00:01
All right guys, we are given the following data for water in rigid tag that is t1 is equal to 200 degree fahrenheit.
00:10
We've got t2 is equal to 20 degrees fahrenheit.
00:14
We've got x1 is equal to 0 .9 and we've got v1 is equal to 6 cubic feet.
00:21
So from superheated water table f7 .1 corresponding to temperature t1, we can obtain the saturation and evaporation -specific volume and internal energy.
00:31
That is vf is equal to 0 .1663 cubic fit per lbm and we've got vfg is equal to 33 .6146 cubic fit per lbm.
00:53
Then we've got uf is equal to 168 .03 btu per lbm and we've got ufg.
01:09
Is equal to 906 .15 btu per lbm.
01:17
Now guys, calculating the initial specific internal energy, that is, u1 is equal to uf plus x1 into ufg, that is equal to 168, 168 .103 plus 0 .9 into 906 .159 into 906 .151 .5 ,000, and 0 .03 plus 0 .9 into 906 .151 .05 .0 .0 .0 .0 .9 into is equal to 983 .5 .56b .m.
01:57
So calculating the initial specific volume, we have v1 is equal to vf plus x1 into vfg, that is equal to 0 .01663 plus 0 .9 into 33 .6463 plus 0 .9 into 33 .646666 .6 .6 .6.
02:19
And that is equal to 30 .27 cubic fit per lbm.
02:27
So calculating mass, we've got m is equal to v1 over small v1, that is the specific volume.
02:38
So we have 6 over 30 .27 is equal to 0 .2lbm.
02:46
So water in rigid tank, water is in rigid tank, so volume remains constant.
02:51
That is v2 is equal to v1 is equal to 30 .27 cubic fit per lbm.
02:59
So in constant volume process, work done is equal to zero.
03:05
So we have w is equal to zero from saturated, solid saturated vapor table f7 .4 corresponding to temperature t2, that is equal to 20 degree fahrenheit.
03:19
Heights, we can obtain saturation solid and evaporation specific volume and internal energy...