 87 308 351.78744 m/sec a 1 a 0 1 C 351.78744 0.4 84694.6498 a 337.4579 m/sec. a RT T a R 337.4579 1.4 87 83.419 K From isentropic relation T 0 T 1 P 0 P P T 1 P 0 T 0 P 83.419 308 Flight Mach number M C a. 337.4579 0.65851 1.4 0.4 105
Fundamentals of Gas Dynamics 1.37 Result (i) Static pressure P 78.481 kpa (ii) Static temperature T 83.491 K (iii) Flight Mach number M 0.65851 Problem 1.8: Determine the velocity of sound in air at 35C what should be the minimum temperature of air if it has to be hypersonic when it has a velocity of 1500 m/sec. [Oct 007, AU] Given Data: T 35 73 308 K, 1.4, R 87 J/kg K (i) Velocity of sound a RT 1.4 87 308 351.7874 m/sec (ii) When the flow is hypersonic, the Mach number M 5 (Assumed) We know that, M C a a C M 1500 5 300 C 1500 m/sec a RT 300 T 300 1.4 87 3.99 K Result (i) Velocity of sound a 351.7874 m/sec (ii) Minimum temperature of air T 3.99 K
1.38 Gas Dynamics and Jet Propulsion Problem 1.9: Air at stagnation condition has a temperature of 800 K. Determine the velocity of sound and the maximum possible fluid velocity. What is the velocity of sound when the flow is at half the maximum velocity. [Apr. 008, AU] Given Data: T 0 800 K (i) Velocity of sound a 0 RT 0 1.4 87 800 (ii) Maximum fluid velocity 566.9567 m/sec C max a 0 1 C max a 0 566.9567 r 1 0.4 167.7539 m/sec when the flow velocity C is at half of the maximum velocity C max i.e., C C max 633.8769 m/sec We know that, C max a 1 C C C max 4 a 1 C max C a 1 3 8 C max
Fundamentals of Gas Dynamics 1.39 3 a 8 1 C max 490.99898 m/sec Result (i) Velocity of sound a 0 566.9567 m/sec (ii) Maximum possible fluid velocity C max 167.7539 m/sec (iii) Velocity of sound when C max C 490.99898 m/sec Problem 1.10: Find the stagnation properties of air by calculation if the static pressure is 50 kpa and static temperature is 15C and velocity is 00 m/s. [Oct, 008, AU] Given Data: P 50 kpa, T 15 73 398 K, C 00 m/sec (ii) T 0 T M C a 1 1 M C RT 00 1.4 87 398 0.50013 T 0 1 0.4 0.50013 398 417.9104 K P 0 P T 0 1 T
1.40 Gas Dynamics and Jet Propulsion P 0 1.4 417.9104 0.4 398 50 96.5789 kpa T 0 T P 0 P r 1 0 1 0 T 0 T 1 1 0.477 kg/m 3 P RT Result (i) Stagnation pressure P 0 96.5789 kpa (ii) Stagnation temperature T 0 417.9104 K (iii) Stagnation density 0.477 kg/m 3 Problem 1.11: A plane travels with a velocity of 1000 km/h at an altitude where the pressure and temperature are 40 kpa and 35C. Find the mach angle and mach number. [Oct 98, Madras], [Apr 004-AU] Given Data: C 1000 1000 3600 77.7778 m/sec, P 40 kpa, T 35 73 38 K Mach number M C RT 77.7778 1.4 87 38 0.8986
Fundamentals of Gas Dynamics 1.41 We know that, mach angle sin 1 1 M When the mach number is less than one, there is no mach angle. Result (i) Mach angle (ii) Mach number M 0.8986 Problem 1.1 The pressure, temperature and mach number at the entry of a flow passage are.45 bar, 7C, and 1.4 respectively. If the exit mach number is.5, determine the stagnation temperature, temperature and velocity of gases at the exit and the flow rate per unit area at inlet. The fluid may be assumed to have adiabatic flow with 1.3 and R 469 J/kg/K. [MKU. MSU, April 96], [Oct 006-AU] [April 010-AU] Given Data: P 1.45 bar, T 1 7 73 300 K, M 1 1.4, M.5, 1.3 (i) From Isentropic table 1.3 and M 1 1.4 T 1 T 01 0.773 T 01 388.0983 K In an adiabatic flow, the stagnation temperature is constant throughout. T 01 T 0 T 03...
