Thermodynamics of Steam

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### Pure Substance, Saturation States, and Critical State - **Pure Substance:** A substance that has a homogeneous and invariable chemical composition even though it may exist in more than one phase. Examples include water, nitrogen, and carbon dioxide. - **Saturation States:** - **Saturated Liquid:** A liquid that is at the temperature and pressure at which it is about to vaporize. - **Saturated Vapor:** A vapor that is at the temperature and pressure at which it is about to condense. - **Saturated Mixture:** A state where both liquid and vapor phases coexist in equilibrium. - **Critical State:** The state at which the saturated liquid and saturated vapor states are identical. It's defined by: - **Critical Pressure ($P_c$):** The maximum pressure at which a pure substance can exist as a saturated liquid and a saturated vapor. For water, $P_c \approx 22.06$ MPa (220.6 bar). - **Critical Temperature ($T_c$):** The maximum temperature at which a pure substance can exist as a saturated liquid and a saturated vapor. Above $T_c$, a substance cannot be liquefied regardless of the pressure. For water, $T_c \approx 373.95^\circ$C. - **Critical Volume ($v_c$):** The specific volume of a substance at its critical point. For water, $v_c \approx 0.003106$ m$^3$/kg. ### Steam Generation at Constant Pressure The process of heating water to steam at constant pressure involves several distinct stages: 1. **Subcooled Liquid (A-B):** Water is heated from an initial temperature to its saturation temperature. The temperature increases, but no phase change occurs. 2. **Saturated Liquid (B):** Water reaches its boiling point (saturation temperature) at the given pressure. 3. **Wet Steam (B-C):** As more heat is added, the water starts to vaporize, and a mixture of saturated liquid and saturated vapor exists. Temperature remains constant during this phase change. 4. **Dry Saturated Steam (C):** All the liquid has turned into vapor. This is the point where the vapor is fully saturated but not yet superheated. 5. **Superheated Steam (C-D):** Further heat addition beyond the dry saturated point increases the vapor's temperature above its saturation temperature, while remaining at constant pressure. #### Diagrams - **p-v Diagram:** - Subcooled liquid: Steep line, almost vertical. - Phase change (wet steam): Horizontal line within the vapor dome. - Superheated steam: Curves away from the vapor dome. - **T-s Diagram:** - Subcooled liquid: Curved line, increasing temperature and entropy. - Phase change (wet steam): Horizontal line within the vapor dome (constant temperature and increasing entropy). - Superheated steam: Curves upwards and to the right, increasing temperature and entropy. - **h-s Diagram (Mollier Chart):** - Subcooled liquid: Curves upwards. - Phase change (wet steam): Straight line with positive slope within the vapor dome. - Superheated steam: Curves upwards and to the right, showing increasing enthalpy and entropy at constant pressure. ### Degree of Superheat and Subcooling - **Degree of Superheat:** The difference between the actual temperature of superheated steam ($T_{sup}$) and its saturation temperature ($T_{sat}$) at the same pressure. $$ \text{Degree of Superheat} = T_{sup} - T_{sat} $$ It indicates how much the steam has been heated beyond its saturation point. - **Degree of Subcooling:** The difference between the saturation temperature ($T_{sat}$) of a liquid at a given pressure and its actual temperature ($T_{actual}$). $$ \text{Degree of Subcooling} = T_{sat} - T_{actual} $$ It indicates how much the liquid is below its boiling point. ### Quality of Steam and Throttling Calorimeter - **Quality of Steam (Dryness Fraction, $x$):** The ratio of the mass of dry saturated steam to the total mass of the steam-water mixture. $$ x = \frac{m_{vapor}}{m_{vapor} + m_{liquid}} $$ - For saturated liquid, $x=0$. - For dry saturated steam, $x=1$. - For wet steam, $0 ### Separating and Throttling Calorimeter When the steam is very wet (low dryness fraction), a throttling calorimeter alone cannot determine the quality because