THERMAL ENGINEERING -2 PROBLEMS
1.Write the complete combustion equation for methane? Ans: CH4+2O2 →2H2O+CO2↑ 2.write the stichiometric combustion equation for methane? Ans: normal air contains 79% of nitrogen and 21% of oxygen So for every 21 moles of oxygen there are 79 moles of nitrogen so,1 mole of oxygen contains 79/27 moles of nitrogen i.e; 3.76 moles of nitrogen.The combustion equation is : CH4+2(O2+(3.76N2))→2H2O+CO2+7.52N2 Mass of fuel :12+4(1)=16g Mass of air=2(2(16)+3.76(2(14))) =274.56g So fuel to air ratio:mass or fuel/mass of air=16/274.56=0.0582 3.Calculate the stichiometric fuel –air ratio of iso octane? Ans: 2C8H8+25(O2+3.76N2)→18H2O+16CO2+94N2 Molecular weight of iso octane=2(8(12)+8(1)) =228g
Molecular weight of air=25(2(16)+3.76(2(14))) =3432g Fuel to air ratio: Mass of fuel/mass of air= 228/3432=0.066 4.calculate the equivalence ratio of iso octane with 120% of theoretical air? Ans: we know that from previous problem the fuel to air ratio of a stichiometric equation is 0.066 Mass of fuel in that stichiometric equation is:228g Mass of air is:3432g In this problem it is given that 120% of stichiometric (or) theoretical air is used so in this actual equation mass of air is:1.2*3432g Mass of air in actual equation is=4118g So the actual fuel air ratio is:228/4118=0.0553 So the equivalence ratio is: actual fuel-air ratio/stichiometric (or)theoretical fuel – air ratio =0.0553/0.066 =0.8378 4. calculate the equivalence ratio of a engine running with propane as fuel if:
a)80% of theoretical air is used b)110% of theoretical air is used c)30% of excess air is used d)10% of efficient air is used ans: the stichiometric combustion equation of propane is: C3H8+5(O2+3.76(N2))→3CO2+4H2O+18.8N2 Mass of fuel used is:3(12)+8(1)=44g Mass of air used:5(2(16)+3.76(2(14)))=686.4g Fuel to air rato is:44/686.4=0.0641 Now we see the special cases given: a)80% of theoretical air is used mass of stichiometric air in above equation is :686.4 80% of the above quantity is =0.8*686.4 =549.12g So the fuel to air ratio inyhis equation is :44/549.12=0.08012 So the equivalence ratio is :0.08012/0.0641=1.25 b)110% of theoretical air: mass of air in this case is:1.1*686.4=755.04g
so the fuel to air ratio is:44/755.04=0.05828 so the equivalence ratio is:0.05828/0.0641=0.9092 c)30% of excess air so 30% of excess air means 130% of stichiometric air:1.3*686.4=892.32g so actual fuel air ratio is:44/892.32=0.0493 so the equivalenca ratio is:0.0493/0.0641=0.7692 d)10% of excess air so multiply 686.4g of stichiometric mass of air with 1.1 so we get mass air is:755.04g so the fuel air ratio is:44/755.04=0.0582 so the equivalence ratio is:0.0582/0.0641=0.9091 6.An engine which is operating with propane as fuel has the dry exhaust analysis as follows:4.9% of CO2, 9.79% of CO, 2.45% of O2, by volume basis then calculate the equivalence ratio? Ans: If the given data constitutes a total of 17.14 of the exhaust gas the the remaing contene would be nitrogen which is obtained by subtracting this value from 100 So the value of nitrogen in the exhaust is 100-17.14=82.86% The stichiometric equation can be wriyyen as follows: XC3H8+Y(O2+3.76(N2))→4.9CO2+9.79CO+2.45O2+82.86N2
Now we equate the total no.of atoms of each element on reactant side and product side: First we equate carbon atoms: No. Of carbon atoms on reactant side:3X No.of carbon atoms on product side:9.79+4.9=14.69 Equating the carbon atoms on both sides: 3X=14.69 X=4.89 Similarly if we equate no.of atoms of nitrogen on both sides: No.of nitrogen molecules on reactant side:3.76Y No.of nitrogen molecules on product side:82.86 Equating both: 3.76Y=82.86 Y=22.03 So the resultant equation is: 4.89(C3H8)+22.03(O2+3.76(N2))→4.9CO2+9.79CO+2.45O2+82. 86N2 The mass of air is:22.03(2(16)+(3.76(2(14)))=3024.27g The mass of fuel is:4.89(3(12)+8(1))=215.16g Fuel to air rato is:215.16/3024.27=0.07114
From the previous problem the stichiometric fuel-air ratio is:0.0641 So the equivalent ratio is:0.07114/0.0641=1.1586 7.A given coal sample contains 85% carbon, 6% O2, 6%H2, by weight and the remaining is inexhaustible then determine the minimum amount air required to burn completely 1 Kg of air? Ans: The actual combustion equation for the given data is: 85/12C+6/32O2+6/2H2+Y(O2+3.76N2)→n2CO2+n3H2O+n4N2 Equating atoms on both sides: n2=85/12 n3=6/2=3 Equating oxygen on both sides: 6/32+Y=n2+n3/2 We get Y=8.3925 So the mass of air is=8.3925(2(16)+3.76(2(14)))=1152.1224Kg That is if we take 100 Kg of fuel the amount of oxygen required to completely burn is 1152.122Kg For 1 Kg of fuel the amount of air required is 11.52Kg 8.Find the chemical formula of fuel if when it is completely burnt the wet analysis gives 80%N2 and 3%H2O?
