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

Correct Answer: 273

Given: Displacement volume, $V_d = 6.0$ L = $0.006 \, {m}^3$ Engine speed, $N = 2300 \, {rpm}$ Mechanical efficiency, $\eta_m = 0.75$ Indicated mean effective pressure, $p_{me} = 800 \, {kPa} = 800{,}000 \, {Pa}$ BSFC = 320 g/h/kW = 0.32 kg/h/kW Calorific value, $CV = 44.6 \, {MJ/kg} = 44.6 \times 10^3 \, {kJ/kg}$ 
Step 1: Calculate Indicated Power (IP) For a four-stroke engine: \[ IP = \frac{p_{me} \cdot V_d \cdot N}{2 \cdot 60} = \frac{800{,}000 \cdot 0.006 \cdot 2300}{120} = \frac{11040{,}000}{120} = 92.0 \, {kW} \] Step 2: Brake Power (BP) \[ BP = \eta_m \cdot IP = 0.75 \cdot 92.0 = 69.0 \, {kW} \] Step 3: Fuel mass flow rate \[ \dot{m}_f = BP \cdot {BSFC} = 69.0 \cdot 0.32 = 22.08 \, {kg/h} \] Step 4: Fuel Equivalent Power \[ {Energy input} = \dot{m}_f \cdot CV = 22.08 \cdot 44.6 \times 10^3 = 985{,}568 \, {kJ/h} \] \[ {Fuel Equivalent Power} = \frac{985568}{3600} \approx \boxed{273.77 \, {kW}} \]