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Fuel Consumption Analysis of I.C. Engines: Measuring Thermal Efficiency

This practical guide focuses on calculating the thermal efficiency of internal combustion engines through fuel consumption measurements. Utilizing an engine test bench, accurate measuring tanks, flow meters, and stopwatches, the procedure involves determining fuel flow rates and conversions from liters to mass based on relative density. Key examples illustrate the calculation of power output and mechanical efficiency. This experiment provides essential insights into engine performance, enabling precise assessments of fuel efficiency in energy production.

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Fuel Consumption Analysis of I.C. Engines: Measuring Thermal Efficiency

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  1. I.C. ENGINES Practical No: 7 (5 May, 2014)

  2. Fuel Consumption • Objective • To calculate thermal efficiency of the engine • Apparatus • Engine test bench • Accurate measuring tank • Flow meter • Stop watch (min time should be not less than two minutes)

  3. Fuel Consumption • Procedure • Use test tank with hook gauges • Or calibrated glass measuring cylinder • Measuring glass shaped so that fuel falls fast at the points of calibration where time finished and finishes • An other method is to use flow meter

  4. Fuel Consumption • Procedure • The scale are not linear since the quantity of fuel flowing in unit time

  5. Fuel Consumption • Calculation • To convert milliliter (ml) or liters ( l) to mass ( kg) • 1000 ml = 1 litre • Since 1 litre of water has a mass of 1 kg • 1 litre of fuel has a mass of 1 x relative density of fuel (kg) • X litre of fuel has a mass of : • X x relative density of fuel ( kg)

  6. Fuel Consumption • Calculation • The quantity of fuel consumed is measured in litre per brake power ( kW) per hour , or kg per brake power ( kW) per hour

  7. Fuel Consumption • Calculation

  8. Example • A Morse test on an engine gave the following results: • Pb All cylinder working 100 kW • Pb No 1 cylinder cut out 69 kW • Pb No 1 cylinder cut out 71 kW • Pb No 1 cylinder cut out 68.5 kW • Pb No 1 cylinder cut out 71.5 kW • Find Pi , Pf and Mechanical efficiency

  9. Example • Formula • Pi = (A x 4) - (B1 +B2+B3+ B4) • Sol • Pi = (100 x4) – ( 69+ 71 + 68.5 + 71.5) • = 400 – 280 • Pi = 120 kW • Pf = Pi - Pb • = 120 – 100 • Pf = 20 kW

  10. Example • Sol • MechEff = 83 %

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