Ans: As per the given data the remaining amount should be carbon di oxide which is:100-(80+3)=17%
Cn1Hn2+Y(O2+3.76N2)→17CO2+3H2O+80N2 When same element atoms are compared on both sides : n1=17 n2=6 So the chemical formula of the fuel is C17H6 9.The following is the complete analysis of a sample of petrol by weight 85% carbon, 15% hydrogen, calculate the equivalence ratio of the mixture if the dry analysis of the fuel Gives the following results : CO2-9.2%, CO-6.44%, O2-1.38% Ans: From the given data excluding the water we get The combustion equation based on the following data; n0(85/12(C)+15(H))+n1(O2+3.76(N2))→9.2CO2+6.44CO+1.38O 2+n2H2O+n3N2 On comparing elements on both sides first carbon: n0(85/12)=9.2+6.44=15.64
n0=2.208 on comparing hydrogen: 15(2.208)=2n2; n2=16.56 on comparing oxygen: 2n1=2(9.2)+6.44+2(1.38)+16.56 2n1=44.16 n1=22.08 mass of air=22.08(2(16)+3.76(28))=3031.1424g mass of fuel:100(2.208)g so the actual fuel air ratio is:220.8/3031.142=0.07284 10.For a sample of hydro carbon fuel ultimate analysis gives 855carbon, 155 hydrogen, by weight calculate the actual fuel-air ratio of the mixture if the dry analysis shows the following results: 8.28% of CO2 , 7.35% of CO, 1.84 %of O2, 82.53% N2 Ans: The combustion equation for the following data is : n0(85/12C+15/1H)+n2(O2+3.76(N2))→8.28CO2+7.35CO+1.84O 2+n3H2O+82.53N2
comparing carbon atoms on both sides we get: 85/12(n0)=8.28+7.35 n0=2.206 comparing hydrogen atoms on both sides: 2.206(15)=2n3 n3=16.545 comparing O2 atoms on both sides: 2n2=2(8.28)+7.35+2(1.84)+16.545 n2=22.06 so mass of air is:22.06(2(16)+3.76(28))=3028.39g mass of fuel:2.206(100)=220.6g so fuel to air ratio is:220.6/3028.39=0.07284
1]A 4-stroke v6 engine of 3L capacity operates at 1500 rpm. If the compression ratio is 9. Calculate for square engine (a) Bore (b) Stroke (c) Vc and Vs (d) Average Piston speed (f) Crank offset (g) Volume at BDC Answer: N=1500rpm vs=3/6 l =0. 5*10^-3 m^3 C. r=vs+vc/vc 9=0.005+vc/vc Therefore, vc=6. 25*10^-5 m^3 As it is square cylinder, B/S=1, So, B=S THEREFORE, vs=3. 14*B*B*S/4 Therefore, B=S=0. 086m And, S*2=a so, a=0. 043m
2]A 4 stroke V6 engine of 4L capacity. It operates at a speed of 1500 rpm. If its compression ratio is 6. Find the following: (A) Bore (B) Stroke (C) Swept volume (D) Volume at BDC (E) Average piston speed (F) Clearence volume (G) Crank offset Answer: Given, engine is ovr-sqrd, and B/S=1.6 we knw, Vs=3.14*(B^2)*S/4 on sub 1st in 2nd eqn, we get Vs=3.14((1.6S)^2)*S so,Vs=2.01(S^3) and givn Vs=4/8 l=0.0005m^3 so eqtn both we get S=62.9mm so,B=1.6S =>B=100.6mm givn, c.r=6 we knw c.r=Vs+Vc/Vc=6
