Gas DynamicsMS Process & Mechanical Engineering 3rd Semester Instructor: Dr. Kamran Rasheed Qureshi

Shock Wave at High Angle of Attack

Normal Shock Waves and Related TopicsFlow over a blunt bodyFlow in a nozzle bodyExamples of Normal Shock Waves

Basic Normal Shock EquationsConsider the continuity equation in the integral form as:Consider a rectangular control volume abcd given by dashed line in the following figure. The shock wave is inside the control volume as shown.Hence for the control volume shown, the above equation becomesContinuity equation for normal shock

Cont.(Basic Normal Shock Equations)Now consider the momentum equationSince the flow is moving only in one direction, so we need to consider only the scalar x component of above equation which isThe surface integrals in the above equation becomeMomentum equation for normal shock waves.

Cont.(Basic Normal Shock Equations)Now consider the energy equation in integral form. For steady, adiabatic, and inviscid flow with no body forces this equation becomes:Evaluating the above equation for the control surface shown in figure, we haveDividing the above equation by 1u1 and 2u2 Energy equation for normal shock waves

Cont.(Basic Normal Shock Equations)So the basic normal shock equations are Continuity:Momentum:Energy:Equation of state:Enthalpy:We have five equations for five unknowns namely, 2, p2, u2, h2 and T2

Calculation of Normal Shock Wave Properties.Basic normal shock wave problem: given the conditions in region 1 ahead of the shock, calculate the conditions in region 2 behind the shock.(1)(2)(3)(4)(5)Dividing (2) by (1), we obtainAlso as we know that then the above equation becomesThis equation is the combination of the continuity and momentum equation.

Cont..The energy equation, equation (3), can be used in one of its alternate forms, namelyThis equation can be written for region 1 and region 2 as:Substituting the above equations into the equation and then after solving we obtain Prandtl RelationThe above equation can also be written as

Cont..The above equation can also be written asSubstituting the above equation in the following equation And after some manipulations, we obtainIt is our first major result for normal shock wave. It states that the Mach number behind the wave is a function only of the Mach number ahead of the wave.

Cont..Similarly we can obtain the ratios of the thermodynamic properties 1/2, p1/p2 and T1/T2 which areImportant questions1) Can we have normal shock if upstream Mach number is less than 1?

2) Why does the entropy increase across the normal shock wave?

Cont..Variation of properties across a normal shock wave as function of upstream Mach number: = 1.4

Measurement of Velocity in a Compressible FlowSubsonic Flow

ContSupersonic FlowRayleigh Pitot Tube Formula

Hugoniot ConditionsAs we know that the relations for the normal shock wave involve velocities and Mach number.Changes across normal shock wave can also be expressed in terms of purely thermodynamic variables as follows:From continuity and momentum equations, we know thatSolving forAlternatively we can solve the continuity and momentum equation for u2

Cont(Hugoniot Conditions)From energy equation we know thatSubstituting the previously derived relations for u1 and u2 , in the above equation, we haveThis simplifies to Hugoniot EquationIt has certain advantages because it relates only thermodynamic quantities across the shock. By definition we have

Cont(Hugoniot Conditions)Now consider a specific shock with a specific value of upstream velocity u1 .

How can we locate a specific point on the Hugoniot curve, point 2, which corresponds tothis particular shock? Hugoniot CurveAs we haveWe know thatOn rearranging we obtain

Oblique Shock and expansion wavesOblique shock and expansion waves are prevalent in two and three dimensional supersonic flows.These waves are inherently two dimensional in nature.Formation of oblique shockFormation of expansion waves

Cont.Why are most waves oblique rather than normal to the upstream flow?Relationship between oblique shock angle and mach wave angleMach angle

Oblique Shock RelationsConsider a oblique shock wave as sketched in the following figure:Applying the continuity equation over the control surfaces of the control volumeContinuity equation

Cont.(Oblique Shock Relations)Momentum equation can be resolved into two components; normal and tangential to the shock wave.First consider the tangential component After control volume analysis we obtainIt is an important result: it states that the tangential component of the flow velocity is constant across an oblique shock.

