Sure! Let's walk through each part **step-by-step**. You’ll need **steam tables** (thermodynamic property tables for water) to look up properties. I’ll explain **how to use the tables** and get the answers. --- ## **Instructions Recap** - For each state (given by pressure, temperature, specific volume, etc.), **find the specific properties**: - \( v \) (specific volume, m³/kg) - \( u \) (specific internal energy, kJ/kg) - \( h \) (specific enthalpy, kJ/kg) - Indicate if the state is compressed liquid, superheated vapor, or mixture. - Mark the state on PV and TV diagrams (do this in your notebook). --- # **a. At P = 1 MPa, v = .001 m³/kg** ### **Step 1: Find Saturation Data at P = 1 MPa** From steam tables (saturated water), at \( P_{sat} = 1 \) MPa: - \( T_{sat} \approx 179.9^\circ C \) - \( v_f \) (sat. liquid) \( \approx .001127 \) m³/kg - \( v_g \) (sat. vapor) \( \approx .1944 \) m³/kg ### **Step 2: Compare Given v to \( v_f \) and \( v_g \)** - \( v = .001 \) m³/kg - \( v_f < v < v_g \)? \( .001127 > .001 < .1944 \) - \( v \) is just slightly less than \( v_f \): **subcooled (compressed) liquid region** ### **Step 3: Approximate Properties** For compressed liquid, properties are close to those of saturated liquid at the same T or P: - \( u \approx u_f @ 1\,\text{MPa} \approx 761.68 \) kJ/kg - \( h \approx h_f @ 1\,\text{MPa} \approx 762.61 \) kJ/kg ### **Step 4: Final Answers** - **State:** Compressed liquid - **\( v \):** .001 m³/kg (given) - **\( u \):** ≈ 761.68 kJ/kg - **\( h \):** ≈ 762.61 kJ/kg --- # **b. At T = 320°C, v = .5 m³/kg** ### **Step 1: Find Saturation Data at T = 320°C** From steam tables: - \( P_{sat} @ 320^\circ C \approx 11.54 \) MPa - \( v_f \approx .001215 \) m³/kg, \( v_g \approx .13871 \) m³/kg ### **Step 2: Compare Given v** - \( v = .5 \) m³/kg > \( v_g \) - **Superheated vapor** ### **Step 3: Go to Superheated Table, Look Up Properties** At \( T = 320^\circ C \), need to know the pressure. But since v is higher than \( v_g \), it's deep in the superheated region. Use superheated tables or ideal gas approximation for large v. (If using the table for \( T = 320^\circ C \) and lowest P, \( v \) values are still much smaller than .5 m³/kg. So this is very low pressure, far into the superheated region.) - **Estimate:** Use \( v = RT/P \) for steam: - \( R = .4615 \) kJ/kg·K - \( T = 320 + 273.15 = 593.15 \) K - \( v = RT/P \rightarrow P = RT/v \) Calculate \( P \): \[ P = \frac{RT}{v} = \frac{.4615 \times 593.15}{.5} \approx 547 \text{ kPa} = .547 \text{ MPa} \] Now use superheated steam tables at \( T = 320^\circ C, P = .5 \) MPa: - \( u \approx 2911 \) kJ/kg - \( h \approx 3245 \) kJ/kg ### **Step 4: Final Answers** - **State:** Superheated vapor - **\( v \):** .5 m³/kg (given) - **\( u \):** ≈ 2911 kJ/kg - **\( h \):** ≈ 3245 kJ/kg --- # **c. At P = 120 kPa, T = 200°C** ### **Step 1: Find Saturation Data at P = 1.2 MPa** At \( P_{sat} = 1.2 \) MPa: - \( T_{sat} \approx 188.15^\circ C \) Given \( T = 200^\circ C > T_{sat} \): - **Superheated vapor** ### **Step 2: Use Superheated Steam Table** At \( P = 1.2 \) MPa, \( T = 200^\circ C \): - \( v \approx .1646 \) m³/kg - \( u \approx 2594. \) kJ/kg - \( h \approx 2855.1 \) kJ/kg ### **Step 3: Final Answers** - **State:** Superheated vapor - **\( v \):** .1646 m³/kg - **\( u \):** 2594. kJ/kg - **\( h \):** 2855.1 kJ/kg --- # **d. At T = 100°C, v = 1.00 m³/kg** ### **Step 1: Find Saturation Data at T = 100°C** At \( T_{sat} = 100^\circ C \): - \( P_{sat} = 101.33 \) kPa - \( v_f \approx .001043 \) m³/kg - \( v_g \approx 1.672 \) m³/kg ### **Step 2: Compare v** - \( v_f < v < v_g \) \( .001043 < 1.00 < 1.672 \) **State:** Mixture (wet vapor) ### **Step 3: Find Quality \( x \)** \[ x = \frac{v - v_f}{v_g - v_f} = \frac{1.00 - .001043}{1.672 - .001043} = \frac{.998957}{1.670957} \approx .598 \] ### **Step 4: Find u and h** From steam tables at 100°C: - \( u_f = 419.04 \) kJ/kg, \( u_g = 2506.5 \) kJ/kg - \( h_f = 419.04 \) kJ/kg, \( h_g = 2676. \) kJ/kg \[ u = u_f + x(u_g - u_f) = 419.04 + .598 \times (2506.5 - 419.04) \approx 419.04 + .598 \times 2087.46 \approx 419.04 + 1247.2 \approx 1666.2 \text{ kJ/kg} \] \[ h = h_f + x(h_g - h_f) = 419.04 + .598 \times (2676. - 419.04) \approx 419.04 + .598 \times 2256.96 \approx 419.04 + 1349.9 \approx 1768.9 \text{ kJ/kg} \] ### **Step 5: Final Answers** - **State:** Mixture (quality \( x = .60 \)) - **\( v \):** 1.00 m³/kg (given) - **\( u \):** 1666.2 kJ/kg - **\( h \):** 1768.9 kJ/kg --- # **e. At P = 4 MPa, T = 160°C** ### **Step 1: Find Saturation Temperature at P = 4 MPa** At \( P = 4 \) MPa: - \( T_{sat} \approx 250.35^\circ C \) Given \( T = 160^\circ C < T_{sat} \): - **Compressed liquid** ### **Step 2: Use Compressed Liquid Approximation** For compressed liquid, use sat. liquid values at 160°C: - \( v_f \approx .001073 \) m³/kg - \( u_f \approx 670.46 \) kJ/kg - \( h_f \approx 670.46 \) kJ/kg ### **Step 3: Final Answers** - **State:** Compressed liquid - **\( v \):** ≈ .001073 m³/kg - **\( u \):** ≈ 670.46 kJ/kg - **\( h \):** ≈ 670.46 kJ/kg --- # **Summary Table** | Case | State | v (m³/kg) | u (kJ/kg) | h (kJ/kg) | |------|----------------------|-------------|-----------|-----------| | a | Compressed liquid | .001 | 761.68 | 762.61 | | b | Superheated vapor | .5 | 2911 | 3245 | | c | Superheated vapor | .1646 | 2594. | 2855.1 | | d | Mixture (x ≈ .60) | 1.00 | 1666.2 | 1768.9 | | e | Compressed liquid | .001073 | 670.46 | 670.46 | --- **Tip:** Always check the phase region by comparing your knowns (\( v \), \( T \), \( P \)) to the saturation properties in the steam tables! Let me know if you need help with the PV or TV diagrams, or any further steps!