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A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec is
- a)1×10−3kg/s
- b)2×10−3kg/s
- c)3×10−3kg/s
- d)4×10−3kg/s
Correct answer is option 'B'. Can you explain this answer?
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A for cylinder engine running at 1200rpm delivers 20kW. The average t...
To find the mass of fuel consumption, we need to follow these steps:
1. Calculate the total power output of the engine:
- Given: Power output = 20 kW
- Since it is a 4-cylinder engine, the total power output will be 4 times the power output of one cylinder.
- Total power output = 4 * 20 kW = 80 kW
2. Calculate the torque of the engine:
- Given: Average torque with one cylinder cut = 110 Nm
- Since the power output of the engine is given in kilowatts and the torque is given in Newton-meters, we need to convert them.
- Power (P) = Torque (T) * Angular velocity (ω)
- Angular velocity (ω) = 2π * (RPM / 60)
- Convert 1200 RPM to radians per second: ω = 2π * (1200 / 60) = 40π rad/s
- Convert 20 kW to Watts: P = 20 kW * 1000 = 20000 W
- Torque (T) = P / ω = 20000 W / (40π rad/s) ≈ 159.155 Nm
- Since the average torque with one cylinder cut is 110 Nm, it means that the torque of one cylinder is 159.155 Nm - 110 Nm = 49.155 Nm
3. Calculate the fuel consumption rate:
- Given: Calorific value of fuel = 43 MJ/kg
- Given: Fuel consumption rate = 360 g/kWh
- Convert 20 kW to kWh: Power (P) = 20 kW * (1 h / 3600 s) = 0.00556 kWh
- Convert fuel consumption rate to kg/s: Fuel consumption rate = 360 g/kWh * (1 kg / 1000 g) * (1 h / 3600 s) = 0.0001 kg/s
4. Calculate the mass of fuel consumption:
- Since the engine is a 4-cylinder engine, the fuel consumption rate will be 4 times the fuel consumption rate of one cylinder.
- Mass of fuel consumption = 4 * 0.0001 kg/s = 0.0004 kg/s = 2 * 10^-3 kg/s
Therefore, the correct answer is option 'B', which is 2 * 10^-3 kg/s.
1. Calculate the total power output of the engine:
- Given: Power output = 20 kW
- Since it is a 4-cylinder engine, the total power output will be 4 times the power output of one cylinder.
- Total power output = 4 * 20 kW = 80 kW
2. Calculate the torque of the engine:
- Given: Average torque with one cylinder cut = 110 Nm
- Since the power output of the engine is given in kilowatts and the torque is given in Newton-meters, we need to convert them.
- Power (P) = Torque (T) * Angular velocity (ω)
- Angular velocity (ω) = 2π * (RPM / 60)
- Convert 1200 RPM to radians per second: ω = 2π * (1200 / 60) = 40π rad/s
- Convert 20 kW to Watts: P = 20 kW * 1000 = 20000 W
- Torque (T) = P / ω = 20000 W / (40π rad/s) ≈ 159.155 Nm
- Since the average torque with one cylinder cut is 110 Nm, it means that the torque of one cylinder is 159.155 Nm - 110 Nm = 49.155 Nm
3. Calculate the fuel consumption rate:
- Given: Calorific value of fuel = 43 MJ/kg
- Given: Fuel consumption rate = 360 g/kWh
- Convert 20 kW to kWh: Power (P) = 20 kW * (1 h / 3600 s) = 0.00556 kWh
- Convert fuel consumption rate to kg/s: Fuel consumption rate = 360 g/kWh * (1 kg / 1000 g) * (1 h / 3600 s) = 0.0001 kg/s
4. Calculate the mass of fuel consumption:
- Since the engine is a 4-cylinder engine, the fuel consumption rate will be 4 times the fuel consumption rate of one cylinder.
- Mass of fuel consumption = 4 * 0.0001 kg/s = 0.0004 kg/s = 2 * 10^-3 kg/s
Therefore, the correct answer is option 'B', which is 2 * 10^-3 kg/s.
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A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer?
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A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer?.
A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A for cylinder engine running at 1200rpm delivers 20kW. The average torque when one cylinder was cut is 110Nm. If the calorific value of the fuel is 43MJ/kg and the engine uses 360gms of gasoline per kWh. Then the mass of fuel consumption in kg/sec isa)1×10−3kg/sb)2×10−3kg/sc)3×10−3kg/sd)4×10−3kg/sCorrect answer is option 'B'. Can you explain this answer?.
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