anna university, chennai. r-2017 me8493 thermal ...volumetric efficiency, isothermal efficiency and...

REGULATION 2017 ACADEMIC YEAR: 2018 - 2019

JIT-JEPPIAAR/MECH/Mr.J.RAVIKUMAR & Ms.S.AROKIYA ANICIA/IInd Yr/SEM 04/ME 8394/THERMAL ENGINEERING-I/UNIT-

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ANNA UNIVERSITY, CHENNAI.

R-2017

ME8493 THERMAL ENGINEERING - I L T P C

3 0 0 3

OBJECTIVES:

✓ To integrate the concepts, laws and methodologies from the first course in thermodynamics into analysis of cyclic processes

✓ To apply the thermodynamic concepts into various thermal applications like IC engines, Steam Turbines, Compressors.

(Use of standard refrigerant property data book, Steam Tables, Mollier diagram and Psychrometric chart permitted)

UNIT I GAS AND STEAM POWER CYCLES 9

Air Standard Cycles - Otto, Diesel, Dual, Brayton – Cycle Analysis, Performance and Comparison – Rankine, reheat and regenerative cycle. UNIT II RECIPROCATING AIR COMPRESSOR 9

Classification and comparison, working principle, work of compression - with and without clearance,

Volumetric efficiency, Isothermal efficiency and Isentropic efficiency. Multistage air compressor with Intercooling.Working principle and comparison of Rotary compressors with reciprocating air

compressors.

UNIT III INTERNAL COMBUSTION ENGINES AND COMBUSTION 9

IC engine – Classification, working, components and their functions. Ideal and actual : Valve and port

timing diagrams, p-v diagrams- two stroke & four stroke, and SI & CI engines – comparison. Geometric,

operating, and performance comparison of SI and CI engines.Desirable properties and qualities of

fuels.Air-fuel ratio calculation – lean and rich mixtures.Combustion in SI & CI Engines – Knocking –

phenomena and control.

UNIT IV INTERNAL COMBUSTION ENGINE PERFORMANCE AND SYSTEMS 9

Performance parameters and calculations.Morse and Heat Balance tests.Multipoint Fuel Injection system

and Common Rail Direct lnjection systems.Ignition systems – Magneto, Battery and

Electronic.Lubrication and Cooling systems.Concepts of Supercharging and Turbocharging – Emission

Norms.

UNIT V GAS TURBINES 9

Gas turbine cycle analysis – open and closed cycle.Performance and its improvement - Regenerative, Intercooled, Reheated cycles and their combinations.Materials for Turbines.

TOTAL:45 PERIODS



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OUTCOMES:

Upon the completion of this course the students will be able to

CO1 Apply thermodynamic concepts to different air standard cycles and solve problems. CO2 Solve problems in single stage and multistage air compressors CO3 Explain the functioning and features of IC engines, components and auxiliaries. CO4 Calculate performance parameters of IC Engines. CO5 Explain the flow in Gas turbines and solve problems.

TEXT BOOKS:

✓ Kothandaraman.C.P.,Domkundwar. S,Domkundwar. A.V., “A course in thermal Engineering", Fifth Edition, ”DhanpatRai& sons , 2016

✓ Rajput. R. K., “Thermal Engineering” S.Chand Publishers, 2017

REFERENCES:

✓ Arora.C.P, ”Refrigeration and Air Conditioning ,” Tata McGraw-Hill Publishers 2008

✓ Ganesan V..” Internal Combustion Engines” , Third Edition, Tata Mcgraw-Hill 2012

✓ Ramalingam. K.K., "Thermal Engineering", SCITECH Publications (India) Pvt. Ltd., 2009. ✓ Rudramoorthy, R, “Thermal Engineering “,Tata McGraw-Hill, New Delhi,2003

✓ Sarkar, B.K,”Thermal Engineering” Tata McGraw-Hill Publishers, 2007

Subject Code : ME 8493 Year/Semester : II/ 04

Subject Name :Thermal Engineering-I Subject Handler : J.Ravikumar & S.A.ArokyaAnicia

UNIT I GAS AND STEAM POWER CYCLE

Air Standard Cycles - Otto, Diesel, Dual, Brayton – Cycle Analysis, Performance and Comparison– Rankine, reheat and regenerative cycle

PART * A

Q.No. Questions

1.

Define thermodynamic cycle? [Oct.1997] BTL1

Thermodynamic cycle is defined as the series of processes performed on the system, so that the

system attains its original state.

2

Mention the assumptions made for air standard cycle analysis? [May 2015&2016] BTL1

(i)The working medium is a perfect gas through i.e., It follows the law pv = mRT

(ii)The working medium does not undergo any chemical change throughout the Cycle.

(iii)The compression and expansion processes are reversible adiabatic i.e.,

There is no loss or gain of entropy.

(iv)The operation of the engine is frictionless.

3

Mention the various processes of dual cycle. [April 1996] BTL1

(i) Isentropic compression.

(ii) Constant pressure heat supplied.

(iii) Isentropic expansion, and



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(iv) Constant pressure heat rejection.

4 Define air standard cycle efficiency. [Dec.2012&May 2014] BTL1

Air standard efficiency is defined as the ratio of work done by the cycle to heat supplied to the

cycle.

5

Define mean effective pressure as applied to gas power cycles. [Dec.2017&May 2016] BTL1

Mean effective pressure is defined as the constant pressure acting on the piston during the working

stroking. It is also defined as the ratio of work done to the stroke volume or piston displacement

volume.

6

Define the following terms (i) Compression ratio (ii) Cut off ratio and (iii) Expansion ratio?

[May 2014] BTL1

(i) Compression ratio is defined as the ratio between total cylinder volumes to clearance volume.

(ii) Cut off ratio is defined as the ratio of volume after the heat addition to volume before the heat

addition.

(iii) Expansion ratio is the ratio of volume after the expansion to the volume before expansion.

7 Which cycle is more efficient with respect to the same compression ratio? [Oct.1995] BTL4

For the same compression ratio, Otto cycle is more efficient than diesel cycle.

8

For the same compression ratio and heat supplied, state the order of decreasing air standard

efficiency of Otto, diesel and dual cycle. [Dec.2013] BTL2

η Otto > η Dual > η Diesel

9 Name the factors that affect air standard efficiency of Diesel cycle. [April 1997] BTL1

Compression ratio and cut-off ratio.

10

What is the effect cut-off ratio on the efficiency of diesel cycle when the compression ratio is

kept constant? [April 2003& Nov.2015] BTL4

When cut-off ratio of diesel cycle increases, the efficiency of cycle is decreased when compression

ratio is kept constant and vice versa.

11

Define Wet and Dry steam. BTL2

The steam which partially evaporated and having water particles in suspension is called Wet Steam.

The steam which is completely in evaporated state without any water particles is called Dry steam.

12

Write the formula for calculating entropy change from saturated water to superheat steam

condition. [April 1999] BTL3

sup

supEntropy of Superheated steam, S logg ps c

s

TS C

T

where

Sg – entropy of dry steam

Tsup-Super heated temperature

Ts-Saturated temperature

Cps-Specific heat of super heated steam



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13

Find the mass of 0.1 m3 of wet steam at a temperature of 1600 and 0.94 dry. [Oct.1998] BTL4

From steam table at 1600C

Vg=0.30676 m3/kg

Specific volume of wet steam = x,vg=0.94x0.30676 m3/kg

=0.2884 m3/kg

Volume of given west steam 0.1 of steam, m=

volume of wet steam 0.2884

0.35

MassSpecific

M kg

14

One kg of steam at 10 bar has an enthalpy of 2500kJ/kg. Find its quality. BTL4

H=2500kJ/kg

H=hr+xhfg

At 10 bar from steam tables

Hf = 762.6kJ/kg;hfg=2013.6kJ/kg

2500=762.6+x+2013.6 2500 762.6

0.8622013.6

x

15

Define the term Efficiency ratio. BTL1

The ratio of actual cycle efficiency to that of ideal cycle efficiency is termed efficiency ratio.

Efficiency ratio = Actual cylcle efficiency

Ideal rankine efficiency

16

Define the term Isentropic efficiency. BTL1

For an expansion process

Isentropic efficiency = Actual work done

Isentropic work done

For an compression process

Isentropic efficiency =Isentropic work done

Actual work done

17

Give the effects of Condenser pressure on the Rankine Cycle. BTL2

By lowering the condenser pressure, we can increase the cycle efficiency. The main

disadvantages are lowering the backpressure is to increases the wetness of steam. Isentropic

compression of wet vapour is very difficult.

18

Mention the improvements made to increase the ideal efficiency of Rankine Cycle.

[June 2014] BTL3

1. Lowering the Condenser pressure.

2. Superheated steam is supplied to the turbine.

3. Increasing the boiler pressure to certain limit.

4. Implementing reheat and regeneration in the cycle.

19

List the advantages of Reheat cycle. [Oct.1997] BTL1

1. Marginal increase in thermal efficiency.

2. Increase in work done per kg of steam which results in reduced size of boiler and auxiliaries

for the same output.

3. It prevents the turbine from erosion.

20 Give the function of feed water heaters in the Regenerative cycle with bleeding. [Oct.1999]

BTL2

The main function of feed water heater is to increase the temperature of feed water to the



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saturation temperature corresponding to the boiler pressure before it enters into the boiler.

PART * B

Q.No. Questions

1

In an air standard Otto cycle the compression ratio is 7, and the compression begins at 35 o C,

0.1 Mpa. The maximum temperature of the cycle is 1100 oC. Find the

(a) the temperature and pressure at the cardinal points of the cycle.

(b) the heat supplied per kg of air,

(c) the work done per kg of air,

(d) the cycle efficiency, and

(e) the m.e.p of the cycle. (13 M) BTL3

Answer : Page No:1.17

Solution:

(2M)

(2M)



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(5M)

(4M)

2

A six cylinder petrol engine has a compression ratio of 5:1. The clearance volume of each

cylinder is 110CC. It operator on the four stroke constant volume cycle and the indicated

efficiency ratio referred to air standard efficiency is 0.56. At the speed of 2400 rpm. It

consumer 10kg of fuel per hour. The calorific value of fuel is 44000KJ/kg. Determine the

average indicated mean effective pressure. (13 M) [April 1995] BTL3


Given:

r = 5

Vc =110 cc

ɳ relative = 0.56

N = 2400rpm

Mf = 1okg

H r= 10/3600 kg/s



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Cv= 44000 KJ/kg

Z =6

Compression ratio:

r = (Vs + Vc)/Vc → 5 = Vs + 110/110 → Vs = 440CC = 44x10−6𝑚3 (2M)

Air standard efficiency:

ɳ = 1- 1 / (𝑟𝛾−1) = 47.47% (𝛄 = 1.4) (2M)

Relative efficiency:

ɳ relative = ɳ actual/ ɳ air- standard → 0.56 = ɳ actual/47.47

ɳ actual = 26.58% (3M)

Actual efficiency = work output/ head input

0.2658 = W/ mf Cv → W = 0.2658 x 10/3600x44000

W = 32.49kW. (3M)

The network output:

W = Pm x Vs x N/60 x Z → 32.49𝑥103 = Pm x 440 x10−6 x 1200/60 x 6

Pm = 6.15 bar (3M)

3

In an air standard diesel cycle, the pressure and temperature of air at the beginning of cycle is

1 bar and 40⁰C. The temperatures before and after the heat supplied are 400⁰C and 1500⁰C.

Find the air standard efficiency and mean effective pressure of the cycle. What is the power

output if it makes 100 cycles / min? (13 M) [Oct.1998] BTL3


Given:

P1 = 1 bar = 100KN/m2

T1 = 40⁰C = 313K

T2 = 400⁰C = 673K

T3 = 1500⁰C = 1773K



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Solution:

1-2 isentropic compression:

T2/T1 = (r) 𝛄-1

Compression ratio:

r = V1/V2 = (T2/T1) 1/𝛄-1 = (673/313) 1/1.4-1 = 6.779 (2M)

2-3 constant pressure heating:

V2/T2 = V3/T3

Cut off ration,

P = V3/V2 = T3/T2 = 1773/673 = 2.634 (2M)

Efficiency:

ɳ = 1- 1/𝛄 (r) 𝛄-1(p𝛄-1/p-1) = 0.4142% (2M)

Mean effective pressure:

Pm = P1 r𝛄 [(p-1) –r 1-( p𝛄-1)]/( 𝛄-1) ( r -1)

=100 x (6.779)1.4[1.4 (2.634-1) – (6.779)1-1.4(2.634 1.4-1)]/ (1.4-1) x (6.779-1)

Pm = 597.77KN/m2 (2M)

Heat supplied:

m x Cp(T3– T2) = 1 x 1.005 (1773 -673)

Qs = 1105.5 KJ /kg (2M)

Work done:

ɳ x Qs = 0.4142 x 1105.5 = 457.89KJ/kg (1M)

Power:

= W x cycle /min = 457.89 x 100 =45 x 10-3 KJ/kg-min = 763.16KJ/kg-sec

= 763.16W/kg (2M)

4 An air standard diesel cycle the compression ratio is 15, and the compression begins at 0.1

Mpa at 40 o C. The heat added is 1.675 MJ/kg. Find (a) the maximum temperature of the

cycle, (b) the work done per kg of air, (c) the cycle efficiency, (d) the temperature at the end of



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isentropic expansion, (e) cut-off ratio, (f) the maximum pressure of the cycle, and (g) the m.e.p

of the cycle. (13 M) BTL3

Answer : Page No: 1.61

(3M)

(3M)

(3M)



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(2M)

(2M)

5

An air standard limited pressure cycle the compression ratio is 15, and the compression

begins at 0.1 MPa at 40 oC, the maximum pressure is limited to 6 MPa. And the heat added is

1.675 MJ/kg. Compute (a) the heat supplied at constant volume per kg of air, (b) the heat

supplied at constant pressure per kg of air, (c) the work done per kg of air, (d) the cycle

efficiency, (e) the temperature at the end of constant volume heating process, (f) cut-off ratio,

and (g) the m.e.p of the cycle. (13 M) [Dec.2013] BTL4




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(2M)

(2M)

(2M)

(2M)



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(2M)

(3M)

6

In a gas turbine plant working on a Brayton cycle the compression ratio is 7, and the

maximum temperature is 800 o C. The compression begins at 0.1 mpa., 35 o C. Find (a) the

heat supplied per kg of air, (b) the net work done per kg of air, (c) the cycle efficiency, and (d)

the temperature at the end of expansion process. (13 M) [May 2003] BTL3




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(2M)

(4M)

(5M)

7

An air standard dual cycle has a compression ratio of 16 and compression begins at 1 bar and

50⁰C. The maximum pressure is 70 bar. The heat transformed to air at constant pressure is

equal to heat transferred at constant volume. Find the temperature at all cardial points, cycle

efficiency and mean effective pressure take Cp= 1.005KJ/kgK, Cv = 0.718KJ/kgK. (13 M)



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BTL3


Given data:

P1 = 1 bar

R = 12

T1 =300K

K = 3% of Vs = 0.03Vs

P3 = 70 bar

D = 25 cm

L = 30cm

Solution:

Specific volumes:

V1 RT1/ P1 = 287 x 300/ 1 x 105

= 0.861 m3 /kg

V3 = V2 = V1/r = 0.861/12

= 0.07175 m3/kg

V4 – V3 = 0.03 (V1 –V2)

V4 = 0.0954275 m3/kg (2M)

Cut off ratio:

P = V4 /V3 = 0.054275/0.07175

P = 1.33 (1M)

1-2 isentropic compression process:

P2 = ( r) 𝛄 x P1 = ( 12) 1.4 x 1

= 32.423 bar

V2 = ( r )𝛄-1 x T1 = ( 12) 1.4 -1 x 300

T2 = 810.57K (2M)

2-3 constant volume heat addition process

P3/T3 = P2/T2

T3 = ( P3/P2) x T2 = ( 70 / 32.423) x 810.57

T3 = 1750K (2M)

3-4 constant pressure heat addition process:

T4 = ( V4/V3) x T3 = ( 0.0954275 / 0.07175 ) x1750



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T4 = 2327.5 K

Pressure ratio, K = (P3/P2) = 70/32.423 = 2.159

Net heat supplied to the cycle:

QS = Cv ( T3 – T2) + Cp ( T4-T3)

= 0.718 ( 1750 -810.57 ) + 1.005( 2327.5-1759)

= 1254.9 KJ/kg (2M)

Efficiency of the cycle:

ɳ = 1- 1/ ( r ) 𝛄-1 [ ( K x P𝛄-1)/(k-1) + K𝛄(p-1)]

ɳ = 61.92%

Net workdone of the cycle:

W = ɳ x Qs

= 0.6192 x 1254.9

= 777.1 KJ/kg (2M)

Mean effective pressure,

Pm = W/ V1 – V2

= 777.1/ 0.361 – 0.07115

= 984.6 KPa

Pm = 9.846 bar (2M)

8

In a gas turbine plant working on a Brayton cycle the air at the inlet is at 27 o C, 0.1 Mpa.

