00:02
For the given problem, first solving for part a.
00:04
In part a, if you solve for the specific volume, let us suppose the specific volume is denoted as gamma.
00:15
So suppose the gamma 1 values equals to rt1 divided upon the pressure 1, that will be equals to r will be 0 .287, multiply it to temperature, that is 50 degrees celsius in kelvin's, it will be at 2 .7.
00:33
Divided upon the ancient pressure is 90k kiosk so this will be meter cube divided upon kilograms therefore the specific volume will come out to be 1 .03 meter cube per kilograms thus if we need to find the mass the mass will be equals to the volume divided upon the specific volume therefore it will be 1 .8 multiplied to 10 to the power minus 3 divided upon the specific volume is 1 .03 hence the value of mass will come up to be 1 .049 multiplied to 10 to the power minus 3 kilograms this is the mass for state 2 if we consider for state 2 in state 2 we can say that the temperature 2 will be equals to the temperature 1 r r is the compression ratio that will be raised to the power k minus 1 that will be equals to 323 in calvin's multiplied to the compression ratio that is 11 and the value of k will be 1 .35 for the given gas minus 1 therefore the value of t 2 comes out to be 747 .6 kelvin similarly for pressure we can calculate p 2 will be equals to p 1r divided or sorry multiply to t 2 divided upon t 1 and the value will come out to be 90 multiplied to the value of r is 11 multiplied t2 is 747 .6 and divided by t1 is 3223 hence the value of p2 will come out to be 2291 .4 kilo pascal it will mean kilopascal similarly if we consider for state 3 for state 3 we can again write that t3 this time will be equal to to t2 plus the heat injected divided upon the mass cv, that's specific heat at constant volume.
02:49
Therefore, it will be t2 is 747 .6 plus q1 is 0 .5 multiplied to 4 divided upon m is 0 .049, multiplied 10 to the power minus 3.
03:05
And for cv, it will be 0 .8 to 1 .1.
03:08
Hence the value of t3 will be 3069 .9 calvin.
03:16
For state 3 again if we say for the pressure, p3 will be equals to p2, t3 divided upon t2.
03:24
Therefore it will be 2291 .4 multiplied to t3 3069 .9.
03:33
Divided upon t2 is 747 .6.
03:37
Hence the value of p3 will be 9409 .3.
03:42
Again, it will be in kilo -pascals.
03:45
So, this is the required pressure for the third stage.
03:52
For part b, for part b, if we consider for the fourth state, we have the value of temperature at fourth state will be equal to t3, r raised to the power 1 minus k.
04:05
If we solve, we have 3069 .9, multiply to the r, that is, compression ratio will be 11, 1 minus 1 .35.
04:17
Hence, the value will come out to be 1326 .3 kelvin.
04:23
Thus, we need to determine the network.
04:26
Therefore, the network will be equal to mcv t3 minus t2 minus times t4 minus t1.
04:36
We have considered this equation because this equation represents the injected heat minus the outward heats...