subs in c.r formula, we get Vc=0.0001m^3 to find avg piston speed,Vp=2SN/60 so, Vp=2*0.0629*1500/60 => Vp=3.145m/s crank offset a=S/2=62.9/2=31.45mm volume at bdc, Vs=Vbdc-vtdc we knw Vs=0.0005m^3 and Vtdc=0.0001m^3 so, Vbdc=0.0006m^3
3]For a 4 stroke petrol square engine whose crank offset a= 40mm, Vc =10% of Vs, N= 1200 rpm. Find : S Vp C.R. Distance between crank axis and pin axis if L=120mm. Number of sparks to be produced in 1 second. Angle is 20. as we knw, S=2a
S=2*40=80mm givn, Vc=10%Vs c.r=Vs+Vc/Vc so, c.r=0.1Vs+Vs/0.1Vs => c.r=11 since it is sq. cylinder,B=S =>B=80mm Vs=3.14(B^2)S/4 since, B=S so, Vs=3.14*(S^3) so, Vs=3.14(0.08^3)=4.02*10^-4 m^3 Vp=2SN/60=2*0.08*1200/60=3.2m/s r=((c^2)-(a^2)+(2*r*a*cosp))^(1/2) here, givn ang.p=20 deg and c=0.12m so on subs in formula we get, r=147.7mm for 2 strokes, 1 spark is produced, so, for 1200 rpm, 600 sparks are produced for 1 min 600 sparks are produced so for 1 sec 10 sparks are produced.
4]A single 4 stroke S.I. Engine operates with an engine on Cv= 40MJ/kg. It consumes 0.01 kg of fuel in one minute. If it runs with 1800 rpm develops 8Nm torque, mechanical efficiency is 90% and Nvol. =92%. Ma/Mf=11. Find: (a) B.P. (b) I.P. (c) Nbth, Niso (d) Vs (e) Break specific fuel consumption (f) Indicated and brake mean effective power. Answer: CV=40MJ/kg m=0.01kg N=1800rpm T=8Nm mech eff.=0.9, vol eff.=0.92, ma/mf=11.1 mf*CV=(0.01*40000*1000)=4*10^5
Brk pwr= BP=Tw= T*2*3.14*N/60= 8*2*3.14*1800/60= 1507.96W= 1.51kW Indctd pwr= IP= BP/mech eff= 1510/.9 = 1.675kW Frictnl powr= FP= IP-BP= 167.56W brk thrml eff= BP/heat lbrtd= 1507.2*60/(4*10^5)= 22.6% indctd thrml eff =1675*60/(4*10^5)= 25.13% vol eff=(mass of actual air entering)/(mass of theoretical air entering) as we know, ma/mf=11 so, ma=11*0.4/60=0.00183kg so, 0.92=0.00183/(Vs*Pa) = 0.00183/(Vs*1.2) so, Vs=0.00166m^3 Indicatd sp fuel consmtn=mf/IP= Break sp fuel consmtn=mf/BP=0.01/(60*1507.96)=1.1*(10^-4) kg/kWhr Indictd mean eff pr Pim=(IP*60)/(Vs*(N/2)*k) here k=no of cylinders=1 so on sub these values in above formula we get Pim=67.269kW simlrly for Break mean effective pr we sub BP instead of IP in formula,
on doing that we get Pib=60.522kW
5]Gasoline 4 cylinder square engine working on 4 stroke develops a B.P. of 2.2 kW. A Morse test was conducted on this engine & B.P. obtained when each cylinder was made inoperative by short circuiting the spark plug are 14.9 kW, 14.3 kW, 14.8 kW and 14.5 kW resp. The test was conducted at constant speed. Find: (a) Ip (b) Nmech (c) Brake mean effective power (D) Indicative mean effective power when all cylinders are firing. Take bore= 80mm speed= 3000rpm. we knw, IP= IP(1)+IP(2)+IP(3)+IP(4) IP(1)=BP(4)-BP(3)=22-14.9=7.1 IIIly, IP(2)=7.7kw, IP(3)=7.2kW, IP(4)=7.5kW so, IP=7.1+7.7+7.2+7.5= 29.5Kw mech eff= BP/IP=0.746= 74.6% Bore=Stroke as it is a sq engine giv.
