Understanding Engine Thermal Efficiency Calculation

This problem requires us to calculate the fuel consumption rate of an engine given its brake power, brake thermal efficiency, and the fuel's calorific value. We need to determine how much fuel the engine burns per hour.

Key Information Provided:

• Brake Thermal Efficiency ($\eta_{th}$): 40% or 0.40
• Brake Power ($P_b$): 20 kW
• Fuel Calorific Value (CV): 60,000 kJ/kg

Calculating Fuel Energy Input Rate

The brake thermal efficiency relates the useful work output (brake power) to the rate at which energy is supplied by the fuel.

The formula for brake thermal efficiency is:

$$ \eta_{th} = \frac{\text{Brake Power}}{\text{Fuel Energy Input Rate}} $$

We can rearrange this formula to find the Fuel Energy Input Rate:

$$ \text{Fuel Energy Input Rate} = \frac{\text{Brake Power}}{\eta_{th}} $$

First, let's ensure the units are consistent. Brake Power is given in kilowatts (kW), which is equivalent to kilojoules per second (kJ/s).

$$ P_b = 20 \text{ kW} = 20 \text{ kJ/s} $$

Now, substitute the values into the formula:

$$ \text{Fuel Energy Input Rate} = \frac{20 \text{ kJ/s}}{0.40} $$

$$ \text{Fuel Energy Input Rate} = 50 \text{ kJ/s} $$

This means the engine consumes 50 kilojoules of energy from the fuel every second.

Calculating Fuel Consumption Rate

The fuel energy input rate is also determined by the amount of fuel consumed and its energy content (calorific value).

The relationship is:

$$ \text{Fuel Energy Input Rate} = (\text{Fuel Consumption Rate in kg/s}) \times (\text{Calorific Value in kJ/kg}) $$

We can rearrange this to solve for the Fuel Consumption Rate in kilograms per second (kg/s):

$$ \text{Fuel Consumption Rate (kg/s)} = \frac{\text{Fuel Energy Input Rate}}{\text{Calorific Value}} $$

Substitute the known values:

$$ \text{Fuel Consumption Rate (kg/s)} = \frac{50 \text{ kJ/s}}{60,000 \text{ kJ/kg}} $$

$$ \text{Fuel Consumption Rate (kg/s)} = 0.0008333... \text{ kg/s} $$

Converting Fuel Consumption to kg/hour

The question asks for the fuel consumption in kilograms per hour (kg/hour). To convert from kg/s to kg/hour, we multiply by the number of seconds in an hour (3600).

$$ \text{Fuel Consumption Rate (kg/hour)} = (\text{Fuel Consumption Rate in kg/s}) \times 3600 \text{ s/hour} $$

$$ \text{Fuel Consumption Rate (kg/hour)} = 0.0008333... \text{ kg/s} \times 3600 \text{ s/hour} $$

$$ \text{Fuel Consumption Rate (kg/hour)} = 3.0 \text{ kg/hour} $$

Final Answer Verification

The calculation shows that the engine consumes 3.0 kg of fuel per hour. This matches one of the options provided.