Let's break down the problem step by step. --- ## **Given Data** - **Atmospheric Pressure (\(P_1\))**: 1 atm = 101325 Pa - **Atmospheric Temperature (\(T_1\))**: \(-20^\circ C = 253.15\,K\) - **Compressor Pressure Ratio (\(r_c\))**: 8 - **Turbine and Compressor Efficiency (\(\eta_c, \eta_t\))**: .88 - **Max Cycle Temperature (\(T_3\))**: \(100^\circ C = 1273.15\,K\) - **Nozzle Efficiency (\(\eta_n\))**: .90 - **Mass Flow Rate (\(\dot{m}\))**: 25 kg/s - **Gas Constant (\(R\))**: 287 J/kg-K (for air) - **Specific Heat at Constant Pressure (\(c_p\))**: 1005 J/kg-K (for air) - **Ratio of Specific Heats (\(\gamma\))**: 1.4 (for air) --- ## **Step 1: Find Compressor Exit Temperature (\(T_2\))** ### **Isentropic Exit Temperature (\(T_{2s}\)):** \[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{(\gamma-1)/\gamma} \] \[ P_2 = r_c \cdot P_1 = 8 \cdot P_1 \] \[ T_{2s} = 253.15 \times 8^{(.4/1.4)} \] Calculate exponent: \[ \frac{.4}{1.4} = .2857 \] \[ 8^{.2857} \approx 1.79 \] \[ T_{2s} = 253.15 \times 1.79 \approx 453.1\,K \] ### **Actual Exit Temperature (\(T_2\)):** \[ \eta_c = \frac{T_{2s} - T_1}{T_2 - T_1} \implies T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c} \] \[ T_2 = 253.15 + \frac{453.1 - 253.15}{.88} \] \[ T_2 = 253.15 + \frac{199.95}{.88} \approx 253.15 + 227.22 = 480.37\,K \] --- ## **Step 2: Turbine Exit Temperature (\(T_4\))** ### **Work Done by Compressor:** \[ W_c = c_p (T_2 - T_1) \] \[ W_c = 1005 \times (480.37 - 253.15) = 1005 \times 227.22 \approx 228,356\,J/kg \] ### **Work Done by Turbine:** \[ W_t = W_c \text{ (since no net shaft work, ignoring losses and bleed)} \] \[ W_t = c_p (T_3 - T_4) \implies T_4 = T_3 - \frac{W_t}{c_p} \] \[ T_4 = 1273.15 - \frac{228,356}{1005} \approx 1273.15 - 227.22 = 1045.93\,K \] #### **Correct for Turbine Efficiency (\(\eta_t\)):** \[ \eta_t = \frac{T_3 - T_4}{T_3 - T_{4s}} \implies T_{4s} = T_3 - \frac{T_3 - T_4}{\eta_t} \] \[ T_{4s} = 1273.15 - \frac{1273.15 - 1045.93}{.88} \] \[ T_{4s} = 1273.15 - \frac{227.22}{.88} = 1273.15 - 258.2 = 1014.95\,K \] --- ## **Step 3: Nozzle Exit Velocity Calculation** ### **Assume Expansion to Atmospheric Pressure (\(P_1\)):** **Isentropic Nozzle Exit Temperature (\(T_{5s}\)):** \[ T_{5s} = T_4 \left( \frac{P_5}{P_4} \right)^{(\gamma-1)/\gamma} \] \(P_4 = P_2 = 8\,\text{atm}\), \(P_5 = 1\,\text{atm}\) \[ T_{5s} = 1014.95 \times (1/8)^{.2857} \] \[ (1/8)^{.2857} = 8^{-.2857} \approx .559 \] \[ T_{5s} = 1014.95 \times .559 \approx 567.5\,K \] **Actual Nozzle Exit Temperature (\(T_5\)):** \[ \eta_n = \frac{T_4 - T_5}{T_4 - T_{5s}} \implies T_5 = T_4 - \eta_n (T_4 - T_{5s}) \] \[ T_5 = 1014.95 - .90 \times (1014.95 - 567.5) \] \[ T_5 = 1014.95 - .90 \times 447.45 = 1014.95 - 402.7 = 612.25\,K \] ### **Nozzle Exit Velocity (\(V_5\)):** \[ \text{Assume } V_1 = \text{ (static test bed)} \] \[ \text{Kinetic energy: } \Delta KE = c_p (T_4 - T_5) \] \[ V_5 = \sqrt{2c_p(T_4 - T_5)} \] \[ V_5 = \sqrt{2 \times 1005 \times (1014.95 - 612.25)} \] \[ V_5 = \sqrt{2 \times 1005 \times 402.7} \] \[ V_5 = \sqrt{2 \times 1005 \times 402.7} = \sqrt{810,427.5} \approx 900\,\text{m/s} \] --- ## **Step 4: Static Thrust Calculation** \[ \text{Thrust} = \dot{m} V_5 \] \[ \text{Thrust} = 25\,\text{kg/s} \times 900\,\text{m/s} = 22,500\,\text{N} \] --- ## **Final Answers** ### **1. Velocity of air leaving the nozzle:** \[ \boxed{900\,\text{m/s}} \] ### **2. Static thrust:** \[ \boxed{22,500\,\text{N}} \] --- ### **Summary Table** | Quantity | Value | |---------------------|-----------| | Nozzle Exit Velocity| 900 m/s | | Static Thrust | 22,500 N |