Unit – III

Flow with Normal Shock Waves

A shock wave is a special kind of steep finite pressure wave and the changes inthe flow properties across the wave are abrupt. When the shock waves are at right anglesto the flow, they are called normal shocks, if it is inclined at an angle, then obliqueshocks.

Normal shocks may be treated as one dimensional hut oblique shocks require twodimensional approach. There will be an increase in static pressure and entropy, but loss instagnation pressure across the shock. The flow changes from supersonic to subsonic withsudden increase in pressure.

The applications of shock waves are

(1 used in shock tubes and supersonic compressors.1i) used in supersonic aircraft to obtain high pressure ratio in an axial compressor.Development of a Normal Shock WaveIn an off-design value of pressure ratio, the variation in fluid properties is sudden and theflow changes from supersonic to subsonic. It occurs in the divergent portion of aconvergent- divergent nozzle is shown in figure in the previous chapter. This is due to theformation of finite shock waves at this section and the thickness of the shock wave is0.00! mm. Hg 3. I Ib) shows a constant area duct which contains gas initially at rest.When the piston moves right the pressure pulses (infinitesimal pressure waves) aretransmitted through the gas to the right. The growth of pressure wave at time tj, 12, 13 isshown in Fig. 3.1 (a) and it travels towards right with the speed of sound relative to thegas. The gas near to the piston moves with a high velocity than the gas in the downstreamregion. Similarly the pressure of the gas nearer to the piston is higher than in the regionsdownstream. Therefore, the pressure waves nearer to the piston travel at higher velocitieson account of higher gas velocity and speed of sound. Thus the upstream waves arecontinuously overtaking the downstream region.On account of the above phenomena the weak pressure wave generated at time i =grows stronger and steeper when it moves towards right. When this growth continuous, atsome stage t = t the form of wave is vertical. This vertical wave front is called NormalShock Wave across which the changes in pressure, density, temperature, velocity andMach number are abrupt.

Fig. 3.2 shows a normal shock wave in a frictionless constant area duct contained in acontrol volume. The governing equations used in normal shock waves are(i) Continuity equation(ii Momentum equation

(iii) Energy equation and(iv) Equation of stateThe properties of gas in the upstream side is x’ and the downstream isy.From continuity equation, the mass flow ratem = Px A Cx = py A C {A = A) A = constant]= p = p...

It is assumed that, there is no heat transfer and the shaft work is zero. Therefore, theadiabatic energy equation for the control volume containing the shock gives= ho = h = constant

From momentum equation(P — P A = iii (c — C

From equation of stateh = f(S,p) 1s = f(P,p) 5

The above equations are used to define the two important curves known as Fanno andRayleigh curves.

Fig. 3.2 shows a normal shock wave in a frictionless constant area duct contained in acontrol volume. The governing equations used in normal shock waves are(i) Continuity equation(ii Momentum equation(iii) Energy equation and(iv) Equation of state

The properties of gas in the upstream side is x’ and the downstream isy.From continuity equation, the mass flow ratem = Px A Cx = py A C {A = A) A = constant]

It is assumed that, there is no heat transfer and the shaft work is zero. Therefore, theadiabatic energy equation for the control volume containing the shock gives= ho = h = constant

From momentum equation(P — P A = iii (c — C

From equation of stateh = f(S,p) 1s = f(P,p) 5

The above equations are used to define the two important curves known as Fanno andRayleigh curves.

Fanno LineFanno line gives an adiabatic flow process in a constant area duct with friction. Sincethere is a friction, therefore the process is irreversible. The stagnation enthalpy and massflow rate per unit area remains constant. The governing equations used for Fanno floware continuity equation, energy equation and equation of state.

By substituting different values of Cy, we will get a line called fanno line on the h — sdiagram as shown in Fig. 3.3. Figure shows the constant pressure lines also. The entropyis maximum at point F’ where the Mach number M = I is derived below. From adiabaticenergy equation

dh + C dC = 0

pdC+Cdp = 0

l’ herefore, at the maximum entropy point (F) on the fanno line is sonic i.e., M = 1. Theupper side of the curve represents subsonic flow whereas the lowerside is supersonic.

Rayleigh LineRayleigh line describes a frictioniess flow process in a constant area duct with heattransfer. The mass flow rate per unit area remains constant. The governing equations usedfor Rayleigh flow are continuity equation, momentum and equation of state. Fromequations

By substituting different values of Cy, we will get a line called Rayleigh line on the h — s diagram as shown in fig. 3.3. The entropy is maximum at point ‘ R’ where theMach number M = I is derived below.

pC = constant

DifferentiatingpdC+Cdp = 0pdC = -Cdp

Prandtl-Meyer Relation

It is a fundamental equation which gives the relation between the gas velocities beforeand after the normal shock and the critical velocity of sound. Praridtl-Meyer equation isthe basis for other equations for shock waves.

