Engine Selection

IntroductionBefore selecting an engine for the SUPRA SAEINDIA 2016 competition, going through the Formula SAE rules it is evident that the restrictions placed on competitors are very precise. The first aim is to gain an understanding of the fundamental principles of internal combustion engines and their design and then relate this understanding back to the Formula SAE competition rules. The short listed engines which will fill the criteria are listed below. Before selecting an optimum engine for the competition, the other factors that affect engine selection were also stated below. The simulation method will be used to select the optimum engine for the competition and the results that the simulation produced.

To obtain a better understanding of the design constraints of the engines some of the most important Formula SAE rules need to be known. The most important rules in regards to engine design include the capacity limit, operating cycle and the type of fuel that is allowed. The engine must not exceed a capacity of 610cc and must utilize a four-stroke operating cycle. The engine must also run on unleaded gasoline, which means that the engine must utilize spark ignition. The introduction of performance boosting agents into the fuel, such as Nitrous Oxide, is also prohibited as no additives are allowed in the fuel (Formula SAE Rules, 2004 pp34-35). Although the rules are strict, they still allow a great degree of freedom in terms of engine design.

Basic factors we consider for an engine1. Application

2. Basic engine design

3. Working cycle

4. Valve or port design and location

5. Fuel

6. Method of mixture preparation

7. Method of ignition

8. Combustion chamber design.

9. Method of cooling

10. Method of load control

Comparing 2 Stroke and 4 Stroke engines:The Formula SAE-A rules specify that the car’s engine must be a four-stroke SI engine. According to the Formula SAE design objectives the car should be aimed at the novice weekend autocross racer. The rules also indicate that the car must have excellent performance characteristics, be inexpensive, reliable, and easily maintained whilst employing common parts. With these factors in mind, two-stroke engines will be analyzed.

With it high lightweight, simple design and high specific power output the two-stroke engine would seem to be the obvious choice for any high performance application. However the two-stroke does not produce double the power of a four-stroke as it theoretically should. Instead the two-stroke produces only about 50% more power. Other drawbacks of two-stroke engines include high fuel consumption, unacceptable polluting emissions and a tendency to be noisy.

Considering the fuel inefficiency and loudness of the two-stroke it still appears a feasible option due to its high performance capabilities. the power curve of a two-stroke engine is quiet irregular. Initially the engine will feel quiet sluggish and then the ‘power-band’ will take effect when the throttle is rolled on. This makes the engine unpredictable and the vehicle hard for the novice to tame. To take full advantage of a two-stroke engine an expert would have to be employed to drive the car.

Engine Selection process: 1. Different types of engines were researched and classified.

2. The parameters that affect the acceleration performance of the vehicle were studied.

3. The list of suitable and available engines made and the relevant data was gathered.

4. Mathematical models of the acceleration performance of the cars were developed.

5. Programs were written in Matlab to perform the above calculations.

6. The vehicle’s performance is obtained by the programs output.

7. Depending upon the financial constraints of the team decision was made.

Engine Option’sThe four-cylinder engines include the:

1. Suzuki GSXR 600;

2. Honda CBR 600;

3. Yamaha YZF r6.

The single cylinder engines include the:

4. RE Bullet 535;

5. KTM Duke 390.

Four cylinder engine specifications:

Manufacturer Suzuki Honda Yamaha

Model GSXR600 CBR600RR YZF600-R6

Operating Cycle 4 -stroke 4 -stroke 4 -stroke

Capacity (cc) 599 599 600

Bore x Stroke (mm) 67 x 42.5 67 x 42.5 65.5x 44.5

Compression Ratio 12.2: 1 12.0: 1 12.0: 1

Cooling System Liquid Liquid Liquid

No. of Cylinders In-line 4 In-line 4 In-line 4

Camshafts DOHC DOHC DOHC

Number of Valves 16 16 16

Four cylinder engine performance:

