Post on 04-Apr-2018
7/31/2019 Thermal Engineering2
1/25
THERMAL ENGINEERING -2 PROBLEMS
1.Write the complete combustion equation for methane?
Ans:
CH4+2O22H2O+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
7/31/2019 Thermal Engineering2
2/25
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:
7/31/2019 Thermal Engineering2
3/25
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
7/31/2019 Thermal Engineering2
4/25
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
7/31/2019 Thermal Engineering2
5/25
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.
86N2The 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
7/31/2019 Thermal Engineering2
6/25
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?
7/31/2019 Thermal Engineering2
7/25
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
7/31/2019 Thermal Engineering2
8/25
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 gives855carbon, 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
7/31/2019 Thermal Engineering2
9/25
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
7/31/2019 Thermal Engineering2
10/25
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/4Therefore, B=S=0. 086m
And, S*2=a so, a=0. 043m
7/31/2019 Thermal Engineering2
11/25
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
7/31/2019 Thermal Engineering2
12/25
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
7/31/2019 Thermal Engineering2
13/25
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.
7/31/2019 Thermal Engineering2
14/25
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
7/31/2019 Thermal Engineering2
15/25
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,
7/31/2019 Thermal Engineering2
16/25
on doing that we get Pib=60.522kW
5]Gasoline 4 cylinder square engine working on 4 strokedevelops 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.
7/31/2019 Thermal Engineering2
17/25
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, weget 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
7/31/2019 Thermal Engineering2
18/25
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= = 0.75cm
7/31/2019 Thermal Engineering2
19/25
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
(air std) = 1 1/(rc(-1))
Cp/Cv = =1.4(air)
RELATIVE EFFICIENCY
(relative)= (brake thermal)/(air std)
NUMERICAL
7/31/2019 Thermal Engineering2
20/25
Q1. A SIX CYLINDER 4-STROKE GASOLINE ENGINE HAVING BORE=90mm,
STROKE=100mm, rC=8, (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*(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
(air std)=1 1/(rc(-1))
(air std)=1 1/80.4
(air std)=0.564
(brake thermal)=BP/HL
(brake thermal)=67593/(CV*0.0564)
0.60=67593/(CV*0.564*0.0564)
C.V.=3.54MJ/kg
HEAT BALANCE QUESTIONS
7/31/2019 Thermal Engineering2
21/25
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=65.8C, TEMP OF EXHAUST GASES=400C, ROOM
TEMP=20.8C. FIND BP, MECH EFFICIENCY, BRAKE SPECIFIC FUEL
CONSUMPTION AND ALSO, DRAW HEAT BALANCE SHEET ON HOUR,
MINUTE, kW AND % BASIS.
A1.
BP=T*
BP=327.4*(2*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
(mech)=BP/IP
(mech)=(59.999/69.999)*100
(mech)=85%
7/31/2019 Thermal Engineering2
22/25
COOLING LOSSES(EC)=mw*CP*T
EC=(16/60)*4.18*103*(65.8-20.8)
EC=50160W
EXHAUST LOSSES(Eg)=me* CP*T
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
kW Hour Minute %
Heat
Supplied
175 630000 10500 100
BP 59.999 215996.4 3599.94 34.28Cooling
Losses50.160 180576 3009.6 28.66
Exhaust
Losses
33.166 119397.6 1989.96 18.95
Unaccounted
Losses
31.674 114026.4 1900.44 18.09
7/31/2019 Thermal Engineering2
23/25
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=25C, AIR FUEL RATIO=32:1. DETERMINE BP,IP, (MECH), INDICATED
THERMAL EFFICIENCY AND DRAW HEAT BALANCE SHEET ON
MINUTE,SECOND AND % BASIS.
A2.
BP=2NT/60
BP=(2*600*120)/60
BP=7539.82W
FP=1500W
IP=BP+FP
IP=7539.82+1500
IP=9039.82W
(INDICATED THERMAL)=IP/(CV*mf)
(INDICATED THERMAL)=(9039.82/(2.5*40.3*106))*100
(INDICATED THERMAL)=32.30%
(mech)=BP/IP
(mech)=(7539.82/9039.82)*100
7/31/2019 Thermal Engineering2
24/25
(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*T
EC=(818/3600)*4.18*103*10
EC=9497.88W
EXHAUST LOSSES(Eg)=me* CP*T
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
Second Minute %
Heat
Supplied
27986.11 1679166.6 100
7/31/2019 Thermal Engineering2
25/25
BP 7539.8 4.52*105 26.94
Cooling
Losses
9497.88 569872.8 33.94
Exhaust
Losses
7728 463680 27.61
Unaccounted
Losses
3220 193200 11.50