Report

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/2 ρa = Inlet density SI = 80 to 85 % CI = 85 to 90 % Volumetric Efficiency • Alternative equivalent definition of Volumetric • = ρ • ρ = 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 pim L A N n K 60 ×1000 = Indicated power [kW] = Indicated mean effective pressure [N/m2 ] = Length of the stock [m] = Area of the piston [m2] = Speed in revolutions per minute [rpm] = Number of stock N/2 & N = Number of cylinder • = 6000 Mean Effective Pressure • indicated mean effective pressure is 6000 • = • Break mean effective pressure is 6000 • = 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 & N K = Number of cylinder Mean Piston Speed • Sp = 2LN L N = Length of the stock [m] = 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) 2 ( ) • • • • • • • • Where b = cylinder diameter Di = inlet valve diameter V = mean piston speed = inlet sonic velocity C = inlet valve average flow coefficient Z = inlet valve Mach index = = = Fuel – Air ( F/A) or Air –Fuel Ratio (A/F) • ϕ= − ℎ − • Where • ϕ = 1 means stoichiometric ( chemically correct) mixture • ϕ ˂ 1 means lean mixture • ϕ ˃ 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:Ip bp 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)