M x M* = 1

Mach Number Downstream of the Normal Shock WaveMT = Mach number before the normal shock (or) upstream Mach numberM = Mach number after the shock (or) Downstream Mach number

Static Pressure Ratio Across the ShockFx = Fy

Temperature Ratio across the Shock

Density Ratio Across the Shock (or) Rankine-Hugoniot EquationThe Rankine-Hugoniot equation gives the relationship between the pressure and densityratios across a shock wave in a perfect gas.From continuity equation,Px C = p C) = pC = constantFrom momentum equation,

From adiabatic energy equation

By substituting this in the adiabatic equation,

We know that,

The above equation can be written as in terms of pressure ratio across the shock.

These Equations are known as Rankine-Hugoniot equations. These equations arecompared with isentropic process equation for the same pressure-density relation .

Stagnation Pressure Ratio across the ShockShock wave is an irreversible one across which there is an increase in static pressure andentropy hut loss in stagnation pressure. This stagnation pressure ratio is derived under asa function of upstream Mach number

We know that

We know that

Change in Entropy across the ShockThe change in entropy for a perfect gas is given by

Therefore the change in entropy across the shock

Impossibility of Rare Faction Shock Wave

The variations of downstream Mach number and change in entropy with upstream Machnumber is shown in the Figure. When the Mach number before the shock is greater, the

Mach number after the shock is very small.

When M, is greater than I. M is less than 1. By substituting ‘ Mi’ value in equation (3.45)for a given value of r, the change in entropy is positive. On the other hand, when M isless than I (subsonic), M will be greater than I (supersonic) and the downstream pressure(P is less than the upstream pressure ( P ). This will be possible only if the shock is anexpansion shock. But the change in entropy is negative. A decrease in entropy in anadiabatic process which violates the second law of thermodynamics. Therefore anexpansion shock wave (Rarefaction shock) is impossible and the shock is always acompression shock.

Strength of a Shock WaveThe strength of a shock wave is defined as the ratio of increase in pressure due to shockto the pressure before the shock. It is used in shock wave analysis.

Thus, the strength of a shock wave is proportional to (M$ — I). When M is greater, theshock waves will be strong shocks.

The strength of the shock wave becomes

It is observed from the above equation, when the density ratio is 6, the strength of theshock is infinity.

Supersonic Wind TunnelsThe convergent divergent section ofa supersonic wind tunnel is shown in Fig. 3.6 and 3.7.It consists of a nozzle, test section and a diffuser. Normal shock takes place at the testsection is shown in fig. 3.6. The supersonic flow leaving the test section is then reversiblydecelerated in the diffuser which raises the gas pressure to a back pressure value.

Due to boundary layer growth, the diffuser throat area is greater than the nozzle throatarea. As a result of shock wave there will be a stagnation pressure loss across the shockand change in critical areas, but mass flow parameter is constant.Applying Fliegner’ s formula across the shock wave.

The above equation shows that, when a shock is at the test section, the diffuser throatMea. is always greater than the nozzle throat area. Therefore for steady runningconditions it is economical to keep the shock at the diffuser throat is shown in the figure.

- change in enthalpv in reversible diffusionDiffuser efficiency 1D = .change in enthalpy in actual diffusion

111(a A gas (r = 1.4, R = 287 J/Kg K) at a Mach number of 1.8, P = 80 KPa and T = 373K passes f/trough a normal shock. Determine its density after the shock. (‘ompare thisvalue in an isentropic compression through the same pressureratio. [ ‘ 95. MS’ Lí j(b) A jet of air at 275 K and 69 KPa has an initial Mach number 2. If it passes through anormal shock wave determine at downstream of the shock the fOllowing Mach number,pressure, temperature, density, speed of sound and jet velocity. [ ‘ 95. Madras, Nov. 95,

MKU Apr. ‘ 96,

2.An air plane having a diffuser designed for subsonic flight, has the normal shockattached to the edge of the diffuser when the plane is flying at a certain Mach number. Ifat the exit oft/ic d?ffuser the mach number is 0.3. What must be the flight mach numberassuming isentropic dfffusion behind the shock ? The area at inlet is 0.29 mand that at exit is 0.44 m [ ‘ 96. Madras]

QUESTIONS AND PROBLEMS1.Describe two practical situations where oblique shock waves are produced. How arestrong and weak shocks generated and how do they affect ?