Model Suzuki Honda Yamaha

Power 124 Bhp @ 13,500rpm 118 Bhp @ 13,500 rpm 122 Bhp @ 14,500 rpm

Torque 69.6Nm @ 11,500 rpm 66 Nm @ 11,210 rpm 65.7 Nm @ 10,500 rpm

Top-speed 255.56 kmph 261.03 kmph 254.27 kmph

Quarter mile 10.87sec @ 126.0mph 10.73sec @ 129.7mph 10.80sec @ 127.8 mph

Comparing Single & Four cylinder engines:Acceleration performance of the car has the largest weighting, in terms of competition points, it is imperative that the engine that is selected has the best acceleration performance in as many events as possible. The suitable single cylinder engines, on average, weigh 30kg less than the four-cylinder engines but have less torque throughout the rev range than the four-cylinder engines. As both of these parameters directly affect the acceleration performance of the vehicle it is not possible to distinguish which engine is the most suitable by only analyzing the engine performance characteristics. In order to determine the engine that would optimize the acceleration performance of the vehicle a mathematical model of the car was developed. To show how the model was created the following section will give a description of the computing equations that were used

Four cylinder engine performance:

Model Bullet Duke

Power 29.1 Bhp @ 5,100rpm 43 Bhp @ 9,500 rpm

Torque 44 Nm @ 4,000 rpm 35 Nm @ 7,250 rpm

Top-speed 145.0 kmph 169.0 kmph

Quarter mile 16.87sec @ 90.0 mph 10.73sec @ 129.7 mph

Single cylinder engine specifications:

Manufacturer Bullet Duke

Model GT 535 KTM 390

Operating Cycle 4 -stroke 4 -stroke

Capacity (cc) 535 373.2

Bore x Stroke (mm) 87 x 90 89 x 60

Compression Ratio 8.5: 1 12.6: 1

Cooling System Air cooled Liquid

No. of Cylinders Single cylinder Single cylinder

Number of Valves 2 4

Design Requirements for an FSAE engine: 1. Acceleration performance

2. Reliability

3. Reproducible

4. Fuel-efficient

5. Cost effective

6. Fuel - Mixture

1. Vehicle Acceleration Performance ModellingGiven that the acceleration performance of the car has the largest weighting, in terms of competition points, it is imperative that the engine that is selected has the best acceleration performance in as many events as possible. The suitable single cylinder engines, on average, weigh 30kg less than the four-cylinder engines but have less torque throughout the rev range than the four-cylinder engines. As both of these parameters

directly affect the acceleration performance of the vehicle it is not possible to distinguish which engine is the most suitable by only analyzing the engine performance characteristics. In order to determine the engine that would optimize the acceleration performance of the vehicle a mathematical model of the car was developed.

Equation of motion

Aerodynamic drag

Rolling resistances forces

Tractive forces

Time to velocity and Distance to velocity

Program construction

Analyzing the acceleration performance

2. ReliabilityWhen considering the reliability of engines for the Formula SAE car the simplest way to find out if any engine had poor reliability was to look at the history of the engine in racing conditions. that each super-sport engine under consideration has had a successful history in the World Super-sport Series. It can be assumed that results of the competition demonstrate an engine’s reliability under racing conditions. Endurance events are very demanding on engines, as the motorcycles are ridden ‘hard’ for days at a time.

3. ReproducibleSince the Formula SAE competition is heavily performance based, it is important to have access to performance enhancing parts. All of the engines that were considered were designed with racing in mind. For this reason a great deal of aftermarket performance enhancing parts are available. The aftermarket parts that can be purchased for each of these motorcycles include mufflers, fuel injection upgrades, camshafts, pistons etc.

4. Fuel efficientComparing fuel consumption is very difficult given each of the engines were originally in two completely different styles of motorcycles. It is possible to calculate the brake specific fuel consumption of each engine. However, this does not take into account the different masses of the vehicle when the car is fitted with each engine.

