00:01
A steam power plant has high pressure as 20 megapascal and low pressure as 10 kilo pascal and an open feed water heater which is operating at 1 megapaascal with an exit as a saturated liquid.
00:18
The maximum temperature is 800 degrees celsius and the turbine has a power output is 5 megawatt assuming the isentropic efficiency of 85 % the physical components and t .s.
00:35
Diagram is shown in the diagram for one feed water heater.
00:40
The same state numbering is used and from the steam table, state s5 is given by pressure and temperature according to which from the steam table h5 value is 4069 .8 kilojoule per kg and entropy is 7 .05 double 4 kilojoules per k .g kelvin.
00:59
Now for state 1 defined by the quality and pressure here h1 equals 191 .8 kilojoule per kg and specific volume in 0 .00101 meter cube per kg.
01:13
Now for the state 3 we have h3 equals to 762 .8 kilojoule per kg and specific volume v3 is equals to 0 .001127 meter cube per kg.
01:31
Now for the pump 1, work done is equals to v1, p2 minus p1, substituting the value 0 .00101 multiply 990 is equal to 1 kilojoule per kg.
01:46
Now from here, value for h2 equals to h1 plus work done by pump 1 is equal to 192 .81 kilojoule per kg.
01:56
For the turbine which has 5 to 6 state, s6 equals, s6 equals to 1.
02:01
To s5 and from here we get h6 equals to 3013 .7 kilojoule per kg...