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

Solution

Solution Steps

Step 1: Understand the Problem

We have a two-phase liquid-vapor mixture of water at the saturation temperature of $100^{\circ} \mathrm{C}$ 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: \[ v = (1-x) \cdot v_f + x \cdot v_g \] where:

$v_f$ is the specific volume of the saturated liquid,
$v_g$ is the specific volume of the saturated vapor.

Step 3: Obtain Specific Volume Values

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

$v_f = 0.001043 \, \mathrm{m}^3/\mathrm{kg}$
$v_g = 1.6720 \, \mathrm{m}^3/\mathrm{kg}$

Step 4: Calculate the Specific Volume of the Mixture

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