5. Cost effectiveIn terms of manufacturing feasibility it was concluded that the four-cylinder super-sport engines and the single cylinder enduro engines would be similar. The reason for this is that both engines are mass-produced so obviously they are both feasible to produce. The costing of the engine and accessories consists of a separate costing for the engine, intake manifold, exhaust manifold, cooling system and mufflers. Given that both engines usually have a single muffler and water-cooling the differences in the cost of these components would be minimal. The cost of the exhaust manifold and intake manifold would also be less on the single cylinder engine as the complexity of these components would less on the single cylinder engine.

6. Fuel MixtureThe type of aspiration, method of fuel mixture, the induction system and exhaust systems are all important factors that play a part in determining the performance of an engine. Brake power is the power measured at the flywheel by an engine dynamometer and is reflective of an engines performance.

Forced air induction is a means of raising the inlet density of the air. The purpose of all forced air induction systems is to provide a pressure boost to the air inducted into the engine. The result of raising the pressure of the incoming air is a proportional increase in the density of the air. There are three general types of forced air induction systems including turbochargers, superchargers and ram-air systems. Ram-air systems do not increase the performance of the engine compared to superchargers and turbochargers and their effectiveness is largely dependent on the vehicle velocity (Haile, J. 2000, p15). Because the speeds that the Formula SAE car travels at are relatively low, ram-air systems will not be considered.

Easily obtainable

Easily tuned

Wide of range parts must be available

Provide the correct fuel/air ratio overall operating conditions

Cost effective

Matlab codes written to check different engines parameters uses mole/mass basis instead of mass of dry air basis

define fuel, compression ratio and initial staten = 8;m = 18; % fuel is C8H18Mf = 114.23; % data for iso-octanephi = 1; % equivalence ratiorc = 12.5; % compression ratioT1 = 350; % KP1 = 101.325; % kPa

xb = 0.08; % burned gas fraction

%% part 1 - compression ...% isentropic process (on a mole basis so we can use universal gas constant)Psi1 = mixprop('Psi',compUU,T1,'mass');fcn = @(T) mixprop('Psi',compUU,T,'mass') - Psi1 + Ruu*log(1/rc);T2 = fzero(fcn,600);P2 = P1*rc*(T2/T1); % get pressure - using ideal gas law

fprintf(' (2) End of compression: %5.1f K %7.2f kPa\n', T2, P2)

%% part 3 - combustion ...% constant volume combustion - same internal energy on mass basis

u2 = mixprop('u',compUU,T2,'mass');

u3 = u2;fcn = @(T) mixprop('u',compBB,T,'mass') - u2;T3 = fzero(fcn,2500);P3 = P2*Rbb/Ruu*T3/T2; % get pressure - using ideal gas law

fprintf(' (3) End of combustion: %5.1f K %7.2f kPa\n', T3, P3)

%% part 4 - expansion ...% isentropic processPsi3 = mixprop('Psi',compBB,T3,'mass');fcn = @(T) mixprop('Psi',compBB,T,'mass') - Psi3 + Rbb*log(rc);T4 = fzero(fcn,1500);P4 = P3/rc*(T4/T3); % get pressure - using ideal gas law

fprintf(' (4) End of expansion: %5.1f K %7.2f kPa\n', T4, P4)

%% part 5 - total work% work (on a per kg mixture basis)W12 = - mixprop('u',compUU,T2,'mass') + mixprop('u',compUU,T1,'mass');W34 = - mixprop('u',compBB,T4,'mass') + mixprop('u',compBB,T3,'mass');work = W12 + W34;

eta = work/Q;

fprintf('\nResults:\n')fprintf(' Compression work: %7.2f kJ/kg\n',W12);fprintf(' Expansion work: %7.2f kJ/kg\n',W34);fprintf(' Net work: %7.2f kJ/kg\n',work);fprintf(' Heat input: %7.2f kg/kg\n',Q);

fprintf(' Efficiency: %7.2f %%\n',eta*100);

% some other parameters12T0 = 298.15; % datumTprop = 1740; % evaluation of mixture properties