The pressure ratio is 6.25 and the maximum temperature is 800 o C. The turbine and the

compressor efficiencies are each 80%. Find (a) the compressor work per kg of air, (b) the

turbine work per kg of air, (c) the heat supplied per kg of air,(d) the cycle efficiency, and (e)

the turbine exhaust temperature. (13 M) [Dec.2016] BTL3




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(5M)

(5M)

(3M)

9

A stationary gas turbine power plant working on a Brayton cycle delivers 20 MW to an

electric generator. The maximum temperature is 1200 K and the minimum 290 K. the

minimum pressure is 95 kPa and the maximum 380 kPa. Determine (a) the power output of

the turbine, (b) the fraction of the output of the turbine used to drive the compressor, (c) the

mass flow rate of air to the compressor, and (d) the volume flow rate to the compressor.

(13 M) [Dec.2017] BTL4




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(3M)

(3M)



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(3M)

(4M)

9

Determine the Rankine cycle efficiency working between 6 bar and 0.4 bar when supplied

with dry saturated steam. By what percentage is the efficiency increased by supplying

superheated steam of 3000C? (13 M) [April/May15] BTL5


At 0.4 bar steam table values (2M)

(1) Rankine efficiency with dry saturated steam: S1=S2, x2=0.86, (3M)

(2) h2= h f2+ x2h fg2= 2312.212 kJ/kg (3M)

(3) Wp= v f2( p2 – p1 ) =0.575 kJ/kg (2M)

(4) Rankine efficiency=(h1- h2) /(h1- h f2)



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ɳ ran =18.16% (3M)

Consider a steam power plant operating on an ideal reheat Rankine cycle. The steam enters

the H.P turbine at 30 bar and 350C. After expansion to 5 bar, the steam is reheated to 350C

and then expanded the L.P turbine to the condenser pressure of 0.075 bar. Determine the

thermal efficiency of the cycle and the quality of the steam at the outlet of the L.P turbine.

(13 M) (Apil/May 15 & May/June14] BTL3


Using steam tables At 30 bar and 3500C, find values

h1=3115.3 kJ/kgK, S1=S2, x2=0.98, h2=2706.56 kJ/kgK, h3=3167.7 kJ/kgK, s3=s4, x4=0.919,

h4=2380.89 kJ/kgK, h5=168.79 kJ/kgK, (6M)

Wp= vf4( p1– p4 ) =3.0164 kJ/kg, (3M)

Efficiency = (h1-h2)/(h1-hf2) = 35% (4M)

Ans: x4 = 0.919, ɳ rh = 35%

PART * C

Q.No. Questions

1

One kg of air taken through, a) Otto cycle, b) Diesel cycle initially the air is at 1 bar and 290

K. The compression ratio for both cycles is 12 and heat addition is 1.9 MJ in each cycle.

Calculate the air standard efficiency and mean effective pressure for both the cycles. (15 M)

[Oct.1995] BTL4


Given data:

P1 = 1 bar = 100KN mᶟ

T1 = 290K

r = 12

Qs = 1.9MJ = 1900KJ

Solution:

a) Otto cycle:



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For process 1-2: isentropic compression:

P₂/P₁ = ( V₁/V₂) → P₂ = P₁ x 𝑟𝛾

P₂ = 3242.3kN/m2

T₂/T₁ = ( V₁/V₂) 𝛾−1 → T₂ = T₁ x ( V₁/V₂) 𝛾−1 = 290 x (12)1.4-1

T₂ = 783.55K (2M)

Heat supplied:

Q = m x Cv ( T₃ - T₂ )

1900 = 1 x 0.718 x (T₃ - 783.55)

T₃ = 3429.79K (1M)

For process 2-3 : constant volume process

P₃/P₂ = T₃/ T₂ → P₃ = P₂ x T₃/ T₂ = 3242.3 x 3429.79/783.55

P = 14196.7KN/m2 (2M)


ɳ = 1- 1 / (𝑟𝛾−1) = 0.6298

ɳ = 62.98% (1M)

Pressure ratio, K = P₃/P₂ = 14196.7/32423 = 4.378

Mean effective pressure,

Pm = p₁ r ( k-1/ 𝛾−1) (𝑟𝛾−1-1/r-1) = 100 x 12 ( 4.378-1/1.4) [ ( 12 1.4-1-1/12-1)]

Pm = 1567.93KN/m2 (1M)

b) Diesel cycle:



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Consider 1-2 isentropic compression process:

T2 = (V1/V2) -1 x T1 = ( r ) 𝛄-1 x T1 = (12) 1.4-1 x 290

T2 = 783.56K (2M)

Consider 2-3 constant pressure heat addition:

Qs = Cp ( T3 – T2 )

1.9 x 10 3 = 1.005 x ( T3 – 783.56 )

T3 = 2674 K (1M)

Cut off ratio:

P = V3/V2 =T3/T2 = 2674/783.56 = 3.413


ɳ = 1-1/𝛄 (r) 𝛄-1 { P𝛄-1/p-1} = 1-1/ 1.4(12) 1.4-1 {3.413 1.4-1/3.413-1}

ɳ = 49.86% (2M)


Pm = P1r𝛄 [ (p-1) –r 1-𝛄 ( p𝛄-1)/ (𝛄-1) ( r-1 )]

100 x (12) 1.4[ 1.4 (3.413-1) –(12)1.4-1( 3.413 1.4-1)]/(1.4-1) (12-1)

Pm = 1241KN/m2 (3M)

2

The pressure, temperature and volume of air at the beginning of dual cycle are 1.03 bar, 35⁰C

and 150 liters respectively. The volume after compression is 10 liters 42KJ of heat is added to

constant volume and 63KJ at constant pressure. Determine air standard efficiency, clearance

and cut off percentage. (15 M) [April 1998] BTL3


Given data:



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P1 = 1.03 bar = 1.03 x 105 N/m2

T1 = 35⁰C = 308K

V1 = 150lit = 0.15 m3

V2 = 10 lit = 0.010 m3 = V3

Qs1 = 42KJ, Qs2 = 63 KJ

Compression ratio (r) = V1/V2 = 0.15/0.01 = 15

For process 1-2 isentropic relation:

T2/T1 = ( V1/V2) 𝛄-1 = ( r ) 𝛄-1

T2 = ( 15) 1.4-1 x 308 = 909.8 K

P2/P1 = (r )

P2 = ( 15) 1.4 x 1.03 x 105

P2 = 45.6 x 105 N/m2 (2M)

PV = mRT

m =P1V1/RT1

= 1.03 x105 x 0.15/287 x 308

= 0.17kg (1M)

R = 287 J/kgk

Qs1 = m x Cv ( T3 –T2 )

42 = 0.17 x 0.718 x ( T3 – 909.8 )

T3 = 1253.89K (1M)

Qs2 = m x Cp ( T4 – T3 )

63 = 0.17 x 1.005 ( T4 – 1253.89 )

T4 = 1622.6K (2M)

For process 2-3 constant volume process:

P2/P3 = T2/T3

P3 = T3/T2 x P2

= 1253.89/909.8 x 45.6 x 105

P3 = 62.9 x 105 N/m2 (2M)

From process 3-4 constant pressure heat addition:

Cut off ratio, P =V4/V3 = T4/T3 = 1622.6/1253.89 = 1.294

Expansion ratio, K = P3/P2 = 62.9/45.6 = 1.37 (1M)




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ɳ dual = 1- 1/ ( r ) 𝛄-1 [ ( K x P𝛄-1)/(k-1) + K𝛄(p-1)]

ɳ dual = 65%

% of clearance volume

V2/V1 x 100 = 0.01/0.15 x 100 =6.6% (1M)

For process 3-4 constant pressure process:

V3/T3 = V4/T4

V4 = T4/T3 x V3

= 1622.6/1253.89 x 0.01

V4 = 0.0189 m3/kg (2M)

Cut off volume = V4 –V3 = 0.0189 – 0.01 = 0.0089m3

% of cut off volume = 0.0089/0.15 x 100 = 5.93 % (3M)

3

A regenerative cycle utilizes steam as the working fluid. Steam is supplied to the turbine at 40

bar and 450ºC and the condenser pressure is 0.03 bar. After expansion in the turbine to 3 bar,

some of the steam is extracted from the turbine for heating the feed water from the condenser

in an open heater. The pressure in the boiler is 40 bar and the state of the fluid leaving the

heater is saturated liquid water at 3 bar. Assuming isentropic heat drop in the turbine and

pumps, compute the efficiency of the cycle. (15 M) BTL5


Use superheated steam tables at 40 bar and 450ºC

h1= 3330.3 kJ/kg, S1= 6.9363 kJ/kgK (2M)

At p 2 = 3 bar taking all the values x2=0.9895, h2 = 2702.65 kJ/kg

x3=0.8, h3 = 2057.63 kJ/ (2M)

h4= hf3 = 101.05 kJ/kg, (2M)

pump Work : (1-m) (h5– h4) = (1-m)*vf3 (p 2 - p 3 )

h5= 101.35 kJ/kg (3M)

Amount of steam bleed m=(hf2– h5)/ (h2– h5) = 0.117 kg (3M)

Wp 6-7 = (h7– h6) = vf2 (p 1 - p 2) = 565.44 kJ/kg (2M)

Regenerative Rankine efficiency= [(h1– h7)-(1-m) (h3– hf3)] /(h1– h7)

= 41.75% (3M)

4 If engine working on dual cycle, the temperature and pressure at the beginning of the cycle

are 90⁰C and 1 bar. The compression ratio is 9. The maximum pressure is limited to 68 bar



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and total heat supplied per kg of air is 1750KJ. Determine air standard efficiency and mean

effective pressure. (15 M) [May 2011&2012] BTL3


Given data:

P1 = 1 bar

T1 = 90⁰C

P3 = P4 = 68 bar

r = 9

Qs = 1750KJ/kg

Solution:

1-2 isentropic compression process:

P2 = ( r ) 𝛄 x P1 = (9) 1.4 x 1

= 21.67 bar

T2 = ( r ) 𝛄 -1 x T1 =(90) 1.4-1 x 363 = 874K (3M)

2-3 constant volume heat addition process:

T3 = (P3/P2) xT2 = (68/21.67) x 874 = 2743K (1M)

3-4 constant volume heat addition process:

Qs = Cv (T3 – T2) + Cp ( T4 – T3)

1750 = 0.718 ( 2743 -874) + 1.005 ( T4 – 2743)

T4 = 3149K

V1 = RT1/P1 = 287 x 363/ 1 x 105 = 1.04181 m3/kg

V3 = V2 =V1 /r = 1.04181/9 = 0.11576m3/kg

V4= (T4/T3) x V3 = ( 3149/2743) x 0.11576

= 0.132894 m3/kg (4M)

Cut of ratio:

P = V4/V3 = 0.132894/0.11576 = 1.148

Pressure ratio K = P3/P2 = 68/21.67 = 3.138

Efficiency of the cycle:

ɳ = 1- 1/ ( r ) 𝛄 -1 [ ( K x P 𝛄 -1)/(k-1) + K 𝛄 (p-1)]

ɳ = 58.19% (3M)

Net work done of the cycle:



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W net = ɳ x Qs

= 0.5819 x 1750

= 1018.33KJ/kg (2M)


Pm = W net /(V1-V2) = 1018.33/1.04181 – 0.11576

Pm = 10.98 bar (2M)

5

Find the efficiency of the prime mover operating on the Rankine cycle between 7 bar and 1

bar for the following initial conditions. (i) The steam has a dryness fraction of 0.8, (ii) The

steam is dry and saturated and (iii) The steam is superheated to 350ºC. Draw the T-s diagram

for each case. Neglect the pump work. (15 M) BTL5


Using Steam tables at 7 bar Ts1 = 164.9ºC and

take all values of hf1, hfg1, sf1, sfg1,sg1 (3M)

At 1 bar Ts2 = 99.63 ºC and hf2, hfg2, sf2, sfg2,sg2 (3M)

(i) if x = 0.8 x2= 0.736, h2=2079.85 kJ/kg,

Rankineefficiency=(h1–h2)/(h1–hf2)=13.93% (3M)

(ii) When steam is dry x2=0.89, h2= 2427 kJ/kg,

Rankine efficiency = 14.28 % (3M)

(iii) x2>1 superheated steam, h2=2272.246 kJ/kg

Rankine efficiency= 16.075 % (3M)



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Subject Name: Thermal Engineering-I Subject Handler: J. Ravikumar &S.A.ArokyaAnicia

UNIT II RECIPROCATING AIR COMPRESSOR

Classification and comparison, working principle, work of compression - with and without clearance,

Volumetric efficiency, Isothermal efficiency and Isentropic efficiency. Multistage air compressor with

Intercooling. Working principle and comparison of Rotary compressors with reciprocating air

compressors.

PART * A

Q.No. Questions

1.

Classify the various types of air compressors. [Dec.2003&Nov.2010] BTL2

1. According to the and principle of operation

a)Reciprocating compressors

b) Rotary compressors.

2) According to the action

a) Single acting compressors

b) Double acting compressors

3) According to the number of stages

a) Single stage compressors

b) Multistage compressors

4) According to the pressure limit

a)Low pressure compressors

b )Medium pressure compressors

c) High pressure compressors

5) According to the capacity

a)Low capacity compressors

b) Medium capacity compressors

c) High capacity compressors

2

Write a short note on single acting compressors? BTL1

In single acting reciprocating compressor, the suction, compression and delivery of air takes place

on one side of the piston

3

What is meant by single stage compressor? BTL1

In single stage compressor, the compression of air from the initial pressure to the final .pressure is

carried out in one cylinder only.

4

Mention a note on double acting compressor? BTL2

In double acting reciprocating compressor, the suction, compression and delivery of air takes place

on both sides of the piston.

5

Indicate the application of reciprocating compressors in industry? [Nov.2004] BTL2

The applications of compressed air as follows:

1) Pneumatic brakes

2) Pneumatic jakes



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3) Pneumatic drills

4) Pneumatic lifts

5) Spray painting

6) Shop cleaning

7) Injecting fuel in diesel engines

8) Supercharging internal combustion engines

9) Refrigeration and air conditioning systems

6

List out the advantages of Multi stage compression with Internal cooling over single stage

compression for the same pressure ratio. [Dec.2013&May.2015&2018] BTL4

1. It improves the volumetric efficiency for the given pressure ratio.

2. It reduces the leakage loss considerably.

3. It gives more uniform torque and hence a smaller size flywheel is required.

4. It reduces the cost of the compressor.

7

Define the terms as applied to air compressors: Volumetric efficiency and isothermal

compression efficiency. [April 2005] BTL1 (or)

Define the mechanical efficiency and isothermal efficiency of a reciprocating air compressor.

Volumetric efficiency:

Volumetric efficiency is defined as the ratio of volume of free air sucked into the compressor per

cycle to the stroke volume of the cylinder.

Volumetric efficiency: Volume of free air taken per cycle/Stroke volume of the cylinder.

Isothermal compression efficiency:

Isothermal efficiency is defined as the ratio between isothermal work to the actual work of the

compressor.

Isothermal efficiency = brake power / Indicated power

8

Define clearance ratio. BTL2

Clearance ratio is defined as the ratio of clearance volume to swept volume (or) stroke volume.

