A 300-m3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, what is the total mass in the tank?
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Answered by
15
Answer:
Mass in Tank = 451.422 kg
Explanation:
Specific Volume of Saturated Steam at 200 kPa
Vg = 0.885735 m³/kg
Specific Volume of Saturated Water at 200 kPa
Vf = 0.00106052 m³/kg
Specific Volume of saturated liquid-vapor mixture of water at 200 kPa
= Vf + (0.75)(Vg - Vf)
= 0.00106052 + (0.75)(0.885735 - 0.00106052)
= 0.00106052 + 0.75 (0.88467448)
= 0.00106052 + 0.66350586
= 0.66456638 m³/kg
Volume of Tank = 300 m³
=> Mass in Tank = Volume of Tank / Specific Volume of saturated liquid-vapor mixture of water
=> Mass in Tank = 300 / 0.66456638 kg
=> Mass in Tank = 451.422 kg
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0
Answer:
what is the physical significance of the two conatants that appear in the van der waals equation of state?
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