00:01
For this problem on the topic of thermodynamic property relations, we are told that saturated water vapor is expanded while its pressure is kept constant to a higher temperature.
00:13
We want to calculate the change in specific enthalpy and entropy using firstly the departure charts as well as the property tables.
00:22
So if we were to use the departure charts, we first made the pressure of water vapor during this process.
00:29
So p1 is equal to p2, the pressure is kept constant, and this is the saturation pressure at our temperature of 400 degrees fahrenheit.
00:44
And from our tables, we know this to be 247 .26 pounds per square inch absolute.
00:57
Now if we go to our ideal gas property table for water vapor, we can find p -chane.
01:03
In enthalpy h bar 2 minus h bar 1 for an ideal water vapor and a motor basis to be 10 ,354 .9 minus 6 ,8955 .6, which gives an enthalpy change during this process of 3 ,459 .3 btu -pan -moe.
01:46
So we have the change in entropy on a molar basis.
01:51
Similarly the change in entropy on a molar basis is s2 bar minus s1 bar again we assume an ideal gas behavior and this is equal to the temperature dependent entropy s2 minus the temperature dependent entropy s1 -0 minus the universal gas constant ru times the natural log of the ratio of pressures p2 over p1.
02:26
And so if we substitute our values into this, this becomes 52 .212 minus 48 .916.
02:37
And obviously the last term becomes zero since p2 and p1 are the same.
02:42
So the natural log of 1 becomes zero.
02:45
This gives us the change in entropy on a molar basis to be 3 .296 btu per pound mole r.
03:03
Now the enthalpy and entropy departures of water vapor at the given states can be determined from the generalized charts.
03:11
And we can do this by first calculating the reduced temperature of the initial state tr1 is the specific temperature t1 over the critical temperature tc.
03:23
And so this is 860 rankin divided by 1 ,164 .8 rankin, which gives us a relative or reduced temperature of 0 .730.
03:41
0.
03:43
The reduced pressure of the first state pr1 is equal to p1 over the critical pressure pc, and that's 247 .26, which we found above, over 3 ,200.
04:02
And simplifying, we get a reduced pressure of 0 .0727.
04:10
So from these values, we find the enthalpy departure, zh1 to be 0 .1342, and our entropy departure of the first state, zs1, to be 0 .1171.
04:32
Now we can do the same for the second state.
04:37
So the reduced temperature of the second state, tr2, is equal to the specific temperature temperature t2 over the critical temperature of water vapor tc and this is 1 ,260 ranken over 1 ,164 .8 which gives us the reduced temperature of 1 .08 of 1 .081 and it reduced pressure pr2 which is p2 over pc and this is 680 80 .56 divided by the critical pressure of 3 ,200.
05:30
So actually no, it's the pressure is constant and so the pressure is 247 .26.
05:43
So the pressure does not change over 3 ,200 which again is 0 .07727.
05:50
So the reduced pressure remains constant as well.
05:56
And this from our generalized charts gives us an entropy departure zh2 of 0 .07213, an entropy departure zs2 of 0 .04595.
06:18
So we can calculate the entropy and entropy changes per most.
06:22
Basis as follows.
06:28
So per mole we get the change in enthalpy h2 bar minus h1 bar to be the change in enthalpy on a molar basis for the ideal gas h2 bar minus h1 bar minus the universal gas constant ru times the critical temperature tc times the change in enthalpy departure is in h2 minus zh1...