C =Vc/Vs

Vc = Clearance volume

Vs = Swept volume

9 Discuss the effect of clearance upon the performance of an air compressor. [Oct.1999] BTL2

The volumetric efficiency of air compressor increases with decrease in clearance of the compressor.

10

Give two merits of rotary compressor over reciprocating compressor. [May 2011] BTL2

1. Rotary compressor gives uniform delivery of air where compared to reciprocating compressor.

2. Rotary compressors are small in size for the same discharge as compared with reciprocating

compressors.

3. Lubricating system is more complicated in reciprocating compressor where as it is very simple in

rotary compressor.

11 Name the methods adopted for increasing isothermal efficiency of reciprocating air

compressor. BTL3

Isothermal efficiency is increased by perfect inter cooling.

12

Why clearance is necessary and what is its effect on the performance of reciprocating

compressor? [June 1999&Dec.2017] BTL4

When the piston reaches top dead center in the cylinder, there is a dead space between piston top

and cylinder head. This space is known as clearance space and the volume occupied by this space is



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known as clearance volume.

13

What is meant by inter cooler? BTL1

An inter cooler is a simple heat exchanger. It exchanges the heat of compressed air from the low-

pressure compressor to the circulating.

14

Give the factors that affect the volumetric efficiency of a reciprocating compressor?

[April 1998] BTL1

1. Clearance volume.

2. Compression ratio.

15 Define compression ratio. BTL1

Compression ratio is defined as the ratio between total volume and clearance volume.

Compression ratio = Total volume/Clearance volume.

16 Define volumetric efficiency of an air compressor. [May & Nov.2015] BTL5

Volumetric efficiency is defined as the ratio of volume of free air sucked into the compressor per

cycle to the stroke volume of the cylinder (Va/Vs).

17

Why clearance is necessary and what is its effect on the performance of reciprocating

compressor? [April 1998&Oct.1999] BTL2

When the piston reaches top dead center in the cylinder, there is a dead space between piston top

and cylinder head. This space is known as clearance space and the volume occupied by this space is

known as clearance volume.

18

List the effects of multi stage compression? [May 2015&2018] BTL1

(i) The work done per kg of air is reduced in multistage compression with intercooler as

compared with single stage compression for the same delivery pressure.

(ii) It improves volumetric efficiency for a given pressure ratio.

(iii) It reduces leakage loss considerably.

(iv) It gives more uniform torque and hence, a smaller size flywheel is required.

(v) It provides effective lubrication because of lower operating temperature

(vi) It reduces the cost of the compressor.

19 Define the term isothermal efficiency. [Nov.2007 & May 2018] BTL1

The ratio of the work required to compress a gas isothermally to the work actually done by the

compressor.

20

Differentiate positive and non-positive displacement compressors. BTL4

Positive displacement compressor is one in which air is compressed adiabatically. The air is

entrapped in between two sets of engaging surfaces. The pressure rise is either by back flow of air

(as in roots blower) or both by variation in the flow and back flow (as in vane blower).

In non-positive displacement compressor, air is not trapped in specific boundaries but it flows

continuously and steadily through the machine (as in centrifugal compressor and axial flow

compressor).

Part*B

Q.No. Questions

1

Explain the working principle of Reciprocating air compressor with neat sketch. (13 M)

[April 2004]

BTL2



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Sketch:

(6M)

Description:

• During the suction stroke the compressor piston starts its downward stroke and

the air underpressure in the clearance space rapidly expands until the pressure falls below

that on the oppositeside of the inlet valve.

• This difference in pressure causes the inlet valveto open into the cylinder until the piston

reaches the bottom of its stroke (Figure 2C).During the compression stroke the piston starts

upward, compression begins, and at point D hasreached the same pressure as the compressor

intake. The spring-loaded inlet valve then closes.

• As the piston continues upward, air is compressed until the pressure in the cylinder

becomesgreat enough to open the discharge valve against the pressure of the valve

springs and the pressure of the discharge line. From this point, to the end of the stroke, the

air compressed within the cylinder is discharged at practically constant pressure. (7M)

2

Derive the expression for work done on reciprocating air compressor with clearance volume.

(13 M) [May 2011,2012&2016] BTL3

Answer: Page No:2.6

Derivation:

Work done in a single stage reciprocating compressor with clearance volume:

p1, V1, T1 = Initial Pressure, Volume, Temperature respectively

p2, V2, T2 = Final Pressure, Volume, Temperature respectively

Vc = Clearance Volume; Vs = Stroke Volume = V1-Vc

n = Polytropic Index (4M)



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(4M)

W= Area 1-2-3-4-1

W= Work done during compression – Work done during expansion

W=n

n−1p1V1 [(

p2

p1)

n−1

n − 1] −n

n−1p4V4 [(

p3

p4)

n−1

n − 1]

With substitution

(5M)

3

Derive the expression for work done by reciprocating air compressor without clearance

volume. (13 M) BTL3


Work done by reciprocating compressor without clearance volume is an ideal condition in

which four process are taken to consideration

a) Isothermal Compression (pV = Constant)

b) Polytropic Compression (pVn = Constant)

c) Isentropic Compression (pVγ= Constant)



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(3M)

a) Isothermal Compression (pV = Constant)

The Process involved are:

Process 1-2: Air is compressed isothermally from Pressure p1 to p2.

Process 2-3: Discharge of air at pressure p2

Process 4-1: Represents the suction of air at pressure p1.

Work done = Area 1-2-3-4-1

W = Wcompression + Wdelivery - Wsuction = p1V1 ln (V1

V2) + p2V2 − p1V1

For Isothermal process p1V1 = p2V2

With substitution W = mRT1 ln (p2

p1) (4M)

b) Polytropic Compression (pVn = Constant)


Process 1-2: Air is compressed polytropically from Pressure p1 to p2.





V2) + p2V2 − p1V1

For polytropicprocevss Wcomp = p2V2−p1V1

n−1

With substitution W = n

n−1mRT1 ln[(

p2

p1)

n−1

n − 1] = n

n−1p1V1 ln[(

p2

p1)

n−1

n − 1] (3M)



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c) Isentropic Compression (pVγ= Constant)


Process 1-2: Air is compressed isentropically from Pressure p1 to p2.





V2) + p2V2 − p1V1

For polytropic processWcomp= p2V2−p1V1

γ−1

With substitution W = n

n−1mRT1 ln[(

p2

p1)

γ−1

γ − 1] = n

n−1p1V1 ln[(

p2

p1)

γ−1

γ − 1] (3M)

4

Explain with neat sketch the working principle of Multistage Reciprocating Compressor.

(13M) [April 2008 & Dec.2013] BTL1

Answer: Page No:2.53

The Multi-stage compressor is serious of cylinder arrangement with the output of first cylinder is

given as the input to the second cylinder with or without intercooling regards to the requirement.

Intercooling:

The compression ratio increases as with the temperature to avoid the heat being

transferred to the second stage to decline the efficiency an intercooler will be installed to

increase the efficiency. (3M)



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(10M)

5

Derive the expression for work done by multistage reciprocating compressor. (13 M)

[Dec.2017] BTL3




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(4M)

Two stage Compressor:

Without Intercooling:

8-1-4-7 - Low Pressure Cycle.

7-4-5-6 – High Pressure Cycle.

Work done W = n

n−1p1V1 [(

p4

p1)

n−1

n − 1] +n

n−1p4V4 [(

p5

p4)

n−1

n − 1] (4M)

With perfect intercooling:

8-1-4-7 - Low Pressure Cycle.

7-2-3-6 – High Pressure Cycle.

W = n

n−1p1V1 [(

p4

p1)

n−1

n − 1] +n

n−1p2V2 [(

p3

p2)

n−1

n − 1] (4M)

Reducing

W = n

n−1p1V1 [(

p2

p1)

n−1

n + (p3

p2)

n−1

n − 2] (1M)

6

Reciprocating air compressor has cylinder with 24 cm bore and 36 cm stroke. Compressor

admits air at 1 bar, 17°C and compresses it up to 6 bar. Compressor runs at 120 rpm.

Considering compressor to be single acting and single stage determine mean effective pressure

and the horse power required to run compressor when it compresses following the isothermal

process and polytrophic process with index of 1.3. Also find isothermal efficiency when

compression is of polytrophic and adiabatic type. (13 M) [Nov.1994] BTL4


Vs = (π/4) D2L = 0.0162 m3

Work done by isothermal Process: Wiso = p1V1 ln (p2

p1)

Wiso= 2.902 kJ [(Wiso x N)/60]

Wiso= 5.80 kW or 7.77 hp (3M)

Work done by polytropic process:



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Wpolyn

n−1p1V1 ln[(

p2

p1)

n−1

n − 1]

Wpoly = 3.594kJ

Wpoly = 7.188 kW or 9.63 hp (3M)

Mean Effective Pressure: MEP = Work done / Swept Volume

MEPiso = 179.13 kPa ;

MEPpoly = 221.85 kPa (3M)

Work done by isentropic,

Wadia = n

n−1p1V1 ln[(

p2

p1)

γ−1

γ − 1] = 3.790 kJ (3M)

Isothermal Efficiency: isothermal work done / Actual work done

Isothermal η for polytropic process = 0.807 or 80.7%

Isothermal η for adiabatic process = 0.765 or 76.5% (1M)

7

A single stage single acting reciprocating air compressor has air entering at 1 bar, 20°C and

compression occurs following polytrophic process with index 1.2 upto the delivery pressure of

12 bar. The compressor runs at the speed of 240 rpm and has L/D ratio of 1.8. The

compressor has mechanical efficiency of 0.88. Determine the isothermal efficiency and

cylinder dimensions. Also find out the rating of drive required to run the compressor which

admits 1 m3 of air per minute. (13 M) [Nov.2002] BTL4


(3M)

(2M)

(3M)

(5M)

8 A reciprocating compressor of single stage, double acting type delivers 20m3/min when

measured at free air condition of 1 bar and 27°C. The compressor has compression ratio of 7



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and the conditions at the end of suction are 0.97 bar and 35°C. Compressor runs at 240 rpm

with clearance volume of 5% of swept volume. The L/D ratio is 1.2. Determine the volumetric

efficiency and dimensions of cylinder and isothermal efficiency taking the index of

compression and expansion as 1.25. Also show the cycle on P-V diagram. (13 M) [Dec.2003]

BTL3


m = pV / RT = 0.387 kg/s

p1V1=mRT1

V1 = 0.352 m3/s (2M)

r = V1/V2

V2 = 0.050 m3/s

p1(V1)1.25 =p2(V2)

1.25

p2 = 9.9 bar (3M)

ηvol= 1 +Vc

Vs−

Vc

Vs[(

p2

p1)

1

n] = 0.729 or 72.9 % (2M)

Dimensions of Cylinder;

Va = Vs x ηvol x N

Vs = 0.1143 m3

Vs = (π/4) D2 x L

D = 0.49 m

L = 0.59 m (3M)

Isothermal Efficiency:

Work done, Wiso = p1V1 ln (p2

p1) = 79.316 kW

Work done, Wpolyn

n−1p1V1 ln[(

p2

p1)

n−1

n − 1] = 100.95 kW

Isothermal Efficiency = Isothermal work done / Actual work done = ηiso = 78.56 % (3M)

9

A two stage double acting reciprocating air compressor running at 200 rpm has air entering

at 1 bar, 25°C. The low pressure stage discharges air at optimum intercooling pressure into

intercooler after which it enters at 2.9 bar, 25°C into high pressure stage. Compressed air

leaves HP stage at 9 bar. The LP cylinder and HP cylinder have same stroke lengths and

equal clearance volumes of 5% of respective cylinder swept volumes. Bore of LP cylinder is 30

cm and stroke is 40 cm. Index of compression for both stages may be taken as 1.2. Determine,

(i) the heat rejected in intercooler, (ii) the bore of HP cylinder, (iii) the hp required to drive

the HP cylinder. (13 M) [Nov.2006] BTL3

Answer: Page No: 2.75

Vs = (π/4) x (DL.P) 2 x LL.P = 11.3 m3/min (2M)

ηvol= 1 +Vc

Vs−

Vc

Vs[(

p2

p1)

1

n] = 0.9285 or 92.85 % (2M)

V1= ηvol x Vs = 10.49 m3/min (1M)

p1V1=mRT

m = 12.26 kg/min (2M) T2

T1= (

p2

p1)(n−1)/n

T2 = 355.86 K (1M)

Heat rejected to the intercooler:



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QR = mCp (T2-T5) = 701.6 kJ/min (1M)

Diameter of HP cylinder

Vs = mRT5/p5 = 3.61m3/min

Vs of H.P cylinder = Vs / ηvol

Vs of H.P cylinder = (π/4) x (DH.P) 2 x LH.P = 3.88 m3/min (2M)

DH.P= 0.175 m

Power Required = (n/n-1) mR(T2-T1) = 20kW. (2M)

PART * C

Q.No. Questions

1

In a two stage compressor in which inter-cooling is perfect, prove that the work done in the

compressor is minimum when the pressure in the inter-cooler is geometric mean between the

initial and final pressure. Draw the P-V&T-S Diagram for two stage compression.

(15 M) [AU Nov.2013] BTL4


(8M)

If we look at compressor work then it shows that with the initial and final pressures p1 and p2

remaining same the intermediate pressure p2 may have value floating between p1 and p2 and

change the work requirement Wc. Thus, the compressor work can be optimized with respect to

intermediate pressure p2. Mathematically, it can be differentiated with respect to p2.



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(7M)

2

Explain the construction and working principle of Multi stage compressor and discuss the

perfect cooling with inter cooler. (15 M) [AU Nov. 2013] BTL2


Multistage compression refers to the compression process completed in more than one stage i.e. a

part of compression occurs in one cylinder and subsequently compressed air is sent to subsequent

cylinders for further compression. In case it is desired to increase the compression ratio of

compressor then multi-stage compression becomes inevitable.

Increasing delivery pressure the volume of air being sucked goes on reducing as evident

from cycles 1234 and 12’3’4’. Let us increase pressure from p2 to p2’ and this shall cause the

suction process to get modified from 4–1 to 4’–1.



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(10M)

Apart from the cooling during compression the temperature of air at inlet to compressor can

be reduced so as to reduce compression work.

In multistage compression the partly compressed air leaving first stage is cooled up to

ambient air temperature in intercooler and then sent to subsequent cylinder (stage) for

compression Intercoolers when put between the stages reduce the compression work and

compression is called intercooled compression.

Intercooling is called perfect when temperature at inlet to subsequent stages of compression

is reduced to ambient temperature. (5M)

3

A single acting reciprocating air compressor has a piston dia of 200mm and a stroke of

300mm and runs at 350 rpm. Air is drawn at 1.1 bar pressure and is delivered at 8 bar

pressure. The Law of compression is PV1.35 =constant and clearance volume is 6% of the

stroke volume. Determine the mean effective pressure and the power required to drive the

compressor. (15 M) [AU May 2013] BTL3


Vs = (π/4) D2L = 0.009424 m3

Vc = 0.06 Vs = 5.65 x 10-4 m3

V1 =Vs+Vc = 0.009898 m3 (5M)

Work done by polytropic process:

Wpoly𝑛

𝑛−1𝑝1𝑉1 ln[(

𝑝2

𝑝1)

𝑛−1

𝑛 − 1]

Wpoly = 1665.30 J [(Wpoly x N)/60] = 9.714 kW (5M)

Mean Effective Pressure:

MEP = Work done / Swept Volume



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= 176.70 kN/m3 (5M)

4

Derive the work done by a 2-stage reciprocating compressor with intercooler and derive the

condition for minimum work input and the expression for minimum work required for 2-

stage reciprocating compressor. (15 M) [AU May 2013] BTL4


(13M)

5

Derive the expression for volumetric efficiency of air compressor. (15 M) [AU May / June

2014] BTL4


n

C

n

P

PV

P

PVV

1

1

2

1

1

234

(3M)

Va = V1 – V4

n

ccs

n

caP

PVVV

P

PVVV

1

1

2

1

1

21



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1

1

1

2n

csaP

PVVV

(5M)

s

n

cs

s

av

V

P

PVV

V

V

1

1

1

2

11

1

1

2n

s

c

P

P

V

V

(5M)

n

P

PCC

1

1

21

C

V

V

s

C

(2M)

6

Explain the working principle of intercooler with neat sketch and explain the working

principle of axial flow compressor. (15 M) BTL2


The basic components of an axial flow compressor are a rotor and stator, the former carrying the

moving blades and the latter the stationary rows of blades. The stationary blades convert the kinetic

energy of the fluid into pressure energy, and also redirect the flow into an angle suitable for entry to

the next row of moving blades. Each stage will consist of one rotor row followed by a stator row,

but it is usual to provide a row of so called inlet g This is an additional stator row upstream of the

first stage in the compressor and serves to direct the axially approaching flow correctly into the first

row of rotating blades. For a compressor, a row of rotor blades followed by a row of stator blades is

called a stage. Two forms of rotor have been taken up, namely drum type and disk type. The disk

type is used where consideration of low weight is most important. There is a contraction of the flow

annulus from the low to the high pressure end of the compressor.

