Brayton Cycle Numericals

Cheatsheet Content

### Brayton Cycle Overview The Brayton cycle is a thermodynamic cycle that describes the operation of **gas turbine engines**. It's an open cycle in practice (air continuously enters and exits), but for analysis, it's often modeled as a closed cycle with the working fluid (air) undergoing a series of processes. It's the foundation for jet engines and gas turbine power plants. **Key Components:** 1. **Compressor:** Increases the pressure of the air. 2. **Combustor:** Adds heat to the high-pressure air (fuel is burned). 3. **Turbine:** Extracts work from the hot, high-pressure gas. 4. **Heat Exchanger (or Atmosphere):** Rejects heat from the gas. ### Cycle Processes and Diagrams The ideal Brayton cycle consists of four internally reversible processes: two isentropic (adiabatic and reversible) and two isobaric (constant pressure). #### P-V Diagram A P-V (Pressure-Volume) diagram shows the relationship between pressure and volume during the cycle. * **1-2: Isentropic Compression** * Air is drawn into the compressor and compressed. * Pressure ($P$) increases significantly, while volume ($V$) decreases. * Temperature ($T$) increases due to compression. * No heat transfer ($Q=0$), entropy ($s$) is constant. * Represented by a steep curve on the P-V diagram. * **2-3: Constant Pressure Heat Addition** * Compressed air enters the combustor where fuel is injected and burned. * Heat is added to the air at constant pressure ($P_2 = P_3$). * Temperature ($T$) and volume ($V$) increase significantly. * Represented by a horizontal line (constant pressure) moving to the right (increasing volume) on the P-V diagram. * **3-4: Isentropic Expansion** * Hot, high-pressure gas expands through the turbine, producing work. * Pressure ($P$) decreases significantly, while volume ($V$) increases. * Temperature ($T$) decreases as work is extracted. * No heat transfer ($Q=0$), entropy ($s$) is constant. * Represented by a steep curve on the P-V diagram. * **4-1: Constant Pressure Heat Rejection** * The gas exits the turbine and releases heat to the surroundings (or through a heat exchanger). * Heat is rejected at constant pressure ($P_4 = P_1$). * Temperature ($T$) and volume ($V$) decrease, returning the air to its initial state. * Represented by a horizontal line (constant pressure) moving to the left (decreasing volume) on the P-V diagram. #### T-s Diagram A T-s (Temperature-Entropy) diagram shows the relationship between temperature and entropy during the cycle. This diagram is particularly useful for visualizing efficiency and heat transfer. * **1-2: Isentropic Compression** * Temperature ($T$) increases from $T_1$ to $T_2$. * Entropy ($s$) remains constant ($s_1 = s_2$). * Represented by a vertical line upwards on the T-s diagram. * **2-3: Constant Pressure Heat Addition** * Temperature ($T$) increases from $T_2$ to $T_3$ (maximum temperature). * Entropy ($s$) increases from $s_2$ to $s_3$. * Heat added ($Q_{in}$) is the area under curve 2-3. * Represented by a curve sloping upwards and to the right on the T-s diagram. * **3-4: Isentropic Expansion** * Temperature ($T$) decreases from $T_3$ to $T_4$. * Entropy ($s$) remains constant ($s_3 = s_4$). * Represented by a vertical line downwards on the T-s diagram. * **4-1: Constant Pressure Heat Rejection** * Temperature ($T$) decreases from $T_4$ to $T_1$ (minimum temperature). * Entropy ($s$) decreases from $s_4$ to $s_1$. * Heat rejected ($Q_{out}$) is the area under curve 4-1. * Represented by a curve sloping downwards and to the left on the T-s diagram. **Net Work ($W_{net}$):** On the T-s diagram, the area enclosed by the cycle (1-2-3-4-1) represents the net work output of the cycle. **Thermal Efficiency ($\eta_{th}$):** $\eta_{th} = W_{net} / Q_{in} = (Q_{in} - Q_{out}) / Q_{in}$. ### Key Formulas for Numerical Solving Assume ideal gas behavior and constant specific heats ($c_p$, $c_v$, $k = c_p/c_v$). * **Gas Properties:** * For air, typically $k \approx 1.4$ and $c_p \approx 1.005 \text{ kJ/(kg} \cdot \text{K)}$. * $c_v = c_p / k$ * $R = c_p - c_v$ (specific gas constant) * **Pressure Ratio ($P_r$):** * $P_r = P_2/P_1 = P_3/P_4$ (for ideal cycle) * **Isentropic Relations (for processes 1-2 and 3-4):** * $T_2/T_1 = (P_2/P_1)^{(k-1)/k} = (P_r)^{(k-1)/k}$ * $T_4/T_3 = (P_4/P_3)^{(k-1)/k} = (1/P_r)^{(k-1)/k}$ or $T_3/T_4 = (P_r)^{(k-1)/k}$ * **Heat Transfer:** * **Heat Added ($Q_{in}$, process 2-3):** $Q_{in} = c_p(T_3 - T_2)$ * **Heat Rejected ($Q_{out}$, process 4-1):** $Q_{out} = c_p(T_4 - T_1)$ * **Work Terms:** * **Work Done by Turbine ($W_T$, process 3-4):** $W_T = c_p(T_3 - T_4)$ * **Work Done on Compressor ($W_C$, process 1-2):** $W_C = c_p(T_2 - T_1)$ * **Net Work Output ($W_{net}$):** * $W_{net} = W_T - W_C$ * Also, from the first law of thermodynamics: $W_{net} = Q_{in} - Q_{out}$ * **Thermal Efficiency ($\eta_{th}$):** * $\eta_{th} = W_{net} / Q_{in} = (Q_{in} - Q_{out}) / Q_{in} = 1 - Q_{out} / Q_{in}$ * For an ideal Brayton cycle, this simplifies to: $\eta_{th} = 1 - 1/(P_r)^{(k-1)/k}$ * This formula shows that efficiency increases with pressure ratio $P_r$. * **Back Work Ratio (BWR):** * BWR = $W_C / W_T$ * A high BWR means a significant portion of the turbine's output is consumed by the compressor, which is characteristic of gas turbines. ### Solving Strategy for Numericals 1. **Understand the Problem & Sketch Diagrams:** * Read carefully to identify if it's an ideal or actual cycle. * Draw the **T-s Diagram** (and P-V if it helps your understanding). Label the four states (1, 2, 3, 4) and the processes. This is crucial for visualizing temperature changes and entropy. 2. **List Given Values and Assumptions:** * Write down all known temperatures ($T_1$, $T_3$ are common inputs), pressures ($P_1$, $P_2$), pressure ratio ($P_r$), and gas properties ($c_p$, $k$). * State assumptions (e.g., ideal gas, constant specific heats, ideal cycle = isentropic processes). 3. **Determine Unknown Temperatures (State Points):** * **State 1 (Compressor Inlet):** Usually $T_1$ and $P_1$ are known (ambient conditions). * **State 2 (Compressor Exit):** Use isentropic relation for 1-2: $T_2 = T_1 \cdot (P_r)^{(k-1)/k}$ (If actual cycle, this would be $T_{2s}$, then calculate $T_{2a}$ using compressor efficiency). * **State 3 (Turbine Inlet):** Usually $T_3$ (maximum cycle temperature) and $P_3 (=P_2)$ are known. * **State 4 (Turbine Exit):** Use isentropic relation for 3-4: $T_4 = T_3 / (P_r)^{(k-1)/k}$ (If actual cycle, this would be $T_{4s}$, then calculate $T_{4a}$ using turbine efficiency). 4. **Calculate Heat and Work Terms (per unit mass):** * **Compressor Work ($W_C$):** $W_C = c_p(T_2 - T_1)$ (always positive for work input) * **Heat Added ($Q_{in}$):** $Q_{in} = c_p(T_3 - T_2)$ * **Turbine Work ($W_T$):** $W_T = c_p(T_3 - T_4)$ (always positive for work output) * **Heat Rejected ($Q_{out}$):** $Q_{out} = c_p(T_4 - T_1)$ 5. **Calculate Net Work Output and Thermal Efficiency:** * **Net Work ($W_{net}$):** $W_{net} = W_T - W_C$ (or $Q_{in} - Q_{out}$) * **Thermal Efficiency ($\eta_{th}$):** $\eta_{th} = W_{net} / Q_{in}$ * **As a check for ideal cycle:** $\eta_{th} = 1 - 1/(P_r)^{(k-1)/k}$ 6. **Handle Non-Ideal Conditions (if applicable):** * **Isentropic Efficiencies:** * **Compressor Efficiency ($\eta_C$):** $\eta_C = (T_{2s} - T_1) / (T_{2a} - T_1)$ * First calculate $T_{2s}$ (isentropic temperature) using the ideal relation. * Then find the actual compressor exit temperature: $T_{2a} = T_1 + (T_{2s} - T_1) / \eta_C$. * Use $T_{2a}$ for subsequent calculations of $Q_{in}$ and $W_C$. * **Turbine Efficiency ($\eta_T$):** $\eta_T = (T_3 - T_{4a}) / (T_3 - T_{4s})$ * First calculate $T_{4s}$ (isentropic temperature) using the ideal relation. * Then find the actual turbine exit temperature: $T_{4a} = T_3 - \eta_T(T_3 - T_{4s})$. * Use $T_{4a}$ for subsequent calculations of $Q_{out}$ and $W_T$. * **Pressure Drops:** If pressure drops in the combustor or heat exchanger are given, adjust the pressures for states 2, 3, 4 accordingly (e.g., $P_3 = P_2 - \Delta P_{combustor}$). This makes $P_2 \neq P_3$ and $P_4 \neq P_1$. ### Example Problem (Ideal Brayton Cycle) **Problem:** An ideal Brayton cycle operates with air from a minimum temperature of $300 \text{ K}$ and a maximum temperature of $1300 \text{ K}$. The pressure ratio of the cycle is $8$. Assume $k=1.4$ and $c_p = 1.005 \text{ kJ/kg} \cdot \text{K}$. Calculate: a) The temperatures at the exit of the compressor ($T_2$) and the exit of the turbine ($T_4$). b) The net work output per unit mass of air. c) The thermal efficiency of the cycle. d) The back work ratio. **Solution:** **1. Given Values:** * $T_1 = 300 \text{ K}$ (Compressor Inlet Temperature) * $T_3 = 1300 \text{ K}$ (Turbine Inlet Temperature / Max Cycle Temperature) * $P_r = P_2/P_1 = P_3/P_4 = 8$ * $k = 1.4$ (Ratio of specific heats for air) * $c_p = 1.005 \text{ kJ/kg} \cdot \text{K}$ (Specific heat at constant pressure for air) **2. Determine Unknown Temperatures:** * **a) For $T_2$ (Compressor Exit, Isentropic Compression 1-2):** Using the isentropic relation: $T_2/T_1 = (P_r)^{(k-1)/k}$ $T_2 = T_1 \cdot (P_r)^{(k-1)/k} = 300 \text{ K} \cdot (8)^{(1.4-1)/1.4} = 300 \text{ K} \cdot (8)^{0.4/1.4} = 300 \text{ K} \cdot (8)^{0.2857}$ $T_2 = 300 \text{ K} \cdot 1.8115 = \mathbf{543.45 \text{ K}}$ * **a) For $T_4$ (Turbine Exit, Isentropic Expansion 3-4):** Using the isentropic relation: $T_3/T_4 = (P_r)^{(k-1)/k}$ $T_4 = T_3 / (P_r)^{(k-1)/k} = 1300 \text{ K} / (8)^{0.2857} = 1300 \text{ K} / 1.8115$ $T_4 = \mathbf{717.64 \text{ K}}$ **3. Calculate Work and Heat Terms:** * **Work Done on Compressor ($W_C$):** $W_C = c_p(T_2 - T_1) = 1.005 \text{ kJ/kg} \cdot \text{K} \cdot (543.45 - 300) \text{ K} = 1.005 \cdot 243.45 = 244.67 \text{ kJ/kg}$ * **Work Done by Turbine ($W_T$):** $W_T = c_p(T_3 - T_4) = 1.005 \text{ kJ/kg} \cdot \text{K} \cdot (1300 - 717.64) \text{ K} = 1.005 \cdot 582.36 = 585.27 \text{ kJ/kg}$ * **Heat Added ($Q_{in}$):** $Q_{in} = c_p(T_3 - T_2) = 1.005 \text{ kJ/kg} \cdot \text{K} \cdot (1300 - 543.45) \text{ K} = 1.005 \cdot 756.55 = 760.33 \text{ kJ/kg}$ * **Heat Rejected ($Q_{out}$):** $Q_{out} = c_p(T_4 - T_1) = 1.005 \text{ kJ/kg} \cdot \text{K} \cdot (717.64 - 300) \text{ K} = 1.005 \cdot 417.64 = 419.72 \text{ kJ/kg}$ **4. Calculate Net Work and Efficiency:** * **b) Net Work Output ($W_{net}$):** $W_{net} = W_T - W_C = 585.27 - 244.67 = \mathbf{340.6 \text{ kJ/kg}}$ (Check: $W_{net} = Q_{in} - Q_{out} = 760.33 - 419.72 = 340.61 \text{ kJ/kg}$ - Matches!) * **c) Thermal Efficiency ($\eta_{th}$):** $\eta_{th} = W_{net} / Q_{in} = 340.6 \text{ kJ/kg} / 760.33 \text{ kJ/kg} = 0.4479 = \mathbf{44.8\%}$ (Check with formula: $\eta_{th} = 1 - 1/(P_r)^{(k-1)/k} = 1 - 1/(8)^{0.2857} = 1 - 1/1.8115 = 1 - 0.5520 = 0.448 = 44.8\%$ - Matches!) * **d) Back Work Ratio (BWR):** BWR = $W_C / W_T = 244.67 \text{ kJ/kg} / 585.27 \text{ kJ/kg} = \mathbf{0.418}$ This means 41.8% of the turbine's work output is consumed by the compressor.