NUMERICAL PROBLEMS INTERNAL COMBUSTION ENGINE

 

NUMERICAL PROBLEMS

INTERNAL COMBUSTION ENGINE

Solution of numerical problems help to increase clarity. It gives the real  inside understanding. This is helpful in the design and construction of Internal Combustion Engines. Solution of two numerical problems on internal combustion engines is given below. One is on two stroke while the other is on four stroke cycle.

PROBLEM 1

 Data recorded during testing of a TWO STROKE gas engine is

Diameter of the piston                  d= 150 mm

Stroke length                                     L= 180 mm

Clearance volume                            Vc = 0.89 liter

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 p L AN/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

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