00:01
All right, guys, we are given the following data for our 410a, that is t1 is equal to 10 degrees centigrade.
00:09
We've got t2 is equal to 10 degrees centigrade, v is equal to 10l, that is equal to 0 .01 cubic meter meter, then we've got a is equal to 0 .006 meter square.
00:26
And then we've got p nod p nod is equal to 100 kilo x is equal to 0 .9 mp is equal to 61 .18 kg so from r 410 a table corresponding to t is equal to 10 degrees centigrade we can obtain specific volume and internal energy as vf is equal to 0 .00886 cubic meters per kg, and then we've got vfg is equal to 0 .0299 cubic meters per kg.
01:33
And then we've got uf is equal to 72 .24 kilojoules per kg.
01:42
And then we've got ufg is equal to 183 .66 kilojouz per kg so initial specific volume is given as v1 plus v1 is equal to vf plus x into vfg and that is equal to 0 .000000 886 plus x into vfg and that is equal to 0 .00806 plus 0 .9 into 0 .02295 and that is equal to 0 .0 .0211 cubic meters per kg.
02:27
So initial specific internal energy is given as u1 is equal to uf plus x into ufg and that is equal to 72 .24 plus 0 .9 into 0 .02295 and the u1 is equal to is equal to 237.
02:55
237 .534 kilojoules per kg.
03:01
So calculating our 410a mass we've got m is equal to v1 over small v1 that is 0 .0 divided by 0 .021541 is equal to 0 .46243 kd.
03:27
So final pressure is given as p2 is equal to p nought plus mp into g divided by a that is equal to 100 ,000 plus 61 .18 into 9 .81 divided by 0 .006, that is equal to 200 kilo pasco.
03:56
And the answer to p2 is equal to 200 kilo pasco...