This is necessary to maintain the axial velocity at a reasonably constant level throughout the length

of the compressor despite the increase in density of air. Figure illustrates flow through compressor

stages. In an axial compressor, the flow rate tends to be high and pressure rise per stage is low. It

also maintains fairly high efficiency.



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The basic principle of acceleration of the working fluid, followed by diffusion to convert

acquired kinetic energy into a pressure rise, is applied in the axial compressor. The flow is

considered as occurring in a tangential plane at the mean blade height where the blade peripheral

velocity is U . This two dimensional approach means that in general the flow velocity will have two

components, one axial and one peripheral denoted by subscript w, implying a whirl velocity.

It is first assumed that the air approaches the rotor blades with an absolute velocity, , at an angle

to the axial direction. In combination with the peripheral velocity U of the blades, its relative

velocity will be at and angle as shown in the upper velocity triangle. After passing through

the diverging passages formed between the rotor blades which do work on the air and increase its

absolute velocity, the air will emerge with the relative velocity of at angle which is less than

.

This turning of air towards the axial direction is, as previously mentioned, necessary to provide an

increase in the effective flow area and is brought about by the camber of the blades. Since is

less than due to diffusion, some pressure rise has been accomplished in the rotor. The velocity

in combination with U gives the absolute velocity at the exit from the rotor at an angle to

the axial direction. The air then passes through the passages formed by the stator blades where it is

further diffused to velocity at an angle which in most designs equals to so that it is



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prepared for entry to next stage. Here again, the turning of the air towards the axial direction is

brought about by the camber of the blades.

(15M)



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Subject Name :Thermal Engineering- I Subject Handler : J. Ravikumar & S.A.ArokyaAnicia

UNIT III INTERNAL COMBUSTION ENGINES AND COMBUSTION

IC engine – Classification, working, components and their functions. Ideal and actual : Valve and port

timing diagrams, p-v diagrams- two stroke & four stroke, and SI & CI engines – comparison. Geometric,

operating, and performance comparison of SI and CI engines. Desirable properties and qualities of fuels.

Air-fuel ratio calculation – lean and rich mixtures. Combustion in SI & CI Engines – Knocking –

phenomena and control.

PART * A

Q.No. Questions

1. Define compression ratio of an IC engine. (Nov.2004 & April 2005) BTL1

It is the ratio of volume when the piston is at BDC to the volume when the piston is at TDC.

2 Define the terms Mean effective pressure. (Nov.2004) BTL1

It is defined as the algebraic sum of the mean pressure acting on during one complete cycle.

3

What is meant by highest useful compression ratio? (April 1997) BTL1

The compression ratio which gives maximum efficiency is known as highest useful compression

ratio.

4

Why compression ratio of petrol engines is low while diesel engines have high compression

ratio? (Oct.1998) BTL1

Since fire point of petrol is less as compared to diesel, petrol engine has low compression ratio.

5

Compare the thermal efficiency of petrol engines with diesel engines. Give reasons. (April

2000) BTL-1

Thermal efficiency of diesel engine is greater than petrol engine this is due to high compression

ratio.

6

Write a short note on scavenging in I.C. Engines. (May 2003) BTL1

The process of removing the burnt gases from the combustion chamber of engine cylinder by using

fresh air fuel mixture is known as Scavenging.

7 Define Cetane number. (April 2003) BTL 1

The property that quantities the ignition delay is called as Cetane number.

8

Which is better efficient two stroke or four stroke engines? (April 1998)BTL 2

Two-stroke engine give always lesser efficiency than four-stroke engine due to incomplete

combustion and poor scavenging.

9

Define delay period with respect to a CI engine. (Nov.2003) BTL2

The physical delay period is the time between the beginning of injection and the attainment of

chemical reaction reaction conditions. During this period fuel is atomized, mixed with air and raised

to its self-ignition temperature.

During the chemical delay reactions start slowly ad then accelerate until ignition takes place.



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10

Give the purpose of providing spark plug in SI engine? BTL2

The function of a spark plug is to produce an electric spark for the ignition of compressed air-fuel

mixture inside the engine cylinder.

11 What are the factors which contribute to knocking in SI engine? (May 2011 & 2012) BTL2

Temperature factor, Density factor, Time factor and Composition factor

12

State the function of Push rod and Rocker arm. BTL2

The push rod and rocker arm actuate valves according to the engine stroke by cams. They allow the

push rods to push up on the rockers arms and therefore, push down on the valves.

13

State the function of Flywheel. (Nov.2010 & May 2015) BTL2

The flywheel is heavy and perfectly balanced wheel usually connected to the rear end of the

crankshaft. Flywheel serves as energy reservoir. It stores energy during power stroke and releases

energy during other strokes. Thus, it gives a constant output torque.

14

Name the basic thermodynamic cycles of the two types of internal combustion reciprocating

engines. (April 2001) BTL2

Otto cycle is used for SI engines and Diesel or dual cycle is used for CI engine.

15

Define the term valve timing diagram. (May 2016) BTL 1

The exact moment at which each of valves open and closes with reference to the position of piston

and crank can graphically be shown in diagram. This diagram known as valve timing diagram.

16 What are the causes of knock in CI engine? BTL4

1. Long ignition delay

2. Auto-ignition delay

17

Brief the term ignition delay. (Nov.2003) BTL2

In the actual engine cylinder, there is a certain time interval between instant of spark and instant of

pressure rise due to combustion. This time interval or time delay is known as “Ignition lag” or

“Delay period”.

18

Describe the phenomenon of detonation in SI engine. (Nov. 2010 & May 2016) BTL 1

If the temperature of the unburnt mixture exceeds the self-ignition temperature during the ignition

delay periods, auto-ignition occurs at various location in the cylinder. It will generate pressure

pulses. These high pressure pulses can causes damage to the engine and quite often are in audiable

frequency range. This phenomenon is often called knocking or detonation.

19

List the factors that increase the knocking in SI engines. BTL 1

1. Low octane number

2. Lean mixture of air-fuel ratio

3. Decreasing atmospheric humidity

4. Low self-ignition temperature

5. Short ignition delay.

20 What is stoichiometric air-fuel ratio? BTL2



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The amount of air required to burn 1 kg of fuel for making complete combustion knows as air-fuel

ratio. It known as stoichiometric or theoretical air-fuel ratio or minimum quantity of air.

Stoichiometric air-fuel ratio = Amount of air required for complete combustion / Amount of fuel

used.

Part *B

Q.No. Questions

1

Summaries the classifications of IC Engines? (13 M) BTL2


Based on working cycle

• Two stroke engines

• Four stroke engines (1 M)

Based on method of ignition

• Compression ignition engines (C.I engines)

• Spark ignition engines (S.I engines) (1 M)

Based on Fuel used

• Light fuel oil engines (Petrol engines)

• Diesel engines

• Gas engines (2 M)

Based on applications

• Stationary engines

• Portable engines

• Automobile engines

• Marine engines

• Aero engines (2 M)

Based on arrangement of the cylinder

• Horizontal engines

• Vertical engines



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• Radial engines

• V-type engines (2 M)

Based on speed of the engine

• Slow speed engines

• Medium speed engines

• High speed engines (2 M)

Based on number of cylinders

• Single cylinder engines

• Multi-cylinder engines (2 M)

Based on method of cooling

• Water cooled engines

• Air cooled engines (1 M)

2

Explain the Components of I.C engines. (13 M) BTL2


• Cylinder block

• Cylinder head

• Piston assembly

• Connecting rod

• Crank shaft

• Crank case

• Valves and Valve operating mechanism



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• Fuel supply system

• Ignition system

• Lubrication system

• Cooling system

• Inlet and Exhaust system

Cylinder:

• The circular cylinders in the engine block in which the pistons reciprocate back and forth.

• The cylinder block is the main structure for the various components. The cylinder head is

mounted on the cylinder block. The cylinder head and cylinder block are provided with

water jackets in the case of water cooled engines or with cooling fins in the case of air

cooled engines.

• The cylinder head is held tight to the cylinder block by number of bolts and studs.

• The bottom portion of the cylinder block is called crank case.

• The piston reciprocates inside the cylinder and the motion of the piston is transmitted to the

crank shaft by connecting rod and crank assembly. (3M)

Piston rings:

• Metal rings that fit into circumferential grooves around the piston and form a sliding surface

against the cylinder walls.

• Inlet and exhaust valves are provided for suction of charge and removal of exhaust gases

• Inlet manifold is provided on suction side which allows the charge entering the cylinder

during suction process.

• Exhaust manifold is provided on exhaust side which allows the exhaust gases letting to

atmosphere during exhaust process. (3M)

Camshaft:

• Rotating shaft used to push open valves at the proper time in the engine cycle, either directly

or through mechanical or hydraulic linkage (push rods, rocker arms, tappets).

• Push rods: The mechanical linkage between the camshaft and valves on overhead valve

engines with the camshaft in the crankcase. (2M)

Combustion chamber:

• The end of the cylinder between the head and the piston face where combustion

occurs.

• The size of combustion chamber continuously changes from minimum volume when

the piston is at TDC to a maximum volume when the piston at BDC. (2M)

Crankshaft:

• Rotating shaft through which engine work output is supplied to external systems.

• The crankshaft is connected to the engine block with the main bearings.

• It is rotated by the reciprocating pistons through the connecting rods connected to the

crankshaft, offset from the axis of rotation. This offset is sometimes called crank

throw or crank radius. (2M)



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Connecting rod:

Rod connecting the piston with the rotating crankshaft, usually made of steel or alloy forging in

most engines but may be aluminum in some small engines. (1M)

3

Discuss the Theoretical and Actual Valve Timing Diagram of Four stroke engine. (13 M) (May

2011 & 2012) BTL4


• The inlet valve opens at TDC and suction takes place from TDC to BDC.

• At BDC the inlet valve closes and the compression takes place from BDC to TDC.

• At TDC the fuel is fired and the expansion takes place from TDC to BDC.

• At the end of expansion (BDC) the exhaust valve opens and exhaust takes place from BDC

to TDC.



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• In actual practice it is difficult to open and close the valve instantaneously.

• The inlet valve is opened 10o to 30o in advance of the TDC position to enable the fresh

charge to enter the cylinder and to help the burnt gases at the same time, to escape to the

atmosphere.

• The suction of the mixture continues upto 30o to 40o or even 60o after BDC position.

• The inlet valve closes and the compression of the entrapped mixture starts.

• The spark plug produces a spark 30o to 40o before the TDC position, thus fuel gets more

time to burn.

• The pressure becomes maximum nearly 10o past the TDC position. The exhaust valve opens

30o to 60o before BDC position and the exhaust gases are driven out of the cylinder by

piston during its upward movement.

• The exhaust valve closes when piston is nearly 10o past TDC position.

(7M)

Valve Timing Diagram of Four stroke diesel engine



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• The inlet valve opens 10o to 25o in advance of TDC position and closes 25o to 50o after the

BDC position.

• Exhaust valve opens 30o to 50o in advance of BDC position and closes 10o to 15o after the

TDC position.

• The fuel injection takes place 5o to 10o before TDC position and continues upto 15o to 25o

near TDC position. (6M)

4

Discuss the Theoretical and Actual Port Timing Diagram of Two stroke engine. (13 M) (May

2013) BTL4


(5 M)

• The expansion of the charge after ignition starts as the piston moves from TDC towards

BDC.

• First the exhaust port opens before the piston reaches BDC and the burnt gases start leaving



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the cylinder.

• After a small fraction of the crank revolution, the transfer port also opens and the fresh

charge enters into the engine cylinder.

• This is done as the fresh incoming charge helps in pushing out the burnt gases. Now the

piston reaches the BDC and then starts moving upwards.

• As the piston moves little beyond BDC, first the transfer port closes and then exhaust port

also closes. This is done to suck fresh charge through the transfer port and to exhaust the

burnt gases through the exhaust port simultaneously.

• Now the charge is compressed with both the ports closed and then ignited with the help of

spark plug (petrol engine) or injector (diesel engine) before the end of the compression

stroke. This is done as the charge requires some time to ignite.

• By the time the piston reaches TDC, the burnt gases push the piston downwards with full

force and expansion of the burnt gases takes place. (8M)

5

Explain the abnormal combustion in IC Engines. (13 M) (Nov.2003 & May 2015) BTL2


➢ In CI engine, auto ignition of first part of droplet which is responsible to produce

knocking and rough running of engine.

➢ Delay period is more, more droplets are accumulated in the combustion chamber

➢ That Accumulated fuels start burning with creates high rate of pressure rise.

➢ Heavy vibration accompanied by a knocking sound.

➢ Causing

✓ Overheating of piston ,

✓ Cylinder head

✓ Drop in power and damage to bearings

✓ Possible to piston seizure.

(4M)



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(4M)

Knocking/detonation prevention

• Detonation can be prevented by any of the following techniques:

• the use of a fuel with high octane rating, which increases the combustion temperature of

the fuel and reduces the proclivity to detonate

• enriching the air-fuel ratio which alters the chemical reactions during combustion,

reduces the combustion temperature and increases the margin above detonation

• reducing peak cylinder pressure by decreasing the engine revolutions (e.g., shifting to a

higher gear, there is also evidence that knock occurs more easily at high rpm than low

regardless of other factors)

• Decreasing the manifold pressure by reducing the throttle opening, boost pressure or

reducing the load on the engine.

(2M)

Comparison of Knocking in S.I. and C.I. Engines



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➢ In S.I. engines, the knocking occurs near the end of combustion whereas in C.I.

engine, this occurs at the beginning of combustion.

➢ The knocking in S.I. engine takes place in a homogeneous mixture; therefore, the

rate of pressure rise is high. In C.I. engines, the mixture is heterogeneous and hence

rate of pressure is lower.

➢ The question of pre-ignition does not arise in C.I. engines as the fuel is supplied only

near the end of compression stroke.

➢ The knocking in S.I. engines is because of auto-ignition of the last part of the charge.

To avoid this, the fuel must have long delay period and high self-ignition

temperature. To avoid knock in C.I. engine, the delay period should be as small as

possible and fuel self –ignition temperature should be as low as possible. (3M)

6

Explain the stages of combustion in SI engine. (13 M) BTL2


(i)In a spark-ignition engine a sufficiently homogeneous mixture of vaporized fuel, air and residual

gases is ignited by a single intense and high temperature spark between the spark plug electrodes (at

the moment of discharge the temperature of electrodes exceeds 10,000°C), leaving behind a thin

thread of flame.

(ii)From this thin thread combustion spreads to the envelop of mixture immediately surrounding it

at a rate which depends primarily upon the temperature of the flame front itself and to a secondary

degree, upon both the temperature and the density of the surrounding envelope. In this manner there

grows up, gradually at first, a small hollow nucleus of flame, much in the manner of a soap bubble.

(iii)If the contents of the cylinder were at rest, this flame bubble would expand with steadily

increasing speed until extended throughout the whole mass.

