Let's break down and solve the problem **step by step**. --- ## **Given Data** - **Volume of water:** \( V = 130 \, \text{m}^3 \) - **Initial Pressure:** \( P_1 = 1000 \, \text{kPa} \) - **Initial Quality:** \( x_1 = 11.59\% = 0.1159 \) - **Final State:** Water is saturated vapor (\( x_2 = 1 \)), and volume is constant. - **Heating rate:** \( \dot{Q} = 14,717 \, \text{kJ/s} \) --- ## **a) Final temperature and pressure of the water** ### **Step 1: Initial State Properties** From steam tables at \( P_1 = 1000 \) kPa: - \( v_f @ 1000\, \text{kPa} = 0.001127 \, \text{m}^3/\text{kg} \) - \( v_g @ 1000\, \text{kPa} = 0.1944 \, \text{m}^3/\text{kg} \) - \( T_{sat} @ 1000\, \text{kPa} = 179.88^\circ \text{C} \) ### **Step 2: Find Initial Mass** Specific volume at initial state: \[ v_1 = v_f + x_1 (v_g - v_f) \] \[ v_1 = 0.001127 + 0.1159 \times (0.1944 - 0.001127) \] \[ v_1 = 0.001127 + 0.1159 \times 0.193273 \] \[ v_1 = 0.001127 + 0.022420 = 0.023547 \, \text{m}^3/\text{kg} \] Total mass: \[ m = \frac{V}{v_1} = \frac{130}{0.023547} \approx 5523.4 \, \text{kg} \] ### **Step 3: Final State (Saturated Vapor, Volume Constant)** At final state, \( v_2 = v_1 \) and \( x_2 = 1 \) (saturated vapor): Set \( v_g = v_1 \) and find the corresponding pressure from steam tables. Looking for \( v_g \approx 0.02355 \, \text{m}^3/\text{kg} \): From steam tables: - At \( P \approx 6300 \, \text{kPa} \), \( v_g \approx 0.02355 \, \text{m}^3/\text{kg} \) - Corresponding temperature: \( T_{sat} @ 6300\, \text{kPa} \approx 275.6^\circ \text{C} \) ### **Final Answers for (a):** - **Final Pressure:** \( P_2 \approx 6300 \, \text{kPa} \) - **Final Temperature:** \( T_2 \approx 275.6^\circ \text{C} \) --- ## **b) Amount of Work Done on the Water** For **constant volume process**, **boundary work** is zero: \[ W = \int_{V_1}^{V_2} P \, dV = 0 \] --- ## **c) Amount of Heat Added to the Water** \[ Q = m(u_2 - u_1) \] **From steam tables:** - At \( P_1 = 1000 \, \text{kPa}, x_1 = 0.1159 \): - \( u_f = 762.81\, \text{kJ/kg} \), \( u_g = 2013.6\, \text{kJ/kg} \) - \( u_1 = u_f + x_1(u_g - u_f) \) - \( u_1 = 762.81 + 0.1159 \times (2013.6 - 762.81) \) - \( u_1 = 762.81 + 0.1159 \times 1250.79 \) - \( u_1 = 762.81 + 145.00 = 907.81\, \text{kJ/kg} \) - At \( P_2 = 6300 \, \text{kPa}, x_2 = 1 \): - \( u_g @ 6300 \, \text{kPa} \approx 2594.0 \, \text{kJ/kg} \) (verify with steam tables) \[ Q = m(u_2 - u_1) = 5523.4 \times (2594.0 - 907.81) \] \[ Q = 5523.4 \times 1686.19 \approx 9,\!316,\!000\, \text{kJ} \] --- ## **d) Time to Reach Final State** \[ Q = \dot{Q} \cdot t \implies t = \frac{Q}{\dot{Q}} \] \[ t = \frac{9,\!316,\!000}{14,\!717} \approx 633.3\, \text{s} \] \[ t \approx 10.56\, \text{min} \] --- ## **Summary Table** | Part | Description | Value | |------|----------------------------------------------|-------------------------------| | a | Final Pressure, Temperature | 6300 kPa, 275.6°C | | b | Work Done | 0 | | c | Heat Added | 9,316,000 kJ | | d | Time to Reach Final State | 633.3 s (10.56 min) | --- **Let me know if you need steam table references or further clarifications!**