energy in agricultue-first lecture

8/4/2019 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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-


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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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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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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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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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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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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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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)

energy in agricultue-first lecture

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