(iv)In the actual engine cylinder, however, the mixture is not at rest. It is, in fact, in a highly

turbulent condition the turbulence breaks the filament of flame into a ragged front, thus presenting a

far greater surface area from which heat is radiated; hence its advance is speeded up enormously.

(v)The rate at which the flame front travels is dependent primarily on the degree of turbulence, but

its general direction of/movement, that of radiating outward from the ignition point, is not much

affected. According to Ricardo the combustion can be imagined as if developing in two stages, one

the growth and development of a semi propagating nucleus of flame called ignition lag or

preparation phase, and the other, the spread of the flame throughout the combustion chamber.



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(vi)The former is a chemical process depending upon the nature of the fuel, upon temperature

and pressure, the proportion of the exhaust gas, and also upon the temperature coefficient of the

fuel, that is, the relationship between temperature and rate of acceleration of oxidation or

burning. The second stage is a mechanical one pure and simple. The two stages are not entirely

distinct, since the nature and velocity of combustion change gradually.

(vii)The starting point of the second stage is where first measurable rise of pressure can be seen

on the indicator diagram, i.e., the point where the line of combustion departs from the

compression line. A shows the point of passage of spark - (say 28° before TDC), B the point at

which the first rise of pressure can be detected (say, 8°before TDC) and C the attainment of

peak pressure. Thus AB represents the first stage (about 20° crank angle rotation) and BC the

second stage.

(viii)Although the point C makes the completion of the flame travel, it does not follow that at

this point the whole of the heat of the fuel has been liberated, for even after the passage of the

flame, some further chemical adjustments due to re-association, etc., and what is generally

referred to as after burning, will to a greater or less degree continue throughout the expansion

stroke.

(ix)The first stage AB, by analogy with diesel engines is called ignition lag, which label is

wrong in principle. In spark ignition there is practically no ignition lag and a nucleus of

combustion arises instantaneously near the spark plug electrodes. But during the initial period

flame front spreads very slowly and the fraction of burnt mixture is small so that an increase of

pressure cannot be detected on the indicator diagram.

(x)The increase of pressure maybe just one per cent of maximum combustion pressure

corresponding to burning of about 1.5per cent of the working mixture, and the volume occupied

by the combustion products may be about 5 per cent of the combustion chamber space. The

stage II is the main stage of combustion.



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(xi)The end of second stage is taken as the moment at which maximum pressure is reached

in the indicator diagram. However, combustion does not terminate at this point and after

burning continues for a rather long time near the walls and behind the turbulent flame front.

(xii)The combustion rate in the stage III reduces, due to surface of the flame front becoming

smaller and reduction in turbulence. About 10 per cent or more of heat is evolved in the after-

burning stage and hence the temperature of the gases continues to increase to point D in Fig.9.

However, the pressure reduces because the decrease in pressure due to expansion of gases and

transfer of heat to walls is more than the increase in pressure due to combustion. (13M)

7

Explain the theoretical and actual p-V diagrams for 4S SI engine. (13 M) (May 2015) BTL2


VARIOUS STROKES IN DIESEL ENGINE:

➢ Suction stroke

➢ Compression stroke

➢ Power stroke

➢ Exhaust stroke (6M)



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1-2 compression process

Air is compressed isentropically with rise in pressure increases and entropy remains constant.

2-3 constant pressure heat addition process

Air is heated with rise in temperature at constant pressure.

3-4 isentropic expansion process

Air expands isentropically with fall in pressure and temperature.

4-1 constant volumeheat rejection

Heat is rejected at constant volume. (7M)

8

Compare the 4 stroke engine and 2 stroke engines. (13 M) (Nov.2010 & May 2012) BTL4


Aspect Four stroke Two stroke

1 Completion of cycle In four strokes of the piston

or in two revolution of the

crankshaft.

In two strokes of the piston or in

one revolution of the crank shaft.

2 Flywheel required Heavier flywheel is required. Lighter flywheel is needed.

3 Power produced One power stroke for two

revolutions.

One power stroke in one

revolution. Double the power as

that developed by four stroke

engine (theoretically).

4 Cooling and

lubrication

requirements

Because of one power stroke

in two revolution, lesser

cooling and lubrication

requirements. Lesser rate of

wear and tear.

Because of one power stroke in one

revolution greater cooling and

lubrication requirements. Great rate

of wear and tear.

5 Valve mechanism Contains valves and

maintenance required.

Contains ports. No valves. Less

maintenance problems.

6 Initial cost Because of heavy weight and

complication of valve

mechanism, initial cost is

high.

Because of light weight and

simplicity due to absence of valves,

initial cost is less.

7 Volumetric More due to more time of Less due to lesser time of



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efficiency induction. induction.

8 Thermal efficiency Higher Lower

9 Part load efficiency Higher Lower

10 Applications Used where efficiency is

important. In cars, buses,

trucks, industrial engines,

power generators, etc.

Used where low cost, compactness

and light weight is required. In

scooters, ships, motor cycles, etc.

9

Compare S.I. engine and C.I. engine. (13 M) (June 2009 & Dec. 2011 & 2012) BTL4


Sl. No Aspect S.I engines C.I engines

1 Fuel used Petrol Diesel

2 Air-Fuel ratio 10 : 1 to 20 : 1 18 : 1 to 100 : 1

3 Compression ratio 7 to 11 12 to 24

4 Combustion Spark ignition Compression ignition

5 Fuel supply By carburetor – cheap. By injector – expensive.

6 Cycle of operation Otto cycle Diesel cycle for slow speed engines.

Dual cycle for high speed engines.

7 Power developed Less More

8 Control of power Quantity governing Quality governing

9 Running cost Higher Lower

10 Applications Used where low cost,

compactness and light

weight is required. In

scooters, ships, motor

cycles, air crafts, etc.

Used where efficiency is important.

In cars, buses, trucks, industrial

engines, power generators, etc.

Part *C

Q.No. Questions



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1

Explain the stages of combustion in CI engine. (15 M) BTL 2


Herry Ricardo has investigated the combustion in a compression ignition engine and divided

the same into the following four stages:

1. Ignition delay or delay period.

2. Uncontrolled combustion.

3. Controlled combustion.

4. After burning.

(4M)

The details of these stages of combustion are given below:

Pressure Vs crank angle of a CI engine in a simplified from is shown in fig. The curved line

ABCG represents compression and expansion of the air charge in the engine cylinder when the

engine is being motored, without fuel injection. This curve is mirror symmetry with respect to TDC

line. The curve ABCDEFH shows the pressure trace of an actual engine.

Delay period (3M)

In an actual engine, fuel injection beings at the point B during the compression stroke. The injected

fuel does not ignite immediately. It takes some time to ignite. Ignition sets in at the point C.

During the crank travel B to C pressure in the combustion chamber does not rise above the



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compression curve. The period corresponding to the crank angle B to C is called delay period or

ignition delay (about 0.001 seconds).

During ignition delay, the following events take place. The injected spray enters the

combustion chamber and slowly (at about 55 m/min) bores hole in the air mass, while the fuel

particles are stripped away. Some of these particles are vapourized. Thus, the main body of the

spray is surrounded by vapour liquid particle air envelope. In small combustion chambers, the

spray body may impinge on the walls. Some of the impinged fuel may bounce off the surface,

while the rest may glide on the walls. Vapourization of fuel particles tends to lower the

compression pressure and temperature slightly. At the same time, the energy released in the

preflame reactions tends to raise the pressure. Now in the outer envelope of the spray, ignition

nuclei are formed. Mostly, the nuclei are cool flame reactions, on the verge of autoignition. By

oxidation or cracking reactions, luminescent carbon particles are formed.

Uncontrolled combustion (3M)

At the end of the delay period i.e. at the point C, fuel starts burning. At this point, a good

amount of fuel would have already entered and got accumulated inside the combustion chamber.

This fuel charge is surrounded by hot air. The fuel is finely divided and evaporated. Majority of

the fuel burns with an explosion like effect. This instantaneous combustion is called uncontrolled

combustion. This combustion causes a rapid pressure rise.

During uncontrolled combustion the following take place. Flame appears at one or more

locations and spreads turbulently, with glowing luminosity. Flame of low luminosity marks regions

of vaporized fuel and air (premixed flame. Flames of higher luminosity mark regions of liquid

droplets and air (diffusion flame). The initial spreading of non-luminous and luminous flame arises

from auto ignition and flame propagation. This is the knock reaction with a high rate of energy

release and correspondingly high rate of pressure rise.

Combustion during crank travel C to D is called uncontrolled combustion. This is because

no control over this combustion is possible by the engine operator. Since this combustion is more

or less instantaneous, it is also called rapid combustion.

If more fuel is present in the cylinder at the end of delay period, and undergoes rapid

combustion when ignition sets in, the rate of pressure rise and the peak pressure attained will be

greater. During this combustion the piston is around TDC, and is almost stand still. Too rapid a

pressure rise and severe pressure impulse at this position of the piston will result in combustion

noise called Diesel Knock.

The severity of the knock reactions is in proportion to the mass enflamed. The regions of

premixed flame are probably hotter (and older) than the regions where liquid droplets are present.

As such, the knock reaction may be propagated mainly in the low luminosity state of the flame.

The rate at which the uncontrolled combustion takes place will depend upon the following:

• The quantity of fuel in the combustion chamber at the point C. This quantity

depends upon the rate at which fuel is injected during delay period and the duration

of ignition delay.

• The condition of fuel that has got accumulated in the combustion chamber at the

point C.



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The rate of combustion during the crank travel C to D and the resulting rate of pressure rise

determine the quietness and smoothness of operation of the engine. (5M)

Controlled combustion

During controlled combustion, following thing happen. The flame spreads rapidly (but less

than 135 m/min), as a turbulent, heterogeneous or diffusion flame with a gradually decreasing rate

of energy release. Even in this stage, small auto ignition regions may be present. The diffusion

flame is characterized by its high luminosity. Bright, white carbon flame with a peak temperature

of 2500o C is noticed. In this stage, radiation plays a significant part in engine heat transfer.

During the period D to E, combustion is gradual. Further by controlling the rate of fuel

injection, complete control is possible over the rate of burning. Therefore, the rate o pressure rise is

controllable. Hence, this stage of combustion is called Gradual combustion or Controlled

combustion. The period corresponding to the crank travel D to E is called the period of controlled

combustion.

The rate of burning during the period of controlled combustion depends on the following:

1. Rate of fuel injection during the period of controlled combustion.

2. The fineness of atomization of the injected fuel.

3. The uniformity of distribution of the injected fuel in the combustion chamber.

4. Amount and distribution of the oxygen left in the combustion space for

reaction of the injected fuel.

At the point E, injection of fuel ends, the period of controlled combustion ends at this point.

When the load on the engine is greater, the period of controlled combustion is also greater.

During controlled combustion, the pressure in the cylinder may increase or remain constant

or decrease. Usually during this period, the combustion is more or less at constant pressure (on a

PV diagram) because the downward movement of the piston (i.e. increase in volume) compensates

for the effect of heat release and the consequent pressure rise.

After burning

At the last stage, i.e. between E and F the fuel that is left in the combustion space when the

fuel injection stops is burnt. This stage of combustion is called after burning (burning on the

expansion stroke). In the indicator diagram after burning will not be visible.

This is because the downward movement of the piston causes the pressure to drop inspired of the

heat that is released by the burning of the last portion of the charge.

Increasing excess air, or air motion will shorten after burning i.e. reduce the quantity

of fuel that may undergo after burning).

2

Explain the working principles of two stroke petrol engine. (15 M) BTL 4




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VARIOUS STROKES IN DIESEL ENGINE:

➢ Suction stroke

➢ Compression stroke

➢ Power stroke

➢ Exhaust stroke

1-2 compression process

Air-fuel mixture is compressed isentropically with rise in pressure increases and entropy

remains constant.

2-3 constant pressure heat addition process

Air is heated with rise in temperature at constant pressure.

3-4 isentropic expansion process



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Air expands isentropically with fall in pressure and temperature.

4-1 constant volume heat rejection

Heat is rejected at constant volume. (10M)

3

Briefly explain the factors affecting knocking in SI engine. (15 M) BTL 5


Fuel-air ratio: (2M)

The composition of the working mixture influences the rate of combustion and the amount of heat

evolved. With hydrocarbon fuels the maximum flame velocities occur when mixture strength is

110% of stoichiometric (i.e., about 10% richer than stoichiometric). When the mixture is made

leaner or is enriched and still more, the velocity of flame diminishes. Lean mixtures release less

thermal energy resulting in lower flame temperature and flame speed. Very rich mixtures have

incomplete combustion (some carbon only burns to CO and not to CO2) that results in production

of less thermal energy and hence flame speed is again low.

Compression Ratio: (2M)

A higher compression ratio increases the pressure and temperature of the working mixture and

decreases the concentration of residual gases. These favorable conditions reduce the ignition lag of

combustion and hence less ignition advance is needed. High pressures and temperatures of the

compressed mixture also speed up the second phase of combustion. Total ignition angle is reduced.

Maximum pressure and indicated mean effective pressure are increased.. Lastly, use of a higher

compression ratio increases the surface to volume ratio of the combustion chamber, thereby

increasing the part of the mixture which after-burns in the third phase. The increase in compression

ratio results in increase in temperature that increases the tendency of the engine to detonate.

Intake temperature and pressure: (1M)

Increase in intake temperature and pressure increases the flame speed.

Engine load: (3M)

With increase in engine load the cycle pressures increase. Hence the flame speed increases. In SI

engines with decrease in load, throttling reduces power of an engine. Due to throttling the initial

and final compression pressures decrease and the dilution of the working mixture due to residual

gases increases. This makes the smooth development of self-propagating nucleus of flame difficult

and unsteady and prolongs the ignition lag. The difficulty can be overcome to a certain extent by

enriching the mixture at low loads (0.8 to 0.9of stoichiometric) but still it is difficult to avoid after-

burning during a substantial part of expansion stroke. In fact, poor combustion at low loads and the

necessity of mixture enrichment are among the main disadvantages of spark ignition engines which

cause wastage of fuel and discharges of a large amount of products of incomplete combustion like

carbon monoxide and other poisonous substances.

Turbulence: (2M)



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Turbulence plays a very vital role in combustion phenomenon. The flame speed is very low in non-

turbulent mixtures. A turbulent motion of the mixture intensifies the processes of heat transfer and

mixing of the burned and unburned portions in the flame front (diffusion). These two factors cause

the velocity of turbulent flame to increase practically in proportion to the turbulence velocity. The

turbulence of the mixture is due to admission of fuel-air mixture through comparatively narrow

sections of the intake pipe, valves, etc. in the suction stroke. The turbulence can be increased at the

end of the compression by suitable design of combustion chamber that involves the geometry of

cylinder head and piston crown. The degree of turbulence increases directly with the piston speed.

If there is no turbulence the time occupied by each explosion would be so great as to make the high

speed internal combustion engines impracticable. Insufficient turbulence lowers the efficiency due

to incomplete combustion of the fuel. However, excessive turbulence is also undesirable.

Engine Speed: (3M)

The higher engine speed, the greater the turbulence inside the cylinder. For this reason the flame

speed increases almost linearly with engine speed. Thus if the engine speed is doubled the time

required, in milliseconds, for the flame to traverse the combustion space would be halved. Double

the original speed arid hence half the original time would give the same number of crank degrees

for flame propagation. The crank angle required for the flame propagation, which is the main phase

of combustion, will remain almost constant at all speeds. This is an important characteristic of

spark ignition engines. However, the increase in engine speed would lead to ignition advance due to

the first phase of combustion. This can be illustrated with a numerical example. Consider a petrol

engine running at 1500rpm. Let us say for the first stage of combustion the ignition lag, the time

required in terms of crank angle, is 8° of crank rotation, and for the second stage, the propagation of

flame through the combustion space, 12oofcrank rotation is required. Thus the total ignition period

is20°of crank rotation. Now if the engine speed is doubled from 1500 to 3000 rpm, the time

required for the second stage will again be 12° of crank rotation (due to doubling of turbulence

intensity time in milliseconds is halved and in terms of crank angle remains constant), but for the

first stage time in milliseconds is constant and hence in terms of crank angle it will be doubled, i.e.,

it would be 16°.This would make the total ignition period of 16 + 12 = 28° crank rotation at

3000rpm compared to 8° + 12°= 20° at .1500 rpm. From this it follows that with increase in engine

speed ignition must be advanced. This is done in practice by automatic ignition advance

mechanism.

