t.y. diploma : sem. v [me/pg/pt] power engineering prelim ...vidyalankar.org/upload/pe_soln.pdf ·...

1114/TY/Pre_Pap/2014/CP/PE_Soln 1

T.Y. Diploma : Sem. V [ME/PG/PT]

Power Engineering Time: 3 Hrs.] Prelim Question Paper Solution [Marks : 100

Q.1(a) (i) Duel Cycle

Fig. : Duel cycle

Q.1(a) (ii) Actual and theoretical value timing diagram of u-stroke petrol engine

(a) Theoretical (b) Actual

Fig. : Valve timing diagram of 4-stroke petrol engine

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arFig. : F Duel cycle Duel cycle

etical value timing diagram of u-se timing diagram

Vidyalankar : T.Y. Diploma PE

2

Q.1(a) (iii) Supercharging The power output of an engine depends upon the amount of air inducted per

unit time. The amount of air inducted per unit time can be increased by increasing the

engine speed or by increasing the density of intake air. To increase the engine speed, the engine construction is required to be

robust. Therefore it has a limit. The second alternative is to increase the density of intake air, is used to

increase the power output. Thus, increasing the engine power by increasing the density of intake air is

called as ‘supercharging’. This can be done by supplying the air at a higher pressure at which the engine naturally aspirates air from atmosphere by using a pressure-boosting device called as ‘supercharger’.

Effects of Supercharging The effects of supercharging are as follows

i) A supercharged engine has air supplied to it at a pressure and density higher than atmosphere. This increase the volumetric efficiency.

ii) The mechanical efficiency of supercharged engine is more than that of naturally aspirated engine when both are running at same speed.

iii) The specific fuel consumption of a supercharged engine is less due to proper combustion achieved by better turbulence.

iv) For supercharged engine, chances of detonation increases due to higher pressure.

Q.1(a) (iv) Turbocharging

About 30% of heat input goes with exhaust gases.

The exact percentage depends upon the type of engine and its operating conditions. This exhaust gas can be used to run a gas turbine.

The gas turbine develops the sufficient power to drive centrifugal compressor, which is used to supply the air to engine. This results in increased power output and better thermal efficiency of engine. Thus, supercharging done by driving compressor with the help of gas turbine utilizing exhaust of engine is called as ‘turbo charging’.

The turbochargers are nothing but centrifugal compressors driven by exhaust gas of engine.

Now a days turbochargers are more popular as they utilize the exhaust energy of engine which otherwise go waste.

In order to supply sufficient energy to the turbo charger, the exhaust valve is opened much before the BDC as compared to naturally aspirated engine. This allows the exhaust gas to escape at higher pressure and temperature giving the turbocharger enough energy to drive the compressor.

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arequired to be to b

ntake air, is used to air, is used to

ng the density of intake air is g the density of intake air by supplying the air at a higher supplying the air at a hig

aspirates air from atmosphere by air from atmosp as ‘supercharger’. percharger’.

e as follows has air supplied to it at a pressure supplied to it

e. This increase the volumetric effic increase the volumciency of supercharged engine is ciency of supercharged eng

d engine when both are running at sgine when both are runnel consumption ofmption of a supercharged a super

bustion achieved by better turbulenc achieved by better tucharged engine, chances of detonad engine, chances

re.

ocharging out 30% of heat input goes with exhout 30% of heat input goes w

The exact percentage depends u exact percentage conditions. This exhaust gas cannditions. This exhaust

The gas turbine develophe gas turbine devecompressor, which is uompressor, which is uincreased power outpincreased powsupercharging donesuperchargingutilizing exhaust outilizin

The turbochar tugas of engi

Now a ener

I

Prelim Question Paper Solution

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Q.1(b) (i) d = 150 mm = 15 cm L = 225 mm = 22.5 cm VC = 1.25 × 10 3 m3

VS = 24d L

= 20.15 0.2254

= 3.976 × 103 m3

Cylinder volume = VC + VS = 1.25 × 10 3 + 3.976 × 10 3 = 5.226 × 10 3 m3

Compression Ratio =

Cylinder volumeClearance Volume

= C S

C

V V

V

= 3

3

5.226 10

1.25 10 = 4.18

ASE = 1 J 1

1

r

= 1 1.4 1

1

4.18 = 0.4356

= 43.56%

Q.1(b) (ii) MPFI (Multi Point Fuel Injection) The MPFI means Multi Point Injection System. In this system each cylinder

has number of injectors to supply/spray fuel in cylinders as compared to one injector located centrally to supply/spray fuel in case of single point injection system.

