# Step-by-Step Solution: Calculating Enthalpy at State 4 (Turbine Outlet) in a Rankine Cycle Given Data: - **Pressure, \( P = 9.7 \) kPa** (Saturated mixture at this pressure) - **Quality, \( x = .82 \)** - **Saturated liquid enthalpy, \( h_f = 191.2 \) kJ/kg** - **Latent heat of vaporization, \( h_{fg} = 2409.6 \) kJ/kg** --- ## 1. Understanding Quality (\( x \)) The quality \( x \) represents the mass fraction of vapor in the mixture at the given pressure. - \( x = \): all liquid - \( x = 1 \): all vapor - \( < x < 1 \): mixture --- ## 2. Formula for Enthalpy of a Saturated Mixture The enthalpy at the given state is calculated as: \[ h = h_f + x \cdot h_{fg} \] Where: - \( h \) = mixture enthalpy (unknown) - \( h_f \) = enthalpy of saturated liquid - \( h_{fg} \) = enthalpy of vaporization - \( x \) = quality --- ## 3. Plug in the Values Given: - \( h_f = 191.2 \) kJ/kg - \( h_{fg} = 2409.6 \) kJ/kg - \( x = .82 \) Substitute into the formula: \[ h = 191.2 + .82 \times 2409.6 \] --- ## 4. Calculate \( x \cdot h_{fg} \): \[ .82 \times 2409.6 = 1975.872 \] --- ## 5. Sum to Get Final Enthalpy \[ h = 191.2 + 1975.872 = 2167.072 \text{ kJ/kg} \] --- ## **Final Answer** \[ \boxed{h_4 = 2167.1 \;\text{ kJ/kg}} \] *(rounded to one decimal place)* --- ### **Summary Table** | Parameter | Value | |---------------------|-------------| | Pressure, \( P \) | 9.7 kPa | | Quality, \( x \) | .82 | | \( h_f \) | 191.2 kJ/kg | | \( h_{fg} \) | 2409.6 kJ/kg| | **Enthalpy, \( h \)** | **2167.1 kJ/kg** | --- ### **Explanation** - The enthalpy for a two-phase (liquid-vapor) mixture depends on how much is liquid and how much is vapor, given by the quality \( x \). - \( h_f \) is the enthalpy if it were all liquid. Each increment of quality \( x \) adds \( h_{fg} \) energy per kg of vapor created. - The result tells you the total energy content per kilogram of the mixture at the turbine outlet. --- **If you have further steps in the Rankine cycle, you can use this \( h_4 \) value for efficiency or work calculations.**