37186719 thermal engineering unit i for be students

B.E-MECHANICAL ENGINEERING-IV SEMESTER

ME1251-Thermal Engineering-

UNIT – I- GAS POWER CYCLES

Syllabus

Otto, Diesel, Dual, Brayton cycles, Calculation of mean effective pressure and air standard

efficiency, Actual and theoretical PV diagram of Four stroke engines, Actual and

theoretical PV diagram of two stroke engines.

PART-A

1. Define Air standard efficiency of a Gas power cycle, and relate the air standard efficiency with actual thermal efficiency.

Air standard efficiency may be defined as “ the ratio of work done to the heat supplied

during a thermodynamic cycle when air is employed as the working substance ”.

Wok done by the system per cycleAir standard efficiency ( ή Air std ) = ------------------------------------------------

Heat supplied into the system per cycle

The actual efficiency of a cycle is always less than the air standard efficiency of that cycle

under ideal conditions. This is due to the practical losses such as friction, heat losses,

variation of specific heats of working substance and nature of working substance.

We can relate actual thermal efficiency with air standard efficiency by introducing a new

term, Relative efficiency.

Relative efficiency (ή Relative ) = Actual thermal efficiency Air standard efficiency

2. State the assumptions made for the analysis of the air standard cycles?

The assumptions made for the analysis air standard cycles are,

(i) Air is employed as the working fluid for the whole cycle and air is assumed to be a perfect gas obeying all gas laws.

(ii) The value of specific heats ( Cp and Cv ) of working fluid remains constant throughout the cycle.

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(iii) The compression and expansion processes mare reversible adiabatic(isentropic).

(iv) The addition of heat and rejection of heat is carried out by making hot body and cold body contacts respectively with the engine cylinder head. Effect of combustion of fuel is neglected.

(v) The cycle is considered as closed one and the air used in the first cycle is used again and again.

(vi) At moderate temperatures, the air has the following properties,Molecular weight = 29Specific heat at constant pressure Cp = 1.005 kJ/kg/KSpecific heat at constant pressure CV = 0.718 kJ/kg/K

3. Define Mean Effective Pressure of a cyclic heat engine.

Mean effective pressure of a cyclic heat engine may be defined as “ the average

constant pressure acting on the piston during the whole cycle which will be able to do the

same amount of work as done by the actual varying pressure during the cycle”

It may also be defined as “ the ratio of work done per cycle to the piston displacement volume or stroke volume per cycle”

Workdone per cycleMean Effective Pressure = -------------------------

Stroke volume

In actual cycle of an engine, the pressure inside the cylinder keeps on changing

with the position of the piston. The cyclic average of this varying pressure acting on the

piston is termed as “ Mean Effective Pressure”

There are two mean effective pressures,

(i) Brake mean effective pressure

(ii) Indicated mean effective pressure

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4. Draw the P-V and T-S diagram for Otto cycle and diesel cycle?

p-V and T-s diagram for Otto cycle or constant volume cycle or Petrol engines

1.2 - --Isentropic compression 2-3 --- Constant Volume Heating3.4 ----Isentropic Expansion 4-1 --- Constant Volume Cooling

p-V and T-s diagram for Diesel cycle or constant Pressure cycle

1.2 --- Isentropic compression 2-3 --- Constant pressure Heating

3.4 --- Isentropic expansion 4-1 --- Constant volume cooling

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5. Draw the p-V and T-s diagram for Dual cycle?

p-V and T-s diagram for Duel cycle or Mixed cycle or Limited pressure cycle

1.2 --- Isentropic Compression 2-3 --- Constant Volume Heating

3.4 --- Constant Pressure Heating4-5 --- Isentropic Expansion

5.1 --- Constant Volume Cooling

6. Draw the layout of a closed cycle Gas turbine power plant and Draw the p-V and T-s

diagram for Brayton cycle?

Layout of a closed cycle Gas turbine power plant

p-V and T-s diagram for Brayton cycle or Joule’s cycle

-2 --- Isentropic compression 2-3 --- Constant pressure heat addition

3-4 --- Isentropic Expansion 4-1 --- Constant pressure heat rejection

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7. Define the terms Stroke volume of an I.C engine

Stroke volume (Vs)

The volume of working fluid swept inside the engine cylinder when the piston

moves from Top Dead Centre (TDC) and Bottom Dead Centre (BDC) is callede as stroke

volume or swept volume or piston displacement volume (Vs).

Stroke volume (Vs) = Cross sectional area of the piston x Stroke length

Stroke volume (Vs) = slxd 2

4

π

Where,

d ---Diameter of the piston or cylinder bore in ‘m’

ls --- Stroke length of the piston in ‘m’

Stroke volume (Vs) = ( Total cylinder volume --- Clearance volume)

Stroke volume )( 211 VVVVV cs −=−=

8. Define (i) Clearance volume (ii) Full volume or Total volume of an IC engine

Clearance volume (Vc)

Clearance volume is the volume occupied by the working fluid when piston is at top dead

centre (TDC). Practically clearance volume is provided for an engine to

(i) accommodate valves and their operations

(ii) avoiding hitting of piston with the cylinder head

(iii) to assist in effective suction

Clearance volume (Vc) = ( Total cylinder volume --- Stroke volume)

Clearance volume )( 12 sc VVVV −==

Full volume or Total volume(V1)

Full volume is the volume occupied by the fluid when piston is at bottom

dead centre.

Full volume sVVV += 21

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9. Define ‘Compression ratio’.

It is the ratio of the initial volume to the final volume during compression. It may

also be defined as “The ratio of total cylinder volume (V1 = Vs + Vc) to the clearance

volume (Vc).

Compression ratio (rc) = 2

1

V

V

V

VV

c

sc =+

(rc) = c

s

V

V+1

Compression ratio (rc) = Expansion ratio (re) -----for Otto cycle

Compression ratio (rc) = Expansion ratio (re) x Cut off ratio (ρ) ---for Diesel cycle

Compression ratio (rc) = Expansion ratio (re) x Cut off ratio (ρ) ---for Duel cycle

10. What is ‘Cut-off Ratio’?

Cut off ratio is defined as “The ratio of volume after the heat addition to before the

heat addition in a thermodynamic gas power cycle”. It is denoted by the letter ‘ρ’ . Also it

may be defined the ratio of the final volume to the initial volume during constant

pressure combustion.

Cut off ratio (ρ) = 2

3

V

V for Diesel cycle

Cut off ratio (ρ) = 3

4

V

V for Duel cycle

• In Otto cycle heat is added during constant volume process therefore

cut off ratio for Otto cycle is 1.

• In diesel cycle heat is added during constant pressure processes ρ >1

• In dual cycle heat is added during both constant volume and constant

Pressure process ρ >1.

11. Define ‘Pressure Ratio’ in Brayton Cycle. State its significance in a gas turbine

power plant.

Pressure Ratio

It is defined as the ratio of the maximum pressure to the minimum pressure in the

cycle. It is denoted as “rp” .

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The efficiency of a Brayton cycle is given by ή = ( ) νν 1

11 −−

pr . If we increase the c

value (by increasing maximum pressure) the efficiency will increase. But the work

required for the compressor also will increase. The “rp” value for the maximum net work

output is derived as (rp)optimum work = )1(2

max

min−

νν

T

T . The minimum temperature involved is

the atmospheric temperature and maximum temperature is limited by the materials of the

gas turbine equipments. Hence we can not increase the Tmax and “rp” beyond certain limit.

Hence by inventing suitable very high temperature material we can increase the rp value

and thus the gas turbine power plant efficiency.

12. Define actual IC engine cycle. Why actual work done is always less than the

theoretical work done of an I.C Engine?

Actual cycle:

The cycle using air-fuel mixture and combustion product instead of only air as

working fluid and consisting irreversible compression and irreversible expansion, heat

losses and rapid heating and cooling is called actual IC engine cycle.