Engine size: (2M)

Engines of similar design generally run at the same piston speed. This is achieved by smaller

engines having larger rpm and larger engines having smaller rpm. Due to the same piston speed, the

inlet velocity, the degree of turbulence, and flame speed are nearly same in similar engines

regardless of the size. However, in small engines the flame travel is small and in large engines

large. But with lower rpm of larger engines the time for flame propagation in terms of crank angle

would be nearly same as in smaller engines. In other words the number of crank degrees required

for flame travel will be about the same irrespective of engine size provided the engines are similar.

4

Mention the Factors affecting knocking in CI engine. (15 M) BTL 4


The diesel combustion process which includes ignition delay, premixed burning due to delay period



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and diffusion burning and injector needle lift and pressure variation with respect to crank angle can

be seen in fig. The premixed burning is responsible for diesel knock.

The following are the factors which influence ignition delay and thereby contribute to knock:

Higher inlet air pressure, air temperature and compression ratio reduce knock. Supercharging

reduces knock. Increased humidity increases knock.

Combustion chamber design and associated air motion influence heat losses from the

compressed air. Tendency to knock will be lesser, with less heat losses. A combustion chamber

with a minimum surface to volume ratio and with lesser intensity of air motion is desirable.

Knocking tendency is lesser in engines where compressed air injects the fuel into the

combustion space. In the case of mechanical injection of fuel, finer the atomization of fuel, lesser is

the tendency to knock.

A fuel with long preflame reactions (i.e. self-ignition possible only at a higher temperature)

will result in the injection of a considerable amount of fuel before the initial part ignites. This in

turn results in a large amount or number of parts of the mixture to ignite at the same time and

produce knock. Thus, a good CI engine fuel should have a short ignition delay and low self-ignition

temperature, if knock is to be avoided.

Ignition delay of fuels is generally measured in terms of cetane number. Fuels of higher cetane

number have shorter ignition delay and thus will have a lesser tendency to knock.

The ignition delay of CI engine fuels may be decreased by the addition of small amounts of certain

compounds (called ignition accelerators or improves). These compounds are ethyl nitrate and

amylthionitrate. These compounds affect the combustion process by speeding the molecular

interactions.

Direct injection engines – These engines have a single, open combustion chamber into which the

entire quantity of fuel is injected directed directly. An open combustion chamber is one in which

the combustion space incorporates no restrictions that are sufficiently small to cause large

differences in pressure between different parts of the chamber during the combustion process.

Indirect injection engines – In these engines the combustion space is divided into two parts and the

fuel is injected into the auxiliary chamber which is connected to the main chamber via a nozzle or

one or more number of orifices. The main chamber is situated above the piston. The restrictions or

throat are so small to cause considerable pressure differences between them during the combustion

process. (13M)



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Subject Name : Thermal Engineering-I Subject Handler : J. Ravikumar & S.A.Arokkiya Anicia

UNIT IV INTERNAL COMBUSTION ENGINE PERFORMANCE AND SYSTEMS

Performance parameters and calculations. Morse and Heat Balance tests. Multipoint Fuel Injection system and Common Rail Direct lnjection systems. Ignition systems – Magneto, Battery and Electronic. Lubrication and Cooling systems. Concepts of Supercharging and Turbocharging – Emission Norms.

PART * A

Q.No. Questions

1.

Classify IC engine? BTL1

According to working cycle: a) Four stroke cycle engine b) Two stroke cycle engine

According to the type of fuel used: a) Petrol Engine b) Diesel Engine c) Gas Engine

According to the method of ignition: a) Spark Ignition (SI) b) Compression Ignition (CI)

According to the cooling system: a) Air cooled Engine b) Water cooled Engine

According to the arrangement of cylinders: a) Horizontal Engine b) Vertical Engine c) V – Type

Engine d) Radial Engine e) In – Line Engine f) Opposite Cylinder Engine

According to the number of cylinders: a) Single cylinder b) Multi cylinder

According to the speed of the Engine: a) Low speed b) High speed c) Medium speed

According the Lubrication system: a) Wet sump Lubrication Engine b) Dry sump Lubrication

Engine

According to the Valve Opining: a) Over head valve Engine b) Side valve Engine

2

Define swept volume in IC Engine. BTL1

The volume swept by the piston during one stroke is called the swept volume (or) piston

displacement. In other words, swept volume is the volume covered by the piston while moving from

TDC to BDC.

3

List the various parameter involved in engine performance. BTL1

✓ Brake Power

✓ Indicated Power

✓ Friction Power

✓ Total fuel consumption

✓ Specific fuel consumption

✓ Thermal efficiency

✓ Mechanical efficiency

✓ Mean effective pressure

4 Define the term Brake power. [April 2008 & May 2014] BTL1

Brake power is the power output of the drive shaft of the engine without the power loss caused by

gears, transmission, friction etc. it is also called as useful power or true power.

5 Differentiate between brake power and Indicated power of an IC engine. [ April 2003] BTL2

Indicated power: Power actually developed engine in the engine cylinder.



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Brake power: Power available at the crank shaft. It is always less than the indicated power.

6

What are the important requirements of fuel injection system? [Nov.2007 & Nov.2015] BTL1

✓ The beginning as well as end of injection should take place sharply

✓ Inject the fuel at correct time in the cycle throughout the speed range of the engine.

✓ The injection of fuel should occur at the correct rate and in correct quantity as required by

the varying engine load.

✓ Atomize the fuel to the required degree.

✓ Distribute the fuel throughout the combustion chamber for better mixing.

7 What is gasoline injection system? [May 2015] BTL1

If the fuel is injected directly into the combustion chamber instead of the intake port.

8 Define “Continuous injection” of petrol engine. [May 2016] BTL1

The injection system which is provides a continuous spray of fuel from each injector at appoint

before the intake valve is known as continuous injection systems.

9

Mention different types of fuel injection systems in C.I. engines. [ Oct.1998 & April 1999]

BTL2

a) Air injection system

b) Airless or Solid injection

(i) Common rail system

(ii) Individual pump system.

10

What are the advantages in MPFI system? [May 2017] BTL1

More uniform air-fuel mixture will be supplied to each cylinder hence the difference in power

developed in each cylinder is minimum. The vibrations produced in MPFI engines is very less, due

to this life of the engine component is increased.

11

What is the necessity of cooling in I.C. Engines? [April 2002] BTL1

When the air-fuel mixture is ignited and the combustion takes place at about 2500ºC for producing

power inside the engine, the temperature of the cylinder, cylinder head, piston and valves,

continuous to raise when the engine runs. If these parts are not cooled by some means then by likely

to get damaged and even melted. The piston may cease inside the cylinder. To prevent this, the

temperature of the parts around combustion chamber is maintained as 200ºC to 250ºC. Too much

cooling will lower the thermal efficiency of the engine. Hence, the purpose of cooling is to keep the

engine at its most efficient operating temperature at all engine speeds and all driving conditions.

12

Why anti freezing solutions are used in IC Engines? Give some examples. BTL5

In order to prevent the cooling water from freezing down, some chemicals known as anti – freezing

solutions are mixed up with water. Ex: Denatured alcohol, Ethylene glycol, Distilled glycerine,

Methanol, Sugar solutions, Calcium or magnesium chloride and Kerosene.

13

What is the purpose of a thermostat in an engine cooling system? [April 2003] BTL2 A

Thermostat valve is used in the water-cooling system to regulate the circulation of water in system

to maintain the normal working temperature of the engine parts during the different operating

conditions.



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14

Compare air cooling and water cooling system. BTL1

Air cooling system Water cooling System

The design is simple Design is comparatively complex

No problem of leakage or freezing of water Both the problem exist

More noise Less noise as water dampens the vibration

Maintenance of the cooling system is easier Maintenance is difficult

The heat transfer rate in the system is less The heat transfer rate in the system is more

15

State any three functions of lubrication? BTL1

a) It reduces friction between moving parts.

b) It reduces wear and tear of the moving parts.

c) It minimizes power boss due to friction.

16

Classify Lubrication System. BTL1

1. Various Lubrication systems used in IC Engine

2. Mist Lubrication system or Petro-oil Lubrication system

3. Wet sump Lubrication system a) Splash lubrication system b) Pressure lubrication system

4. Dry sump lubrication system

17

What is Morse test? BTL1

It is a performance test conducted on multi cylinder engines to measure indicated power without

using indicator diagram

18

What is meant by motoring test? [Nov 2016] BTL1

It is a method of engine testing used measure the power output of the engine. The temperature

of heat engine’s pistons and cylinder walls, together with other working parts and also the engine

oil, falls below that of normal working temperature during the motoring tests, and with the lack of

exhaust gases, etc., the frictional and pumping losses are somewhat modified.

19

What are the classifications of an ignition system? BTL1

1. Coil ignition system (or) battery ignition system

2. Magneto ignition system

3. Electronic ignition system

4. Transistorized ignition system

20

What is supercharging? [May 2009] BTL1

It the process of supplying the air fuel mixture to the engine above the atmospheric pressure. A

supercharger increases the pressure of the air fuel mixture from the carburetor before it enters the

engine.

PART * B

Q.No. Questions

1

Write down the performance calculation parameters for IC engines. BTL1

Answer Page No:4.9

1. Brake power



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B.P = 2𝜋𝑁𝑇

60 kW

Where,

N= Engine speed in rpm

T = Torque, T = WgR

W = Dead weight added in kg

W = Wmax

R = Brake drum radius in cm

R= (RD + Rrope) / 2 = 0.21 m

g = gravitational force = 9.81 m/s2 (2 M)

2. Total Fuel consumption

T.F.C = 𝑐𝑐

𝑡𝑓 × Specific gravity ×

3600

1000

𝐾𝑔

ℎ𝑟

Where,

tf= Time taken to consume 10cc of fuel in seconds

cc = Amount of fuel consumption measured in cc (2 M)

3. Specific fuel consumption

S.F.C = kg / kW- hr (1 M)

4. Friction power

Values taken from graph

5. Indicated power

I.P = B.P + F.P kW (1 M)

6. Mechanical efficiency

ηmech = (1 M)

7. Indicated thermal efficiency



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ηI.ther. = % (2 M)

8. Brake thermal efficiency

ηB.th. = % (2 M)

9. Heat Input = 𝐶𝑉 ×𝑇𝐹𝐶

3600 kW (2 M)

2

Write down the morse test calculation parameters for IC engines. BTL1

Answer Page No:4.8

Maximum Load Calculation:

BP = (W×N) / 2000

38 = (WMAX 2000) / 4200

WMAX =--------- (3 M)

1) Brake Power

Brake Power= When all cylinders are in working condition

BP (BP1234) = (W× × 0.736 kW

BP1 (BP1234) = (W× × 0.736 kW

BP2 (BP1234) = (W× × 0.736 kW

BP3 (BP1234) = (W× × 0.736 kW

BP4 (BP1234) = (W× × 0.736 kW (3 M)

2) Indicated Power:

Indicated Power= Indicated Power of engine

IP1 = BP BP1 kW

IP2 = BP BP2 kW

IP3 = BP BP3 kW

IP4 = BP BP4 kW

IP = IP1 + IP2 + IP3 + IP4 kW (2 M)



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3) Friction Power:

FP1 = IP BP1 kW

FP2 = IP BP2 kW

FP3 = IP BP3 kW

FP4 = IP BP4 kW

FP = FP1+ FP2+ FP3+ FP4 kW (2 M)

4) Mechanical Efficiency:

= ×100 % (3 M)

3

Write down the heat balance test calculation parameters for IC engines. BTL1

Answer Page No:4.11

1. Brake Power

P= kW

Where,

V = Voltmeter reading in volts

I = Ammeter reading in amps

η = Generator efficiency = 0.85 (2 M)

2. Total Fuel Consumption

T.F.C = 𝑐𝑐

𝑡𝑓 × Specific gravity ×

3600

1000

𝐾𝑔

ℎ𝑟

Where,

tf= Time taken to consume 10cc of fuel in seconds

cc = Amount of fuel consumption measured in cc (1 M)

3. Total heat supplied

T.H.S = 𝑇.𝐹.𝐶 ×𝐶𝑉

60

𝑘𝐽

𝑚𝑖𝑛

Where,

CV = Calorific value of fuel in kJ (1 M)

4. Heat equivalent to break power

H.E.B.P = Brake power x 60 kJ/min

Where,



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Brake power in kW (1 M)

5. Mass of air entering the cylinder

= 𝐶𝑑 × a (√2𝑔 ℎ𝑤𝜌𝑤 𝜌𝑎 ) kg

min (2 M)

Where,

Cd - Coefficient of discharge of orifice meter = 0.62

a- Area of orifice meter in m2

g- Acceleration due to gravity in m / sec2

hw - difference in manometer reading in m

ρw- Density of water in kg / m3

ρa -Density of air in kg / m3 = 1.23

6. Mass of exhaust gas

Mg = Ma + Mf kg / min (2 M)

Where,

Ma = Mass of air consumed per minute

Mf = Mass of fuel consumed per minute

7. Heat carried by exhaust gas

= mg x Cpg (Te - Ta) kJ / min (2 M)

Where,

Ta &Te= Temperature of air inlet & Temperature of exhaust gas

Mg = Mass of exhaust gas

Cpg = Specific heat capacity of exhaust gas = 1.001 KJ/Kg-K

8. Heat carried by cooling water

= mw x Cpw (Tout - Tin) kJ / min (2 M)

Where,

Mw = Mass of cooling water circulated per minute.

Cpw = Specific heat capacity of water = 4.19 KJ/Kg-K

Tout = Temperature of outlet water

Tin = Temperature of water inlet.

9. Unaccounted heat loss



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= Total heat - (Heat to B.P + Heat carried by cooling water + Heat carried by exhaust gas)

(1 M)

4

A single cylinder 4-stroke oil engine works on Diesel cycle. The following readings were taken

when the engine was running at full load: Area of the indicator diagram = 3 cm2, length of the

diagram = 4 cm, spring constant = 10 bar/cm, speed of the engine = 400 rpm, load on the brake =

380 N, spring balance readings = 50 N, diameter of the brake drum = 120 cm, fuel consumption =

2.8 kg/h, calorific value of the fuel = 42000kJ/kg, diameter of the cylinder = 16 cm, stroke of the

piston = 20 cm. Find (a) frictional power of the engine, (b) mechanical efficiency, (c) brake thermal

efficiency and (d) brake mean effective pressure. [Nov.2010] BTL1

Answer Page No:4.34

Given Data:

Single cylinder 4-stroke oil engine

Area of the indicator diagram, A = 3 cm2

length of the diagram , L= 4 cm

spring constant , S= 10 bar/cm,

speed of the engine, N = 400 rpm

load on the brake = 380 N,

spring balance readings = 50 N

diameter of the brake drum = 120 cm

fuel consumption = 2.8 kg/h

calorific value of the fuel = 42000kJ/kg

diameter of the cylinder = 16 cm

stroke of the piston = 20 cm

(2 M)

(2 M)



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(3 M)

(2 M)

(2 M)

(2 M)

5

Explain the multipoint injection system (MPFI). [Nov.2007, Dec.2008 & 2009 &May 2011]

BTL2

Answer Page No:4.55

In multipoint port injection systems the fuel is injected into the intake port of each engine

cylinder. Thus these systems require one injector per cylinder (plus, in some systems, one or

more injectors to supplement the fuel flow during starting and warm-up). There are both

mechanical injection systems and electronically controlled injection systems. Most modern

automobile SI engaines have multipoint port fuel injectons. In this type of system, injectors

spray fuel into the region directly behind the intake valve, sometimes directly onto the back of

the valve face. Contact with the relatively hot valve surface enhances evaporation of the fuel

into the stationary air just before the intake valve is open, there is a momentary pause in the air

flow, and the air velocity does not promote the needed mixing and evaporation enhancement.

