00:01
Okay, so we know the critical properties occurs when mark number equal 1.
00:04
So now we need to use the table a32 at the back of the book to find the critical properties for nitrogen gas.
00:15
So at mark number equal 1, if you take a look at the table, we know p over p0 is equal 0 .5283, which means that critical pressure is equal to 0 .583 p0.
00:27
And row over row 0 is given as 0 .6339.
00:31
So therefore the critical density is equal to 0 .6339, row 0.
00:39
We have t over t0 is equal to 0 .833.
00:42
So we have a critical temperature is equal to 0 .8333 t0.
00:47
So now that's trying to determine the critical pressure first.
00:50
So when we know the initial pressure, or in other words, stagnation pressure is given at 700 kiosk.
00:58
So now that's plugging back to the equation to determine the critical pressure.
01:03
So the critical pressure is equal to 0 .52 .83 multiplied by 700kcal, which is equal to 370kcal.
01:27
So now that's determining the critical temperature.
01:30
So when the initial temperature is given as 400 kelvin.
01:33
So now that's plugging back to the equation to determine the critical temperature, which is equal to 0 .8.
01:40
3, 3, 3, multiplied by 400 kelvin.
01:44
And this will give us the critical temperature is about 333 kelvin.
01:52
So now we can determine the critical velocity, which is eventually equal to mark number times square root krt asterisk.
02:01
So when the mark number is given as 1, the specific ratio for the nitrogen gas is 1 .4...