Actual work done is always less than the theoretical work done because of the

following reasons,

(i) Losses due to friction . In theoretical cycle compression and expansion

process are considered as reversible processes. But in the actual case

compression and expansion process are irreversible process.

(ii) Pumping losses

(iii) Chemical equilibrium losses

(iv) Losses due to variation of specific heats with temperature

(v) Direct heat losses because perfect insulation is not possible Therefore, that

actual work done is always less than the theoretical work done of an I.C

Engine.

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13. What is the condition for maximum work of an otto cycle?

1) Temperature at the end of compression 312 TTT =

2) Compression ratio =( ))12

1

2

3−

=

χ

T

Tr

Where,

T1-Temperature at the initial condition of the compression process in ‘K’

T3-Temperature at the final condition of the combustion process in ‘K’

14 Which cycle will be efficient for the same compression ratio and heat input?

All the cycles have same heat rejection and same compression ration. For the same

compression ratio the maximum pressure attained reduced in duel and diesel cycle than

otto cycle. We know that tconsQQ

Qs

S

R tan1 =−=η

If QR decreases, Efficiency (η) increases and vice versa.

Since all the cycles reject their heat at the same specific volume, On T-s diagram, area

below the process line from state 4 to 1 , 4’-1 and 4” -1 gives the heat rejected in Diesel,

dual and Otto cycle. Heat rejected otto < Heat rejected dual < Heat rejected diesel

There fore dieseldualotto ηηη >>

15. Which cycle will be efficient for the maximum pressure, and same heat supplied?

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To reach same maximum pressure the compression ratio is to be reduced in Dual and Otto

cycles than diesel cycle.

We know that tconsQQ

Qs

S

R tan1 =−=η

If QR decreases, Efficiency (η) increases and vice versa.

On T-s diagram, Area below the curve 4-1 < 4’-1 < 4”-1

Heat rejected diesel < Heat rejected dual < Heat rejected otto

Therefore ottodualdiesel ηηη >> at maximum pressure and same heat supplied.

16. Which cycle will be efficient for the maximum pressure and temperature?

To reach same maximum pressure and temperature in all cycles , the compression ratio are to be

reduced in dual and otto cycle than diesel cycle. No change in heat rejection in all cycles. So that

efficiency is proportional to the amount of heat supplied (QS).

We know that tconsQSinceQ

tCons

Q

QR

SS

R tantan

11 =−=−=η

The area below 2” -3 > 2’-3’-3 > 2-3

Heat supplied diesel > Heat supplied dual > Heat supplied otto

Therefore Ottodualdiesel ηηη >>

17. Compare Otto cycle with Diesel cycleS.No Otto cycle Diesel cycle

1 Otto cycle is used in petrol engines Diesel cycle is used in diesel engines.

2Otto cycle contains two isentropic processes two constant volume processes.

Diesel cycle contains two isentropic processes, one constant pressure one constant volume pressure.

3 Heat is added at constant volume. Heat is added at constant procure.

4Compression ratio and expansion ratio are same.

Compression ratio and expansion ratio are not same.

5 Air standard efficiency of the cycle depends on compression ratio.

Air standard efficiency of the cycle depends compression ratio and cut off

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ratio.6 Heat rejected is less Heat rejected is more

7Efficiency is more than diesel cycle for the same compression ratio

Efficiency is less than otto cycle for the same compression ratio

18.Compare Diesel cycle with Duel cycleS.No Diesel cycle Duel cycle

1 Diesel cycle is used in diesel engines. Duel cycle is used in crude oil engines.

2Diesel cycle contains two isentropic processes, one constant pressure and one constant volume pressure.

Duel cycle contains two isentropic processes, two constant volume and one constant pressure process

3Heat is added at constant pressure process.

Partially heat is added at constant volume process and remaining is in constant pressure process

4Compression ratio and expansion ratio are not equal.

Compression ratio and expansion ratio are not equal.

5Air standard efficiency of the cycle depends compression ratio and cut off ratio.

Air standard efficiency of the cycle depends compression ratio, cut off ratio and pressure ratio.

6 Heat rejected is more Heat rejected is less

7Efficiency is less than duel cycle for the same compression ratio

Efficiency is more than duel cycle for the same compression ratio

19. Draw the Theoretical and Actual P-v diagram for a four stroke Otto engine.

20. Draw the Theoretical and Actual P-v diagram for a four stroke engine

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2

3

4

5 1

3

4

5

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20. Draw the Theoretical and Actual P-v diagram for a four stroke diesel engine.

5-1 ----Suction 1-2 ----Compression

2-3 ---- Constant pressure combustion 3-4 ---- Expansion

4-1 ---- Sudden fall in pressure 1-5 ---- Exhaust

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PART - B

1. Derive the efficiency for OTTO cycle

OTTO CYCLE PETROL AND GAS ENGINES – CONSTANT VOLUME CYCLE

1.3 Isentropic compression 2-3 Constant Volume Heating 3.5 Isentropic Expansion4.1 Constant Volume Cooling

Since combustion takes place at constant volume the Otto cycle is also called as Constant Volume Cycle.

Heat Supplied = Cv ( T3 – T2) ------- (1)

Heat Rejected = Cv ( T4 – T1) ------- (2)

. . . Work done = Heat Supplied – Heat Rejected

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= Cv ( T3 – T2) - Cv ( T4 – T1)

ηcycle = workdone

Heat Supplied = p 3 2 v 4 1

p 3 2

C ( T - T ) - C ( T - T )

C ( T - T )

ηcycle = 4 1

3 2

1T T

T T

−− −

41

1

32

2

1

1

1

TT

T

TT

T

−

= − −

------- (3)

34 4 2

1 3 2 1

TT T T

T T T T= × ×

[ ]

1 1

3 34 1

1 4 2 2

343 2 4 1

1 2

&

V TT V

T V T V

TTV V V V

T T

γ γ− −

= × ×

= = =Q

Substituting 34

1 2

TT

T T= in equation (3)

1

2

1cy

T

Tη = −

2

1

11

TT

= −

cyη ( ) 1

11

rγ −

= −

1

12 1

1 2

T Vr

T V

γγ

−−

= =

Q

where1

2

Vr compression ratio

V= =

Compression ratio ( r) : It is the ratio of the initial volume to the final volume during compression.

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2. Derive the efficiency for DIESEL cycle.

DIESEL CYCLE

DIESEL ENGINES – CONSTANT PRESSURE CYCLE :

1.3 Isentropic compression2.3 Constant pressure Heating3.5 Isentropic expansion4.1 Constant volume Cooling

Since combustion takes place at constant pressure the Diesel cycle is also called as Constant Pressure Cycle.

Heat Supplied = Cp ( T3 – T2) -------(1) Heat Rejected = Cv ( T4 – T1) -------(2)

. . . Work done = Heat Supplied – Heat Rejected = Cp ( T3 – T2) - Cv ( T4 – T1)

ηcycle =workdone

Heat Supplied =p 3 2 v 4 1

p 3 2

C ( T - T ) - C ( T - T )

C ( T - T )

4 1

3 2

4 1

3 2

( )1

( )

( )1

( )

vcy

p

cy

C T T

C T T

T T

T T

η

ηγ

−= −−

−= − −

p

v

Cratio of specific Heats

Cγ= =Q

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4 41

1 1

13 32

2 2

1 1

1 1

1 1cy

T TT

T T

T TT r

T Tγ

ηγ γ −

− −

= − = −

− −

--------(3)

where 1

1

2 1

1 2

T Vr

T V

γγ

−−

= =

34 4 2

1 3 2 1

TT T T

T T T T= × ×

[ ]

1 1

3 3 1

4 2 2

1

3 34 1

2 2

V T V

V T V

V TV V

V T

γ γ

γ

− −

−

= × ×

= × =

Q

1

3 3

2 2

V V

V V

γ −

= ×

3 3

2 2

, 2-3 constant pressure processT V

T V

=

Q

3

2

3

2

c

c

Vr

V

Vwhere r cutoff ratio

V

γγ

= =

= =

Substituting 4

1

T

T= cr

γ and 3

2

T

T= cr in equation (3),

[ ]1

11

1c

cyc

r

r r

γ

γηγ −

− = − −

Cutoff ratio (rc) : It is the ratio of the final volume to the initial volume during Constant pressure combustion.

