# Turbofan Engine Performance Calculation Let's solve step-by-step for **thrust**, **air-fuel ratio**, and **specific fuel consumption** using the given data. --- ## **Given Data** - Core mass flow (\(\dot{m}_{a}\)): **1. kg/s** - Overall pressure ratio: **12** - Ambient pressure (\(P_\)): **35.6 kPa** - Ambient temperature (\(T_\)): **236.15 K** - Altitude: **800 m** - Mach number (\(M_\)): **.92** - Max turbine inlet temp (\(T_{t4}\)): **140 K** - Diffuser efficiency: **99%** - Fan efficiency: **85%** - Compressor efficiency: **87%** - Turbine efficiency: **89%** - Fan nozzle & core nozzle efficiency: **100%** - Combustion chamber pressure loss: **6%** - Fuel heating value (\(h_{PR}\)): **44,400 kJ/kg** - Gas constant air (\(R_a\)): **1. kJ/kg·K** - \(C_p\) air: **1. kJ/kg·K** - \(\gamma\) air: **1.4** - Gas constant exhaust (\(R_g\)): **.287 kJ/kg·K** --- # **Step 1: Calculate Flight Velocity (\(V_\))** \[ V_ = M_ \cdot \sqrt{\gamma \cdot R_a \cdot T_} \] Plug in values: \[ V_ = .92 \cdot \sqrt{1.4 \cdot 1. \cdot 236.15} \] \[ V_ = .92 \cdot \sqrt{330.61} \] \[ V_ = .92 \cdot 18.18 \approx \boxed{16.73 \text{ m/s}} \] --- # **Step 2: Diffuser (Total Conditions Before Compression)** Isentropic total temperature after diffuser: \[ T_{t} = T_ \left[1 + \frac{\gamma-1}{2} M_^2\right] \] \[ T_{t} = 236.15 \left[1 + .2 \cdot (.92)^2\right] \] \[ T_{t} = 236.15 \left[1 + .2 \cdot .8464\right] \] \[ T_{t} = 236.15 \left[1 + .1693\right] = 236.15 \cdot 1.1693 \approx \boxed{276. \text{ K}} \] Isentropic total pressure after diffuser: \[ \frac{P_{t}}{P_} = \left[1 + \frac{\gamma-1}{2} M_^2\right]^{\frac{\gamma}{\gamma-1}} \] \[ = [1 + .2 \cdot .8464]^{3.5} \] \[ = [1.1693]^{3.5} \approx 1.685 \] \[ P_{t} = 1.685 \cdot 35.6 \approx \boxed{60. \text{ kPa}} \] With diffuser efficiency, actual \(T_{t}\) is nearly same (since efficiency is 99%)—so stick with above values. --- # **Step 3: Compressor Exit Conditions** **Compressor pressure ratio:** 12 \[ P_{t3} = 12 \cdot P_{t} = 12 \cdot 60. = \boxed{720 \text{ kPa}} \] **Compressor temperature rise (account for efficiency):** Isentropic temp rise: \[ \frac{T_{t3s}}{T_{t}} = \left(\frac{P_{t3}}{P_{t}}\right)^{\frac{\gamma-1}{\gamma}} = 12^{.286} \approx 1.95 \] \[ T_{t3s} = 1.95 \cdot 276. = 538.2 \text{ K} \] Actual temp rise (using efficiency \(\eta_c = .87\)): \[ \eta_c = \frac{T_{t3s} - T_{t}}{T_{t3} - T_{t}} \Rightarrow T_{t3} = T_{t} + \frac{T_{t3s} - T_{t}}{\eta_c} \] \[ T_{t3} = 276. + \frac{538.2 - 276.