A separate-exhaust (unmixed) turbofan engine operates at sea-level static conditions. The core stream has an exhaust jet velocity $V_{jc} = 600\text{ m/s}$ and core mass flow rate $\dot{m}_c = 25\text{ kg/s}$. The bypass stream has an exhaust jet velocity $V_{jf} = 300\text{ m/s}$ and bypass mass flow rate $\dot{m}_f = 125\text{ kg/s}$. The bypass ratio $\alpha$ and the total static thrust $F$ (in $\text{kN}$) produced by the engine are:

(A) $\alpha = 5.0$ and $F = 52.5\text{ kN}$

(B) $\alpha = 5.0$ and $F = 37.5\text{ kN}$

(C) $\alpha = 6.0$ and $F = 52.5\text{ kN}$

(D) $\alpha = 0.2$ and $F = 60.0\text{ kN}$

Answer Key:

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### 1. Strategy & Theoretical Principles 1. The bypass ratio $\alpha$ is the ratio of bypass mass flow rate to core mass flow rate: $$\alpha = \frac{\dot{m}_f}{\dot{m}_c}$$ 2. At static sea-level conditions (flight speed $V_0 = 0$), with fully expanded nozzles ($p_e = p_a$), total engine thrust is the sum of core momentum thrust and bypass momentum thrust: $$F = \dot{m}_c V_{jc} + \dot{m}_f V_{jf}$$ ### 2. Analytical Formulation & Calculation Given: - $\dot{m}_c = 25\text{ kg/s}$ - $\dot{m}_f = 125\text{ kg/s}$ - $V_{jc} = 600\text{ m/s}$ - $V_{jf} = 300\text{ m/s}$ 1. Bypass ratio: $$\alpha = \frac{125}{25} = 5.0$$ 2. Core thrust: $$F_c = \dot{m}_c V_{jc} = 25 \times 600 = 15{,}000\text{ N} = 15.0\text{ kN}$$ 3. Bypass thrust: $$F_f = \dot{m}_f V_{jf} = 125 \times 300 = 37{,}500\text{ N} = 37.5\text{ kN}$$ 4. Total static thrust: $$F = F_c + F_f = 15.0 + 37.5 = 52.5\text{ kN}$$ ### 3. Distractor Forensics - **Option A:** Correct bypass ratio and thrust. - **Option B:** Considers only the fan bypass thrust ($37.5\text{ kN}$), omitting the core. - **Option C:** Miscounts total mass flow ratio. - **Option D:** Inverted bypass ratio (core/fan = 0.2). ### 4. Exam Trap & Standard Reference - **Trap Alert:** In separate-exhaust turbofans, both streams contribute to total thrust. - **Standard Reference:** Mattingly, J.D., *Elements of Propulsion*, Chapter 5.

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