Runiv = 8.314; % gas constant

%% MIXTURESfprintf('\n--\n');fprintf('MIXTURE COMPOSITION\n\n');% mixture[compBB,compUU,nBB,nUU,xB,xU,fu] = burnedgas(n,m,phi,Tprop,xb);fprintf('Including residual gas (xb=%0.4f)\n',xb);fprintf(' Unburned: %s\n',compUU);

fprintf(' Burned: %s\n',compBB);

%% HEATING VALUES% sanity check: calculate lower heating value (heywood lists 44.3 MJ/kg)hB = mixprop('h',compBB,T0,'mole');hU = mixprop('h',compUU,T0,'mole');deltaHf0 = hB*nBB - hU*nUU; % per mole of O2 (we don'LHV_fuel = -deltaHf0/fu/Mf;

fprintf('\nLHV = %2.5f MJ/kg fuel\n',LHV_fuel/1000);

% actual heating value of mixture (on a mass basis)[hB,MwUU] = mixprop('h',compBB,T0,'mass');

[hU,MwBB] = mixprop('h',compUU,T0,'mass');

deltaHf0 = hB - hU; % per mole of O2Q = -deltaHf0;

fprintf('Q = %2.5f kJ/kg mixture\n',Q);

% gas constantsRuu = Runiv/MwUU;

Rbb = Runiv/MwBB;

%% DO CYCLE ANALYSISfprintf('\n--\n');fprintf('FUEL/AIR CYCLE (constant volume combustion)\n\n');

fprintf(' (1) Initial state: %5.1f K %7.2f kPa\n', T1, P1)

%% PLOT fuel/air cycle and analyze air-standard cycle for comparison% create axesfiga = figure;subplot(1,2,1), axP = gca; hold on, box on,subplot(1,2,2), axT = gca; hold on, box on

col = get(gca,'colororder');

% create some volume valuesV = linspace(1,1/rc,50);%% compression - fuel/air% solve for temperatures / pressures between points 1 and 2Psi1 = mixprop('Psi',compUU,T1,'mass');T12fu = zeros(size(V)); % create empty solution arrayfor i=1:length(V),

fcn = @(T) mixprop('Psi',compUU,T,'mass') - Psi1 + Ruu*log(V(i));T12fu(i) = fzero(fcn,600);end

P12fu = P1./V.*T12fu/T1;

% plot pressure and temperaturehFu(1,1) = plot(axP,V,P12fu);

hFu(1,2) = plot(axT,V,T12fu);

%% do a standard air cycle in comparison% evaluate gamma valueTair = 600;[cv,MwAir] = mixprop('cv','O2:1,N2:3.776',Tair,'mass');Rair = Runiv/MwAir;

gamma = (cv+Rair)/cv;

% print headerfprintf('\n---\n');fprintf('AIR STANDARD CYCLE (const. volume combustion,gamma=%5.3f)\n\n',gamma);

fprintf(' (1) Initial state: %5.1f K %7.2f kPa\n', T1, P1)

%% compression - standard air% standard isentropic compression with gamma=constT12air = T1*V.^(1-gamma);P12air = P1*V.^-gamma;T2air = T12air(end);P2air = P12air(end);

fprintf(' (2) End of compression: %5.1f K %7.2f kPa\n', T2air, P2air)

% plot pressure and temperaturehAir(1,1) = plot(axP,V,P12air);

hAir(1,2) = plot(axT,V,T12air);

%% combustion - fuel/air% plot pressure and temperaturehFu(2,1) = plot(axP,V(end)*[1 1],[P2 P3]);

hFu(2,2) = plot(axT,V(end)*[1 1],[T2 T3]);

%% combustion - standard air% add heatT3air = T2air + Q/cv;P3air = P2air*T3air/T2air;

fprintf(' (3) End of combustion: %5.1f K %7.2f kPa\n', T3air, P3air)

