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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-
1/QB+Keye/Ver 1.0
6-1
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
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-
1/QB+Keye/Ver 1.0
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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
Answer : Page No:1.25
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
Answer : Page No:1.57
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
Answer : Page No: 1.61
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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
Answer : Page No:1.121
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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
Answer : Page No:1.20
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
Answer : Page No:1.125
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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
Answer : Page No:5.45
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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
Answer : Page No:3.68
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
Answer : Page No:3.75
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
Answer : Page No:1.52
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)
Air standard efficiency:
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
Air standard efficiency:
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)
Mean effective pressure:
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
Answer : Page No:1.87
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)
Air standard efficiency:
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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
Answer : Page No:3.89
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
Answer : Page No:1.89
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)
Mean effective pressure:
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
Answer : Page No:3.63
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 Code : ME 8493 Year/Semester : II/ 04
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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Answer : Page No:2.2
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
Answer : Page No:2.6
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)
The Process involved are:
Process 1-2: Air is compressed polytropically 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 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)
The Process involved are:
Process 1-2: Air is compressed isentropically 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 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
Answer: Page No:2.54
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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
Answer : Page No:2.15
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
Answer: Page No:2.21
(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
Answer: Page No:2.49
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
Answer: Page No:2.52
(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
Answer: Page No: 2.52
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
Answer: Page No: 2.19
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 inter- cooler 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
Answer: Page No: 2.53
(13M)
5
Derive the expression for volumetric efficiency of air compressor. (15 M) [AU May / June
2014] BTL4
Answer: Page No:2.24
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
Answer: Page No:2.92
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 Code : ME 8493 Year/Semester : II/ 04
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
Answer : Page No:2.1
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
Answer : Page No:2.5
• 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
Answer : Page No:2.16
• 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
Answer : Page No:2.29
(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
Answer : Page No: 2.64
➢ 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
Answer : Page No:3.52
(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
Answer : Page No:2.20
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
Answer : Page No:2.30
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
Answer : Page No:2.32
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
Answer : Page No:3.60
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
Answer : Page No:2.22
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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
Answer : Page No:3.55
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
Answer : Page No:3.62
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 Code : ME 8493 Year/Semester : II/ 04
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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(8 M)
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Subject Code : ME 8493 Year/Semester : II/ 04
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
Answer Page No: 5.55
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
Answer Page No: 5.50
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
Answer Page No: 5.52
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
Answer Page No: 5.15
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
Answer Page No: 5.51
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
Answer Page No: 5.52
12
What are all modifications carried out in Brayton cycle? Why? BTL2
Answer Page No: 5.52
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
Answer Page No: 5.53
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
Answer Page No: 5.52
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
Answer Page No: 5.53
When the pressure ratio is very high, then the intercooler is provided between compressors.
16
What is the effect of reheat cycle? BTL1
Answer Page No: 5.54
(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
Answer Page No: 5.54
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
Answer Page No: 5.55
(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
Answer Page No: 5.53
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
Answer Page No: 5.54
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
Answer Page No: 5.15
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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
Answer Page No: 5.11
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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
Answer Page No: 5.23
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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Answer Page No: 1.121
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
Answer Page No: 1.118
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
Answer Page No: 5.26
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
Answer Page No: 5.45
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
Answer Page No: 5.39
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(4 M)
(4 M)
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(4 M)
(3 M)