Cont.(Oblique Shock Relations)The normal component of the integral momentum equation isAfter control volume analysis the above equation becomesSimilarly, we can write the energy equation after control volume analysisWe know that from the fig thatHence

Cont.(Oblique Shock Relations)Lets now gather our resultsContinuity equationNormal component of Momentum eqEnergy equationThese equation involve the normal components of the velocity that is u1 and u2 . The tangential component does not appear in these equations.Same algebra as applied to the normal shock equations when applied to above equations will lead to identical expression for changes across oblique shock in terms of the normal component of the upstream Mach number Mn,1

Cont.(Oblique Shock Relations)Note that Hence The temperature ratio follows from the equation of stateM2 can be found from Mn,2 as

Cont.(Oblique Shock Relations)The above equation introduces deflection angle in to oblique shock analysis; we need to be able to calculate M2. is a parameter which is function of M1 and as follows:From the geometry of figureDividing the above two equations and invoking the continuity equation we have, , M relation

Cont.(Oblique Shock Relations)This relation is vital to the analysis of oblique shock and the results obtained from it are plotted in the following figure for =1.4.This plot provides wealth of physical phenomena associated with oblique shock waves. For example

Cont.(Oblique Shock Relations)1). For a given upstream Mach number M1, if > Max

Cont.(Oblique Shock Relations)2). For any given less than Max , there are two oblique shock solutions for a given upstream Mach number.

For example, if M1= 2 and =15o then can equal either 45.3 or 79.8.3). If =0, then equals either 90o or 4).Effect of increasing the Mach number

Cont.(Oblique Shock Relations)5). Effect of increasing the deflection angle.

Supersonic Flow over Wedges and ConesFlow over a wedgeFlow over a cone

Shock Interactions and ReflectionsRegular reflection of shock from a solid boundaryMach reflection

Cont(Shock Interactions and Reflections)Intersection of left and right running shock waves

Cont(Shock Interactions and Reflections)Intersection of two left running shock waves

Detached Shock Wave in Front of a Blunt Body

Shock Polar Graphical explanations go a long way towards the understanding of supersonic flow with shock waves. One such graphical representation of oblique shock properties is given by shock polar described below: Physical planeHodograph plane The graph of velocity components Vx and Vy is called Hodograph plane.

Cont(Shock Polar)Shock polar for a given V1Geometric construction using shock polarShock polar for different Mach number

Prandtl-Meyer Expansion Waves An expansion wave emanating from sharp convex corner as sketched in above figure, is called centered expansion wave. The problem of expansion wave is as follows: Given the upstream flow (region 1) and the deflection angle, calculate the downstream flow (region 2).

Cont(Prandtl-Meyer Expansion Waves)Consider a very weak wave produced by an infinitesimal small flow deflection d as sketched in the following figure: We consider the limit of this picture as d 0; hence the wave is essentially a Mach wave at the angle to the upstream flow. examining the geometry in the above figure, from the law of sine applied to the triangle ABC we see that

Cont(Prandtl-Meyer Expansion Waves)From trigonometric identities, we know thatSubstituting the above two equations in the previous equation, we haveFor small d, the above equation becomesExpanding a power series and ignoring the high order terms, we have

Cont(Prandtl-Meyer Expansion Waves)And finally we obtain This is the differential equation which precisely describe the flow inside the expansion wave.Let us integrate the above equation

Cont(Prandtl-Meyer Expansion Waves)In the above equation the integral is called the Prandtl-Meyer function. Carrying out the integration the above equation becomes

Reflection and Intersection of Expansion WavesRefection of expansion wave from flat wallIntersection of expansion waves

1) A diaphragm at the end of 4m long pipe containing air at a pressure of 200 kPa and a temperature of 30oC suddenly ruptures causing an expansion wave to propagate down the pipe. Find the velocity at which the air is discharged from the pipe if the ambient air pressure is 103 kPa. Also find the velocity of the front and the back of the wave and hence find the time taken for the front of the wave to reach the end of pipe.2) A shock tube essentially consists of a long tube containing air and separated into two sections by a diaphragm. The pressures on the two sides of the diaphragm are 300 kPa and 30 kPa and the temperature is 15o C in both sections. If the diaphragm is suddenly ruptured, find the velocity of the air between the moving shock wave and the moving expansion wave that are generated. Problems