00:01
Now from a given question we have first drawn the ts diagram of the rakeen cycle with reheat.
00:11
Now we will obtain the properties of steam at pressure value 40 bar, temperature value of 350 degrees centigrade from superheated water tables.
00:25
Now here we have the value from superheated water tables of h1 to be equal to 3091 .8 kilojoules per kg, s1 being 6 .581 kilojoules per kg kelvin.
00:43
We have to note that the properties of steam at state 2 when steam expands and it is just dry saturated.
00:51
We obtain the properties at s2 being equal to s1 being equal to s of g which is equal to 6 .581 kilojoules per kg from saturated water tables.
01:04
We also get the value h2 being equal to 277.
01:07
78 .3 kilojoule per kg, and p2 being equal to 1014 .4k.
01:17
Kaspel.
01:18
Next, we've obtained the properties at p2 pressure being equal to p3, which is equal to 1014 .4kcal and at temperature 350 degree centigrade from superheated water tables.
01:34
We also got the value, h3 being equal to 3157 .1 kilojoule per kg.
01:41
S3 being equal to 7 .294 kilojoule per kg.
01:49
And next we'll find the isentropic enthalpy at the exit of the first turbine by using isentropic efficiency of the turbine, which is given as h1 minus h2 by h2s.
02:09
Substituting the value of efficiency being 0 .84, h1 being 3091 .8.
02:19
Kilojouze per kg minus 2778 .3 h2 value kilojoues per kg divided by rearranging and solving which we get h of s to be equal to 2718 .585 .85 kilojoues per k.
02:42
Therefore the isyentropic enthalpy at the exist of the first turbine by using isentropic efficiency was determined to be 2718 .58 .85 kilojoules per kege.
02:54
Now for the process 3 to 4 s, it is an isotropic expansion in the second turbine, which is given as s3 being equal to s4 of s.
03:05
Let us obtain properties of steam at pressure 4 being equal to 0 .035 bar or 3 .5 kilo -pascal from saturated water tables.
03:17
We get h of f to be equal to 1 .1 .8 kilojoums per kg.
03:24
M .s.
03:25
Of f to be equal to 0 .390 kilojoule per k .g.
03:32
Calvin.
03:33
F to be equal to 0 .001 meter cube per k.
03:37
G to be equal to 2549 .3 kilojoule per kg.
03:45
Nose of g being equal to 8 .52 kilojoule per kg calvin.
03:52
Next we'll obtain the dryness fraction of steam at pressure for being 0 .035 .0.
04:00
And 4 is being 7 .294 kilojoule per k .g.
04:09
Calving by using the following equation that is equal to at 3 .5 kilopascal plus at 3 .5 kilopascal.
04:32
Substituting the values we get 7 .294 equal to 0 390 plus 8 .52 minus 0 .390 plus 8 .52 minus 0 .3...