When the valve then opens, the fuel vapor and liquid droplets are carried into the cylinder by

the onrush of air, often, with the injector continuing to spray.

Any backflow of hot residual exhaust gas that occurs when the intake valve opens also

enhances the evaporation of fuel droplets. Each cylinder has its own injector or set of injectors



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which give a fairly constant fuel input cycle-to-cycle and cylinder-to-cylinder, depending on the

manufactured quality of the injector parts. Even with perfect control of the fuel flow, there

would still be variations in air/ fuel ratio due to the imperfect air flow cycle-to-cycle and

cylinder-to-cylinder. Multipoint injector systems are better than carburetors or throttle body

injector systems at giving consistent air/fuel delivery. Some multipoint systems have an

additional auxiliary injector or injectors mounted upstream in the intake manifold to give added

fuel when rich mixtures are needed for startup, idling, WOT acceleration, or high speed

operation. The amount of fuel injected each cycle and injection pressure are controlled by the

Electronic Management System (EMS). Injection pressure is generally on the order of 200 to

300 kPa absolute, but can be much higher. Engine operating conditions and information from

sensors in the engine and exhaust system are used to continuously adjust air/fuel ratio and

injection pressure.

Sensing the amount of oxygen in the exhaust is one of the more important feedbacks in

adjusting injection duration for proper air-fuel ratio. This is done by measuring the partial

pressure of the oxygen in the exhaust manifold. Other feedback parameters include engine

speed, temperatures, air flow rate, and throttle position. Engine startup when a richer mixture is

needed is determined by coolant temperature and the starter switch. The advantages of port fuel

injection are increased power and torque through improved volumetric efficiency and more

uniform fuel distribution, more rapid engine response to changes in throttle position, and more

precise control of the equivalence ratio during cold start and engine warm-up. Fuel injection

allows the amount of fuel injected per cycle, for each cylinder, to be varied in response to inputs

derived from sensors which define actual engine operating conditions. (9 M)



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(4 M)

6

Explain the Battery or Coil ignition system. [Nov.2006 & April 2006] BTL4

Answer Page No:4.63

(5 M)



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Most of the modern spark ignition engines use battery ignition system. This system

consists of the following components.

(i) Battery (6 to 12 volts)

(ii) Ignition switch

(iii) Induction coil

(iv) Circuit/Contact breaker

(v) Condenser

(vi) Distributor

• One terminal of the battery is ground to the frame of the engine and other is connected

through the ignition switch to one primary terminal of the ignition coil (consisting of a

few turns of thick wire).

• The other primary terminal is connected to one end of the contact points of the circuit

breaker and through closed points to ground. The primary circuit of the ignition coil thus

gets completed when contact points of the circuit breaker are together and switch is

closed.

• The secondary terminal of the coil is connected to the central contact of the distributor

and hence to distributor rotor.

• The secondary circuit consists of secondary winding (consisting of large number of

turns of fine wire) of the coil, distributor and four spark plugs.

• The contact breaker is driven by a cam whose speed if half the engine speed (for four

stroke engines) and breaks the primary circuit one for each cylinder during one complete

cycle of the engine.

To start with,

• the ignition switch is made on and the engine is cranked the contacts touch, the current

flows from battery through the switch, primary winding of the induction coil to circuit

breaker points and the circuit is completed through the ground.

• A condenser connected across the terminals of the contact breaker points prevent the

sparking at these points.

• The rotating cam breaks open the contacts immediately and breaking of this primary

circuit brings about a change of magnetic field, due to which a very high.

• Due to high voltage the spark jumps across the gap in the spark plug and air-fuel



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mixture is ignite voltage to the tune of 8000 V to 12000 V is produced across the

secondary terminalsed in the cylinder.

• On account of its combined cheapness, convenience of maintenance, attention and

general suitability, it has been adopted universally on automobiles. (8 M)

7

Explain the Magneto-Ignition system. [April 2010 & Dec.2017] BTL2

Answer Page No:4.66

(5 M)

• The magneto ignition system has the same principle of working as that of coil ignition

system, except that no battery is required, as the magneto acts as its own generator.

• It consists of either rotating magnets in fixed coils or rotating coils in fixed magnets.

• The current produced by the magneto is made to flow to the induction coil which works

in the same way as that of coil ignition system.

• The high voltage current is then made to flow to the distributor which connects the

sparking plugs in rotation depending upon the firing order of the engine.

This type of ignition system is generally employed in small spark ignition engines such

as scooters, motor cycles and small motor boat engines.

Cooling System

• In an I.C engine, the temperature of the gases inside the engine cylinder may vary from

35oC or less to as high as 2750oC during the cycle.

• If an engine is allowed to run without external cooling, the cylinder walls, cylinder and



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piston will tend to assume the average temperatures of the gases to which they are

exposed, which may be of the order of 1000 to 1500oC.

• Obviously at such high temperatures, the metals will loose their characteristics and

piston will expand considerably and seize the liner.

• If the cylinder wall temperature is allowed to rise above a certain limit, about 65oC, the

lubricating oil will begin to evaporate rapidly and both cylinder and piston may be

damaged.

• In view of this, part of the heat generated inside the engine cylinder is allowed to be

carried away by the cooling system. (8 M)

8

Explain the Cooling system of IC Engines. [Nov.2007 & Dec 2008] BTL2

Answer Page No:4.81

(i) Air cooling system.

• In this method, heat is carried away by the air flowing over and around the engine

cylinder.

• It is used in scooters, motor cycles, etc. Here fins are cast on the cylinder head and

cylinder barrel which provide additional surface for heat transfer.

The fins are arranged such a way that they are at right angles to the cylinder axis.

Advantages of Air Cooling System

(i) Simple in design and cheap

(ii) Absence of cooling pipes, radiator, etc. makes the cooling system simpler

(iii) No damage of coolant leakage

(iv) The engine is not subjected to freezing troubles, etc. usually encountered in water cooled



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engines

(v) The weight per unit power output is less than that of water cooled engines

(vi) Easy installation

Disadvantages of Air Cooling System

➢ Their movement is noisy

➢ Non-uniform cooling

➢ The output of air cooled engine is less than that of water cooled engines

➢ Maintenance is not easy

➢ Smaller useful compression ratio (4 M)

(ii) Liquid Cooling

• In this method of cooling system, the cylinder walls and heads are provided with jackets

through which the cooling liquid can circulate.

• The heat is transferred from the cylinder walls to the liquid by convection and

conduction.

• The liquid becomes heated in its passage through the jackets and is itself cooled by

means of an air-cooled radiator system.

• The heat from liquid in turn is transferred to air. (3 M)

Thermostat water cooling system

• Too lower cylinder temperature may result in severe corrosion damage due to

condensation of acids in cylinder walls.



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• To avoid such situation, it is customary to use a thermostat (a temperature controlling

device) to stop flow of coolant below a pre-set cylinder wall temperature.

• Most modern cooling systems employ a thermostat device which prevents the water in

the engine jacket from circulating through the radiator for cooling until its temperature

has reached to a value suitable engine operation.

• The water cooling system as shown in fig is used in the engines of cars, busses, trucks,

etc.

• In this system, the water is circulated through water jackets around each of the

combustion chambers, cylinders, valve seats and valve stems. The water is kept

continuously in motion by a centrifugal pump which is driven by a V-belt from the

pulley on the engine crankshaft.

• After passing through the engine jackets in the cylinder block and heads, the water is

passed through the radiator.

• In the radiator the water is cooled by air drawn through the radiator by a fan. Usually fan

and water pump are mounted and driven on a common shaft.

• After passing through the radiator, the water is drained and delivered to the water pump

through a cylinder inlet passage. The water is again circulated through the engine

jackets. (4 M)

Advantages liquid cooling

(i) Compact design of engine is possible

(ii) The fuel consumption of high compression liquid cooled engine is lower than air-

cooled engine

(iii) Uniform cooling of cylinder barrels (walls) and heads

(iv) Installation is not necessary at the front of vehicles as in the case of air-cooled

engines

Disadvantages of liquid cooling

1. This is dependent system in which supply of water for circulation in the jacket is

required

2. Power absorbed by the pump for water circulation is considerably higher than that for

cooling fans

3. In the event of failure of cooling system serious damage may be caused to the engine

4. Cost of the system is considerably high



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5. Requires more maintenance (2 M)

9

Explain the Lubrication System. [Dec.2011, May 2014 & 2015] BTL2

Answer Page No:4.76

Purpose of lubrication

1. Reduce the friction and wear between the parts having relative motion

2. Cool the surface by carrying away heat generated due to friction

3. Seal a space adjoining the surfaces such as piston rings and cylinder liner

4. Clean the surface by carrying away the carbon and metal particles caused by wear

5. Absorb shock between bearings and other parts and consequently reduce noise

Wet sump lubrication

• These systems employ a large capacity oil sump at the base of crank chamber, from

which the oil is drawn by low pressure oil pump and delivered to various parts.

• Oil there gradually returns back to the sump after serving the purpose.

The general arrangement of wet sump lubrication system is shown in fig. In this case oil

is always contained in the sump which is drawn by the pump through a strainer.

(4 M)

Dry sump lubrication system

• In this system, the oil from the sump is carried to a separate storage tank outside the engine

cylinder block.

• The oil from sump is pumped by means of a sump pump through filters to the storage tank.



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• Oil from storage tank is pumped to the engine cylinder through oil cooler. Oil pressure may

vary from 3 to 8 bar.

• Dry sump lubrication system is generally adopted for high capacity engines.

(4 M)

Mist lubrication system

• This system is used for two stroke cycle engines.

• Most of these engines are crank charged, i.e. they employ crankcase compression and thus,

are not suitable for crankcase lubrication.

• These engines are lubricated by adding 2 to 3 % lubricating oil in the fuel tank.

• The oil and fuel mixture is induced through the carburettor.

• The gasoline is vapourised, and the oil in the form of mist, goes via crankcase into the

cylinder.

• The oil which impinges on the crankcase walls lubricates the main and connecting rod

bearings and rest of the oil which passes on the cylinder during charging and scavenging

periods, lubricates the piston, piston rings and the cylinder.

Advantages of mist lubrication

1. System is simple

2. Low cost (because no oil pump, filter, etc. are required)

Disadvantages

1. A portion of lubrication oil burns in the combustion chamber leading to increasing

exhaust emission and formation of deposits on the piston



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2. Lubricating oil loses its anti-corrosion properties as it comes in contact with the acidic

vapours produced during combustion.

3. There should be thorough mixing of lubricants and fuels for effective lubrication. This

requires separate mixing prior to use or some additives to give the oil good mixing

characteristics.

4. Because of burning of some lubricating oil in the combustion chamber, there will be

excess consumption of 5 to 15 % lubricant.

5. Since there is no control over the lubricating oil, once introduced with fuel, most of the

two stroke engines are over-oiled most of the time. (3 M)

Properties of lubricants

1. Viscosity

2. Flash point and Fire point

3. Cloud point

4. Pour point

5. Oiliness

6. Corrosion

7. Emulsification

8. Physical and Chemical stability

9. Neutralization number

10. Adhesiveness

11. Film strength

12. Specific gravity

Main Parts of an engine to be lubricated

1. Main crankshaft bearing

2. Big end bearing

3. Small or gudgeon pin bearing

4. Piston rings and cylinder walls

5. Timing gears

6. Camshaft and camshaft bearings

7. Valve mechanism

8. Valve guides, valve tappets and rocker arms (2 M)



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10

Explain the Common Rail Direct Injection Systems. BTL2

Answer Page No:4.61

CRDI is an intelligent way of controlling a diesel engine with use of modern computer systems.

CRDI helps to improve the power, performance and reduce harmful emissions from a diesel engine.

Conventional Diesel Engines (non-CRDI engines) are sluggish, noisy and poor in performance

compared to a CRDI engine.

CRDI or common rail direct injection system is also sometimes referred to by many similar or

different names. Some brands use name CRDe / DICOR / Turbojet / DDIS / TDI etc. All these

systems work on same principles with slight variations and enhancements here and there.

CRDI system uses common rail which is like one single rail or fuel channel which contains diesel

compresses at high pressure. This is a called a common rail because there is one single pump which

compresses the diesel and one single rail which contains that compressed fuel. In conventional

diesel engines, there will be as many pumps and fuel rails as there are cylinders.

As an example, for a conventional 4 cylinder diesel engine there will be 4 fuel-pumps, 4 fuel rails

each feeding to one cylinder. In CRDI, there will be one fuel rail for all 4 cylinders so that the fuel

for all the cylinders is pressurized at same pressure.

The fuel is injected into each engine cylinder at a particular time interval based on the position of

moving piston inside the cylinder. In a conventional non-CRDI system, this interval and the fuel

quantity was determined by mechanical components, but in a CRDI system this time interval and

timing etc., are all controlled by a central computer or microprocessor based control system.



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To run a CRDI system, the microprocessor works with input from multiple sensors. Based on the

input from these sensors, the microprocessor can calculate the precise amount of the diesel and the

timing when the diesel should be injected inside the cylinder. Using these calculations, the CRDI

control system delivers the right amount of diesel at the right time to allow best possible output with

least emissions and least possible wastage of fuel.

The input sensors include throttle position sensor, crank position sensor, pressure sensor,

lambda sensor etc. The use of sensors and microprocessor to control the engine makes most

efficient use of the fuel and also improved the power, fuel-economy and performance of the engine

by managing it in a much better way.

One more major difference between a CRDI and conventional diesel engine is the way the fuel

Injectors are controlled. In case of a conventional Engine, the fuel injectors are controlled by

mechanical components to operate the fuel injectors. Use of these mechanical components adds

additional noise as there are many moving components in the injector mechanism of a conventional

diesel engine. In case of a CRDI engine, the fuel injectors are operated using solenoid valves which

operate on electric current and do not require complex and noisy mechanical arrangement to operate

the fuel Injection into the cylinder. The solenoid valves are operated by the central microprocessor

of the CRDI control system based on the inputs from the sensors used in the system. (5+8 M)



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PART * C

Q.No. Questions

1

Explain the working of Turbocharging in IC engines. BTL1

Answer Page No:4.92

In turbo charging, the supercharger or blower is being driven by a gas turbine which uses the energy

in the exhaust gases. In this case, there is no mechanical linkage between the engine and the

supercharger. The major parts of a turbocharger are turbine wheel, turbine housing, turbo shaft,

compressor wheel, compressor housing and bearing housing.

During engine operation, hot exhaust gases blow out through the exhaust valve opening into

the exhaust manifold. The exhaust manifold and the connecting tubing route these gases into the

turbine housing. As the gases pass through the turbine housing, they strike on the fins or blades on

the turbine wheel. When the engine load is high enough, there is enough gas flow and this makes

the turbine wheel to spin rapidly. The turbine wheel is connected to the compressor wheel by the

turbo shaft. As such, the compressor wheel rotates with the turbine. Compressor wheel rotation

sucks air into the compressor housing. Centrifugal force throws the air outward. This causes the air



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to flow out of the turbocharger and into the engine cylinder under pressure.

In the case of turbocharging, there is a phenomenon called turbo lag. It refers to the short

delay period before the boost or manifold pressure increases. This is due to the time the

turbocharger assembly takes the exhaust gases to accelerate the turbine and compressor wheel to

speed up.

If the supercharger is driven directly by the engine, part of the power developed by the engine will

be used in running the supercharger.

Comparative of the heat balanced of the naturally aspirated and supercharged diesel engines.

If is found that the gain in the power output of an engine due to supercharging will be many

time the power required to drive the supercharger. Of course, this is possible only with increased

fuel supply to the engine. It is to be noted that at full loads, the compression of the supercharger is

not fully utilized. This will result in greater loss. Therefore, the specific fuel consumption of a

mechanically driven supercharged engine will be more at part loads when compared to that of a

naturally aspirated engine.

