Question 1 (10 point) A 350000 L rigid tank is filled with saturated liquid-vapor mixture of water at 250 kPa . If 35 percent of the mass is liquid and 65 percent of the mass is vapor, what is the total mass in the tank? Question \( 2\left(20\right. \) point) Two air flows are combined to a single flow. One flow is \( 1 \mathrm{~m}^{3} / \mathrm{s} \mathrm{at} 200^{\circ} \mathrm{C} \), and the other is \( 2 \mathrm{~m}^{3} / \mathrm{s} \) at \( 200^{\circ} \mathrm{C} \), both at 100 kPa . They mix without any heat transfer to produce an exit flow at 100 kPa . Neglect kinetic energies and find the volume flow rate at the exit. Assuming constant specific heat
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To find the total mass in the tank, you can use the relationship between the volumes of saturated liquid and vapor at the given pressure. By consulting steam tables, you can determine the specific volumes (\(v_f\) and \(v_g\)) at \(250\) kPa, then apply the quality formula. For the mixture, multiply the tank volume by the mass fractions (35% liquid, 65% vapor) to establish individual masses and add them together for total mass. When mixing two air flows, you simply add the volumetric flow rates to find the exit flow rate. Since both flows are at \(200^\circ C\) and are at a consistent pressure of \(100\) kPa, the exit volume flow rate is the sum of the individual flow rates: \(1 \, \text{m}^3/\text{s} + 2 \, \text{m}^3/\text{s} = 3 \, \text{m}^3/\text{s}\). Thus, the exit flow rate is \(3 \, \text{m}^3/\text{s}\).