so B=S=0.08m as we knw,V=3.14*(B^2)*s= 3.14*(0.08^3)= 4.021*(10^-4) m^3 Pbm=BP/(k*V*0.5*(N/60)) here BP=22000W, N=3000rpm on sub foll values in formula, we get Pbm=5.47bar IIIly to get Pim, we substitue BP instead of IP in same formula, on doing so we get, Pim=7.34bar 6) A cylinder engine running at 1250 rpm delivers 21 kW. The average torqe when one cylinder is cut off 110Nm. Cv of fuel is 43 MJ/kg. The engine uses 360g of fuelfor delivering 1kWh power. Find indictd thrml effncy: (a) B.P.= 21kW…… N=1250 rpm, T(when cylinder is cut) = 110Nm, CV=43MJ/kg= 43000kJ/kg BP of 3 cylinders = T of 3 cylinders * W T*2*3.14*N/60= 110*2*3.14*1250/60= 14399W IP(1 cylinder) = BP(4 cylinders)-BP(3 cylinders)= 21000-14399= 6601W So, IP(4 cylinders)= 4* IP(1 cylinder)= 4*6601= 26.404kW We knw, brk sp fuel consmtn=mf/BP So,(360*10^-3)/(3600*1000)= mf/21000 so, mf=0.0021kg/S
As we knw, indctd thrml eff= IP/(mf*CV)= 26.404/(0.0021*43000)=0.2924 Indctd thrml eff=29.24%
7)A single cylinder 2 stroke I.C. engine when tested following observations are available: Area of indicated diagram = 3cm2 Length of indicated diagram= 4cm Brake drum diameter= 120 cm Dead weight on brake= 380N Spring balance reading= 50N Fuel consumption= 2.8 kg/h Calorific value of fuel= 42000 kJ/kg Cylinder diameter= 16 cm Piston stoke =20 cm Calculate frictional power, mechanical efficiency , specific fuel consumption efficiency. Mean efftv pr= ¾= 0.75cm
So, in bars mean efftv pr= 0.75*spring const= 0.75*10= 7.5 bars= 750000W Pim=(IP*60)/(V*N) V=3.14*(B^2)*S/4= 3.14*(0.2^2)*(0.16)/4= 0.0040212 m^3 750000= (IP*60)/(0.0040212*400)= 20.1kW T=(Wd-Ws)*D/2= (380-50)*1.2/2= 198Nm BP=T*2*3.14*N/60 = 198*2*3.14*400/60= 8.293kW Frictnl pwr= IP-BP=20.1-8.293= 11.807kW Mech eff= BP/IP= 8.293/20.1= 0.4126= 41.26% Brk spcfc fuel consm=mf/BP =2.8/20.1= 0.139kg/kWh Indctd thrml eff=IP/(mf*CV)= (20.1*1000)/(2.8*42000)= 0.1709= 17.09%
AIR STANDARD EFFICIENCY ƞ(air std) = 1 – 1/(rc(ɣ-1)) Cp/Cv = ɣ =1.4(air) RELATIVE EFFICIENCY
ƞ(relative) = ƞ(brake thermal)/ƞ(air std) NUMERICAL
Q1. A SIX CYLINDER 4-STROKE GASOLINE ENGINE HAVING BORE=90mm, STROKE=100mm, rC=8, ƞ(RELATIVE)=60%,WHEN BRAKE SPECIFIC FUEL CONSUMPTION(BSFC)=3009g/kWh. ESTIMATE CALORIFIC VALUE(C.V.) OF FUEL, FUEL CONSUMPTION IF, PBM=8.5bar, N=2500r.p.m. A1.
