energy in agricultue-first lecture
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Heat En ineHeat Engine:A machine which converts heat energy into mechanical energy (work) as a result ofcombustion
Combstion:A process in which fuel combines with oxygen or air and liberates heat. As a result of
, .These types of reactions are exothermic and significant amount of heat is evolved whichincreases temperature and pressure.
2C8H18 +25O2 = 16CO2 + 18H2O + Heat
2C16H34 +49O2 = 32CO2 + 34H2O + Heat
CH4 + O2 = CO2 + H2O + Heat
Hydrocarbons (Fuel) + O2 = C02 + H2O + Heat
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Heat Engines
External combustion Engines
ReciprocatingRotaryReciprocating Rotary
Piston type engine
Steam turbine(Power Plants) Gas Turbine
JPP, Air planes, Power plants
Wankel
(some(Locomotive+
old tractors)-
movers)
C.I.Engines S.I.Engines
Four Stroke cycle,
Tractors, Cars, buses, etc
Two Stroke cycle
Very rare, Motor cycles,CHP
Four Stroke cycle
Cars, tractors, CHPTwo Stroke cycle
(Very rare)
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Compression
Heat engine
additionHeatrejection
cycle
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It is a machine/prime moverwhich converts the
reciprocating motion intorotary mot on as a resu t othermal expanion causedby combustible gases
As the combustion takesplace inside the enginecylinder, so the engine iscalled internal combustion
engine
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Com onents of Internal combustion En ines
The nucleus of all activity,bur but principally for
receiving and burning the fuel
Piston:Receives power generated in the cylinder
Connecting Rod:Transmits power to crankshaft and assists inchanging to and fro motion to rotary motion ofcrankshaft
Crankshaft:Receives power from the piston throughconnecting rod and converts the reciprocatingmotion into rotary motion
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:
Covers the top of the cylinder and
houses other components
Crankcase:Covers the bottom of the cylinder and holds engine lubricating oil andother components
nlet and Exhaust valves:Inlet receives fuel (petrol engine) or
air (diesel engine) while exhaustre eases urn gases.
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ams a t:
Operates the valves
Fly wheel:
Helps keep engine running, by its inertia
during idle strokes (inlet, compression,
exhaust)
En ine block
Foudation block to which all the above
components are directly or indirectly
attached
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Engine Terminology
Bore (d):The internal diameter of the cylinder CV
Stroke (l):The maximum length of travel of piston from oneextreme position to other extreme position in l P.D
Top dead cenre(TDC):The osition of the iston at the end of its travel
d
Piston
when moving towards the cylinder head
Bottom Dead Centre (BDC):e pos on o e p s ona e en o s rave
when movingtowards the crankcase
Piston dis lacement PD :
Connecting rod
The volume displaced or covered by the pistonwhen it moves from TDC to BDC
PD = d2/4*l
crankshaft
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Clearance volume (CV) CV
piston and the engine cylinder when thepiston is at TDC. It is also called combustion
chamber l P.D
Total cylinder volume (TCV)The volume designated by the sum of thepiston displacement and the clearance
d
Piston
BDC
volume
TCV = P.D + C.V
The ratio of total cylinder volume to clearancevolume
C.R = TCV/CV
Connecting rod
Engine SizeIt is the product of diameter of piston and
crankshaft
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Four Stroke( cycle) SI Engine
A four stroke cycle engine one which completes its cycle (suction, compression, power,exhaust) in four stroke of the piston resulting two complete revolution (360*2 = 720) ofcrankshaft
The piston moves from TDC to BDC. The inlet valve is opened to allow the fuel mixtureto enter and fill the partial vacuum created by the movement of the piston. The exhaustvalve remaim closed.