1.4 Gas Dynamics and Jet Propulsion (ii) From isentropic table 1.3 and M.5 T T 0 0.516 T 00.5873 K We know that, M C RT C M RT.5 1.3 469 00.58 873.56 m/sec (iii) The flow rate/m section of the inlet area of cross We know that, m 1 A 1 C 1 A C m A 1 1 C 1 P 1 M RT 1 RT 1 1 5 1.4 1.3 P 1 M 1.45 10 RT 1 469 300 104.604 kg/sec m Result (i) Stagnation temperature of the gas T 0 388.0983 K (ii) Temperature of gas at exit T 00.5873 K
Fundamentals of Gas Dynamics 1.43 (iii) Velocity of gas at exit C 873.56 m/sec (iv) Flow rate per unit area at inlet m A 1 104.604 kg /m sec Problem 1.13 An aircraft is flying at an altitude of 1000 m at a mach number of 0.8. The cross sectional area of the inlet diffuser before the L.P compressor stage is 0.5 m. Determine (a) The mass of air entering the compressor per second (b) The speed of the aircraft (c) The stagnation pressure and temperature of air at the diffuser entry. [MSU, Nov, 95] [Oct 006-AU] Given Data: Altitude Z 1000 m, M 1 0.8, A 0.5 m From table (page number 19) at an altitude of 1000 m, the properties are T 1 16.65 K, P 1 0.193 bar, 1 0.311 kg/m 3 and a 95. m/sec From isentropic table 1.4 and M 1 0.8 T T 0 0.881, P P 0 0.643 T 0 45.913 K and P 0 0.300155 bar We know that, M 1 C 1 a 1 C 1 M 1 a 1 0.8 95.5
1.44 Gas Dynamics and Jet Propulsion 4.064 m/sec 3600 1000 871.43 km/hr. Mass flow rate m 1 A 1 C 1 0.311 0.5 4.064 37.64 kg/sec Result (a) Mass of air entering the compressor per sec 37.64 kg/sec (b) Speed of the air craft C 871.43 km/hr (c) Stagnation pressure at diffuser entry P 0 0.300155 bar Stagnation temperature at diffuser entry T 0 45.913 K Problem 1.14 (a) Air expands isentropically from 0 bar and 100C to 1 bar. Determine the temperature and density at the final state. Also find the ratio of initial to final acoustic velocity. (b) Air at a temperature of 33C and a pressure of 1.1 bar is flowing with a velocity of 30 m/s. Determine the total pressure, temperature and density. [MSU. Nov 96] [Apr. 007-AU] Given Data: (a) P 1 0 bar, T 1 100 73 373 K, P 1 bar. a 1 RT 1 1.4 87 373 387.1377 m/sec In an isentropic flow T P 1 T 1 P 1
Fundamentals of Gas Dynamics 1.45 T 0.4 1 1.4 0 373 3.34688 K P RT 1 10 5 87 3.34688 1.97107 kg/m 3 a RT 359.8874 m/sec ratio of initial to final acoustic velocity a 1 a 1.0757 Result (i) Temperature and density at the final state T 3.34688 K, 1.97107 kg/m 3 (ii) Ratio of initial to final acoustic velocity a 1 a 1.0757 (b) P 1 1.1 bar T 1 33 73 306 K, C 1 30 m/s M 1 C 1 RT 1 T 01 1 1 30 1.4 87 306 M 1 T 1 T 0 1 0.4 0.91608 306 356.9706 K 0.91608
1.46 Gas Dynamics and Jet Propulsion T 0 T P 0 P P 0 356.9706 306 1 P 0 P T 0 T 1.4 P 0 1.88617 bar. 0.4 1.1 1 0 P 0 1.88617 105 1.841055 kg/m3 RT 0 87 365.9706 Result (i) Total pressure P 0 1.88617 bar (ii) Total temperature T 0 356.9706 K (iii) Total density 0 1.841055 kg/m 3 Problem 1.15 A gas whose 1.658 and molecular weight = 39.94 is stored in a reservoir at 98 K, determine stagnation enthalpy and velocity of sound. Given T o 98 K 1.658 M molecular weight 39.94 To find (i) Stagnation enthalpy h o Solution (ii) Stagnation velocity of sound a o We know that R R u 8.314 M 39.94