% plot pressure and temperaturehAir(2,1) = plot(axP,V(end)*[1 1],[P2air P3air]);

hAir(2,2) = plot(axT,V(end)*[1 1],[T2air T3air]);

%% expansion - fuel/air% solve for temperatures / pressures between points 3 and 4Psi3 = mixprop('Psi',compBB,T3,'mass');T34fu = zeros(size(V)); % create empty solution arrayfor i=1:length(V),fcn = @(T) mixprop('Psi',compBB,T,'mass') - Psi3 + Rbb*log(rc*V(endi+1));T34fu(i) = fzero(fcn,1600);end

T4fu = T34fu(end);

P34fu = P3/rc./V(end:-1:1).*T34fu/T3;

P4fu = P34fu(end);

%% adjust line styles etc - P-v / T-v diagrams% add labels (P-v)title(axP,'Pressure vs. Volume')hl(1) = xlabel(axP,'V/V_{max} [-]');hl(2) = ylabel(axP,'P [kPa]');

% add labels (T-v)

title(axT,'Temperature vs. Volume')hl(3) = xlabel(axT,'V/V_{max} [-]');hl(4) = ylabel(axT,'T [K]');hl(5) = legend(axT,[hFu(2),hAir(2),hFu2(2)], ...{'Fuel/Air - thermo data','Air - standard','Fuel/Air - approx.'}, ...'location','northeast');% adjust font & line stylesset(axP,'fontsize',14)set(axT,'fontsize',14)set(hl,'fontsize',12);set([hFu,hAir,hFu2],'linewidth',2)set(hFu,'color',col(1,:));set(hAir,'color',col(2,:),'linestyle','--')set(hFu2,'color',col(3,:),'linestyle','-.')

% plot pressure and temperaturehFu(3,1) = plot(axP,V(end:-1:1),P34fu);hFu(3,2) = plot(axT,V(end:-1:1),T34fu);%% expansion - standard air% standard isentropic expansion with gamma=constT34air = T3air*(rc.*V(end:-1:1)).^(1-gamma);P34air = P3air*(rc.*V(end:-1:1)).^-gamma;T4air = T34air(end);P4air = P34air(end);

fprintf(' (4) End of expansion: %5.1f K %7.2f kPa\n', T4air, P4air)

% plot pressure and temperaturehAir(3,1) = plot(axP,V(end:-1:1),P34air);

hAir(3,2) = plot(axT,V(end:-1:1),T34air);%% close cycles% plot pressure and temperaturehFu(4,1) = plot(axP,V(1)*[1 1],[P4fu P1]);hFu(4,2) = plot(axT,V(1)*[1 1],[T4fu T1]);hAir(4,1) = plot(axP,V(1)*[1 1],[P4air P1]);

hAir(4,2) = plot(axT,V(1)*[1 1],[T4air T1]);

%% analyze work% work (on a per kg mixture basis)W12air = -cv*(T2air-T1);W34air = -cv*(T4air-T3air);workAir = W12air + W34air;etaAir = workAir/Q;fprintf('\nResults:\n')fprintf(' Compression work: %7.2f kJ/kg\n',W12air);fprintf(' Expansion work: %7.2f kJ/kg\n',W34air);fprintf(' Net work: %7.2f kJ/kg\n',workAir);fprintf(' Heat input: %7.2f kg/kg\n',Q);

fprintf(' Efficiency: %7.2f %%\n',etaAir*100);

%% FUEL AIR cycle with adjusted gamma% plot internal energiesT = (300:100:3000)';Tref = 1000;% unburned mixtureTsU = [500,1000];polyU = polyfit(TsU,mixprop('u',compUU,TsU,'mass'),1); % linear fitcvU = polyU(1);gammaU = (cvU+Ruu)/cvU;

uFU = polyval(polyU,Tref); % reference value

%% burned mixtureTsB = [1500,3000];polyB = polyfit(TsB,mixprop('u',compBB,TsB,'mass'),1); % linear fitcvB = polyB(1);

gammaB = (cvB+Rbb)/cvB;uFB = polyval(polyB,Tref); % reference value% show results in a figure (corresponds to Heywood Fig. 4-2)figb = figure;hold on, ax2 = gca; box onh = plot(ax2,T, [mixprop('u',compBB,T,'mass'), polyval(polyB,T), ...mixprop('u',compUU,T,'mass'), polyval(polyU,T)]);

h(5) = plot(Tref*[1 1],get(gca,'ylim'),'k:');

