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I.C. ENGINES

I.C. ENGINES. Practical No: 6 (5 May, 2014). Morse Test. Date Objective This is one method of obtaining the indicated power P i Mechanical efficiency Friction pumping losses of an engine. Morse Test. Apparatus Engine Indicator. Morse Test. Procedure

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I.C. ENGINES

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  1. I.C. ENGINES Practical No: 6 (5 May, 2014)

  2. Morse Test • Date • Objective • This is one method of obtaining the indicated power Pi • Mechanical efficiency • Friction pumping losses of an engine

  3. Morse Test • Apparatus • Engine • Indicator

  4. Morse Test • Procedure • Engine will be coupled to a dynamometer • All cylinders should keep working at certain speed and under load • Deprive of power of one cylinder by cutting ignition of fuel • Engine speed falls

  5. Morse Test • Procedure • Dynamometer load must be reduced till returns to its original figure • Pb is again measured but fond to be lower

  6. Morse Test • Procedure • This reduction in Pbis approximately the Pi of the cylinder because the working cylinders have to overcome the friction pumping losses of the idle cylinder as well as their own; thus the reduction in Pb must be the Pi of the idle engine

  7. Morse Test • Procedure • To calculate the Pi and friction power ( Pf ) of a four cylinder engine • Five test are required to be made • One with all cylinders working • The others with each of the cylinders cut out in turn

  8. Morse Test • Calculation • Let A = Pb when all cylinders are working • B = Pbwhen one cylinders is not working • Then A – B = Pi of the idle cylinder • Therefore • Pi of all the cylinder = (A – B1) + (A – B2) + (A – B3) + (A – B4)

  9. Morse Test • Calculation • Pi = (A x 4) - (B1 +B2+B3+ B4) • Friction power = Pi - Pb • Friction covers all losses like friction, pumping, fan, dynamo and water pump

  10. Morse Test • Graph Pi (kW) Mecheff Power ( kW) Pb (kW) Pf (kW) Rev/min

  11. Example • Agave 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

  12. 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

  13. Example • Sol • MechEff = 83 %

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