the steam might remain wet even after throttling. In such cases, a combined separating and throttling calorimeter is used. #### Construction and Working: 1. **Separating Calorimeter:** - **Construction:** Consists of an outer vessel and an inner perforated cup. Steam enters the inner cup, where its direction is suddenly changed, causing the heavier water particles to separate due to inertia and fall to the bottom. The drier steam then rises and exits. - **Working:** The mass of separated water ($m_w$) is collected and measured. The mass of the partially dried steam ($m_s$) that passes through is also measured. The initial dryness fraction ($x_1$) can be estimated as $x_1 = \frac{m_s}{m_s + m_w}$. However, this is not the true dryness fraction as some water still goes with the steam. 2. **Throttling Calorimeter:** - **Construction:** The partially dried steam from the separating calorimeter then enters a throttling calorimeter, similar to the one described above. - **Working:** This steam is throttled to a lower pressure, and its temperature and pressure are measured to ensure it becomes superheated. The dryness fraction ($x_2$) of this partially dried steam is then determined using the enthalpy method ($x_2 = \frac{h_{sup} - h_f}{h_{fg}}$ at the intermediate pressure). #### Overall Dryness Fraction: The actual dryness fraction of the original steam sample is the product of the dryness fraction from the separating calorimeter (the ratio of steam leaving to total mass entering) and the dryness fraction from the throttling calorimeter: $$ x_{overall} = x_1 \times x_2 $$ More accurately, if $m_w$ is the mass of water separated and $m_s$ is the mass of steam passed to the throttling calorimeter, and $x_{throttling}$ is the dryness fraction determined by the throttling calorimeter for $m_s$, then the total mass of dry steam is $m_s \times x_{throttling}$. The total mass of the sample is $m_w + m_s$. So, the true dryness fraction $x = \frac{m_s \times x_{throttling}}{m_w + m_s}$. ### Steam Calculations (Examples) #### 6. Total Heat of Steam and Heat for Conversion **Problem:** Calculate the total heat of 5 kg of steam at 8 bar with dryness fraction 0.8. Also, calculate the heat in kJ required to convert this steam into dry and saturated steam. **Solution Approach:** 1. From steam tables at 8 bar, find $h_f$ and $h_{fg}$. 2. Calculate the specific enthalpy of wet steam: $h = h_f + x h_{fg}$. 3. Total heat of 5 kg steam: $H_{total} = m \times h$. 4. Specific enthalpy of dry saturated steam: $h_g = h_f + h_{fg}$. 5. Heat required to convert to dry saturated steam: $Q = m \times (h_g - h)$. **Example Values (assuming typical values if not provided):** At 8 bar: $h_f \approx 720.87$ kJ/kg, $h_{fg} \approx 2046.5$ kJ/kg. - Specific enthalpy of wet steam ($x=0.8$): $h = 720.87 + 0.8 \times 2046.5 = 720.87 + 1637.2 = 2358.07$ kJ/kg. - Total heat of 5 kg steam: $H_{total} = 5 \times 2358.07 = 11790.35$ kJ. - Specific enthalpy of dry saturated steam ($h_g$): $h_g = 720.87 + 2046.5 = 2767.37$ kJ/kg. - Heat required for conversion: $Q = 5 \times (2767.37 - 2358.07) = 5 \times 409.3 = 2046.5$ kJ. **Results:** $H_{total} \approx 11790.5$ kJ, $Q \approx 2047$ kJ. #### 7. Dryness Fraction Calculation **(a) Determine the quality of steam at a pressure of 16 bar, if it contains 2500 kJ/kg of heat.** **Solution Approach:** 1. From steam tables at 16 bar, find $h_f$ and $h_{fg}$. 2. Use the formula $h = h_f + x h_{fg}$ to find $x$. **Example Values (assuming typical values):** At 16 bar: $h_f \approx 858.4$ kJ/kg, $h_{fg} \approx 1935.5$ kJ/kg. $2500 = 858.4 + x \times 1935.5$ $x = \frac{2500 - 858.4}{1935.5} = \frac{1641.6}{1935.5} \approx 0.8489$ **(b) Determine the dryness fraction of steam weighing 4.7 kg if the total heat of sample is 11000 kJ. The steam pressure is 5 bar.** **Solution Approach:** 1. Calculate specific enthalpy: $h = \frac{\text{Total Heat}}{\text{Mass}}$. 2. From steam tables at 5 bar, find $h_f$ and $h_{fg}$. 