# Calculating Saturation Pressure and Specific Volume of Water at 256°C To find the **saturation pressure** and **specific volume at saturation** for water at \( 256^\circ\text{C} \), we refer to the thermodynamic property tables for water (commonly called steam tables). --- ## 1. **Saturation Pressure at 256°C** ### Step 1: Understand the Problem - **Saturation pressure** at a given temperature is the pressure at which water boils at that temperature. - \( T = 256^\circ\text{C} \) is not typically listed directly in basic steam tables, so **interpolation** is needed between 255°C and 260°C. ### Step 2: Find Table Values From the steam tables: | Temp (\( ^\circ\)C) | \( P_\text{sat} \) (MPa) | |---------------------|-------------------------| | 255 | 3.605 | | 260 | 3.801 | ### Step 3: Linear Interpolation Let: - \( T_1 = 255^\circ\text{C} \), \( P_1 = 3.605\,\text{MPa} \) - \( T_2 = 260^\circ\text{C} \), \( P_2 = 3.801\,\text{MPa} \) - \( T = 256^\circ\text{C} \) **Interpolation formula:** \[ P = P_1 + \frac{(P_2 - P_1)}{(T_2 - T_1)} \times (T - T_1) \] Plug in values: \[ P = 3.605 + \frac{3.801 - 3.605}{260 - 255} \times (256 - 255) \] \[ P = 3.605 + \frac{.196}{5} \times 1 \] \[ P = 3.605 + .0392 \] \[ P \approx 3.644\,\text{MPa} \] --- ## 2. **Specific Volume at Saturation Pressure (at 256°C)** Specific volume at saturation pressure refers to the specific volume of **saturated vapor (steam)** at that temperature and pressure. ### Step 1: Get Values from Table From steam tables: | Temp (\( ^\circ\)C) | \( v_g \) (m³/kg) | |---------------------|-------------------| | 255 | .06310 | | 260 | .06115 | ### Step 2: Interpolate Let: - \( v_1 = .06310 \,\text{m}^3/\text{kg} \) at \( T_1 = 255^\circ\text{C} \) - \( v_2 = .06115 \,\text{m}^3/\text{kg} \) at \( T_2 = 260^\circ\text{C} \) \[ v = v_1 + \frac{(v_2 - v_1)}{(T_2 - T_1)} \times (T - T_1) \] \[ v = .06310 + \frac{.06115 - .06310}{5} \times 1 \] \[ v = .06310 - \frac{.00195}{5} \] \[ v = .06310 - .00039 \] \[ v \approx .06271\,\text{m}^3/\text{kg} \] --- ## 3. **Final Answers** - **Saturation pressure at 256°C:** \[ \boxed{3.644\,\text{MPa}} \] - **Specific volume of saturated vapor at 256°C:** \[ \boxed{.06271\,\text{m}^3/\text{kg}} \] --- > **Note:** For most engineering purposes, linear interpolation is sufficiently accurate for small temperature intervals in steam tables. > The specific volume of saturated liquid (\( v_f \)) is much smaller and often negligible compared to \( v_g \) (the vapor value) for steam calculations.