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.

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 $\mathrm{m}^{3} / \mathrm{kg}$.
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Solution

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Solution Steps

Step 1: Understand the Problem

We have a two-phase liquid-vapor mixture of water at the saturation temperature of 100C100^{\circ} \mathrm{C} with a quality of 50%50\%. We need to determine the specific volume of the mixture in the tank.

Step 2: Define Quality and Specific Volume

The quality xx of a mixture is defined as the mass fraction of the vapor phase in the mixture. For a quality of 50%50\%, x=0.5x = 0.5.

The specific volume vv of the mixture can be calculated using the formula: v=(1x)vf+xvg v = (1-x) \cdot v_f + x \cdot v_g where:

  • vfv_f is the specific volume of the saturated liquid,
  • vgv_g is the specific volume of the saturated vapor.
Step 3: Obtain Specific Volume Values

At 100C100^{\circ} \mathrm{C}, the specific volume values from steam tables are:

  • vf=0.001043m3/kgv_f = 0.001043 \, \mathrm{m}^3/\mathrm{kg}
  • vg=1.6720m3/kgv_g = 1.6720 \, \mathrm{m}^3/\mathrm{kg}
Step 4: Calculate the Specific Volume of the Mixture

Substitute the values into the specific volume formula: v=(10.5)0.001043+0.51.6720 v = (1 - 0.5) \cdot 0.001043 + 0.5 \cdot 1.6720 v=0.50.001043+0.51.6720 v = 0.5 \cdot 0.001043 + 0.5 \cdot 1.6720 v=0.0005215+0.8360 v = 0.0005215 + 0.8360 v=0.8365m3/kg v = 0.8365 \, \mathrm{m}^3/\mathrm{kg}

Final Answer

The specific volume of the mixture in the tank is 0.8365m3/kg\boxed{0.8365 \, \mathrm{m}^3/\mathrm{kg}}.

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