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A vessel of volume 1.0 m3 contains a mixture of liquid water and steam in equilibrium at 1.0 bar. Given that 90% of the volume is occupied by the steam, find the fraction of the mixture. Assume at 1.0 bar, vf = 0.001 m3/kg and vg = 1.7 m3/kg.
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- 5.266 × 10-3
- 5.266 × 103
- 3.442 × 10-3
- 3.442 × 10-7
Correct Option: A
vf = 0.001 m3 /kg
vg = 1.7 m3 /kg
| mass of mixture = | + | |||
| vg | vf |
| mass of mixture = | + | = 100.53 kg | ||
| 1.7 | 0.001 |
| Specific volume v = | = 9.947 × 10-3 m3 / kg | |
| m |
v = vf + x(vf - vg)
9.947 × 10-3 = 0.001 + x(1.7 - 0.001)
x = 5.266 × 10-3