The inlet valve is closed. The exhaust valve also remain closed. The piston moves fromBDC to TDC compressing the fuel mixture into the clearance volume; thereby raising thetemperature
Power trokeWhen the piston approaches TDC during the compression stroke, a spark ignites the fuelmixture and expanding gases, drive the piston from TDC to BDC
The piston returns from BDC to TDC, sweeping the the burnt fuel/gases through theexhaust valve which has been opened, while the intake valve remain closed
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Four stroke cycle-SI Engine
or wo s ro e an ma on, rowse e n
http://www.keveney.com/otto.html
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Four stroke cycle-CI engines
In case of diesel engine, the same number of piston events take placeas in 4-troke etrol en ine but the method of i nitin the fuel isdifferent. The following sequence of operation occurs
Only air is drawn into the cylinder
Compression Strokee a r s compresse nto t e cy n er w c ra se ts temperature
very high
Power strokeJust at the end of compression stroke, a fine spay of diesel is injectedinto the hot compessed air which ignites the fuel instantly and
expanding gases, drive the piston for the power stroke.Exhaust StrokeSimilar to that of petrol engine
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Two stroke Engine
This type of engines completes one cycle in only two strokes of thepiston.
There are no definite intake and exhaust valves but instead, there
are openings or ports located in the cylinder wall which get covered.
When the piston moves from TDC to BDC, a fresh fuel mixture,
though the exhaust port which is also uncovered.Due toscavenging, some of the unburnt fuels also escapes with exhaustases which not onl reduce the thermal efficien but also cause
environmental pollution. Now a days, these engines are rarely used
The com ression and ower stroke are similar to those for four stroke engines.
However to com lete one revolution of the crankshaft revolution ofthe piston, makes only two strokes which is equivalent to 2*180 =360 or one revolution of the crankshaft.
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Two stroke cycle engine
For two stroke animation, browse the linkhttp://www.youtube.com/watch?v=8G4is0-ZFZs&feature=related
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Two stroke cycle engine
For two stroke animation, browse the linkhttp://www.keveney.com/twostroke.html
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Example 1: An engine has bore of 100 mm and stroke of 124
mm. If the compression ratio is 6. Calculate
a- s on sp acemen , - earance vo ume
c- Total Cylinder Volume (TCV)
Solution
Given dataCV
cylinder bore = diameter of piston(d) = 100 mm
Stroke length (l) = 124 mm
Compression Ration (CR) = 6 l P.D
(a) Diston Displacement(PD) =??
We know that
PD = d2/4 *l
d
Piston
= [(3.14)(100)2
/4]*124 = 973,400 mm3
(b) Clearance volume (CV) = ??
= v ng y
CR = PD/CV + 1
PD = (CR-1)CV
CV = PD/ CR -1
Connecting rod
= 973,400 / (6-1) = 194680 mm3
Cylinder Volume (TCV) = ??
=
= 973400 + 194680 =1168080 mm3 crankshaft
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.
stroke of 127 mm. Find the clearance volume (CV) whenthe compression ratio (CR) is 18.5
CV
o u on
Given data
Cylinder bore (d) = 91.44 mmStroke l = 127 mm l P.D
127 mm
Compression ration (CR) = 18.5
Clearance volume (CV) =?? d
Piston91.44 mmor
PD =( d2/4)*l
= 3.14 91.44 2/4 * 127
= 833,577 mm3
We know that
Connecting rod
. = -
CV = PD/(CR-1)
CV = 833,577/17.5
CV = 50 490 mm3
crankshaft
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Example 3
An engine has clearance volume 53.763 cm3 and
compress on ra o . an s ro e eng mm.
Calculate
(a) Piston displacement
b BoreCV
Solution
Given data l P.Dearance vo ume = . cm
= 53,763 mm3
Compression ratio (CR) = 16.5
Stroke l = 127mm
d
Piston
(a)We know that
PD =(CR-1)CV
= .
PD = 833,326 mm3
(b) Connecting rod
PD = (d2 /4).l
d2 = (4*PD)/(*l)
= (4*833326)/(3.14*127)
d = 91.42 mmcrankshaft
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Example 4: Calculate CV, CR and engine capacity of a 4-
cylinder tractor engine when total volume of one cylinder is
One cylinder
Other cylinder
. . .
respectivily.
