Question

A four-stroke diesel engine has a displacement volume of 6.0 L and it operates at 2300 rpm with 75% mechanical efficiency. The indicated mean effective pressure is 800 kPa. If the engine has brake-specific fuel consumption of 320 g.h\(^{-1}\).kW\(^{-1}\), considering calorific value of the fuel as 44.6 MJ.kg\(^{-1}\), the fuel equivalent power is _________ kW. (Rounded off to 2 decimal places)

Hint:

Fuel equivalent power is calculated by considering brake power, brake-specific fuel consumption, and calorific value. Make sure the units are consistent when using the formula.

Answer 1

Given:

• Displacement volume, \( V_d = 6.0 \, \text{L} = 0.006 \, \text{m}^3 \)
• Engine speed, \( N = 2300 \, \text{rpm} \)
• Mechanical efficiency, \( \eta_m = 0.75 \)
• Indicated mean effective pressure, \( p_{me} = 800 \, \text{kPa} = 800{,}000 \, \text{Pa} \)
• BSFC = 320 g/h/kW = 0.32 kg/h/kW
• Calorific value, \( CV = 44.6 \, \text{MJ/kg} = 44.6 \times 10^3 \, \text{kJ/kg} \)

Step 1: Calculate Indicated Power (IP)

\( IP = \frac{p_{me} \cdot V_d \cdot N}{2 \cdot 60} = \frac{800{,}000 \cdot 0.006 \cdot 2300}{120} = \frac{11{,}040{,}000}{120} = 92.0\, \text{kW} \)

Step 2: Brake Power (BP)

\( BP = \eta_m \cdot IP = 0.75 \cdot 92.0 = 69.0\, \text{kW} \)

Step 3: Fuel Mass Flow Rate

\( \dot{m}_f = BP \cdot \text{BSFC} = 69.0 \cdot 0.32 = 22.08\, \text{kg/h} \)

Step 4: Fuel Equivalent Power

\( \text{Energy input} = \dot{m}_f \cdot CV = 22.08 \cdot 44.6 \times 10^3 = 985{,}568\, \text{kJ/h} \)

\( \text{Fuel Equivalent Power} = \frac{985{,}568}{3600} \approx \boxed{273.77\, \text{kW}} \)