00:03
So, allusion of this question is for part a.
00:10
Now here given, eccentric efficiency is equal to 0 .7 multiplied by efficiency of compact.
00:23
So now efficiency of compact is equal to t2s minus t1 divide by t22.
00:39
Minus t 1 so now m eta com is equal to 0 .7 equal to 550 .68 minus 330 3 30 divide by t 2 minus 3 30 so in solving this t 2 is equal to 645 .1 5 4 kelvin now 3 divided by t 3 divided by t 4 s is equal to r p raised to par gamma minus 1 divided by gamma for centropic process centropic process so now 1200 divided by t4s equal to 6 raise to power 0 .4 divided by 1 .4.
02:07
So on solving this, t4s is equal to 719 .20 kelvin.
02:18
So now for eccentricropic efficiency, is eta t is equal to 0 .8 given.
02:31
So now eta t is equal to t3 minus t4 divide by t3 minus t 4 s.
02:45
Now substituting values in this equation 0 .8 equal to 0 .8 is equal to 0 .8 is equal to 1200 minus t 4 divide by 1200 minus t 4 s which is 719 t 4 s now this equation can be written as 1200 minus t 4 divide by 1200 minus 700 minus 700 minus 719 0 .20 so on solving the equation the value of t4 is equal to 815 .36 kelvin so now part a temperature at exit of turbine is given by temperature for at exit of turbine of turbine is given by temperature for at exit of turbine of turbine is equal to 815.
04:41
This is 815 .36 kelvin...