Oblique Shock Homework Help, Assignment Help, Online Tutor, Tutoring

Submit Assignment

* *

Payment Online Assignment Help

Metallurgical Engineering

Homework Help

sfsf

Oblique Shock

Introduction

We have studied the one dimensional compressible flow. The shock encountered by such the flow is the normal shock. Herewith we are going to deal the compressible flows in two dimensions, 2D. There are two entities which are going to tackle the expansion and compression of the 2D compressible flow. Such situations are two situations shown in Fig. 1a and Fig. 1b which the flow might encounter. If the flow turns in to itself then the compression of flow takes place through shock as shown in Fig. 1a. Since this shock makes certain angle with the flow, it is called as the oblique shock. However if the flow turns away from itself then the expansion of the flow takes place through an expansion fan as shown in Fig. 1b. Deriving various relations for angles or properties ratios for the two dimensional compressible flows in the presence of shock or expansion fan is the central objective here onwards.

The main reason of having shock or expansion fan at the deflection corners is to deflect the flow accordingly so as to maintain its parallelism with the wall in order to avoid the direct collision. Therefore the angle of shock with the wall or free stream velocity will be dependent on flow deflection angle and free stream velocity.

Mach Wave

Let’s consider a railway travelling along its track from location A to location B. If it makes a beep sound while leaving station A and at intermediate locations C and D then the sound wave travels spherically out with the velocity (say) 'a' m/s. The distance travelled by the sound waves originated from A, C and D locations are given by the circles drawn, considering respective points as centres. Now if the speed of the train is less than acoustic velocity (a), say v m/s, then the train will always remain inside the circles. This statement also means that the distance travels by the train in some time interval is always lower than the distance travelled by the acoustic wave in the same time interval (Fig. 2a). Now if we consider the speed of train to be more than speed of sound then the train will always cross the respective circle of acoustic speed in that time interval. Now if we draw the tangent to all circles, representing position of acoustic wave, from the end location of train (B), then we can see in Fig. 2b that such a tangent makes an angle μ with the train track. The expression for this angle is as,

This angle (μ) is called as Mach angle and it becomes the reference angle for our following discussions.

Shock angle which we are going to understand is always more than the Mach angle.

Oblique Shock

Consider the flow taking place along a wedge as shown in Fig. 3. Let θ be the wedge angle and β be the shock angle with the wall which is parallel to the approaching free stream.

As we have already proved that shock exists only for supersonic flows, consider a supersonic flow of Mach number M₁ approaching the wedge. In the presence of the shock, flow deflects by an angle θ which is the wedge angle.

Let’s solve the mass, momentum and energy equations for this flow. Consider the control volume as shown in Fig. 3. In this special control volume, inlet and outlet are parallel to the shock. Other two faces of the control volume are parallel to the streamline hence these faces will not contribute to the mass, momentum and energy fluxes.

Let u be the velocity normal to the shock and w be the velocity parallel to the shock. Graphical demonstration of these velocities is given in Fig. 3. Station 1 corresponds to inlet or preshock conditions while station 2 corresponds to outlet or post shock conditions.

Let’s assume the flow to be steady, hence,

Hence,

(ρu)₂ = (ρu)₁ or ρ₁u₁ = ρ₂u₂ (1)

This is the mass conservation equation for oblique shock conditions expressed in terms of velocities normal to the shock.

Now consider the momentum conservation equation for the same flow. Since momentum is the vector equation,

we have to consider, two equations, viz, normal and parallel to the shock. Let’s initially consider the momentum equation in integral form for inviscid flow.

For steady flow, this equation becomes,

Now consider the momentum equation in the direction parallel to the shock wave. Since there is no pressure difference in this direction, the right hand side will be zero. Hence,

(ρuw)_pre = (ρuw)_post

but using mass conservation Eq (1) we can re-write it as,

(w)_pre = (w)_post or w₁ = w₂

This expression clearly suggests that velocity parallel to the shock remains conserved.

Now consider the momentum equation normal to the shock.

(p + ρu²)_post = (p + ρu²)_post

or

p₁ + ρ₁u₁² = p₂ + ρ₂u₂² (2)

We can clearly see that the momentum equation looks exactly same as that for the normal shock relations.

Here u is the velocity normal to the shock. Therefore only velocity component normal to the shock wave is responsible for the change in momentum since momentum and velocity tangential to shock are conservervative.