Question

A vessel contains 10 kg of water with a quality of 0.85 at a pressure of

a) 150 kPa

b) 200 kPa

c) 2 MPa

For the above pressures, determine the volume of liquid and volume of vapor. Also, what is the temperature of mixture.

Answer 1

Data at 100 C:
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T = 100.0 C
Psat = 101.42 kPa
vf = 0.001043 cu m / kg uf = 419.06 kJ / kg
vfg = 1.670957 cu m / kg ufg = 2087.0 kJ / kg
vg = 1.6720 cu m / kg ug = 2506.0 kJ / kg

v1 = vf1 + ( x1 ) ( vfg1 )
v1 = 0.001043 + ( 0.1230 ) ( 1.670957 ) = 0.2055 cu m / kg

u1 = uf1 + ( x1 ) ( ufg1 )
u1 = 410.06 + ( 0.1230 ) ( 2087.0 ) = 666.8 kJ / kg

m = V / v1 = ( 10 L ) ( 1.0cu m / 1000 L ) / ( 0.2055 cu m / kg )

Since rigid tank and a closed tank, v2 = v1 .

Examining Steam Table at 150 C shows : vf < v2 < vg2 Therefore it is still
a two-phase mixture at 150 C. v2 and data at 150 C give x2 and u2 :
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m = 0.04866 kg

v2 = v1 = 0.2055 cu m / kg

Need saturated steam table data at 150 C:
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T = 150
Psat = 476.16 kPa
vf = 0.001091 uf = 631.66 kJ / kg
vfg = 0.391389 cu m / kg ufg = 1927.4 kJ / kg
vg = 0.39248 cu m / kg ug = 2559.1 kJ / kg

x2 = ( v2 - vf2 ) / ( vfg2 )

x2 = ( 0.2055 - 0.001091 ) / ( 0.391389 ) = 0.5223

u2 = uf2 + ( x2 ) ( ufg2 )

u2 = 631.66 + ( 0.5223 ) ( 1927.4 ) = 1638.3 kJ / kg

delta u = u2 - u1 = 1638.3 - 666.8 = 971.5 kJ per kg

delta U = ( m ) ( delta u )

delta U = ( 0.04866 kg ) ( 971.5 kJ / kg )

delta U = 47.27 kJ

Q = delta U = 47.27 kJ <---------------------------------------...