Fundamentals of Gas Dynamics 1.47 0.0816 kj/kg K 08.16 J/kg K We know that Stagnation velocity of sound a o R T o 1.658 08.16 98 30.7 m/s We know that C p R 1 1.658 08.16 1.658 1 54.51 J/kg K Stagnation enthalpy h o C p T o 54.51 98 156304.7 J/kg h o 156.304 kj/kg Problem 1.16 In a setting chamber air is at P o 5 bar, T o 498 K. Determine the value of h o, a o, c max, T, C and a. Given data P o 5 bar
1.48 Gas Dynamics and Jet Propulsion T o 498 K To find (i) Stagnation enthalpy h o Solution (ii) Stagnation velocity of sound a o (iii) Maximum velocity of fluid C max (iv) Critical temperature (v) Critical velocity of fluid (or) sound (i) We know that, stagnation enthalpy, h o C p T o Since the fluid is air 1.4, C p 1.005 kj/kg K, R 87 J/kg K h o 1.005 498 h o 500.49 kj/kg K (ii) Stagnation velocity of sound a o a o R T o 1.4 87 498 a o 447.3 m/s (iii) Maximum velocity of fluid C max We know that C max h o 500.49 10 3
Fundamentals of Gas Dynamics 1.49 C max 1000.48 m/s (iv) Critical temperature Assuming isentropic flow, for critical condition, from gas table, corresponding to M 1, 1.4 T T o 0.834 T 0.834 498 T 415.33 K (v) Critical velocity of fluid (or) sound c a We know that c a R T 1.4 87 415.33 c a 408.5 m/s 1.8 BERNOULLI EQUATION Starting from the adiabatic energy equation, derive (i) (ii) P C 1 P 0 1 P 0 P 0 P C 0 P 0 1 0 [for reversible incompressible flow] [for reversible compressible flow] From adiabatic energy equation, h 0 h C constant... (1)
1.50 Gas Dynamics and Jet Propulsion By differentiating this equation, we get dh CdC 0 dh CdC 0... (1.41) In an isentropic flow dh dp and if the flow is assumed to be incompressible; constant. Integrating the equation (1.41) 1 dp CdC constant P C constant... (1.4) When the flow is isentropically decelerated to zero velocity at zero elevation, the resultant pressure is stagnation pressure. when C 0 0, P P 0, 0 P C P 0 0 In an incompressible flow; constant 0 P C P 0... (1.43) It is the well known Bernoulli equation and it is valid only when the flow is isentropic and incompressible The adiabatic energy equation can be expressed in terms of pressures for compressible flows also.
Fundamentals of Gas Dynamics 1.51 For a perfect gas, h C p T 1 RT 1 P From adiabatic energy equation h 0 h C P 1 C P RT RT P 1 P C h 0 P 0 1 We know that T 0 T P 0 P 1 0 P P 1 P 0 0 P 0... (1.44) 1 0 P 0 P Multiply Numerator and Denominator by P 0 P P 0 P 0 P 1 P 0 P 0 1 P 0 P 1 P 0 P 0 P 0 P 1 0 P 0 P 0 1 1 1
1.5 Gas Dynamics and Jet Propulsion By substituting this in equation (1.44), we get 1 1 P 0 P 0 P C 0 P 0 1 0... (1.45) This is the another form of Bernoulli equation for isentropic compressible flow. 1.9 EFFECT OF MACH NUMBER ON COMPRESSIBILITY From Bernoulli equation for incompressible flow, the value of pressure co-efficient (or) compressibility factor is unity. i.e., P 0 P C P 0 P C 1... (1.46) For compressible flow the value of pressure co-efficient deviates from unity and the magnitude of deviation increases with the mach number of the flow. We know that T 0 T 1 P 0 1 1 M P P 0 P 1 1 M 1 This can be expanded by Binomial expansion
Fundamentals of Gas Dynamics 1.53 where x i.e., 1 x n n n 1 1 nx x! P 0 P 1 1 1 M and n 1 1 M n n 1 n 3! x 3 1 1 1 1 4 M4 1 1 1 1 13 M 6 6 8 P 0 P 1 M M4 M6 8 48 P 0 P P M M4 M6 8 48 Divide both sides by M P 0 P P M 1 M 4 M6 48 M P 0 P 1 P 0 P P 1 M 4 M4 4... (1.47) We know that, P RT M C a C RT M C RT