}{.87} = 276. + \frac{262.2}{.87} = 276. + 301.38 = \boxed{577.4 \text{ K}} \] --- # **Step 4: Combustor Exit (Turbine Inlet) Conditions** Given \(T_{t4} = 140 \text{ K}\) Combustor pressure loss: \(6\%\) of inlet pressure \[ P_{t4} = .94 \cdot P_{t3} = .94 \cdot 720 = \boxed{676.8 \text{ kPa}} \] --- # **Step 5: Turbine Expansion (Core Nozzle Entry)** The turbine must provide enough work to drive compressor and fan. For this question, as only core flow is given, assume all work is for core compressor: Turbine work required: \[ W_{comp} = C_p \cdot (T_{t3} - T_{t}) \] \[ = 1. \cdot (577.4 - 276.) = 301.4 \text{ kJ/kg} \] **Turbine efficiency (\(\eta_t = .89\))** Isentropic expansion: \[ W_{turb,iso} = C_p (T_{t4} - T_{t5s}) \] \[ \eta_t = \frac{T_{t4} - T_{t5}}{T_{t4} - T_{t5s}} \Rightarrow T_{t5} = T_{t4} - \eta_t (T_{t4} - T_{t5s}) \] But, for power balance: \[ C_p (T_{t4} - T_{t5}) = W_{comp} \Rightarrow T_{t5} = T_{t4} - \frac{W_{comp}}{C_p} = 140 - 301.4 = 1098.6 \text{ K} \] Now, find the isentropic \(T_{t5s}\): \[ T_{t5} = T_{t4} - \eta_t (T_{t4} - T_{t5s}) \Rightarrow T_{t5s} = T_{t4} - \frac{T_{t4} - T_{t5}}{\eta_t} \] \[ T_{t5s} = 140 - \frac{140 - 1098.6}{.89} = 140 - \frac{301.4}{.89} = 140 - 338.6 = 1061.4 \text{ K} \] --- # **Step 6: Core Nozzle Exit Conditions** Nozzle efficiency: **100%** Ambient pressure: \(P_ = 35.6 \text{ kPa}\) \[ \frac{P_{t5}}{P_} = \frac{P_{t4}}{P_} \] Assuming total pressure after turbine is \(P_{t5} = P_{t4}\) (if no pressure loss in turbine): \[ \frac{P_{t4}}{P_} = \frac{676.8}{35.6} = 19.02 \] Isentropic expansion to ambient: \[ \frac{T_{e}}{T_{t5}} = \left(\frac{P_}{P_{t5}}\right)^{\frac{\gamma-1}{\gamma}} = (1/19.02)^{.286} = (.0526)^{.286} \approx .476 \] \[ T_e = .476 \cdot 1098.6 \approx 523 \text{ K} \] Exit velocity: \[ V_e = \sqrt{2 C_p (T_{t5} - T_e)} = \sqrt{2 \cdot 1. \cdot (1098.6 - 523)} = \sqrt{2 \cdot 575.6} = \sqrt{1151.2} \approx \boxed{33.93 \text{ m/s}} \] --- # **Step 7: Air-Fuel Ratio (AFR)** Energy balance in combustor: \[ \dot{m}_a C_p (T_{t4} - T_{t3}) = \dot{m}_f h_{PR} \eta_{comb} \] Assume \(\eta_{comb} = 1\): \[ \frac{\dot{m}_f}{\dot{m}_a} = \frac{C_p (T_{t4} - T_{t3})}{h_{PR}} = \frac{1. \cdot (140 - 577.4)}{44,400} = \frac{822.6}{44,400} = .01853 \] \[ AFR = \frac{\dot{m}_a}{\dot{m}_f} = \frac{1}{.01853} \approx \boxed{54.} \] --- # **Step 8: Thrust Calculation** \[ F = \dot{m}_a (V_e - V_) + \dot{m}_f (V_e - V_) \] But \(\dot{m}_f \ll \dot{m}_a\), so: \[ F \approx (\dot{m}_a + \dot{m}_f)(V_e - V_) \] \[ = (1 + .01853) \cdot (33.93 - 16.73) = 1.01853 \cdot 17.20 = \boxed{17.5 \text{ N}} \] --- # **Step 9: Specific Fuel Consumption (SFC)** \[ SFC = \frac{\dot{m}_f}{F} = \frac{.01853}{17.5} = .00106 \text{ kg/s/N} \] Convert to \(\text{mg/s/N}\) or \(\text{kg/hr/kN}\): \[ SFC = .00106 \times 360 \div 1 = 3.8 \text{ kg/hr/kN} \] --- # **Summary Table** | Quantity | Value | |---------------------------|------------------| | **Thrust** | 17.5 N | | **Air-Fuel Ratio (AFR)** | 54. | | **Specific Fuel Consumption (SFC)** | .00106 kg/s/N or 3.8 kg/hr/kN | --- ## **Final Answers** - **Thrust:** \(\boxed{17.5 \text{ N}}\) - **Air-Fuel Ratio:** \(\boxed{54.}\) - **Specific Fuel Consumption:** \(\boxed{.00106 \text{ kg/s/N}}\) or \(\boxed{3.8 \text{ kg/hr/kN}}\) --- If you need more detail on any step, let me know!