00:01
Hello everyone.
00:02
In the present problem it is given that air enters a converging diverging nozzle at pressure pi is equal to 1mpa.
00:11
Air enters a converging diverging nozzle at temperature t .i.
00:16
Is equal to 300 kelvin.
00:21
Air enters a converging diverging nozzle at negligible velocity and mac number at the exit plane of the nozzle at the upstream.
00:32
Of the shock ma1 is 2 .4.
00:36
Coming to the solution we have from ideal gas properties of air specific heat ratio of air k is equal to 1 .4 gas constant of air are is equal to 0 .287 kilojoule per kg kelvin.
01:41
The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible then p .o1 is equal to pi is equal to 1 mp a 8th.
03:01
T .o .1 is equal to t .o .1 is equal to t .i.
03:05
Is equal to 300 kelvin.
03:14
Then inlet of air at the upstream of the shock is t1 is equal to t01 multiplied by 2 divided by 2 plus k minus 1 multiplied by m a 1 square t1 is equal to 300 multiplied by 2 divided by 2 plus 1 .4 minus 1 multiplied by 2 .0 multiplied by 2 .4 square.
04:25
Solving we have t1 is equal to 139 .4 kelvin and p1 is equal to 139 .4 kelvin and p1 is equal to to p01 multiplied by t1 divided by t not whole raised to k divided by k minus 1 so p1 is equal to 1 multiplied by 139 .4 divided by 300 whole raised to 1 .4 divided by 100 whole raised to 1 .4 divided by 0 .4 divided by 0 .4.
05:19
So p1 is equal to 0 .06840 mpa.
05:39
The fluid properties after the shock are related to those before the shock through the functions listed in.
06:33
Table which gives one dimensional shock functions for an ideal gas.
07:07
So, for m .a .1 equal to 2 .4.
07:15
We read m .a2 is equal to 0 .5 .5.
07:24
2 .31, p02 divided by p .01 is equal to 0 .5401.
07:37
P2 divided by p1 is equal to 0 .5401.
07:40
P2 divided by p1 is equal to 6 .5533 and t2 divided by t1 is equal to 2 .0403.
07:58
Therefore, mac number after the shock wave is 0 .5231...