3. Derive the efficiency for semi diesel cycle or dual cycle.

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SEMI DIESEL CYCLE OR DUAL CYCLE

CRUDE OIL ENGINE - LIMITED PRESSURE CYCLE

1.3 Isentropic Compression2.3 Constant Volume Heating3.5 Constant Pressure Heating4.5 Isentropic Expansion5.2 Constant Volume Cooling

Since part of the combustion takes place at constant volume and part of the combustion takes place at constant pressure the semi diesel cycle is also called as Dual Cycle.

Heat Supplied = [Cv(T3 –T2) + Cp(T4 –T 3)] ------(1)

Heat Rejected = Cv(T5 –T1) ------( 2)

Work Done = [Cv(T3 –T2) + Cp(T4 –T 3)] - Cv(T5 –T1)

v 3 2 p 4 3 v 5 1cy

v 3 2 p 4 3

[C (T -T ) + C (T -T )] - C (T -T )

[C (T -T ) + C (T -T )]η =

v 5 1cy

v 3 2 p 4 3

C (T -T )1

[C (T -T ) + C (T -T )]η = −

5 1

3 2 4 3

(T -T )1

[(T -T ) + (T -T )]γ= − ------(3)

p

v

C

Cγ

=

Q

5

1

1 3 3 4

2 1 3

T -1

T1

T T T1 + 1

T T Trγ γ−

= −

− −

1

12 1

1 2

T Vr

T V

γγ

−−

= = Q

5 5 34 2

1 4 3 2 1

T T TT T

T T T T T= × × ×

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1 1

34 4 1

5 3 2 2

PV V V

V V V P

γ γ− −

= × × ×

[ ]3 2 5 1;V V V V= =Q

= 34

3 2c p

PVr r

V P

γγ

γ × = where 3

2p

Pr

P= = Pressure ratio

( ) [ ]1 3

1

11

1 1

c p

cy

c p c

r r

Tr r r

T

γ

γη

γ−

− = − − + −

13 3 2

1 2 1p

T T Tr r

T T Tγ −= × =

( ) [ ]1 1

11

1 1

c p

cy

c p p c

r r

r r r r r

γ

γ γη

γ− −

− = − − + −

( ) [ ]{ }1

11

1 1

c p

cy

p p c

r r

r r r r

γ

γη

γ−

− = − − + −

Pressure ratio (rp) : It is the ratio of the final pressure to the initial pressure during constant volume combustion.

Mean Effective Pressure ( ) ( ) ( ){ }

( )

1 12 1 1 1

1 ( 1)

p p c c pP r r r r r r r r

r

γ γ γγ

γ

− −− + − − −=

− −

4. OTTO CYCLE- PROBLEMS1. In an ottto cycle air at 15oc and 1.02 bar is compressed until the pressure is 12.5 bar.

Heat is added at constant volume until the pressure rises to 35 bar absolute. Calculate the compression ratio, air standard efficiency and the mean effective pressure γ = 1.4. Given:

P1= 1.02 bar;

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T1=15oC+ 273=288K P2= 12.5 bar; P3= 35 bar;To find: compression ratio, air standard efficiency and the mean effective pressure.Compression ratio r:

( ) 98.502.1

5.12

02.1

5.12 4.1

1

4.1

2

1

1

2 =⇒

=⇒=⇒

= rrr

v

v

p

pγ

Compression ratio: r = 5.98Air standard efficiency ottoη :

( ) ( )%51

51.098.5

11

11

14.11

=

=−=−= −−

otto

ottor

η

η γ

Mean effective pressure Pm:

322

2

1

315

1

11111

135.098.5

810.098.5

810.01002.1

2882871

mvvv

v

mvp

mRTvmRTvp

mestrokevolu

workdonepm

=⇒=⇒=

=⇒×××==⇒=

=

At adiabatic process1-2

( )

( ) ( ) 94.58828898.598.5

98.5

14.11

14.12

14.1

1

2

1

1

2

=×=×=

=

=

−−

−−

TT

v

v

T

Tγ

T2 = 588.94K

At constant volume heat addition:

94.5885.12

35

94.5885.12

353

3

2

3

2

3 ×=⇒=⇒= TT

T

T

p

p

T3 = 1649.03K

( ) ( ) KJTTmCworkdone

pliedheatworkdonepliedheat

workdoneefficiency

v 3865881649717.0151.0

supsup

23 =−××=−×=

×=⇒=

η

ηη

53

21

1073.5135.0810.0

10386 ×=−×=

−==

vv

workdone

mestrokevolu

workdonepm

Mean effective pressure Pm=5.73 bar

2. An engine equipped with a cylinder having bore of 150mm and a stroke of 450mm operated on an otto cycle. If clearance volume is 2000cm3compute the air standard efficiency.Given:Bore D = 150 mm = 0.15 mStroke L = 450 mm = 0.45mClearance volume = 2000 cm 3 = 2000x10-6m3

So Vc = 0.002m3

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Stroke volume ( )

33

3322

1094.7

1094.745.04

15.014.3

4

mV

mLD

V

s

s

−

−

×=

×=××=×= π

Compression ratio r = 9.4002.0

1094.7002.0 3

=×+=+ −

c

sc

V

VV

Air standard efficiency ottoη :

( ) ( )%47

47.09.4

11

11

14.11

=

=−=−= −−

otto

ottor

η

η γ

3. In an air standard otto cycle the compression begins at 35oC, 0.1 MPa. The maximum temperature of the cycle is 1100oC. Find the temperature and pressure at all corner points of p-v diagram, the heat supplied per Kg of air, the work done per Kg of air, air standard efficiency and mean effective pressure.Given:

73

4

2

1 ===v

v

v

vr P1= 1. bar;

T1=35oC+ 273=308K T3= 1100+273=1373K At adiabatic process1-2

( )

( ) ( ) barPrP

rv

v

P

P

2.1517 4.112

2

1

1

2

=×==

=

=

γ

γγ

P2 = 15.2 bar

( )

( ) ( ) 79.67030898.598.5 14.11

14.12

1

1

2

1

1

2

=×=×=

=

=

−−

−−

TT

rv

v

T

T γγ

T2 = 670.79K

At constant volume heat addition 2-3:

barPTT

Tp

T

T

c

p11.31

79.670

2.15137332

2

33

2

33 =×=⇒×=⇒=

P3 = 31.1 bar

At adiabatic expansion 3-4:

barPr

P

rv

v

P

P

04.211.317

11

1

4.1

34

4

3

3

4

=×

=

=

=

=

γ

γγ

P2 = 2.04bar

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( ) ( ) KTT

rv

v

T

T

4.630137398.598.5

1

14.13

14.12

11

4

3

3

4

=×=×=

=

=

−−

−− γγ

T2 = 630.4K

Heat supplied:( ) ( ) KJTTmCQ vs 4.5327.6701373717.0123 =−×=−=

Qs = 532.4 KJHeat rejected ( ) ( ) KTTmCQ vR 16.2313084.630717.0114 =−×=−=

Work done:W = heat supplied –heat rejected = (532.4-231.16)=301.23 KJW=301.23 KJ

Cycle efficiency:= pliedheat

workdone

sup = 56.04.532

23.301 =

%56=ottoηMean effective pressure:

322

2

1

315

1

11111

128.07

88.07

88.01002.1

3082871

mvvv

vr

mvp

mRTvmRTvp

=⇒=⇒==

=⇒×××==⇒=

2

21

/64.401128.088.0

23.301mKN

vv

workdone

mestrokevolu

workdonepm =

−=

−==

Pm=4.01 bar

5. DIESEL CYCLE- PROBLEMS1. In a diesel cycle engine the compression ratio is 13:1 and the fuel is cut off at 8% of

the stroke. Find the air standard efficiency of the engine. Take γ = 1.4.