%% adjust line styles etc - u vs. T diagram% add labelstitle(ax2,'Simple Analytic Ideal Gas Model');hll(1) = legend(ax2,h,{'burned - exact','burned - fit', ...'unburned - exact','unburned - fit','T_{ref}'}, 'location','southeast');hll(2) = xlabel(ax2,'Temperature [K]');hll(3) = ylabel(ax2,'Internal Energy [kJ/kg]');% adjust font & line stylesset(ax2,'fontsize',14)set(hll,'fontsize',12);

set(h(1:2),'color',col(1,:))

set(h(3:4),'color',col(2,:))set(h(2:2:end),'linestyle','--')

set(h(1:4),'linewidth',2)

%% run cycle analysisfprintf('\n-----------------------------------------------------------------\n');fprintf('FUEL/AIR CYCLE (const. vol. comb., gammaU=%5.3f,gammaB=%5.3f)\n\n',gammaU,gammaB);

fprintf(' (1) Initial state: %5.1f K %7.2f kPa\n', T1, P1)

%% compression - standard air% standard isentropic compressionT12fu2 = T1*V.^(1-gammaU);

P12fu2 = P1*V.^-gammaU;

T2fu2 = T12fu2(end);P2fu2 = P12fu2(end);fprintf(' (2) End of compression: %5.1f K %7.2f kPa\n', T2fu2, P2fu2)hFu2(1,1) = plot(axP,V,P12fu2);

hFu2(1,2) = plot(axT,V,T12fu2);

%% combustion - standard air% equate internal energies (see Eq. 4.21a/4.22a, in book)T3fu2 = T2fu2*cvU/cvB + (uFU-uFB);P3fu2 = P2fu2*Rbb/Ruu*T3fu2/T2fu2;fprintf(' (3) End of combustion: %5.1f K %7.2f kPa\n', T3fu2, P3fu2)hFu2(2,1) = plot(axP,V(end)*[1 1],[P2fu2 P3fu2]);

hFu2(2,2) = plot(axT,V(end)*[1 1],[T2fu2 T3fu2]);

%% expansion - standard airT34fu2 = T3fu2*(rc.*V(end:-1:1)).^(1-gammaB);P34fu2 = P3fu2*(rc.*V(end:-1:1)).^-gammaB;T4fu2 = T34fu2(end);P4fu2 = P34fu2(end);fprintf(' (4) End of expansion: %5.1f K %7.2f kPa\n', T4fu2, P4fu2)hFu2(3,1) = plot(axP,V(end:-1:1),P34fu2);

hFu2(3,2) = plot(axT,V(end:-1:1),T34fu2);

%% close cycleshFu2(4,1) = plot(axP,V(1)*[1 1],[P4fu2 P1]);hFu2(4,2) = plot(axT,V(1)*[1 1],[T4fu2 T1]);%% analyze work% work (on a per kg mixture basis)W12fu2 = -cvU*(T2fu2-T1);W34fu2 = -cvB*(T4fu2-T3fu2);workFu2 = W12fu2 + W34fu2;

etaFu2 = workFu2/Q;

fprintf('\nResults:\n')fprintf(' Compression work: %7.2f kJ/kg\n',W12fu2);fprintf(' Expansion work: %7.2f kJ/kg\n',W34fu2);fprintf(' Net work: %7.2f kJ/kg\n',workFu2);fprintf(' Heat input: %7.2f kg/kg\n',Q);

fprintf(' Efficiency: %7.2f %%\n',etaFu2*100);