1.54 Gas Dynamics and Jet Propulsion P M RT C C RT... (1.48) P 0 P C For 1.4 1 M 4 M4 4 P 0 P C 1 M 4 M4 40... (1.49) Equation (1.49) gives the percentage deviation of the pressure co-efficient from its incompressible flow value with the Mach number. By substituting different values of M, we get the following table. Table 1.1 Effect of Compressibility Sl.No. Mach Number M % deviation 1 0.1 0.3 0. 1.0 3 0.3.3 4 0.4 4.1 5 0.5 6.4 6 0.6 9.3 7 0.7 1.9 8 0.8 17 9 0.9 10 1.0 7.5 From the table it is observed that, percentage deviation increases when Mach number increases. When the mach number is unity, the percentage deviation is 7.5%
Fundamentals of Gas Dynamics 1.55 Problem 1.17: Air C p 1.05 kj/kg K, 1.38 at P 1 3 10 5 N/m and T 1 500 K flows with a velocity of 00 m/sec in a 30 cm diameter duct available. Calculate, (a) mass flow rate (b) stagnation temperature (c) Mach number (d) Stagnation pressure values assuming the flow as compressible and incompressible respectively. [MSU. Nov. 95] [April 005-AU] Given Data: C p 1.05 kj/kg K, 1.38, P 1 3 10 5 N/m, T 1 500 K C 1 00 m/s, Area 4 [ 0.3 ] 0.0706858 m (a) Mass flow rate m 1 A 1 C 1 P 1 RT 1 A 1 C 1 3 105 0.0706858 00 89.1304 500 [R 1 9.337 kg/sec C p 0.8913 kj kg K ] (b) T 0 T C C p Velocity temperature 00 500 1050 519.047 K (c) M C a 00 1.38 89.1304 500 0.44777
1.56 Gas Dynamics and Jet Propulsion (d) Stagnation pressure (i) Flow is compressible T 0 T P 0 P 1 P 0 P T 0 T 1.4 P 0 519.047 500 0.4 3 3.436 bar (ii) Flow is incompressible Result P 0 P C 1 3 10 5 3 10 5 00 89.1304 500 3.415 10 5 N/m 3.415 bar (a) Mass flow rate m 9.337 Kg/sec (b) stagnation temperature T 0 519.047 K (c) Mass number M 0.44777 (d) Stagnation pressure when the flow is (i) compressible P 0 3.436 bar (ii) Incompressible P 0 3.415 bar Problem 1.18: Air flowing in a duct has a velocity of 300 m/sec, pressure 100 kpa, temperature 90 K. Taking 1.4 and R 87 J/kg K. Determine (i) Stagnation pressure and temperature. (ii) Velocity of sound in the dynamic and stagnation conditions. (iii) Stagnation pressure assuming constant density. [Oct 006, AU]
Fundamentals of Gas Dynamics 1.57 Given Data: C 1 300 m/sec, P 1 100 kpa, T 1 90 K, 1.4, R 87 J/kg K C p R 1 1.4 87 0.4 1004.5 J/kgK We know that T 01 T C C p We know that, T 0 1 P 0 T 1 P P 01 T 01 T 1 90 334.7984 90 1 P1 165.389 kpa 300 1004.5 334.7984 K 1.4 0.4 100 Velocity of sound at dynamic condition a RT 1.4 87 90 341.35318 m/sec
1.58 Gas Dynamics and Jet Propulsion Velocity of sound at stagnation condition a 0 RT 0 366.774 m/sec At constant density the stagnation pressure P 0 P C P 100 103 1.014898 kg/m3 RT 87 90 P 0 100 10 3 1.014898 300 154.067 kpa Result (i) Stagnation pressure and temperature P 01 165.389 and T 01 334.7984 K (ii) Velocity of sound in the dynamic and stagnation condition a 341.35318 m/sec and a 0 366.774 m/sec (iii) Stagnation pressure assuming constant density P 0 154.067 kpa Problem 1.19: Steam at a section of a pipe has a pressure = 10 bar, temperature = 600 K, velocity = 10 m/sec, datum head = 10 m (a) Taking C p.15 kj/kg K, C v 1.615 kj/kg K, determine mach number, stagnation pressure and temperature. (b) compare the stagnation pressure value with that obtained from the Bernoulli equation and comment on the difference.
