Questions: A 1.5 m^3 tank holds a two-phase liquid-vapor mixture of water at the saturation temperature, 100°C. The quality of the mixture is 50%.
1. Determine the specific volume of mixture in the tank, in m^3 / kg.
Transcript text: A $1.5 \mathrm{~m}^{3}$ tank holds a two-phase liquid-vapor mixture of water at the saturation temperature, $100^{\circ} \mathrm{C}$. The quality of the mixture is $50 \%$.
1. Determine the specific volume of mixture in the tank, in $\mathrm{m}^{3} / \mathrm{kg}$.
Solution
Solution Steps
Step 1: Understand the Problem
We have a two-phase liquid-vapor mixture of water at the saturation temperature of 100∘C with a quality of 50%. We need to determine the specific volume of the mixture in the tank.
Step 2: Define Quality and Specific Volume
The quality x of a mixture is defined as the mass fraction of the vapor phase in the mixture. For a quality of 50%, x=0.5.
The specific volume v of the mixture can be calculated using the formula:
v=(1−x)⋅vf+x⋅vg
where:
vf is the specific volume of the saturated liquid,
vg is the specific volume of the saturated vapor.
Step 3: Obtain Specific Volume Values
At 100∘C, the specific volume values from steam tables are:
vf=0.001043m3/kg
vg=1.6720m3/kg
Step 4: Calculate the Specific Volume of the Mixture
Substitute the values into the specific volume formula:
v=(1−0.5)⋅0.001043+0.5⋅1.6720v=0.5⋅0.001043+0.5⋅1.6720v=0.0005215+0.8360v=0.8365m3/kg
Final Answer
The specific volume of the mixture in the tank is 0.8365m3/kg.