Question #128452

A rigid vessel with a volume of 10 m3 contains a water-vapor mixture at 400 kPa. If the quality is 60 percent, find the mass. The pressure is lowered to 300 kPa by cooling the vessel; find mKand m,. Am. 35.98 kg. 16.47 kg, 19.51 kg

Expert's answer

a) The pressure (p)=400kPa=0.400MPa

Mass of mixture =M

quality (X)= 0.60

Volume of mixture (V)=10 m3

From steam table at P=0.400MPa

Specific volume of saturated water (vf)=0.00108355 m3/kg

Specific volume of saturated steam (vg)=0.46238 m3/kg

Therefore, the volume of steam having X=0.6 is given by

V=M[vf+X(vg-vf)]

10= M[0.00108355+0.6(0.46238-0.00108355)]

M=36.04748 kg

If pressure is lowered to 300kPa

p=0.300MPa

From steam table we get,

Specific volume of saturated water (vf)=0.00107317 m3/kg

Specific volume of saturated steam (vg)=0.60576 m3/kg

Therefore, the volume of steam having X=0.6 is given by

V=M[vf+X(vg-vf)]

10= M[0.001007317+0.6(0.60576-0.00107317)]

M= 26.4825 kg

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