Given:

Compression ratio = r = 2

1

v

v =13

Let us assume the clearance volume v2 = 1 m3

And stroke volume v1-v2 = 13-1 = 12m3

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Volume at cut off v3 = v2 + 8 % of stroke volume

( )

( ) 3

212

96.1113100

81

100

8

m

vvv

=−+=

−+=

We know cut off ratio 96.11

96.1

2

3 ===v

vρ

Air standard efficiency dieselη :

( ) ( ) ( ) ( )

%3.58

%3.58583.0417.01

196.14.1

196.1

13

11

1

111

4.1

14.11

===−=

−−−=

−−−= −−

diesel

diesel

dieselr

ηη

ργρη

γ

γ

2. An ideal diesel engine has a diameter 150mm and stroke 200mm. the clearance volume is 10% of the swept volume. Find the compression ratio and air standard efficiency of the engine if cut of takes place at 6% of the stroke.Given:D = 150 mm = 0.15mL = 200 mm = 0.2 mvc = 10% of vs= 0.1 vs

Swept volume 3322

1053.32.04

15.014.3

4mL

Dvs

−×=××=×= π

333 1035.01053.31.01.0 mvv sc−− ×=××==

Compression ratio r: = 3

33

10353.0

10353.01053.3−

−−

××+×=

+

c

cs

v

vv = 11

r = 11

Air standard efficiency dieselη :Cut off takes place at 6% of stroke volumeCut off volume ( ) ( )33

23 1053.306.010353.006.0 −− ××+×=×+= svvv33

3 10565.0 mv −×=

cut off ratio 6.110353.0

10565.03

3

2

3 =××== −

−

v

vρ

( ) ( ) ( ) ( )

%5.57

%5.575753.04246.01

4246.0116.14.1

16.1

11

11

1

111

4.1

14.11

===−=

−=

−−−=

−−−= −−

diesel

diesel

dieselr

ηη

ργρη

γ

γ

3. The compression ratio of an ideal air standard efficiency of a diesel cycle is 15. The heat transfer is 1465 KJ/kg. Find the pressure and temperature at the end each process and determine the cycle efficiency. What is the mean effective pressure of the cycle if the inlet conditions are 300K and 1 bar?Given:

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Compression ratio r = 152

1 =v

v Heat supplied (2-3) Qs= 1465 KJ

T1 = 300K; P1= 1 bar

Adiabatic compression (1-2)

( ) ( ) ( ) barPrPrv

v

P

P3.44115 4.1

122

1

1

2 =×==⇒=

= γγ

γ

P2 = 44.3 bar P2=P3=44.3bar

( ) ( ) ( ) KTTrv

v

T

T2.8863001515 14.1

114.1

21

1

2

1

1

2 =×=×=⇒=

= −−−

−γ

γ

T2 =886.2KConstant pressure heat addition (2-3):

( )

( ) KTT

TTmCQ ps

2.2351886005.1

1465886005.111465 33

23

=+=⇒−×=

−=

T3 =2351.2K3

22

33

2

3

2

3 1523.00574.0886

3.2351mv

T

Tv

T

T

v

v=×=×=⇒=

v3 = 0.1574 m3

At adiabatic expansion 3-4:

322

2

1

315

1

11111

0574.015

861.015

861.010.1

3002871

mvvv

vr

mvp

mRTvmRTvp

=⇒=⇒==

=⇒×

××==⇒=

KTT

v

v

T

T

7.11752351861.0

1523.0

861.0

1523.0

861.0

1523.0

14.1

3

14.1

2

14.11

4

3

3

4

=×

=×

=

=

=

−−

−−γ

T2 = 1175.7KConstant volume heat rejection (4-1)

barpT

Tpv

T

T

p

p92.31

300

7.11754

1

443

1

4

1

4 =×=×==⇒=

P4 = 3.92 bar

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Heat rejected ( ) ( ) KJTTmCQ vR 4.6233001175005.1114 =−×=−=

Air standard efficiency dieselη =S

RS

Q

QQ

pliedheat

edheatrejectpliedheat −=−

sup

sup

= 5745.01465

4.6231465 =−

dieselη = 57.45%

Mean effective pressure:

2

21

/3.10470574.0861.0

4.6231465mN

vv

QQ

mestrokevolu

workdonep RS

m =−−=

−−

==

Pm=10.47 bar

4. An engine with 200 mm cylinder diameter and 300 mm stroke works on theoretical diesel cycle. The initial pressure and temperature of air used 1 bar and 27oC the cut off is 8% of stroke determine

a) pressure and temperature at all salient pointsb) theoretical air standard efficiencyc) mean effective pressured) Power of the engine if the working cycles per minute are 380 assume compression r

= 15.Given:

D = 200 mm = 0.2mL = 300 mm = 0.3 mvc = 8% of vs= 0.08 vs

Compression ratio r = 152

1 =v

v

T1 = 300K; P1= 1 bar

Swept volume 322

00942.03.04

2.014.3

4mL

Dvs =××=×= π

33

323

001426.0

001426.000942.008.00067.008.0

mv

mvvv s

=

=×+=+=

Compression ratio r: =c

c

c

cs

c

cs

v

v

v

vv

v

vv +=⇒

+=⇒

+ 00942.01515

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3121

2

132 0101.015150067.0

00942.015

mvvvv

vrmvv

vv

c

cc

=⇒×=⇒====

=−

Adiabatic compression (1-2)

( ) ( ) ( ) barPrPrv

v

P

P3.44115 4.1

122

1

1

2 =×==⇒=

= γγ

γ

P2 = 44.3 bar P2=P3=44.3bar

( ) ( ) ( ) KTTrv

v

T

T2.8863001515 14.1

114.1

21

1

2

1

1

2 =×=×=⇒=

= −−−

−γ

γ

T2 =886.2K

Constant pressure heat addition (2-3):

KTv

vT

T

T

v

v1878886

00067.0

001426.02

2

33

2

3

2

3 =×=×=⇒=

T3 = 1878KAt adiabatic expansion 3-4:

KTT

v

v

T

T

858300861.0

1523.0

861.0

1523.0

0101.0

001426.0

14.1

3

14.1

2

14.11

4

3

3

4

=×

=×

=

=

=

−−

−−γ

T2 =858K

3.440101.0

001426.0

0101.0

001426.0

0101.0

001426.04.1

4343

4 ×

=⇒

=⇒

= ppp

p

pγγ

P3 = 2.85 bar

We know cut off ratio 12.200067.0

001426.0

2

3 ===v

vρ ρ =2.12

Air standard efficiency dieselη :

( ) ( ) ( ) ( ) 598.0112.24.1

112.2

15

11

1

111

4.1

14.11=

−−−=

−−−= −− ργ

ρηγ

γrdiesel

%8.59=dieselη

Mean effective pressure:( ) ( )

00942.01423 TTmCTTmC

v

QQ

mestrokevolu

workdonep vp

s

RSm

−−−=

−==

=( ) ( ) 2/741

00942.0

300858717.0117.08861878005.1117.0mN=−×−−×

Pm= 7.41 bar

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Power of the engine:

cycleKJpervPworkdone

mestrokevolu

workdonep

sm

m

99.600942.01041.7 5 =××=×=

=

power = work done / cycle x no of cycles per second

= 6.99 x 380/60 = 44.27 KW so power = 44.27 KW

6. DUAL CYCLE (SEMI DIESEL CYCLE)

1. An oil engine working on the dual combustion cycle has a compression ratio 10 and cut off takes place at 1/10 of the stroke. If the pressure at the beginning of compression is 1 bar and max. pressure 40 bar, determine air standard efficiency of the cycle. Take γ = 1.4.Given:

r = 10 = 2

1

v

v;

Cut off = 10

1 of stroke;

P1 = 1 bar, P4 = 40 bar,P3 = 40 bar.

