Steam enters an adiabatic turbine at 5 MPa and 600 °C with a...

Steam enters an adiabatic turbine at 5 MPa and 600 °C with a velocity of 50 m/s. The steam leaves the turbine as a saturated liquid-vapor mixture with a velocity of 120 m/s at 0.5 MPa and a quality of 0.7 which driving by a power output of 6000 kW. The elevation change between turbine inlet and exit is 4 m. a) Determine the changes in enthalpy (?h), kinetic energy (?ke) and potential energy (?pe), in kJ/kg. b) Determine the work done per unit mass of the steam flowing through the turbine. c) Calculate mass flow rate of the system.

Transcript

00:01 Here in this question at 5 megapascal and 600 degrees celsius from stream table we get h1 is equal to 0 .03107 kilojou per kg and v1 is equal to 50 per second at e2 equal to 0 .5 megapar and x equal 0 .7 hf is equal 640 .1853 kilojou per kg and hg is equal to 2748 .1076 kilojoules per kg now for calculating the change in enthalpy change in kinetic energy and change in potential energy we just take this values from steam table and now now first we are going to calculate the value of h2.

02:22 So h2 is equal to hf plus x multiply by hg minus hf.

02:37 Now we substitute the values from above.

02:42 So we get h2 is equal to 640.

02:50 01853 plus 0 .7 multiply by 2748.

02:59 Multiply by 2748.

03:03 1076.

03:06 1 minus 640.

03:16 1858.

03:18 Now, now on simplifying this we get h2 is equal to 211 .5 .730 kilojoule per kg.

03:37 Now here we have v2 is equal to 120 meter per second.

03:46 Now first in part a we are going to calculate the change in lphalp and change in enthalpy is given by delta h equal to h2 minus h1 now on substitute the values of h2 and h1 we get delta h is equal to two minus h one now on substitute the values of h2 and h1 we get delta h is equal to two 1115.

04:20 0 .730 minus 3666 6 .6 0 .8310 now on subtracting this we get delta is equal to minus 1551 .1 .10 per kjee.

05:04 Now here the negative sign shows the decrease in enthalpy.

05:16 So as negative sign shows decrease in enthalpy therefore delta h is equal to 1551 0 0 below to per kg now again we are going to calculate the change in kinetic energy which is given by delta k e equal to v2 square minus v1 square divided by 200 now here we substitute 128 124 v2 and 50 for v1 divided by 200 now again we simplify this so here on simplifying this we get delta k e is equal to 5 .95 kilojou per kg now this is the change in kinetic energy.

07:16 Now we are going to calculate the change in potential energy...

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