Answer to Question #140100 in Mechanical Engineering for Kalmic

Question #140100
Determine the density of steam at a pressure of 5.5 bar and dryness fraction 0.85
1
Expert's answer
2020-10-28T08:45:15-0400

Given as:

Pressure of steam(P)=5.5 bar=0.55MPa

Dryness fraction(x)=0.85

Specific volume(v)="V_f+x(V_g-V_f)"

From steam stable: For saturation pressure P=0.55MPa

Vf=0.00109648 "\\frac{m^3}{kg}" , Vg=0.33972 "\\frac{m^3}{kg}"

V=0.00109648+0.85"\\times" (0.33972-0.00109648)=0.28979 "\\frac{m^3}{kg}"

Density of steam("\\rho" )="\\frac{1}{Specific volume(V)}= \\frac{1}{0.28979}" =3.45071 "\\frac{m^3}{kg}"


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