Answer to Question #140100 in Mechanical Engineering for Kalmic

Question

Determine the density of steam at a pressure of 5.5 bar and dryness fraction 0.85

Answer 1

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 $\frac{m^3}{kg}$