Pressure ratio β = 2

3

P

P;

Adiabatic compression 1-2

1

2

P

P =

γ

2

1

v

v= ( )γr ⇒ P2 = ( ) γr P1 = ( ) 4.110 X 1 ⇒ P2 = 25.1 bar.

β = 2

3

P

P =

1.25

40= 1.59 ⇒ β = 1.59.

Cut off ratio ρ = 3

4

v

v ⇒ v4 = v2 +

10

1 of vs

= v2 + 10

1 (v1-v2)

Dividing by v2 on both sides +=

1

2

4

v

v

10

1

−1

2

1

v

v

ρ = 1 + 10

1 (10-1) ⇒ ρ = 1.9.

Air standard efficiency of dual cycle.

( )[ ]

( ) ( )[ ]

−+−

−−= − 11

111

1 ργ βββ ρη

γ

γr

= 1- ( )( )

( ) ( )[ ]

−×+−

−− 19.159.14.1159.1

19.159.1

10

1 4.1

14.1 =

+−−

259.0

19.31

= 1-0.446 = 0.554.

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4.55=η %

2. The swept volume of a diesel cycle engine working on dual cycle is 0.0053m3 and clearance volume is 0.00035 m3. The max. Pressure is 65 bar, Fuel injection ends at 5% of stroke. The temperature and pressure at the shaft of the compression are 80°C and 0.4 bar. Determine the air standard efficiency of the cycle. Take γ = 1.4.

Given:

Swept volume vs = 0.0053 m3 ;

Clearance volume vc = v3 = v2 = 0.0035 m3 ;

Max. pressure P3 = P4 = 65 bar

Initial temp T1 = 80+273 = 353 K

P1 = 0.9 bar Find =dualη ?

Compression ratio (r) = c

cs

v

vv + =

00035.0

00035.00053.0 + = 16.4

Cut off ratio = 2

4

v

v=ρ =

( )

2

2 100

5

v

vv s+=

00035.0

)0053.0(05.000035.0 +

.7757.1=ρ

Pressure ratio 2

3

p

p=β

( ) ( ) ( ) .9.04.16 4.11

4.12

4.1

2

1

1

2 ×=×=⇒=

= pp

v

v

p

p γγγ

22.442 =p bar.3.The compression ratio for single cylinder engine operating on dual cycle is 9. The max.

pressure in the cylinder is limited to 60 bar. The pressure and temperature of the air at the beginning of the cycle are 1 bar and 30°C and heat added at constant pressure process upto 4% of the stroke. D=250 mm, L=300mm.

i. Determine the air standard efficiencyii. The power developed if the number of working cycles are 3 per

second.

Given:

D = 250mm = 0.25m r=9

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L=300mm=0.3mInitial pressure 11 =p bar. Initial temperature T1 = 30+273 = 303 K;Max. pressure == 43 pp 60 bar.No. of working cycles = 3/s.

(i). Air standard efficiency.

Swept volume 322 0147.03.025.044

mLDvs =××== ππ

Compression ratio (r)

r =c

cs

v

vv +

c

c

v

v+=⇒ 0147.09 30018.0

8

0147.0mvc ==⇒

30018.0 mvc =3

1 0165.00018.00147.0 mvvv cs =+=+=

For adiabatic process 1-2

( ) ( ) 67.2119 4.112

2

1

1

2 =×=×=⇒

= prp

v

v

p

p γγ

bar.

67.212 =p bar.

( ) ( ) ( ) =×=×=⇒=

= −−

−

3039 4.01

12

1

1

2

1

1

TrTrv

v

T

T ϕϕγ

729.6 K

Constant Volume process 2-3

.202067.21

606.7292

2

33

2

3

2

3 KTp

pT

T

T

p

p=×=×=⇒=

T2 = 2020 KAlso,

804.0104.019

104.0

100

4

1

1 ×=−⇒=−−⇒==

−− ρρρ

r32.11)804.0( =+×=ρ

32.1=ρFor constant pressure process 3-4

KTTT

T

v

v4.2666202032.134

3

4

3

4 =×=×=⇒== ρρ

KT 4.26664 =For adiabatic expansion process 4-5

4

1

5

1

5

2

2

4

1

5

4

4

5 Tr

Tv

v

v

v

v

v

T

T×

=⇒

×=

=

−−− γγγρ

KT 12374.26669

32.14.0

5 =×

=

KT 12375 =

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08.4609

32.14.0

455

4

4

5 =×

=×

=⇒

= p

rp

v

v

p

pγγ

ρbar.

08.45 =p bar.Total heat supplied = Qs at constant volume + Qs at constant pressure

= m Cv (T3-T2) + m Cp (T4-T3)= +1 X1.005(2666-2020)

Qs = 1562.58 kJ/kg.Heat rejected Qr = m Cv (T5-T1)

=1 X 0.717 (1237-303)= 663.14 kJ/kg.

5756.058.1562

14.66385.1562 =−=−

=s

rs

Q

QQη

efficiency 56.57=η %

Power developed by the engine

Mass of air m = .0189.0303287

0165.0101 5

1

11 kgRT

vp=

×××=

Work done = ( ) kJQQm rs 999.16)14.6631562(0189.0 =−=−Work done = 16.999 kJ.Power = Work done per cycle X No. of cycles per second.

= 16.999 X 3 = 50.99 kW. Power = 50.999 kWRESULT:

Efficiency of the dual cycle = 57.56%Work done = 16.999 kW.Power developed P = 50.99 kW.

7. Explain working principle of four stroke S.I. engine with the help of neat sketches.(Or)

Explain working principle of four stroke petrol engine with the help of neat sketches.

In four stroke petrol engine, the cycle of operation is completed in four strokes of piston or two revolution of crank. The four strokes are

i. Suction strokeii. Compression strokeiii. Power stroke iv. Exhaust stroke.

The air fuel is ignited by means of an electric spark at the end of compression stroke and hence petrol engines are known as spark ignition engines.

Working principle:Suction stroke

During suction stroke, the inlet valve remains open and the exhaust valve is closed and the piston moves from the top dead centre (TDC) to bottom dead centre (BDC). The fresh charge of fuel air mixture is sucked into the cylinder until the piston reaches the BDC. At this position the inlet valve is closed. Here the crank rotates through an angle of 180°.

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Compression strokeDuring compression stroke, both the inlet and exhaust valves remain closed and the piston

moves from the BDC to TDC. When the piston moves upwards the charge drawn in the enclosed cylinder gets compressed. The pressure and temperature of the charge increase. When the piston reaches the TDC, the spark plug gives a spark by means of which the charge is ignited and hence combustion takes place. Due to combustion the pressure and temperature of gases inside the cylinder increased. Here the crank rotates through an angle of 180°. This completes one revolution of crank shaft.Power stroke

During the power stroke, both the inlet and exhaust valves remain closed. Due to the increase in pressure and temperature a great force is exerted by the gases which push the piston downwards. This expansion continues till the piston reaches BDC. This is also known as expansion stroke. At this position the exhaust valve is opened. Here the crank rotates through an angle of 180°. Exhaust stroke

As soon as the exhaust valve is opened the pressure falls to atmospheric pressure. This makes the piston to move from BDC to TDC by pushing all the combustion products through exhaust valve. Here the crank rotates through an angle of 180° and this completes second revolution of crank shaft and constitutes one cycle of operation. As soon as the piston reaches, the inlet valve opens and cycle gets repeated. 8. Explain working principle of four stroke C.I. engine with the help of neat sketches.

