00:01
In this question, the area ratio is given as 2 .9, the back pressure is given as 18 .05 kilo pascal, the reservoir pressure is given as 199 .6 kilo pascal.
00:19
So area ratio will be given by 1 by m a 2 by gamma plus 1 multiplied by 1 plus gamma minus 1 by 2 m a square to the power gamma plus 1 by 2 gamma minus 1.
00:41
So specific heat ratio will be 1 .4.
00:44
So this will be 2 .9 is equal to 1 by m a 2 by 1 .4 plus 1 1 plus 1 .4 minus 1 by 2 m a square to the power 1 .4 plus 1 by 2 1 .4 minus 1.
01:04
So mach number 8 .8 comes out as 2 .6.
01:11
The exit pressure will be given by p naught divided by 1 plus gamma minus 1 by 2 m a square to the power gamma by gamma minus 1.
01:26
So this will be 199 .6 divided by 1 plus 1 .4 minus 1 by 2 into 2 .6 square to the power 1 .4 by 1 .4 minus 1.
01:43
So exit pressure comes out as 10 kilo pascal.
01:49
The ratio of back pressure to the exit pressure is given by 1 plus 2 gamma by gamma plus 1 multiplied by m a square sin square beta a minus 1.
02:09
Here beta is the oblique shock wave.
02:12
This will be 18 .05 by 10 is equal to 1 plus 2 into 1 .4 by 1 .4 plus 1 2 .6 square sin square beta a minus 1...