Question

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08 Question 2.7 Two possibilities are considered for the design of a two-dimensional jet engine inlet as shown in the figures (a-b) below. The inlet is to operate at the Mach number Mi, shown in the table below. In the first design configuration, shown in figure (a), the deceleration of the flow takes place through a normal shock. For the second case, shown in figure (b), a wedge-shaped diffuser is used to decelerate the flow through two oblique shock waves followed by a normal shock wave, where the wedge turning angles are both, θ, as shown in the figure and specified in the table. Demonstrate the advantage of the "weak shock diffuser" design [shown in (b)] over the 209mple inlet [shown in (a)], by evaluating the loss of stagnation pressure for each case. 210 Design Data Value Unit 212Mach number (M) 213 Wedge turning angles (e) deg 214 215 Calculate: Your answers a) Ratio of stagnation pressures (Po Po) for simple inlet 216 [1 mark] b) Angle of first shock for weak shock diffuser deg [l mark] 2 marks] deg 1 mark] [2 marks] 217 c) Ratio of stagnation pressures (Po Po) for weak shock diffuser 218 d) Angle of second shock for weak shock diffuser e) Ratio of stagnation pressures (Pos Po) for weak shock diffuser 220 f) Ratio of stagnation pressures (Po Po:) for weak shock diffuser g) Overall stagnation pressure ratio (Po Po) for weak shock diff 221 [1 mark] [1 mark]

Answer 1

(a) Considering the first case where the flow is decelerated by means of generation of normal shock wave.

Given upstream mach number ;

M = 2.8

To find the stagnation pressure ratio we have to use the below formula :

:
Poi Pi
and
Po2 P2

Dividing the above two equations would give us the stagnation pressure ratio. But as we can see due to lot of questions involved in the process let's make use of shock tables.

From normal shock table at
y1.4 and Mi- 2.8
we can obtain

$M_{2}=0.488$

P2 A = 8.98 Pi

r02-0.389 Po2 Po1

Therefore for simple inlet, the stagnation pressure ratio is
r02-0.389 Po2 Po1

(b)To find angle and stagnation pressure ratio
Po2
for first shock of weak shock diffuser:

Given
M = 2.8
and wedge angle $\theta =9^{0}$

From oblique shock tables for
1.4 . θ-90 and M-28
, it can be observed that

The first shock wave angle
β = 28.07750
(note in tables if the value of wedge angle is not given then interpolate between the given values of wedge angle)

To find the value of stagnation pressure ratio for oblique shock, the mach numbers needs to be resolved perpendicular to the shock and can be solved by treating the same as normal shock.

Mn1 = Misin 3 and Mn2= M, sin(3-0
where $\beta \ represents \ shock \ wave \ angle \ and \ \theta \ represents \ wedge \ angle.$

Now using normal shock tables with value of $M_{n1}$:

$M_{n1}=2.8*sin(28.0775)$

$M_{n1}=1.32$

Again returning to normal shock tables , at $M_{n1}=1.32$ and $\gamma =1.4$,

$\frac{P_{02}}{P_{01}} = 0.976$

$M_{n2}=0.776$

Therefore $M_{2}=\frac{M_{n2}}{sin(\beta -\theta )}$

0.776 sin(28.0775 -9) เงิ

$M_{2}=2.374$

Therefore the stagnation pressure ratio for the first shock for weak shock diffuser is  $\frac{P_{02}}{P_{01}} = 0.976$.

To find angle and stagnation pressure ratio $\frac{P_{03}}{P_{02}}$ for second shock of weak shock diffuser:

To find the value of stagnation pressure ratio for oblique shock, the mach numbers needs to be resolved perpendicular to the shock and can be solved by treating the same as normal shock.

$M_{n2} = M_{2}sin\beta \ and \ M_{n3} = M_{3}sin(\beta -\theta )$where $\beta \ represents \ shock \ wave \ angle \ and \ \theta \ represents \ wedge \ angle.$

Given $M_{2}=2.374$ and wedge angle $\theta =9^{0}$

From oblique shock tables for
1.4 . θ-90 and M-28
, it can be observed that

The second shock wave angle $\beta = 32.9^{0}$ (note in tables if the value of wedge angle is not given then interpolate between the given values of wedge angle)

Now using normal shock tables with value of $M_{n2}$:

Mn22.374 sin(32.9)

$M_{n2}=1.289$

Again returning to normal shock tables , at $M_{n2}=1.289$ and $\gamma =1.4$,

$\frac{P_{03}}{P_{02}} = 0.979$

$M_{n3}=0.793$

Therefore $M_{3}=\frac{M_{n3}}{sin(\beta -\theta )}$

$M_{3}=\frac{0.793}{sin(32.9-9)}$

$M_{3}=1.957$

Therefore the stagnation pressure ratio for the first shock for weak shock diffuser is  $\frac{P_{03}}{P_{02}} = 0.979$

To find the stagnation pressure ratio across the normal shock between stations 3 and 4 :

Now the flow passes through a normal shock.

We have obtained the value of 1.957;

from normal shock tables ,

$\frac{P_{04}}{P_{03}} = 0.744$

Hence the third stagnation pressure ratio is $\frac{P_{04}}{P_{03}} = 0.744$

Now , $\frac{P_{04}}{P_{01}} = \frac{P_{04}}{P_{03}}*\frac{P_{03}}{P_{02}}*\frac{P_{02}}{P_{01}}$

$\frac{P_{04}}{P_{01}} = 0.976*0.979*0.744$

$\frac{P_{04}}{P_{01}} = 0.7109$

Therefore overall stagnation pressure ratio is $\frac{P_{04}}{P_{01}} = 0.7109$

RESULTS OBTAINED:

(a)For simple inlet, the stagnation pressure ratio is

r02-0.389 Po2 Po1

(b)The first shock wave angle

β = 28.07750

(c)The stagnation pressure ratio for the first shock for weak shock diffuser is  $\frac{P_{02}}{P_{01}} = 0.976$

(d)The second shock wave angle $\beta = 32.9^{0}$

(e)The stagnation pressure ratio for the first shock for weak shock diffuser is  $\frac{P_{03}}{P_{02}} = 0.979$

(f)The third stagnation pressure ratio is $\frac{P_{04}}{P_{03}} = 0.744$

(g) The overall stagnation pressure ratio is $\frac{P_{04}}{P_{01}} = 0.7109$

(Important note: Shock tables have been used so that the required values can be found right away.Using the data in formulae takes lot of time as already mentioned above. Also note that the values from tables have been interpolated for the required solution as the direct values for wedge angle of  $9^{0}$ were not mentioned in the tables

COMMENT :

As it is evident from the stagnation pressures ratios of of both the cases the simple case results in drastic loss in stagnation pressure. In other words the ability to extract useful work is being destroyed due to excessive loss in stagnation pressure. Whereas the second case i.e; the wedge shaped diffuser minimizes the stagnation pressure loss to a greater extent. Hence this gives an edge over the simple diffuser.