NUMERICAL PROBLEMS ON INTERNAL COMBUSTION ENGINES

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NUMERICAL PROBLEMS ON INTERNAL COMBUSTION ENGINES

Two numerical problems have been solved on internal combustion engines. One is on two stroke while the other is on four stroke cycle.

PROBLEM 1

The following data was recorded during testing of a TWO STROKE gas engine:

Diameter of the piston                  d= 150 mm

Stroke length                                     L= 180 mm

Clearance volume                            Vc = 0.89 litre

RPM of the engine                            N = 300

Indicated mean effective pressure pm= 6.1 bars

Gas consumption                                   m. = 6.1 m3/h

Calorific value of the gas (fuel)          CF = 17000 kJ/m3 

Determine the followings:

  • Air Standard Efficiency
  • Indicated power (IHP)developed by the engine
  • Indicated thermal efficiency of the engine

SOLUTION

Swept volume  Vs = πd2L/4 = π(0.150)2 x 180/4 = 0.00318 m3

Clearance volume Vc= 0.00089 m3

Total volume = Swept volume + clearance volume

                  VT       = 0.00318 + 0.00089 = 0.00407 m3

Compression ratio γ = Total volume/Clearance volume

                                     =        0.00407/0.00089 = 4.573

  • Air standard Efficiency η = 1 –1/(r)γ—1 = 1—1/(4.573)4—1

       =  0.456 = 45.6 %

  • Indicated power IHP = 100 pLAN/60 when p is in bars

     = 100 x 6.1x 0.180 x [π(0.150)2/4] 300/60 = 9700  W

(c ) Indicated Thermal Efficiency

ηIT = Indicated power in (kJ/s)/Heat supplied in (kJ/s)

     = (9700/1000)/(6.1 x 17000/3600) = 0.3367 = 33.67 %

 

PROBLEM 2

Following data is available for a FOUR STROKE petrol engine:

Air fuel ratio 15.5 : 1

Calorific value of fuel 16000 kJ/kg

Air Standard Efficiency: 53

Mechanical Efficiency: 80 %

Indicated Thermal Efficiency: 37 %

Volumetric Efficiency: 80 %

Stroke/bore ratio: 1.25

Suction pressure: 1 bar

Suction Temperature: 270C

RPM: 2000

Brake Power: 72 kW

Calculate the followings:

  • Brake specific fuel consumption
  • Bore and stroke

SOLUTION

Find compression ratio from air standard efficiency

η = 1 –1/(r)γ—1

0.53 = 1 –1/(r)1.4—1

   r = 6.6

IHP = BP/Mech efficiency = 72/0.80 = 90 kW

ηIT =IHP/(Sp.Fuel Consp x Cal value)

0.37 = 90/sfc x 16000

sfc is specific fuel consumption

sfc = 0.0152 kg/s

Brake sfc =  sfcIHP/BP = 0.0152/72=0.00021 kg/s /kW

Brake sfc = 0.7601 kg/kWh

  • Bore and stroke of the engine

Mass of air fuel mixture/kg of fuel = 15.5 +1 = 16.5

Mass of fuel supplied to the engine = 0.0152 x 16.5 = 0.2508

Volume of air fuel mixture = mRT/p=0.2508x 287×300/(1×105)

V = 0.2159 m3/s

Swept volume = volume of mixture supplied/vol efficiency

Vs = 0.2159/0.80= 0.2699

Vs=( πd2L/4) n x (rpm/2) /60

Where n is the number of cylinders

= (πd2 x 1.25 d/4)n rpm/120

    d= 0.152 m= 152 mm

     L = 190 mm