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

Answer:

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

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