Liquid water as a compressed liquid at 65 degree F and 20 ps
Solution
SOLUTION:
1) given-
first stream of cold water at 65 farenhite or 18.33 degree celsius has property as
t1=18.33+273=291.33 degree kelvin
Cpa=1.005 and x1=dryness fraction=0,Hliq1=1.005*291.33=292.78 KJ/Kg
second stream of vapor has property as
DBT=100 degree celsius and Hliq2=456.47 KJ/Kg and Hg2=2689.31 KJ/Kg and x2=1 from steam table
2) condition of mixture of vapor and liquid turn to steam, hence enthalpy at exit is given as
m1*Hliq1+m2*Hg2=m3*H3
AS m1=m2,hence
m1+m2=m3 on arranging we get,
H3=(Hliq1+Hg2)/2=1491.04 KJ/Kg
3) temperature at exit is given as
m1*t1+m2*t2=m3*t3
t3=(t1+t2)/2=65+212/2=138.5 degree farenhite
4) condition at exit is given as
m1*x1+m2*x2=m3*x3
x3=(x1+x2)/2=(0+1)/2=.5
hence dryness fraction of steam would be 0.5 or .5 kg of vapor in kg of wet steam