(Or)Explain working principle of four stroke diesel engine with the help of neat sketches.

In four stroke diesel engine, the cycle of operation is completed in four strokes of piston or two revolution of crank. The four strokes are

i. Suction strokeii. Compression stroke

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iii. Power stroke iv. Exhaust stroke.

In C.E. engine the combustion takes place due to high pressure and temperature generated during compression stroke. So, spark plug is absent, instead fuel injectors are used.

Working principle:Suction stroke

During suction stroke, the inlet valve remains open and the exhaust valve is closed and the piston moves from the top dead centre (TDC) to bottom dead centre (BDC). The fresh air at atmospheric pressure is sucked into the cylinder until the piston reaches the BDC. At this position the inlet valve is closed. Here the crank rotates through an angle of 180°.

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Compression strokeDuring compression stroke, both the inlet and exhaust valves remain closed and the piston

moves from the BDC to TDC. When the piston moves upwards the air drawn in the enclosed cylinder gets compressed. The pressure and temperature of the air increase. Since the compression ratio is high in this air, the air is compressed to very high pressure and temperature. Fuel is injected as fine spray into the cylinder and combustion starts instantaneously. Due to combustion the pressure and temperature of gases inside the cylinder increased. Here the crank rotates through an angle of 180°. This completes one revolution of crank shaft.

Power strokeDuring the power stroke, both the inlet and exhaust valves remain closed. Due to the

increase in pressure and temperature a great force is exerted by the gases which push the piston downwards. This expansion continues till the piston reaches BDC. This is also known as expansion stroke. At this position the exhaust valve is opened. Here the crank rotates through an angle of 180°.

Exhaust strokeAs soon as the exhaust valve is opened the pressure falls to atmospheric pressure. This

makes the piston to move from BDC to TDC by pushing all the combustion products through exhaust valve. Here the crank rotates through an angle of 180° and this completes second revolution of crank shaft and constitutes one cycle of operation. As soon as the piston reaches, the inlet valve opens and cycle gets repeated.

9. Explain working principle of two stroke S.I. engine with the help of neat sketches.(Or)

Explain working principle of two stroke petrol engine with the help of neat sketches

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In two stroke petrol engine, the cycle of operation is completed in two strokes of piston or one revolution of crank. The two strokes are

i. Compression strokeii. Power stroke Two stroke petrol engine uses petrol air mixture as a fuel charge and spark plug to ignite

the compressed charge. Fig shows the operation cycle of a two stroke petrol engine using deflection shape piston.

i) Compression strokeFigure shows the position at the end of the compression stroke. the ignition starts before TDC and continues after TDC through some crank rotation. The high pressure gases push the piston downwards with great force and this movement produces the power stroke. This process continues until piston reaches BDC position. While moving the piston from TDC to BDC, the piston first uncovers the exhaust port and burnt gases expanded in the cylinder escapes out through it. A little later, the piston uncovers the transfer port, and the cylinder is directly connected with the crank case in that position. So, the partially compressed charge (air fuel) is transferred from crank case to the engine cylinder through transfer port, so, the charge entering into the cylinder pushes out the remaining burnt gases through exhaust port and thus complete scavenging is achieved. At the same time, the charge inside the crank case gets compressed by the down side of the piston and continues till the piston reaches the BDC position. So, during downward stroke.

Power is developed during expansion stroke.

Exhaust gases are pushed out completely by scavenging.

The charge inside the crank case is compressed to a pressure of about 1.4 bar.

The charge is transferred from the crank case to the engine cylinder through transfer port.

ii) Expansion stroke (Upward stroke of the piston)

As soon as the piston reaches the BDC position, the piston moves upward and hence it first covers the transfer port, so, the flow of charge from the crank case to the engine cylinder is stopped. A little later, the piston covers the exhaust port and thus the two ports are in closed position. Now the charge inside the cylinder is compressed by the upside piston and is continued until the piston reaches the TDC position. At the same time, the pressure inside the crank case becomes low and thus creating the partial vacuum inside it. Due to this action, the charge is drawn inside the crankcase through inlet port as it is uncovered by the piston. Shortly before the piston reaches the TDC position, a spark is produced by spark plug by which the compressed charge is ignited, at the end of compression stroke the pressure and temperature of the gases are very high. The ignition continues through same degree of crank rotation after TDC position and thus the cycle gets repeated. So, the cycle of the engine is completed within two strokes of the piston and makes the crank to turn one revolution.

So, during upward stroke.

Partial scavenging takes place The fresh charge is drawn into the crank case through inlet port

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The compression of the charge by upside of the piston while it moves from BDC to TDC position.

Ignition starts at a little before the piston reaches TDC position.

10. Explain working principle of two stroke C.I. engine with the help of neat sketches.(Or)

Explain working principle of two stroke diesel engine with the help of neat sketches

Compression strokeFigure shows the position at the end of the compression stroke. The ignition starts before TDC and continues after TDC through some crank rotation. The high pressure burnt gases pushes the piston downwards with a great force. and this movement produces the power stroke. This process continues until piston reaches BDC position. While moving the piston from TDC to BDC, the piston first uncovers the exhaust port and burnt gases expanded in the cylinder escapes out through it. A little later, the piston uncovers the transfer port, and the crank case is directly connected with the cylinder. So, the partially compressed air is drawn into the engine cylinder from the crank case through transfer port. so, the air entering into the cylinder pushes out the remaining burnt gases through exhaust port and thus complete scavenging is achieved. At the same time, the air inside the crank case gets compressed by the down side of the piston and continues till the piston reaches the BDC position.So, during downward stroke.

Power is developed during expansion stroke. Exhaust gases are forced out for complete scavenging. The charge inside the crank case is compressed to a pressure of about 1.4 bar. The charge is transferred from the crank case to the engine cylinder through

transfer port.

ii) Expansion stroke (Upward stroke of the piston)

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As soon as the piston reaches the BDC position, the piston moves upward and hence it first covers the transfer port. So, the flow of charge from the crank case to the engine cylinder is stopped. A little later, the piston covers the exhaust port and thus the two ports are in closed position. Now the air inside the cylinder is compressed by the upside piston and is continued until the piston reaches the TDC position. At the same time, the pressure inside the crank case becomes low and thus creating the partial vacuum inside it. Due to this action, the air is drawn inside the crankcase through inlet port as it is uncovered by the piston. Shortly before the piston reaches the TDC position, the fuel oil is injected into the engine cylinder in the form of very fine spray through the fuel nozzle known as fuel injector valve. The pressure and temperature of the air at the end of compression is very high and is enough for better combustion when the fuel oil(diesel) comes into contact with these gases. So, when the fuel is injected into the engine cylinder, immediately it starts burn and continuous through some crank rotation after TDC position. thus the cycle gets repeated and the cycle of operation is completed witin two strokes of the piston. This makes the crank to turn one revolution. So, during upward stroke.

Partial scavenging takes place The fresh air is drawn into the crank case through inlet port The compression of the air by upside of the piston while it moves from BDC to TDC

position. Fuel injection starts at a little before the piston reaches TDC position

11. Distinguish Between 2 Stroke and 4 Stroke Engines

Four Stroke Cylinder Engine Two Stroke Cylinder Engine1. For every two revolution of the crank shaft, there is one power stroke.2. Because of the above, turning moment is not so uniform and hence heavier flywheel is needed.3. For the same power more space is required.4. Because of one power stroke in two revolutions, lesser cooling and lubrication requires. Lower rate of wear and tear.5. Valves are required – inlet and exhaust valves.6. Because of heavy weight, complicated valve mechanism and water cooled, making it complicated design and difficult to maintain.7. The air-fuel mixture is completely utilized thus efficiency is higher.