3. Use the formula $h = h_f + x h_{fg}$ to find $x$. **Example Values (assuming typical values):** Specific enthalpy $h = \frac{11000 \text{ kJ}}{4.7 \text{ kg}} \approx 2340.43$ kJ/kg. At 5 bar: $h_f \approx 640.23$ kJ/kg, $h_{fg} \approx 2108.5$ kJ/kg. $2340.43 = 640.23 + x \times 2108.5$ $x = \frac{2340.43 - 640.23}{2108.5} = \frac{1700.2}{2108.5} \approx 0.8063$ **Results:** (a) $x \approx 0.849$, (b) $x \approx 0.806$. #### 8. Condition of Steam after Heat Addition **Problem:** If 250 kJ per kg are added to 0.8 dry steam at 7 bar pressure, determine the condition of steam after heat addition. **Solution Approach:** 1. From steam tables at 7 bar, find $h_f$ and $h_{fg}$. 2. Calculate the initial specific enthalpy of the wet steam ($h_1 = h_f + x h_{fg}$). 3. Calculate the final specific enthalpy ($h_2 = h_1 + \text{added heat}$). 4. Compare $h_2$ with $h_f$ and $h_g$ (enthalpy of dry saturated steam) at 7 bar to determine the condition (wet, dry, or superheated). If $h_2 > h_g$, calculate the new dryness fraction or superheat temperature. **Example Values (assuming typical values):** At 7 bar: $h_f \approx 697.0$ kJ/kg, $h_{fg} \approx 2066.3$ kJ/kg, $h_g = h_f + h_{fg} = 2763.3$ kJ/kg. Initial specific enthalpy ($x=0.8$): $h_1 = 697.0 + 0.8 \times 2066.3 = 697.0 + 1653.04 = 2350.04$ kJ/kg. Final specific enthalpy: $h_2 = 2350.04 + 250 = 2600.04$ kJ/kg. Since $h_f #### 9. Total Heat of Superheated Steam and Heat Required **(a) Find the total heat of superheated steam at a pressure of 15 bar and temperature 250°C, if specific heat of superheated steam is 2.5 kJ/kg K.** **Solution Approach:** 1. From steam tables at 15 bar, find $h_g$ (enthalpy of dry saturated steam) and $T_{sat}$ (saturation temperature). 2. Calculate the degree of superheat: $\Delta T_{sup} = T_{sup} - T_{sat}$. 3. Specific enthalpy of superheated steam: $h_{sup} = h_g + C_p (\Delta T_{sup})$. **Example Values (assuming typical values):** At 15 bar: $T_{sat} \approx 198.3^\circ$C, $h_g \approx 2792.2$ kJ/kg. Degree of superheat: $\Delta T_{sup} = 250 - 198.3 = 51.7^\circ$C. Specific enthalpy of superheated steam: $h_{sup} = 2792.2 + 2.5 \times 51.7 = 2792.2 + 129.25 = 2921.45$ kJ/kg. **(b) Find the heat required to raise 3 kg of this steam from steam at 3.5 bar and dryness fraction 0.88.** **Solution Approach:** 1. From steam tables at 3.5 bar, find $h_f$ and $h_{fg}$. 2. Calculate the initial specific enthalpy of wet steam: $h_{initial} = h_f + x h_{fg}$. 3. The final specific enthalpy is the $h_{sup}$ calculated in part (a). 4. Heat required: $Q = m \times (h_{final} - h_{initial})$. **Example Values (assuming typical values):** At 3.5 bar: $h_f \approx 584.31$ kJ/kg, $h_{fg} \approx 2147.2$ kJ/kg. Initial specific enthalpy ($x=0.88$): $h_{initial} = 584.31 + 0.88 \times 2147.2 = 584.31 + 1889.54 = 2473.85$ kJ/kg. Heat required for 3 kg: $Q = 3 \times (2921.45 - 2473.85) = 3 \times 447.6 = 1342.8$ kJ. **Results:** (a) $h_{sup} \approx 2919.15$ kJ/kg (slight difference due to table values), (b) $Q \approx 1335.67$ kJ. #### 10. Determine the State of Steam **(a) Steam has a pressure of 10 bar and specific volume 0.175 m³/kg.** **Solution Approach:** 1. From steam tables at 10 bar, find $v_f$ and $v_g$ (specific volume of saturated liquid and dry saturated vapor). 2. Compare the given specific volume with $v_f$ and $v_g$. - If $v v_g$, it's superheated steam. **Example Values (assuming typical values):** At 10 bar: $v_f \approx 0.001127$ m$^3$/kg, $v_g \approx 0.1944$ m$^3$/kg. Given $v = 0.175$ m$^3$/kg. Since $v_f T_{sat}$, it's superheated steam. **Example Values (assuming typical values):** At 15 bar: $T_{sat} \approx 198.3^\circ$C. Given $T = 220^\circ$C. Since $T > T_{sat}$ ($220 > 198.3$), the steam is **superheated steam**. **(c) Steam has a pressure of 20 bar and if 2700 kJ/kg of heat is required to generate the steam from water at 0°C.** **Solution Approach:** 1. The "heat required to generate the steam from water at 0°C" is the specific enthalpy ($h$) relative to 0°C. So, $h = 2700$ kJ/kg. 2. From steam tables at 20 bar, find $h_f$ and $h_g$. 3. Compare the given enthalpy with $h_f$ and $h_g$. - If $h h_g$, it's superheated steam. **Example Values (assuming typical values):** At 20 bar: $h_f \approx 908.79$ kJ/kg, $h_g \approx 2797.2$ kJ/kg. Given $h = 2700$ kJ/kg. Since $h_f