SolutionCV
Given data
Total cylinder volume(TCV) = 1027.33 cm3
= 1027330 mm3 l P.D.
Stroke (l) = 127 mm
(a) We know that
PD = d2 / 4 *(l)
d
Piston= (3.14* (98.4)2/4)(127)
= 965,303 mm3
Clearanve volime (CV) = =??
= 1027.33 -965.303 = 62.027 mm3
Compression ratio (CR) = ??
(CR-1)CV = PD
Connecting rod
CR -1 = 965303/62028
CR = 15.56 +1 = 16.56
Engine capacity = ??
* .
= 965302*4
= 3861,208 mm3
=3861.208 CC
crankshaft
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used ??For a single cylinder engine, there is one
power stroke in two revolutions of the fly. ,
cylinder is not smooth and quite. So, inorder to make the engine running smooth,multi cylinder engines are used
Firing orderIt the the order inwhich the power stroke ineach cylinder takes place
Firing interval(FI)
It the interval between succesive power
FI = Angle turned during one cycle / No. ofcylinders
For 4-cylinder 4-stroke engine
Crankshaft
= =
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Firing order- 4 and 6 cylinder engine
Crank- Cylinder Cylinder Cylinder Cylinder Firing Firing
-angle
180 P E C S 1 1
Firing interval = 720/4
= 180
540 S C E P 4 4
720 C P S E 2 3
Firing order 1 = 1,3,4,2
Firing order 2 = 1,2,4,3
1,6
For 6-cylinder Engine
120120
Firing interval =720/6
=120
120Firing Order 1 =1,5,3, 6,2,4
Firing order 2 =1,4, 2, 6,3,5
2,5 3,4
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General gas equation
Boyle`s lawTemperature remaining constant, volume of a given mass of a gas is inversly proportionalo s pressure
Mathematically
PV = constant if and only if T = constant
Charle`s lawPressure remaining constant, volume of a given mass of a gas is directly proportional to
V/T = Constant if P = Constant
enera gas equa onCombining Boyles law and Charles law, we get
PV/T = constant
1 1 1 2 2 2 .....
Also
PV = mRT
Where
P = Pascal, V = m3, m = kg R = J / kg.K T = K
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Units of HeatCalorie (Cal)
It is the amount of heat energy required to raise the temperatre ofone
gram of water through 1C
I kCal = 1000 Calories (1 kg through 1C)
r s erma n
It is the amount of heat energy required to raise the temperature of1
Centigrade heat unit (CHU)
It is the amount of heat energy required to raise the temeperature of
one pound of water through 1C
Joule
It is the amount of heat energy required to raise the temperature of
1/4.187 gram of water through 1C
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Work and PowerWork
Product of force into displacement; Work = force*displacement (F*d)
Joule
When a force ofone Newton is acted u on a bod and roduces a
displacement ofone meter. Then it is called one joule work
1 joule = 1N.m; and 1 kiloJoule (kJ) = 1 KN.m
Other units of work are erg and foot pound
Power
It is the unit of power. Power =Work/time
Watts W
The rate of workdone @1Joule/sec
Other units are horse power. 1 horsepower = 33000 ft-lb/min or 550 ft-lb/sec
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Laws of thermodynamics
First law of thermodynamics
Heat and work are mutually convertable
Second Law of thermodynamics
Heat flows from a hotbody to a cold body unaided but
it is impossible for the heat to flow from a cold body to
o o y w ou e a o ex erna wor
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The Pressure-Volume diagram
Graphical representation ofpressure and volume of a gas or
-
vapor
From figure
Workdone = Worce =
Pressure = P
Cross-sectional area = Assure
(P)
,
W= F*d(But as we know P =F/A)
* *
Pre
Work done =Area under
the curvedv
W = P*(A*d)
W = P*VVolume (V)
diagram
Work done by gasF
)1(........