2.Derive the energy equation for flow through an oblique shock:

State the assumptions used. Why is it same for the normal and oblique shocks?

3. Starting from the general energy equation for flow through an oblique shock obtain thePrandtl’ s equation:

Deduce from this the corresponding relations or the normal shock and an infinitesimalpressure wave.

4 Derive the Rankinc-Heugonoit relation for an oblique shock.

Compare graphically the variation of density ratio with the initial Mach number inisentropic flow with oblique shock.

5. Using the normal shock relations obtain the explicit expressions for the followingquantities in terms of the initial Mach number and the wave angle, for oblique shocks:

7 Show graphically the variations of the following quantities for flow through an obliqueshock, with the initial Mach number at two different values of the wedge angle:

8.a What is a shock polar? (h) Derive the shock polar equation

b. Sketch shock polar diagrams at two values of the initial Mach number. Show thepositions of the strong and weak oblique shocks, the normal shock and the infinitesimalwave on the diagrams.9.Sketch the Mach waves in a flow field over a convex corner of a curved wall. Theinitial flow is sonic and the final Mach number is 2.0. Draw Mach waves indicating theMach angles at

10. The stagnation pressure and temperature of air at the entry of a nozzle are 5 bar and500 K respectively. The exit Mach number is 2.0 where a normal shock occurs. Calculatethe following quantities before and after the shock: Static and stagnation pressures andtemperatures, air velocities and Mach numbers. What are the values of stagnationpressure loss and increase in entropy across the shock?

11. A Mach-2 aircraft engine employs a subsonic inlet diffuser of area ratio 3. A normalshock is formed just upstream of the diffuser inlet. The free-stream conditions upstreamof the diffuser are: p = 0.10 bar, T = 300 K. Determine(a) Mach number, pressure and temperature at the diffuser exit.(b) Diffuser efficiency including the shock.Assume isentropic flow in the diffuser downstream of the shock.

12. (a) Derive the following relations for normal shock waves:

13. Depict graphically the variation of the following quantities with the Mach numberupstream of the normal shock:

14. A supersonic is provided with a constant diameter circular duct at its exit. The ductdiameter is same as the nozzle exit diameter. Nozzle exit cross-section is three times thatof its throat. The entry conditions of the gas (t’ = 1.4, R=0.287 kJ/kg K) are p 10 bar, T600 K. Calculate the static pressure, Mach number and the velocity of the gas in the duct:(a) when the nozzle operates at its design condition,(b) when a normal shock occurs at its exit, and(c) when a normal shock occurs at a section in the diverging part where the area ratioA/A* = 2.0

15. A nozzle is designed for superheated steam ( = 1.3) with a pressure ratio P/Po =0.185; this operates at an off-design condition with a pressure ratio of 0.754. Determinethe area ratio (A/A*) at the section where the normal shock occurs. What are the valuesof design and off-design Mach numbers at the nozzle exit?

16. Determine the pressure ratio and exit Mach number for a normal shock at the arearatio A/A* = 1.305 in the nozzle of problem 6.15. What is the maximum value of theMach number occurring in the nozzle?

17(a) Explain how strong compression and expansion waves arc formed in acompressible fluid.(b) Under what conditions a compression wave changes into a shock wave(C) Why are expansion shock impossible?

18(a) Write down the four basic equations which satisfy the static points before andafter a normal shock wave.(b) Using the above equations prove that at the maximum entropy on the enthalpy-entropy diagrams the Mach number is unity.

19 Starting from the energy equation for flow through a normal obtain the followingrelations:

20.Derive the following relations for flow through a normal shock:

21. (a) Show that the stagnation pressure ratio and the change in entropy across a normalshock are given by

22. (b) If a diffuser achieves compression of air through a normal shock wave at an initialMach number of 1.5 determine (I) its efficiency (ii) stagnation pressure loss and (iii) theincrease in entropy.

23.A jet of air at 275 K and 0.69 bar has an initial Mach number of 2.0. If it passesthrough a normal shock wave d (a) Mach number (b) pressure (c) temperature (d) density(e speed of Sound and ( jet velocity downstream of the shock.

24. M aircraft flies at a Mach number of 1.2 at an altitude of 160 metres(p = 103 mbar, T= 216.65 K). The compression in its engine is partlyachieved by a normal shock wave standing at the entry of its diffuser.Determine immediately downstream of the shock(a) Mach number(b) temperature of the air(c) pressure of the air and(d) stagnation pressure loss across the shock