00:01
Here we have water at 1 megapascal initially and 250 degrees c and it is brought to a saturated vapor, so x2 is 1, in a piston cylinder with an adiabatic process, so q12 is 0.
00:22
So we know our energy equation now is this change in specific internal energy is minus equals minus, the change in specific work.
00:38
And the entropy equation, because again we're assuming that we're going to assume that this is reversible, gives us that the entropy in each state has to be the same.
00:49
So there's no change in entropy during this.
00:53
Let's see here.
00:54
So now we know two properties in state one, so we can get whatever else we need.
00:59
And i've got the specific volume, the specific internal energy, and the specific entropy.
01:06
Now we know that s2, is equal to s1 so that gives us two properties at the second state and we can figure out the temperature is roughly 141 degrees celsius and this is actually on the on the saturated vapor boundary here so and then we have the pressure at that state and the specific volume and the specific internal energy and they asked us to estimate to get the let's see here to estimate specific specific work from the area in the pv diagram and compare it to the correct value with the correct value we can get from here and that's 159 .35 kilojoules per kilogram now we can estimate it again we can have our pv diagram here and we have let's see here um pb diagram we have state one is like up here and state two is on the boundary here so we can um get this work, estimate the area by assuming that there's a linear approximation between the two states...