PBM= (60*BP)/(VS*6*N/2) (8.5*105*Π(0.1)(0.090)2*2500*6)/(4*2*60)=BP B.P.=67.593Kw
BSFC=mf/BP 3.009*67.593=mf mf=0.0564kg/s
ƞ(air std)=1 – 1/(rc(ɣ-1)) ƞ(air std)=1 – 1/80.4 ƞ(air std)=0.564
ƞ(brake thermal)=BP/HL ƞ(brake thermal)=67593/(CV*0.0564) 0.60=67593/(CV*0.564*0.0564) C.V.=3.54MJ/kg
HEAT BALANCE QUESTIONS
Q1. FOLLOWING OBSERVATIONS WERE RECORDED DURING A TRIAL ON A FOUR STROKE DIESEL ENGINE. POWER ABSORBED BY NON-FIRING ENGINE WHEN DRIVEN BY ELECTRIC MOTOR IS 10kW. SPEED=1750rpm, T=327.4Nm, mf=15kg/h, CV=42MJ/kg. AIR SUPPLIED =4.75kg/min. COOLING WATER SUPPLIED=16kg/min. OUTLET TEMP OF COOLING WATER=65.8°C, TEMP OF EXHAUST GASES=400°C, ROOM TEMP=20.8°C. FIND BP, MECH EFFICIENCY, BRAKE SPECIFIC FUEL CONSUMPTION AND ALSO, DRAW HEAT BALANCE SHEET ON HOUR, MINUTE, kW AND % BASIS. A1. BP=T*ɯ BP=327.4*(2Π*1750/60) BP=59.999kW HEAT LIBERATED(HL)=mf*CV HL=(15/3600)*42*103 HL=175kJ/s BSFC=mf/BP BSFC=15/59.999 BSFC=0.250kg/kWh INDICATED POWER(IP)=BP+FP IP=59.999+10000 IP=69.999kW Ƞ(mech)=BP/IP Ƞ(mech)=(59.999/69.999)*100 Ƞ(mech)=85%
COOLING LOSSES(EC)=mw*CP*ΔT EC=(16/60)*4.18*103*(65.8-20.8) EC=50160W EXHAUST LOSSES(Eg)=me* CP*ΔT Eg=me*1.05*103*(400-20.8) me=ma+mf me=(4.75/60)+(15/3600) me=0.0833kg/s Eg=0.0833*1.05*103*(400-20.8) Eg=33166.728W UNACCOUNTED LOSSES=Qs-(BP+EC+Eg) UNACCOUNTED LOSSES=175*10-3-(59.999*103+50160+33166.72) UNACCOUNTED LOSSES=31674.28W
HEAT BALANCE SHEET Type of Energy Heat Supplied BP Cooling Losses Exhaust Losses Unaccounted Losses
kW
Hour
Minute
%
175
630000
10500
100
59.999 50.160
215996.4 180576
3599.94 3009.6
34.28 28.66
33.166
119397.6
1989.96
18.95
31.674
114026.4
1900.44
18.09
Q2. A 2 STROKE ENGINE WAS MOTORED WHEN METER READING WAS 1.5kW. THEN, THE TEST ON THE ENGINE WAS CARRIED OUT FOR 1 HOUR AND THE FOLLOWING OBSERVATIONS WERE RECORDED. BRAKE TORQUE=120Nm, SPEED=600rpm, FUEL USED=2.5kg/h, CV=40.3MJ/kg, COOLING WATER USED=818kg/h. RISE IN TEMP OF COOLING WATER=25°C, AIR FUEL RATIO=32:1. DETERMINE BP,IP, ƞ(MECH), INDICATED THERMAL EFFICIENCY AND DRAW HEAT BALANCE SHEET ON MINUTE,SECOND AND % BASIS. A2. BP=2ПNT/60 BP=(2П*600*120)/60 BP=7539.82W FP=1500W IP=BP+FP IP=7539.82+1500 IP=9039.82W Ƞ(INDICATED THERMAL)=IP/(CV*mf) Ƞ(INDICATED THERMAL)=(9039.82/(2.5*40.3*106))*100 Ƞ(INDICATED THERMAL)=32.30% Ƞ(mech)=BP/IP Ƞ(mech)=(7539.82/9039.82)*100
Ƞ(mech)=83% TOTAL HEAT SUPPLIED=mf*CV TOTAL HEAT SUPPLIED=(2.5/3600)*40.3*106 TOTAL HEAT SUPPLIED=27986.11W COOLING LOSSES(EC)=mw*CP*ΔT EC=(818/3600)*4.18*103*10 EC=9497.88W EXHAUST LOSSES(Eg)=me* CP*ΔT Eg=me*1.05*103*(345-25) ma/mf=32 ma=2.5*32 ma=80kg/h ma=0.022kg/s me=ma+mf me=0.023kg/s Eg=0.023*1.05*103*(400-20.8) Eg=7728W UNACCOUNTED LOSSES=Qs-(BP+EC+Eg) UNACCOUNTED LOSSES=3220W HEAT BALANCE SHEET Type of Energy Heat Supplied
Second
Minute
%
27986.11
1679166.6
100
BP Cooling Losses Exhaust Losses Unaccounted Losses
7539.8 9497.88
4.52*105 569872.8
26.94 33.94
7728
463680
27.61
3220
193200
11.50
