00:01
A simple rancine cycle uses water as the working fluid.
00:08
The boiler operates at 6 ,000 kioskle and the condenser operates at 50 kiosk at the entrance to the turbine.
00:23
Entrance to turbine temperature is 450 degrees celsius the isotropic efficiency of the turbine is 94 percent the pressure and pump losses are negligible and the water leaving the condenser water leaving condenser has the temperature which is sub -cooled 6 .3 degrees celsius the boiler is sized by for a mass flow rate of 20 kg per second then determine the rate at which heat is added to the boiler.
01:09
Next the power required to operate the pumps and the net power produced by the cycle.
01:21
And one more is thermal efficiency.
01:27
For the pressure of 50 kylpascal and initial temperature t1, say to be 6 .3 degrees celsius, we have saturation temperature 81 .3 degrees celsius and according to which the value of enthalpy at 1 state is 314 kilojoule per k .g.
01:49
For the enthalpy h2, we will add the work done by the pump.
01:53
H2 equals h1 plus, worked done by pump is v1 into ph minus pl substituting the values 314 plus 0 .00103 into 6 ,000 minus 50.
02:15
On solving we have h2 value is 320 .13 kilojoules per kg.
02:22
To determine the h3 value we have temperature 450 degrees celsius according to which we have from the steam table h3 is 330 0 .0 kilojoule per kg and entropy at the same state s3 is 6 .720 kilojoule per kg kelvin is equals to the entropy at 4th state also and the low pressure with the appropriate to determine the 4th enthalpy h4n is 2340 340 k kilojoule per kg will use the efficiency as entropy efficiency 94 % to determine the value of h4 is h3 minus efficiency multiplied by h3 minus h4 and substituting the value from above we have h4 equals 2397 .6 kilojoule per kg to determine the rate of heat q .b dot b is to mass flow rate multiplied by h3 minus h2...