A turboprop engine produces a shaft brake power $P_{\text{shaft}} = 2{,}500\text{ kW}$ and an exhaust jet thrust $F = 4.5\text{ kN}$ at a flight speed $V_0 = 150\text{ m/s}$. The propeller efficiency is $\eta_p = 0.80$. The **equivalent shaft power** ($\text{ESHP}$ in $\text{kW}$) delivered by the engine is:

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### 1. Strategy & Theoretical Principles In turboprop engines, residual thrust from the core exhaust contributes additional thrust power. The Equivalent Shaft Power (ESHP) credits this jet thrust by converting it to equivalent shaft power via the propeller efficiency $\eta_p$: $$\text{ESHP} = P_{\text{shaft}} + \frac{F \cdot V_0}{\eta_p}$$ ### 2. Analytical Formulation & Calculation Given: - $P_{\text{shaft}} = 2{,}500\text{ kW} = 2{,}500{,}000\text{ W}$ - $F = 4.5\text{ kN} = 4{,}500\text{ N}$ - $V_0 = 150\text{ m/s}$ - $\eta_p = 0.80$ 1. Thrust power produced by the jet exhaust: $$P_{\text{thrust}} = F \cdot V_0 = 4{,}500 \times 150 = 675{,}000\text{ W} = 675\text{ kW}$$ 2. Equivalent shaft power of the jet: $$P_{\text{equiv, jet}} = \frac{P_{\text{thrust}}}{\eta_p} = \frac{675\text{ kW}}{0.80} = 843.75\text{ kW}$$ 3. Total Equivalent Shaft Power: $$\text{ESHP} = 2{,}500 + 843.75 = 3{,}343.75\text{ kW}$$ ### 3. Distractor Forensics - Candidate multiplying by $\eta_p$ instead of dividing gets $2500 + 675 \times 0.80 = 3040\text{ kW}$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always divide jet thrust power by $\eta_p$ to find how much engine shaft power would have been required to produce that same thrust. - **Standard Reference:** Mattingly, J.D., *Elements of Propulsion*, Chapter 5.

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