Question 5 3AE01 - Third Assistant Engineer

The Correct Answer is B This problem requires calculating the Indicated Power (IP) developed by the engine using the standard formula: $$IP (W) = P_m \cdot L \cdot A \cdot n \cdot N_{cyl}$$ Where: * $P_m$ = Mean Effective Pressure (in Pascals, Pa) * $L$ = Stroke length (in meters, m) * $A$ = Piston Area (in square meters, $m^2$) * $n$ = Power strokes per second * $N_{cyl}$ = Number of cylinders ### 1. Calculation of Parameters **A. Unit Conversion and Pressure ($P_m$):** The pressure is given as $30 \text{ kg}/\text{cm}^2$. We must convert this to the SI unit, Pascals (Pa). $$1 \text{ kg}/\text{cm}^2 \approx 98,066.5 \text{ Pa}$$ $$P_m = 30 \text{ kg}/\text{cm}^2 \cdot 98,066.5 \text{ Pa} / (\text{kg}/\text{cm}^2) = 2,941,995 \text{ Pa}$$ **B. Stroke (L):** $$L = 1400 \text{ mm} = 1.4 \text{ m}$$ **C. Piston Area (A):** Bore $D = 650 \text{ mm} = 0.65 \text{ m}$ $$A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.65 \text{ m})^2 \approx 0.33183 \text{ m}^2$$ **D. Power Strokes per Second (n):** The engine is four-stroke, meaning there is one power stroke per two revolutions. $$n = \frac{N_{RPM}}{60} \cdot \frac{1}{2} = \frac{100 \text{ RPM}}{60 \text{ s/min}} \cdot \frac{1}{2} = \frac{100}{120} \text{ Hz} \approx 0.83333 \text{ s}^{-1}$$ ### 2. Calculation of Indicated Power (IP) $$IP (W) = P_m \cdot L \cdot A \cdot n \cdot N_{cyl}$$ $$IP (W) = (2,941,995 \text{ Pa}) \cdot (1.4 \text{ m}) \cdot (0.33183 \text{ m}^2) \cdot (0.83333 \text{ s}^{-1}) \cdot 8$$ $$IP (W) \approx 9,111,100 \text{ W}$$ ### 3. Conversion to Kilowatts (kW) $$IP (\text{kW}) = \frac{9,111,100 \text{ W}}{1000} \approx 9,111 \text{ kW}$$ This result matches Option B. *** ### Why Other Options Are Incorrect **A) 1,689 kW:** This value is significantly too low and does not result from common errors like miscounting the stroke cycle or the number of cylinders. **C) 12,388 kW:** This could arise from using a slightly incorrect conversion factor for pressure (e.g., using a common but less precise factor for metric horsepower) or an error in calculating the area or the $n$ term, resulting in a number about 36% higher than the precise calculation. **D) 24,776 kW:** This value is extremely high, suggesting a major error, most likely treating the engine as a **two-stroke** cycle engine ($n = 100/60 \text{ Hz}$) while also incorporating other calculation mistakes (as doubling the precise power, $9,111 \text{ kW} \times 2 = 18,222 \text{ kW}$, still does not reach this option). This represents a gross overestimation of the developed power.