8. Volumetric efficiency is high due to more time for induction.9. Lower fuel consumption per horse power.

10. Used in heavy vehicles, e.g. Buses,

1. For every one revolution of the crank shaft, there is one power stroke.2. Because of the above, turning moment is more uniform and hence a lighter flywheel is used.3. For the same power less space is required.

4. Because of one power stroke for every revolution, greater cooling and lubrication requirements. Higher rate of wear and tear.5. Ports are made in the cylinder walls – inlet, exhaust, and transfer port.6. Simple in design, light weight and air cooled and easy to maintain.

7. As inlet and exhaust port open simultaneously, some times fresh charges may escape with exhaust gases. The exhaust gases are not always completely removed. This cause lower efficiency.8. Volumetric efficiency is low due to lesser time for induction.9. The fuel consumption per horse power is more because of fuel dilution by the exhaust gas.10. Used in light vehicles, e.g. Motor cycle,

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lorries, trucks etc.11. The engine cost is more.12. The exhaust is less noisy.

scooter, etc.11. The engine cost is less.12. The exhaust is noisy due to short time available for exhaust.

12. Distinguish Between S.I. Engines and C.I. Engines Engines

S.I. Engines C.I. Engines1. The fuel used is gasoline (Petrol).2. Air + Fuel mixture is taken during suction.3. For mixing air and fuel a separate device called carburettor is required.4. Since homogeneous mixture is produced in carburettor, no need of injector. 5. Pressure at the end of compression is about 10 bar.6. A spark plug is used to ignite the air fuel mixture.7. Self ignition temperature of fuel is not attained. In other words, the fuel is not self ignited.8. S.I. Engines works on otto cycle (i.e) combustion takes place at constant volume.9. Compression ratio is around 6 to 10.10. Cold starting of engine is easy.11. These are very lighter.12. Cost is comparatively low. 13. Running cost is high.14. Less maintenance.15. η thernal is about 25%.16. Over heating trouble is more.17. Spark plug needs frequent maintenance.18. These are high speed engines.19. Noiseless operation due to less compression ratio.20. Engine weight / kW is less.21. Vibration is less.22. Generally employed for light duty vehicles e.g. two wheeler, otto etc.

1. Fuel used is Diesel.2. Only air taken during suction.3. No need of carburetor.

4. For atomizing and spraying the fuel inside the cylinder, fuel injector is necessary.5. Pressure at the end of compression is about 35 bar.6. Spark plug is not necessary.

7. The fuel get ignited due to the high temperature of compressed air.

8. C.I. Engines works on diesel cycle (i.e) combustion takes place at constant pressure.9. Compression ratio is around 15 to 25.10. Cold starting of engine is diffucult.11. Heavier engine.12. Cost is high.13. Running cost is not high.14. High maintenance is needed.15. η thernal is about 35 to 45%.16. Over heating trouble is less.17. Fuel injector needs less maintenance.18. These are low speed engines.19. Very noisy operation due to high compression ratio.20. Engine weight / kW is more.21. More vibration is there.22. Generally employed for heavy duty vehicles e.g. truks, buses, etc.

13. Explain Brayton Cycle with Regeneration

The temperature of the exhaust gases of the turbine is higher than the temperature of the

air after compression. If the heat energy is used to heat the air after compression in the heat

exchanger called “regeneration”. It will reduce the energy requirement from the fuel thereby

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increasing the efficiency of the cycle. Fig (a) shows the single stage regenerative gas turbine cycle

and fig (b) is the corresponding cycle represented on T-s diagram.

Fig (a). Bryton cycle with regenerator

Air is drawn from the atmosphere into the compressor and is compressed isentropically to

state 2. It is then heated at constant pressure in the regenerator to state 3 by the exhaust gases from

the turbine. Since the temperature of air is increased before its reaches the combustion chamber,

less amount of fuel will be required to attain designed turbine inlet temperature of the products of

combustion.

After combustion at constant pressure in the combustion chamber, the gas enters the

turbine at stage 4 and expands to 5. It then enters the regenerator as stated earlier, where it gives

up a portion of its heat energy to the compressed air from the compressor and leaves the

regenerator at state 6.

In ideal regenerative cycle, the temperature of the air leaving the regenerator to

combustion chamber is equal to the temperature of exhaust gases leaving the turbine. i.e.,

T3=T5.

But in actual cycle, the temperature of the air leaving the generator is less than T5. i.e.,

T3<T5

The effectiveness of the regenerator is given by the ratio of the actual temperature rise to

the maximum possible rise.

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Effectiveness, 25

23

65

23

TT

TT

TT

TT

−−

=−−

∈=

The regenerative cycle, for unit mass flow rate

Heat supplied, Qs=Cp (T4-T3)

(For process 2-3 heat supplied by regenerator)

Heat rejected, )( 16 TTCQ pR −=

(For process 5-6 heat is rejected by regenerator)

Turbine work, )( 54 TTCW PT −=

Compressor work, )( 12 TTCW PC −=

Efficiency,S

R

Q

Q−=1η

Efficiency for regenerative Brayton cycle,

γγ

η1

4

1 )(1−

−= PRT

T

From the above formula, it is obvious that the efficiency of the regenerative Brayton cycle

depends not only on the pressure ratio but also on the ratio of the two extreme temperatures.

14. Explain Brayton Cycle with Inter Cooling

The thermal efficiency of the Brayton cycle may further be increased by providing

multistage compression with intercooler between the compressors and multistage expansion with

reheater between the turbines. The work required during multistage compression with intercoolers

is less than the single stage compression. Similarly, the work output from the turbine is increased

by multistage expansion with reheating. As a result, the net work output from the plant increases.

Fig (a) shows an ideal gas plant operated by Brayton cycle with two-stage compression.

Fig (b). Shows the Vp − and sT − diagram of the multistage compression Brayton

cycle with intercooler.

Initially the air is compressed in the Low Pressure (L.P.Compressor) and it is passed to an

intercooler, which reduces the temperature of the air to its original temperature at constant

pressure. After that, the compressed air is again compressed in the High-Pressure Compressor

(H.P.Compressor). Then the compressed air is passed through the heating chamber were heat is

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added to the air. Now, the air is expanded through the turbine. Finally the air is cooled in the

cooling chamber to its original temperature.

In fig (b). The ideal cycle without inter cooling is represented by 1-2’-5-6-1 and the cycle

with intercooler is represented by 1-2-3-4-5-6-1. The area under the Vp − diagram is increased

by the amount 2-3-4-2’-2; therefore the net work output is increased.

∴The work done by turbine per kg of air, )( 65 TTCW pT −=

Work required by the compressor per kg of air

)( 12 TTCW pC −= + )( 34 TTC p −

Note:

For perfect cooling 31 TT = and 42 TT =

The intermediate pressure for perfect cooling is

654123 pppppp ∗=∗==

15. Explain Brayton Cycle with Reheater

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As stated earlier, the work output can be increased by multistage expansion with reheating

between stages. Fig (a) shows the ideal Brayton cycle with reheating and fig (b) shows the Vp −

and sT − diagrams for the same.

The air is first compressed in the compressor, passed into the heating chamber, and then to

the first turbine. The air is once again passed into the heating chamber called reheater and then to

the second turbine. The area under the Vp − diagram is increased by the amount 4’-4-5-6-4’.

Therefore the network is increased. In Fig (b), the ideal cycle without reheater is shown the

process 1-2-3-4’-1, and the cycle with reheater is shown by 1-2-3-4-5-6-1.

Fig

Work done required by the compressor per kg of air

)( 12 TTC pc −=W

Work done by two turbines per kg of air

)()( 6543 TTCTTCW ppT −+−=

Network, CT WWW −=

Note:

For obtaining maximum work 6

5

4

3

p

p

p

p=

216354 pppppp ∗=∗== {32

61

pp

pp

==

and}

16. Derive the Expression of Optimum Pressure Ratio For Maximum Net Work Output In

An Ideal Brayton Cycle. What Is The Corresponding Cycle Efficiency?