Fundamentals of Gas Dynamics 1.59 Given Data: R C p C v 0.534 kj/kg K; C p C v 1.3317 Z 10 m ; P 1 10 bar; T 1 600 K ; C 1 10 m/sec T 01 T C C p 10 600 150 603.349 K T 0 T 1 1 M T 0 T 1 1 M 603.349 600 1 0.3317 M M 0.18357 We know that T 0 1 P 0 for compressible flow T 1 P P 0 T 0 T 1 P P 0 10.6 10 5 N/m
1.60 Gas Dynamics and Jet Propulsion (b) The stagnation pressure from Bernoulli equation is P 0 P C 10 10 5 10 10 5 1 10 105 535 600 10 10 535 100 10.49 10 5 N/m The difference in pressure is 190.093 N/m 3 This difference is because we assumed from Bernoulli equation, the density is constant. It means that the fluid is incompressible. But in practical cases all fluids have compressible nature only. Result (a) Mach number M 0.18357 Stagnation pressure P 0 10.6 bar Stagnation temperature T 0 603.349 K (b) Stagnation pressure obtained from Bernoulli equation P 0 10.49 bar The difference in pressure P 190.093 N/m
Fundamentals of Gas Dynamics 1.61 Questions and Problems 1.1 Define gas dynamics and give some applications. 1. Differentiate between compressible and incompressible flows. 1.3 Derive an adiabatic energy equation for nozzles and diffusers. 1.4 Define stagnation state and stagnation properties. 1.5 Derive an expression for sonic velocity in a medium having and R as ratio of specific heats and difference in specific heats respectively. Ans : a RT 1.6 Explain how small disturbance travels in a fluid under the following condition with C as the fluid velocity and a as sonic velocity. C a 1, C a 1 and C a 1. 1.7 Show that the sound wave movement pattern when the object producing sound waves with (i) subsonic velocity and (ii) sonic velocity. 1.8 Derive an expression for accoustic velocity in terms of temperature of a fluid. Ans : a RT 1.9 From first principles, obtain the relations between the stagnation properties P 0 P and T 0 T in terms of Mach number for a reversible one dimensional steady flow of an ideal gas through a nozzle.
1.6 Gas Dynamics and Jet Propulsion 1.10 Show that for a perfect gas flowing at Mach number M and pressure P, the stagnation pressure P 0 is P 0 P 1 1 M 1 where is the ratio of specific heats. 1.11 What are the five conclusions that can be drawn from the expression for Mach number? [Ans : Five regions of flow] 1.1 Define the following terms : (i) Mach angle, (ii) Mach cone, (iii) zone of action and (iv) zone of silence. 1.13 Show that the Mach angle is equal to sin 1 1 M. 1.14 Derive an expression for T T 0 in terms of Mach number. 1.15 Explain the sound wave propagation with sketch in two dimension when the object producing sound travels a supersonic velocity and hence show the mach cone, zone of silence and zone of action. 1.16 What are the advantages of using M instead of M and show that M M 1 M 1 1/ 1.17 Define Crocco number and prove that
Fundamentals of Gas Dynamics 1.63 C r 1 T T 0 1/ 1.18 What is the effect of Mach number on the compressibility? Prove for = 1.4. P 0 P C 1 M 4 M4 40... 1.19 Derive the energy equation a r 1 C C max a 0 1 h 0 constant stating the assumptions used. 1.0 An aircraft flies at a velocity of 700 kmph in an atmosphere where the pressure 75 kpa and temperature is 5C. Calculate the Mach number and stagnation properties. [Nov. 95, BDU, Nov. 96, MKU] [Ans : M 0.5817 ; P 0 94.3186 kpa ; T 0 96.8136 K and 0 1.107 kg /m 3 ] 1.1 oxygen at 00 kpa flows at a velocity of 50 m/s. Find the Mach number at a point where its density is.9 kg /m 3. Molecular weight of O is 3. [Nov. 95, BDU and Nov. 96, MKU] [Ans : M 0.1609 ] 1. A diffuser has an area ratio of 1.5 to 1.0. The inlet pressure and temperature are 1 bar and 15C. Assuming the flow to be isentropic, calculate the following for the exit air.
1.64 Gas Dynamics and Jet Propulsion (i) the final pressure, (ii) the exit temperature and (iii) the exit Mach number. [Nov. 95, BTU] [Ans : P 1.358 bar ;T 314.39 K ; M 0.41 ] 1.3 A stream of air flows with a velocity of 50 m/sec in a duct of 10 cm diameter. Its temperature and pressure at that point are 5C and 40 kpa. What will be its stagnation pressure and temperature? What is the mass flow rate? [Nov. 96, BTU] [Ans : P 0 57.976 kpa; T 0 309.094 K ; m 0.98438 kg/ sec] 1.4 (a) Air at a temperature of 33C and a pressure of 1.1 bar is flowing with a velocity of 30 m/sec. Determine the total pressure, temperature and density. [Ans : T 0 356.945 K ; P 0 1.8857 bar ; 0 1.8407 kg /m 3 ] (b) An aeroplane travels at an altitude where the temperature is 37C with a Mach number of 1.. Determine the velocity of the plane in km/hr. [Nov. 96, MSU] [Ans:C 1330.85 km hr]