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Question 56
Question Statement:
A diesel engine operating with B20 biodiesel blend (20% biodiesel, 80% diesel by volume) develops 10 kW brake power at brake-specific fuel consumption 0.26 kg/kWh. Density of biodiesel = 880 kg/m³, diesel = 850 kg/m³. Find the volume of biodiesel required to run the engine for 3 hours (L). (Rounded to 2 decimal places)
Official Answer Key:
1.80 to 1.84
Step-by-Step Solution & Explanation
• **Concept & Given Data:**
- Brake power $BP = 10\text{ kW}$, duration $t = 3\text{ h}$, $bsfc = 0.26\text{ kg/kWh}$.
- Total fuel mass consumed $m_{fuel} = BP \times bsfc \times t = 10 \times 0.26 \times 3 = 7.8\text{ kg}$.
- Blend B20 by volume: $20\%\text{ biodiesel}$ ($\rho_b = 880\text{ kg/m}^3 = 0.880\text{ kg/L}$) and $80\%\text{ diesel}$ ($\rho_d = 850\text{ kg/m}^3 = 0.850\text{ kg/L}$).
• **Average Blend Density:**
$\rho_{blend} = 0.20 \times 880 + 0.80 \times 850 = 176 + 680 = 856\text{ kg/m}^3 = 0.856\text{ kg/L}$
• **Total Fuel Volume:**
$V_{total} = \frac{m_{fuel}}{\rho_{blend}} = \frac{7.8}{0.856} = 9.1121\text{ L}$
• **Biodiesel Volume Required:**
$V_{biodiesel} = 0.20 \times V_{total} = 0.20 \times 9.1121 = 1.8224\text{ L}$
• **Conclusion:** Biodiesel required = $1.82\text{ L}$ (Official range: 1.80 to 1.84).
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