An automobile engine consumes fuel at a rate of 22 L/h and delivers 65 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm³, determine the efficiency of this engine. The efficiency of the engine is %.

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Problem Statement:**

An automobile engine consumes fuel at a rate of 22 L/h and delivers 65 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm³, determine the efficiency of this engine.

**Objective:**

Calculate the efficiency of the engine.

**Solution Approach:**

1. **Convert fuel consumption to kg/h:**

   \[
   \text{Density} = 0.8 \text{ g/cm}^3 = 0.8 \text{ kg/L}
   \]

   \[
   \text{Fuel consumption in kg/h} = 22 \text{ L/h} \times 0.8 \text{ kg/L} = 17.6 \text{ kg/h}
   \]

2. **Calculate the energy content of the fuel consumed per hour:**

   \[
   \text{Energy content (kJ/h)} = 17.6 \text{ kg/h} \times 44,000 \text{ kJ/kg} = 774,400 \text{ kJ/h}
   \]

3. **Convert power to kJ/h:**

   \[
   \text{Power (kJ/h)} = 65 \text{ kW} \times 3600 \text{ s/h} = 234,000 \text{ kJ/h}
   \]

4. **Determine the efficiency:**

   \[
   \text{Efficiency} = \left(\frac{\text{Power output}}{\text{Energy input}}\right) \times 100
   \]

   \[
   \text{Efficiency} = \left(\frac{234,000}{774,400}\right) \times 100 \approx 30.21\%
   \]

**Conclusion:**

The efficiency of the engine is approximately **30.21%**.
Transcribed Image Text:**Problem Statement:** An automobile engine consumes fuel at a rate of 22 L/h and delivers 65 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm³, determine the efficiency of this engine. **Objective:** Calculate the efficiency of the engine. **Solution Approach:** 1. **Convert fuel consumption to kg/h:** \[ \text{Density} = 0.8 \text{ g/cm}^3 = 0.8 \text{ kg/L} \] \[ \text{Fuel consumption in kg/h} = 22 \text{ L/h} \times 0.8 \text{ kg/L} = 17.6 \text{ kg/h} \] 2. **Calculate the energy content of the fuel consumed per hour:** \[ \text{Energy content (kJ/h)} = 17.6 \text{ kg/h} \times 44,000 \text{ kJ/kg} = 774,400 \text{ kJ/h} \] 3. **Convert power to kJ/h:** \[ \text{Power (kJ/h)} = 65 \text{ kW} \times 3600 \text{ s/h} = 234,000 \text{ kJ/h} \] 4. **Determine the efficiency:** \[ \text{Efficiency} = \left(\frac{\text{Power output}}{\text{Energy input}}\right) \times 100 \] \[ \text{Efficiency} = \left(\frac{234,000}{774,400}\right) \times 100 \approx 30.21\% \] **Conclusion:** The efficiency of the engine is approximately **30.21%**.
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