2
1
EqPdVW V=
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Thermodynamics processes
Constant volume Process
Heat
1........2
EPdVWV
=For constant volume process
ssure
2
It meansPr
1Change in volume
dV = 0Volume
Put in Eq.(1)
W = 0
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Thermodynamics processes
Costant pressure Process
Heat
1........2
EPdVWV
= 1
As ressure is constant
ssure
1 2
It meansP = C
Pre
Volume changes from
V to V variable .Volume
Then
W = P V V
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Costant Temperature Process
T2
OR
Isothermal processHeat
)1(........2
1
EqPdVWV=
For constant temperature process
sure
1P
1V
1= P
2V
2= PV = constant(C)
Or
P = C/V
Pre
2Substitute C/V in Eq. (1)
=
2
)/(
V
dVVCW
Volume=C (lnV2lnV1)
=C ln V /V
= P1V1 C ln rv (Where rv is volume ratio)
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General law PVn = Constant
T2
)1(........2
1
EqPdVWV
V= H
eat
P1V1n =P2 V2
n = PVn = C
Ort
P= C /Vn
=2
)/(V n
dVVCW sure
PVn = Constant
=
2
1
)(
V
V
n
dVVCW
Pre
2
=
+ 1V
CWn V2
Volume
+nV1
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++ 11 nn
T2
+
=
1
12
nCW
Heat
=
++ 11
12 CVCV
Wnn
=
++ 1111
1222 VVPVVP
Wnnnn
sure 1
PVn = Constant
+ n
1122 VPVP
Pre
2
+ 1n
2211 VPVPVolume
=
1n
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Thermodynamics processes
Recersible adiabatic Process
PV = Constant
For reversible adiabatic process= . or a r
= Cp /Cv
ssure 1
equation
Pre
2
=
1
2211
nW
Volume
=1
2211
W
s s e equa on or revers e a a a c process
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Law of conservation of Ener
Energy can neither be created nor be destroyed, however it may be changed
from one form to another
In thermodynamics
Total heat supplied = Work done + increase in internal energy
(Internal Energy (U=Cv ( T2- T1) is the rise in energy due to rise in temperature
Mathematically
and Cv & Cp are specific heat at constant volume & pressure, = Cp / Cv)
Q = W + U
Sign convension
work is done by gases on piston = +W+Q
work is done by piston on gases = - W
heat absrbed by the system = +Q
heat rejected from the system = - Q+ Un erna energy as ncreae = +
internal energy has decreased = - U-Q
-
Th d i
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Thermodynamics processes
Costant volume Process
Heat
OR
Isochoric process
Workdone
W = 0
ssure
2Internal Energy
Pre
1
U =Cv(T2-T1)
Volume
Q = U= Cv(T2-T1)
Th d i
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Thermodynamics processes
Costant pressure Process
Heat
Isobaric process
Workdone
W = P(V2-V1)
ssure
1 2Internal EnergyU =cv(T2-T1)
Pr
Heat absorbed
= W + UVolume
= P(V2-V1) + cv(T2-T1)
=Cp(T2-T1)
Th d i
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Thermodynamics processes
T2Costant Temperature Process
Heat
Isothermal process
Workdone
W = P V ln(V /V )
ssure 1
Internal Energy
Pr
2U =0
VolumeHeat absorbed
Q = W = P1V1ln(V2/V1)
Th d i
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Thermodynamics processes
T2General law PVn = Constant
Heat
Workdone
1 1 2 1
ssure 1
n erna nergy
U =Cv ( T2-T1)
Pre
2Heat absorbed
Volume= 1 1 n 2 1 + v 2- 1
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OR
Constant Entropy process
Workdone
W = (P1V1-P2V2)/(n-1)
sure 1Internal Energy
U =Cv(T2-T1)
Pre
2Heat absorbed
Q =W+UVolume
0 = W + U
W = -U
- U = W
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OR
y
OR
ThrottlingWhen a gas a allowed to expand from an aperture of minute dimension.
called irreversible adiabatic .It can not be reversed as the frictionalreheating between the fluid and the walls of the container.Due to this
re-heatin it is also called as constant enthal rocess Enthal =total heat)
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