Question

insulated rigid tank 3kg saturated liquid vapor at 200kpa. 3/4 mass is liquid. heated until all is vaporized. determine quality of states nd entropy change of the steam

Answer 1

from the steam table

we get to know the values

`p_(1) =200Kpa\\x_(1) =0.25`

`therefore\\v_(i) =v_(f) +x_(1) v_(fg) \\v_(i) =0.001061+(0.25)(0.88578-0.001061)\\v_(i) =0.22224m^3/kg`

we have

`s_(i) =s_(f) +x_(1) s_(fg) \\s_(i) =1.5302+(0.250)(5.5968)=2.9294 Kj/kg.K`

to find s 2

`v_(2) =v_(1) =0.22224m^3/kg\\x_(2) =1\\s_(2) =6.6335Kj/kg.K\\`

so the entropy change is

`S=m(s_(2)- s_(1) )`

put values

`S=(3kg)(6.6335-2.9294)Kj/kg.K\\S=11.1123Kj/K`