Thermal Engineering2

7/31/2019 Thermal Engineering2

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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


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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:


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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


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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


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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


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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?


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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


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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


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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


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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


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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


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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


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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.


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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


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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,


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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.


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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


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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


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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


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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


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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%


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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


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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


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(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


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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

Thermal Engineering2

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