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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### **Question 1**

**Given:**
- **Volume of tank, $$ V $$** = 350,000 L = 350 m³
- **Pressure, $$ P $$** = 250 kPa (saturated mixture)
- **Mass fractions:** 35% liquid ($$ x_f = 0.35 $$) and 65% vapor ($$ x_g = 0.65 $$)

**Approach:**
1. **Determine Specific Volumes:**
   - From steam tables at **250 kPa**:
     - **Saturated Liquid Specific Volume ($$ v_f $$)** ≈ 0.001156 m³/kg
     - **Saturated Vapor Specific Volume ($$ v_g $$)** ≈ 0.1269 m³/kg
2. **Calculate Mixture Specific Volume ($$ v $$):**
   $$
   v = x_f \cdot v_f + x_g \cdot v_g = 0.35 \times 0.001156 + 0.65 \times 0.1269 \approx 0.08289 \, \text{m}³/\text{kg}
   $$
3. **Determine Total Mass ($$ m $$):**
   $$
   m = \frac{V}{v} = \frac{350 \, \text{m}³}{0.08289 \, \text{m}³/\text{kg}} \approx 4217 \, \text{kg}
   $$

**Answer:**  
The total mass in the tank is approximately **4,200 kg**.

---

### **Question 2**

**Given:**
- **Flow 1:** $$ V_1 = 1 \, \text{m}³/\text{s} $$ at $$ T_1 = 200°C $$ and $$ P_1 = 100 \, \text{kPa} $$
- **Flow 2:** $$ V_2 = 2 \, \text{m}³/\text{s} $$ at $$ T_2 = 200°C $$ and $$ P_2 = 100 \, \text{kPa} $$
- **Exit Pressure:** $$ P_{\text{exit}} = 100 \, \text{kPa} $$
- **Assumptions:** Adiabatic mixing, no heat transfer, constant specific heats, and neglecting kinetic energies.

**Approach:**
1. **Since both incoming air streams are at the same temperature and pressure:**
   - **Mass Flow Rates:**
     $$
     m_1 = \frac{P V_1}{R T}, \quad m_2 = \frac{P V_2}{R T}
     $$
   - **Total Mass Flow Rate:**
     $$
     m_{\text{exit}} = m_1 + m_2 = \frac{P (V_1 + V_2)}{R T}
     $$
2. **Determine Exit Volume:**
   - Since temperature and pressure remain unchanged, specific volume remains the same.
   - Therefore, **Volume Flow Rates Add Up:**
     $$
     V_{\text{exit}} = V_1 + V_2 = 1 \, \text{m}³/\text{s} + 2 \, \text{m}³/\text{s} = 3 \, \text{m}³/\text{s}
     $$

**Answer:**  
The exit volume flow rate is **3 m³/s**.
**Question 1:** The total mass in the tank is approximately **4,200 kg**. **Question 2:** The exit volume flow rate is **3 m³/s**.

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