In the case of the exhaust gas turbine driven supercharger, the engine is not required to

supply any power to run the supercharger turbine. This type of supercharging is called turbo

charging. The turbo charging gives about 5% higher thermal efficiency at full load. This increase

in efficiency results in reduced fuel consumption compared to that of a naturally aspirated engine

for the same power output. (5+10 M)

2

What are the Effects of turbo charging? BTL2

Answer Page No:4.92

The following are the effects of supercharging engines. Some of the points refer to CI engines:

1. Higher power output

2. Mass of charge inducted is greater

3. Better atomization of fuel

4. Better mixing of fuel and air

5. Combustion is more complete and smoother



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6. Can use inferior (poor ignition quality) fuels.

7. Scavenging of products is better

8. Improved torque over the whole speed range

9. Quicker acceleration (of vehicle) is possible

10. Reduction in diesel knock tendency and smoother operation

11. Increased detonation tendency in SI engines

12. Improved cold starting

13. Eliminates exhaust smoke

14. Lowers specific fuel consumption, in turbocharging

15. Increased mechanical efficiency

16. Extent of supercharging is limited by durability, reliability and fuel economy

17. Increased thermal stresses

18. Increased turbulence may increase heat losses

19. Increased gas loading

20. Valve overlap period has to be increased to about 60 to 160 degrees of crank

angle

21. Necessitates better cooling of pistons and valves. (15 M)

3

Explain the engine emission norms. BTL2

Answer Page No:4.95

Federal exhaust emission test procedures for light duty vehicles less than 6000 lb GVW covering

the period 1972 to 1975 assess hydrocarbon, carbon monoxide and nitric oxide emissions in terms

of mass of emission emitted over a 7.5 mile chassis dynamometer driving cycle. Results are

expressed as grams of pollutant emitted per mile.

There are two procedures in using the same test equipment which assess vehicle emissions. One

is CVS-1 (constant volume sampling), employs a single bag to collect a representative portion of

the exhaust for subsequent analysis. This single bag system applied to testing of 1972, 1973 and

1974 vehicles. Based on this test, emission standards for vehicles have been set at



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Hydrocarbons 3.4 g/mile (1972 to 1974)

Carbon monoxide 3.9 g/mile (1972 to 1974)

Oxide of nitrogen 3.0 g/mile (1973 to 1974)

The second test procedure, termed CVS-3 uses three sampling bags and is designed to give a

reduced and more realistic weighing to cold start portion of the test. This three bag system applies

to testing of 1975 to 1976 vehicles. Exhaust emission standards based on this test are

Hydrocarbons 0.41 g/mile (1975 to 1976)

Carbon monoxide 3.4 g/mile (1975 to 1976)

Oxide of nitrogen 3.0 g/mile (1975)

One of the latest U.S standards ( 1982) for passenger cars and equivalents are

Hydrocarbons 0.41 g/mile

Carbon monoxide 3.4 g/mile

Oxide of nitrogen 1.5 g/mile

These are measured by following a prescribed test procedure. (15 M)

4

The following data refer to a single cylinder four stroke petrol engine:

Compression ratio = 5.6

Mechanical efficiency = 80%

Brake specific fuel consumption = 0.37 kg/kWh

Calorific value = 44000 kJ/kg

Adiabatic index of air = 1.4

Find the (i) brake thermal efficiency (ii) indicated thermal efficiency (iii) air stander efficiency

(iv) relative efficiency with respect to indicate thermal efficiency and (v) relative efficiency

with respect to brake thermal efficiency. [Nov.2007] BTL4

Answer Page No:4.38

Given data:

Compression ratio, r = 5.6

Mechanical efficiency = 80% = 0.8



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Brake specific fuel consumption, BSFC = 0.37 kg/kWh

Calorific value, CV = 44000 kJ/kg

Adiabatic index of air, γ = 1.4

Solution:

(10 M)

(5 M)

5

The following details were noted in a test on a four-cylinder, four stroke engine, diameter =

100 mm; stroke = 120mm; speed of the engine = 1600 rpm; fuel consumption = 0.2 kg/min;

fuel calorific value = 44000 kJ/kg; difference in tension on eigther side of the brake pulley = 40

kgf; brake cicumference is 300 cm. If the mechanical efficiency is 80%, Calculate the:

(i) Brake thermal efficiency

(ii) Indicated thermal efficiency

(iii) Indicated mean effective pressure and

(iv) Brake specific fuel consumption [Dec.2008 & May 2017] BTL4

Answer Page No:4.41

Given data:



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Solution:

(7 M)



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6-93

(8 M)



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Subject Name : Thermal Engineering-I Subject Handler : J. Ravikumar & S.A.Arokkiya Anicia

UNIT V GAS TURBINES

Gas turbine cycle analysis – open and closed cycle. Performance and its improvement - Regenerative, Intercooled, Reheated cycles and their combinations. Materials for Turbines.

PART * A

Q.No. Questions

1.

Define open cycle gas turbine. BTL1

Answer Page No: 5.5

In the open cycle gas turbine, air is drawn into the compressor from atmosphere and is

compressed. The compressed air is heated by directly burning the fuel in the air at constant

pressure in the combustion chamber. Then the high pressure hot gases expand in the turbine and

mechanical power is developed.

2

Define closed cycle gas turbine. [May 2011] BTL1

Answer Page No: 5.50

The compressed air from the compressor is heated in a heat exchange (air heater) by some

external source of heat (coal or oil) at constant pressure. Then the high pressure hot gases expand

passing through the turbine and mechanical power is developed. The exhaust gas is then cooled to

its original temperature in a cooler before passing into the compressor again.

3

List the various factors which influence the performance of gas turbine. [Dec.2010] BTL1

Answer Page No:5.51

i) Air temperature

ii) Humidity

iii) Inlet and exhaust losses

iv) Fuels

v) Fuel heating

vi) Diluent injection

vii) Air extraction

4

Write the major field of application of gas turbines. BTL1


The major fields of application of gas turbines are:

i) Aviation

ii) Power generation

iii) Oil and gas industry

iv) Marine propulsion.

5

Define Gas turbine plant and write the working medium of this gas turbine. BTL1

Answer Page No: 5.5

A gas turbine plant may be defined as one “in which the principal prime-mover is of the

turbine type and the working medium is a permanent gas.

6 What are the components of gas turbine plant? [Dec.10 & 13] BTL1


A simple gas turbine plant consists of the following:



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i) Turbine

ii) Compressor

iii) Combustor

iv) Auxiliaries.

A modified plant may have in addition and intercooler, a regenerator, a reheater etc.

7

What are the methods to improving the thermal efficiency in open cycle gas turbine plant?

[May 2012&Dec.2012] BTL1


Methods for improvement of thermal efficiency of open cycle gas turbine plant are :

i) Inter cooling

ii) Reheating

iii) Regeneration

iv) Combination of intercooling, Reheating and Regenerator.

8

Name two combined power cycles. BTL1

Answer Page No: 5.5

(i) Combined cycle of gas turbine and steam power plant.

(ii) Combined cycle of gas turbine and diesel power plant.

9

Define Reheat cycle. BTL1


If the dryness fraction of steam leaving the turbine is less than 0.88, then, corrosion and

erosion of turbine blades occur. To avoid this situation, reheat is used.

10 Why is power generation by gas turbines attractive these days? [May 2011] BTL2


Gas turbines are attractive because of their ability to quickly ramp up power production.

11

Sketch the schematic diagram of open cycle gas turbine plant and name the component.

[April 2004] BTL2


12

What are all modifications carried out in Brayton cycle? Why? BTL2


In Brayton cycles, the following devices can be incorporated to increase its thermal efficiency such

as, (i) Regenerator, (ii) Reheater and (iii) Intercooler.



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13

What is intercooling and why it is done? [May 2014] BTL1 & BTL2


The process of reducing the temperature of the compressor gas which reduces its volume is known

as intercooling. It is done to reduce the work done by the compressor with less volume which will

reduce the input power.

14

When will be the gas turbine cycle efficiency reaches maximum? BTL3


The gas turbine cycle efficiency reaches maximum when pressure ratio, Rp= 1 and is equal to

(T3-T1/ T3).

15 When will intercooler be provided between tow compressors? BTL2


When the pressure ratio is very high, then the intercooler is provided between compressors.

16

What is the effect of reheat cycle? BTL1


(i) Thermal efficiency is less since the heat supplied ids more.

(ii) Turbine output is increased for the same expansion ratio.

17

How dose regeneration improve the thermal efficiency of gas turbines cycle? [Dec.2014]

BTL2


Regeneration reduces the energy requirement from the fuel thereby increasing the efficiency of the

cycle.

18

How the gas turbine blades are cooled? [May 2011&Nov.2008] BTL2


(i) Drawing cooling air from compressor

(ii) Injection of coolant onto blade surface

(iii) Creating of an insulating sub layer

(iv) Lowering the effective gas temperature in the boundary layer.

19

What is reheating and regeneration of gas turbine? [Nov.2016] BTL2


Reheating: The process of supplying additional between two turbines by adding fuel is called

reheating.

Regeneration: The process of preheating the air which is entering the combustion chamber to

reduce the fuel consumption and to increase the efficiency is known as regeneration. It is done by

the heat of the hot exhaust gases coming out of the turbine.

20

What is the condition for maximum work in the case of reheater employed in the gas turbine

cycle? BTL3


For optimum work, the pressure ratio is equal for all stages.

Rp1 = Rp2 = ……= (Rp)1/n



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Where, Rp = pressure ratio

N = number of stages.

PART * B

Q.No. Questions

1

What are the classifications of the gas turbine? BTL2

Answer Page No: 5.1

1.According to the Process of Combustion:

(a) Constant pressure or continuous combustion type. The cycle working on this principle

is called Joule or Brayton cycle

(b) Explosion or constant volume type. The cycle working on this principle is called

Atkinson cycle.

2. According to Action of Expansion:

(a) Impulse turbine

(b) Impulse-reaction turbine

3. According to Path of Working Fluid:

(a) Open cycle gas turbine

(b) Closed cycle gas turbine

(c) Semi closed gas turbine

4. According to the Direction of Flow:

(a) Axial flow

(b) Radial flow (13 M)

2 Explain the working principles of an open cycle gas turbine with neat sketch. BTL2

Answer Page No: 5.5



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(5 M)

• In the open cycle gas turbine, air is drawn into the compressor from atmosphere and is

compressed.

• The compressed air is heated by directly burning the fuel in the air at constant pressure in

the combustion chamber.

• Then the high pressure hot gases expand in the turbine and mechanical power is developed.

• Part of the power developed by the turbine (about 66%) is used for driving the compressor.

The remaining is available as useful output.

• The working fluid, air and fuel, must be replaced continuously as they are exhausted into the

atmosphere. Thus the entire flow comes from the atmosphere and is returned to the

atmosphere. (8 M)

3

Explain the working principles of an open cycle gas turbine with reheater. BTL2




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(5 M)

• A reheater is basically a superheater that superheats steam exiting the high-pressure stage

of a turbine. The reheated steam is then sent to the low-pressure stage of the turbine.

• Reheating is applied in a gas turbine in such a way that it increases the turbine work without

increasing the compressor work or melting the turbine materials.

• When a gas turbine plant has a high pressure and low pressure turbine a reheater can be

applied successfully.

• Reheating can improve the efficiency up to 3 %.

• A reheater is generally is a combustor which reheat the flow between the high and low

pressure turbines (8 M)

4

Explain the working principle of gas turbine cycle with regenerator. BTL2




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(5 M)

• Regenerative air standard gas turbine cycles shown in figure

• A regenerator (counter flow heat exchanger) through which the hot turbine exhaust gas and

comparatively cooler air coming from compressor flow in opposite directions.

• Under ideal conditions, no frictional pressure drop occurs in either fluid stream while

turbine exhaust gas gets cooled from 4 to 4’ while compressed air is heated from 2 to 2’.

• Assuming regenerator effectiveness as 100% the temperature rise from 2 – 2’ and drop from

4 to 4’ is shown on T-S diagram. (8 M)

5

What are the factor affecting performances of combustion chamber? BTL2


The following factors affect the performance of a combustion chamber

1. Pressure loss

2. Outlet temperature distribution

3. Combustion stability

4. Combustion efficiency

5. Combustion intensity.

The pressure loss in a combustion chamber is due to 2 reasons

(a) Skin friction and turbulence

(b) Rise in temperature due to combustion.

A uniform outlet temperature distribution helps in reducing hot spots and thermal stresses in the

blades.

Stability of combustion refers to smooth burning and ability of flame to sustain over a wide

operating range. Beyond certain limits of air- fuel ratios (rich and weak), the flame becomes

unstable.

Combustion efficiency is defined as the ratio of actual temperature rise to theoretical temperature

rise. Its value is about 98%. (13 M)

6

In a gas turbine plant working on a Brayton cycle the compression ratio is 7, and the

maximum temperature is 800 o C. The compression begins at 0.1 mpa., 35 o C. Find (a) the

heat supplied per kg of air, (b) the net work done per kg of air, (c) the cycle efficiency, and (d)

the temperature at the end of expansion process. [May 2003] BTL4



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Solution:

(3 M)

(5 M)

(5 M)

7 In a gas turbine plant working on a Brayton cycle the air at the inlet is at 27 o C, 0.1 Mpa.

The pressure ratio is 6.25 and the maximum temperature is 800 o C. The turbine and the



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compressor efficiencies are each 80%. Find (a) the compressor work per kg of air, (b) the

turbine work per kg of air, (c) the heat supplied per kg of air,(d) the cycle efficiency, and (e)

the turbine exhaust temperature. [Dec.2016] BTL4


Solution:

(3 M)

(4 M)



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(6 M)

PART * C

Q.No. Questions

1

What are the applications of gas turbines? BTL2


The applications are:

1. Aviation: Compared to reciprocating I.C. engines, for same power, gas turbines are smaller in

size and lower in weight. So, widely used in aviation.

2. Central Stations: For electric generating stations, gas turbines are used for both base load

regeneration and for peak loads. Low initial cost, quick starting and possibility of remote control

make gas turbines very useful. In case of total breakdown of electrical supply, gas turbines are still

capable of starting - They can operate completely independent of main electric supply. An open

cycle gas turbine doesn't require a source of water and so they have very useful application in power

stations having limited water resources.

3. Combination with Steam Plant: Exhaust of gas turbine is about 400°C and temperature of flue

gases in steam plants is about 200°C. The energy can be recovered to

(a) Generate low pressure steam for different purposes

(b) To preheat air or feed water for the boiler.

By combining with steam plant, the overall efficiency can be increased.

4. Industry : Gas turbines have been used for transport of natural gas, crude oil pumping, chemical

processing, refineries, power supply for laboratories, blast furnace air etc.

5. Transportation: Gas turbines can be used on locomotives and cars. Advantages of gas turbines

in vehicles;

(a) Uniform torque and absence of vibrations. Smooth operation and better comfort.

(b) Compact and light

(c) Reduced pollution due to complete combustion

(d) Cheaper fuels can be employed.

(e) Easy to control and less maintenance

(f) Easy cold starting and less lubrication.

Disadvantages are - Poor part load efficiency, braking is not easy, initial cost, delay in acceleration

due to high inertia of parts.

6. Marine: In marine field, gas turbines have limited applications. War ships are powered by gas

turbines. Although the specific fuel consumption is poor in this field, they give higher speeds.



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7. Space: Gas turbines can be used in turbojet and turbo prop for providing thrust. (15 M)

2

A stationary gas turbine power plant working on a Brayton cycle delivers 20 MW to an

electric generator. The maximum temperature is 1200 K and the minimum 290 K. the

minimum pressure is 95 kPa and the maximum 380 kPa. Determine (a) the power output of

the turbine, (b) the fraction of the output of the turbine used to drive the compressor, (c) the

mass flow rate of air to the compressor and (d) the volume flow rate to the compressor.

[Dec.2017] BTL4


Solution:

(3 M)



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(6 M)

(3 M)



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(4 M)

3

In air stander Brayton cycle, The air is enters the compressor at a 1 bar and 15 0 C. The

pressure leaving the compressor is 5 bar and the maximum temperature in the cycle is 900 o C.

Find the following:

a) Compressor and expander work per kg of air

b) The cycle efficiency

If an ideal generator is incorporated into the cycle, determine the percentage change in

efficiency. [April 1996] BTL2




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(4 M)

(4 M)



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(4 M)

(3 M)

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