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Braton cycle is a constant pressure cycle for a perfect gas. It is also called Jule cycle. The hea

tanfers are achieved in reversible constant pressure heat exchangers. An ideal gas turbine plant

would perform the processes that make up a brayton cycle. The cycle is shown in the figure and it

is represented on p-v and T-s diagrams as shown in figure.

The various operations are as follows:

Operation 1-2. the air is compressed isentropically from the lower pressure p1 to the upper

pressure p2, the temperature rising from T1 to T2. No heat flow occurs. Operation 2-3. heat flows

into the system increasing the volume from V2 to V3 and temperature from T2 to T3 whilst the

pressure remains constant at p2.

heat received = mcp(T3-T2)

Operation 3-4. The air is expanded isentropically from p2 to p1, the temperature falling from T3 to

T4. no heat flow occurs.

Operation 4-1. Heat is rejected from the system as the volume decreases from V4 to V1 and the

temperature from T4 to T1 whilst the pressure remains constant at p1.

heat rejected = mcp (T4-T1)

heat received/cycle- heat rejected/cycle

heat

η air-standard = receivedheat

workdone

= cyclereceivedheat

cyclerejecteHeatcyclereceivedheat

/

// −

= 23

14

23

1423 1)(

)()(

TT

TT

TTmc

TTmcTTmc

p

pp

−−

−=−

−−−

now from iisentropic expansin,

γγ 1

1

2

1

2

−

=

p

p

T

T

( ) γγ 1

12

−= prTT where pr = pressure ratio

similarly

γγ 1

1

2

4

3

−

=

p

p

T

T or ( ) γγ 1

43

−= prTT

η air-standard = 1- ( ) ( ) ( ) γγ

γγ

γγ 11

1

1

4

14 11

)(−−− −=

−

−

ppp rrTrT

TT

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pressure ration for maximum work:

now we shall prove that the pressure ration for maximum work is a function of the limiting

temperature ratio.

Work output during the cycle

= cyclerejecteHeatcyclereceivedheat // −

= )()( 1423 TTmcTTmc pp −−−

= )()( 1243 TTmcTTmc pp −−−

=

−

−

1

21

3

43 1

T

TT

T

TTmc p

in case of a given turbine the minimum temperature T1 being the temperature of the atmosphere

and T3 the maximum temperature which the metals of turbine would withstand. Consider the

specific heat at constant pressure cp to be constant. Then

( )1

21

4

3

T

Tr

T

Tp ==

−γ

γ

using constant ‘z’ = γγ 1−

we have, work output/cycle

W= K ( )

−−

− 1

11 13

zpz

p

rTr

T

( ) 011

1 13 =

−−

−= z

pzpp

rTr

TKdr

dW

( ) ⇒= −+

)1(11

3 zpz

p

rzTr

zT

1

32

T

Tr z

p =

⇒

=

z

p T

Tr

21

1

3)1(2

1

3−

=

γγ

T

Trp

17. Problem on brayton cycle

Air enters to the compressor of a gas turbine plant operating on brayton cycle at 101.325

Kpa, 27°C.the pressure ratio in this cycle is 6.calculate the maximum temperature in the

cycle and the cycle efficiency. Assume WT = 2.5 Wc where WT & Wc are the turbine and

compressor work respectively. Take γ = 1.4.

Solution:

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Pressure of intake air P1 = 101.325 kPa

Temperature of intake air T1 = 27+273 = 300K

Pressure ratio rP= 6

(i). Maximum temperature in the cycle, T3:

( ) ( ) 668.16 4.1

14.11

1

1

2

1

2 ===

=

−−−

γγγ

γ

prp

p

T

T

T2 = 1.668 T1 = 1.668 X 300 = 500.4 K

( ) ( ) 668.16 4.1

14.11

4

3 ===−−

γγ

prT

T

668.13

4

TT =

WT = 2.5 Wc

mCp(T3-T4) = 2.5 mCp(T2-T1)

KTT

T 1251501)3004.500(5.2668.1 3

33 =⇒=−=−

Cycle efficiency:

KT

T 750668.1

1251

668.13

4 ===

)(

)()(

23

1243

TTmC

TTmCTTmC

addedheat

worknet

p

ppcycle −

−−−==η

4.0)4.5001251(

)3004.500()7501251( =−

−−−= or 40%

PART-B1. Derive an expression for the air standard efficiency of Otto cycle in terms of compression ratio. Show the variation of air standard efficiency with compression ratio.

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2. Derive an expression for the air standard efficiency of the diesel cycle in terms of the compression ratio and cutoff ratio.

3. Derive an expression for the air standard efficiency of the dual combustion cycle in terms of the compression ratio, explosion ratio and cut off ratio.

4. Derive an expression for the air standard efficiency, work ratio and optimum pressure ratio for maximum work of the Brayton cycle.

5. Derive an expression for the mean effective pressure of the otto cycle in terms of the compression ratio and pressure ratio.

6. Derive an expression for the mean effective pressure of the diesel cycle in terms of the compression ratio and cut off ratio.

7. Derive an expression for the mean effective pressure of the dual cycle in terms of the compression ratio, pressure ratio and cut off ratio.

8. Explain haw the actual indicator diagram deviate from the ideal indicator diagram for a four stroke diesel engine.

9. Explain how the actual indicator diagram deviates from the ideal indicator diagram for a four stroke petrol engine.

Problems:

1. The diameter and stroke of the cylinder of an engine working on ideal Otto cycle are 180mm and 320 mm respectively the clearance volume is 0.002 m3. Calculate the air standard efficiency of the engine.

2. An ideal Otto cycle operator with a compression ratio of 7. The initial conditions are 30 oC and 1.03 bars. The maximum temperature of the cycle as 1147 oC.Determine,(i) Net work per kg of air. (iv) Cycle thermal efficiency.(ii) Peak pressure in the cycle. (v) Heat rejected per kg of air and(iii) Mean effective pressure.

3. What is the mean effective pressure in an Otto cycle having a compression ratio of 6, pressure ratio of 3 and the pressure at the beginning of section is 1.0 bar.

4. The compression ratio in a diesel cycle is 15.If the air before compression is at 15oC and 1atm and 1860 kJ/kg of heat is added to the gas per cycle. Calculate,(i) Pressure and temperature at each point of the cycle.(ii) The heat that must be removed. (iii) The thermal efficiency of the cycle. (iv) The cut off ratio.

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5. A diesel engine has a bore of 150mm and a stroke of 250mm. Its clearance volume is 400 cm3 . Fuel injection takes place for duration of 5% of the stroke. Calculate the air standard efficiency of the cycle.

6. A mixed cycle has a compression ratio of 12 and uses air as the working fluid at the initial conditions of 1.03 bar and 30oC. The maximum pressure of the cycle is 41.4 bar and the maximum temperature is 1393 oC. Calculate,(i) Cycle thermal efficiency. (ii) The mean effective pressure.

7. An engine working on dual cycle has cylinder bore of 20cm, and stroke of 40 cm. The compression ratio is 14.5 and the pressure ratio of the constant volume process is 1.5. The constant pressure process is cut off at 4.9% of the stroke. Determine the temperature at the end of heat addition and at the end of expansion. Also calculate the air standard efficiency. The suction conditions are 1 atm and 17oC.

8. An ideal Brayton cycle operates with air at the initial condition of 15oC and 1.03 bar. The pressure ratio is 7 and the maximum temperature is 816 oC. The efficiency of compressor and turbine is 85% respectively. Determine (i) the net work of the cycle (ii) the work ratio (iii) the thermal efficiency of the cycle (iv) mass flow rate of air for an output of 3750kw.

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37186719 thermal engineering unit i for be students

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