I.C. ENGINES

LECTURE NO: 13(28 Apr 2014)

ENGINE PERFORMANCE PARAMETERS

• Indicated Thermal Efficiency • Brake Thermal Efficiency • Mechanical Efficiency • Volumetric Efficiency • Relative Efficiency • Mean Effective Pressure• Mean Piston Speed• Specific Power Output• Specific Fuel Consumption

ENGINE PERFORMANCE PARAMETERS

• Inlet Value Mach Index• Fuel Air Ratio• Calorific Value of the Fuel

Atmospheric Pressure

Work

• Work is the movement of a body against an opposing force.

Torque

Conversion of Work into Torque

Power

• Rate of doing Work is called power• Through testing, it was found that the average

horse could lift a 200-lb. weight to a height of 165 ft in 1 minute. The equivalent of one horsepower can be reached by multiplying 165 ft by 200 lb. (work formula) for a total of 33,000 ft lb. per minute

Power

Dynamometer

• Pony Brake

Dynamometer

Classic Dynamometer

Indicated Thermal Efficiency

• Indicated thermal efficiency is the ratio of energy in the indicated power ip, to the input fuel energy in appropriate units

ηith = ip [kJ/s] energy in fuel per second [kJ/s]

ηith = ip [kJ/s] mass of fuel/s x calorfic value of fuel

Brake Thermal Efficiency

• Brake thermal efficiency is the ratio of energy in the brake power bp, to the input fuel energy in appropriate units

ηbth = bp [kJ/s] energy in fuel per second [kJ/s]

ηith = ip [kJ/s] mass of fuel/s x calorific value of fuel

Mechanical Efficiency

• Mechanical efficiency is the ratio of bp (delivered power) to indicated power ( power provided to the piston)

ηm = bp [kJ/s] = bp ip [kJ/s] bp+fp

fp = ip - bp

Volumetric Efficiency

• Volumetric efficiency is the ratio actual volume flow rate of air into the intake system to the rate at which the volume is displaced by the system

. ηv = ma / ρa

V disp N/2ρa = Inlet density

SI = 80 to 85 %CI = 85 to 90 %

Volumetric Efficiency

• Alternative equivalent definition of Volumetric

• = is atmospheric air density • = the pumping performance of the entire

inlet system• And if it is taken as the air density in the

manifold then is the pumping performance of the inlet port and valve only

Relative Efficiency or Efficiency Ratio

• Relative efficiency is the ratio thermal efficiency of actual cycle to ideal cycle

. ηrel = Actual Thermal Efficiency

Air Standard Efficiency

Mean Effective Pressure

• ip = Indicated power [kW]pim = Indicated mean effective pressure [N/m2 ]L = Length of the stock [m]A = Area of the piston [m2]N = Speed in revolutions per minute [rpm]n = Number of stock N/2 & NK = Number of cylinder•

Mean Effective Pressure• indicated mean effective pressure is•• Break mean effective pressure is

ip = Indicated power [kW]pim = Indicated mean effective pressure [N/m2 ]L = Length of the stock [m]A = Area of the piston [m2]N = Speed in revolutions per minute [rpm]n = Number of stock N/2 & NK = Number of cylinder

Mean Piston Speed

• Sp = 2LN

L = Length of the stock [m]N = Crank Speed in revolutions per minute

[rpm]

Specific Power Output

• Ps = bp/A = constant x pbm x Sp

Specific Fuel Consumption

• • sfc = Fuel Consumption per unit Time

Power

Inlet – Valve Mach Index (Z)

• • u = Gas velocity through the inlet valve• Ap = piston area

• Ai = nominal intake valve operating area

• Ci = inlet valve flow coefficient

Inlet – Valve Mach Index (Z)

• Where• b = cylinder diameter• Di = inlet valve diameter• V = mean piston speed• 𝛼 = inlet sonic velocity• C = inlet valve average flow coefficient• Z = inlet valve Mach index

Fuel – Air ( F/A) or Air –Fuel Ratio (A/F)

• Where• = 1 means stoichiometric ( chemically correct)

mixture• ˂ 1 means lean mixture• ˃ 1 means rich mixture

Calorific Value ( CV)

• Calorific value of a is the thermal energy released per unit quantity of the fuel when the fuel is burning completely and the products of combustion are cooled back to the initial temperature of the combustible mixture.

PROBLEMS

PROBLEM NO 1

• The Mechanical effencicy of a single cylinder four stroke engine is 80 %. The frictional power is esitmated to be 25 kW. Calculate:-

Ipbp

SOLUTION

• ηm = bp/ip

• fp = ip - bp

PROBLEM NO 2

• A 42.5 kW has a mechanical efficiency of 85%. Find the ip and fp. If the frictional power is assumed to be constant with load, What will be mechanical efficiency at 60 % of the load?

SOLUTION

• ηm = bp / ip

• fp = ip – bp

• ηm = bp / (bp + fp)