Advantages i) Superior fuel combustion. ii) Better fuel management. iii) Better engine performance. iv) Refused pollution. Simple Carburettor Simple carburetor consists of a float chamber, nozzle with metering orifice,

venture and throttle valve. The float and needle valve system maintains a constant height of petrol in

the float chamber. If the amount of fuel in float chamber drops below the designed level, the

float lowers, thereby opening the needle of fuel supply valve. When designed level has been reached the float closes and needle valve

stops the additional flow of fuel from supply system. The float chamber is vented to the atmosphere.

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ar = 4.18

1

an1.4 11.

1 = 0.4356 = 0

= 43.56% = 43.56%

Fuel Injection) ections Multi Point Injection Point Injection System. In thSyste

injectors to supply/spray fuel in cylors to supply/spray futed centrally to supply/spray trally to supply/spra fuel in

antages Superior fuel combustion. Superior fuel combustion.

ii) Better fuel management. Better fuel manage iii) Better engine performance. Better engine perfor iv) Refused pollution. ) Refused pollution.

Simple Carburettor Simple CarbureSimple carburetor coSimple carburventure and throttlventu

The float and The fthe float ch

If the floa


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During suction stroke, air is drawn through the venturi. Venturi has minimum cross-section at throat. Venturi tube is also known as ‘chock tube’ and is so shaped that it gives minimum resistance to airflow.

The air passing through the venture increases in velocity and pressure

decreases. From the float chamber, fuel is led to discharge jet, the tip of which is located in the throat of the venturi.

Because of pressure difference in throat and float chamber, the fuel from

float chamber is discharged at throat. This pressure difference is called as ‘carburettor depression’.

The rate of flow is controlled by the size of the smallest section in the fuel

passage. This is provided by discharge jet and size of jet is designed to give required engine performance.

Fig. 1 : Simple carburettor The pressure difference at throat for full open throttle is about 4 to 5 cm of Hg

below atmosphere. To avoid wastage of fuel, the level of the liquid in the float chamber is kept at short distance below the tip of discharge jet.

The quantity of fuel discharged is regulated by butterfly type throttle valve. As the throttle is closed, less air flows through the venture and less air-fuel

mixture is delivered to the cylinder and less power is developed. As throttle is opened, more air flows through the venture and the power

developed increases.

Limitations of Simple Carburettor i) A simple carburetor gives the correct mixture at only one engine speed

and load. Therefore it is suitable only for engine running at constant speed and load.

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arhe tip

er, the fuel from l fromerence is called as is called as

mallest section in the fuel section in the fuel size of jet is designed to give size of jet is designed to g

Fig. 1 : S

The pressure difference at t The pressure difference abelow atmosphere. To avelow atmosphere. To avchamber is kept at shochamber is kep

The quantity of fuThe qu

As the thrott mixture is

As thde


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ii) At very low speed, the mixture supplied is so weak that it will not ignite properly. For this, arrangement must be made so that at starting it should give richer mixture.

iii) In simple carburetor the mixture is weakened when the throttle is suddenly opened because of inertia effect of the fuel, when mixture demand is of rich mixture.

iv) The working of simple carburetor is affected by changes in atmospheric temperature and season.

Q.2(a) (i) Mechanical efficiency

It is defined as the ratio of brake power to the indicated power of engine.

m =

Break Power

Indicatedpower

(ii) Relative efficiency It is the ratio of the indicated thermal efficiency to the cross pointing air

standard efficiency.

r = Break thermal efficiencyAir standard efficiency

(iii) Thermal efficiency It is the ratio of power output to the heat energy of the fuel supplied during

the same duration.

Break thermal efficiency The efficiency based on the break power is known as break thermal

efficiency, and is given by

btn =

Heat equivalent of B.P.

Heat supplied

Indicated thermal efficiency Efficiency based on indicated power is termed as indicated thermal efficiency

and is given by

ith =

Heat equivalent of I.P.

Heat supplied

(iv) BSEC (Brek Specific Fuel Consumption) It is the height of fuel required to develop 1 kw of break power for a period of

1 hr. and it is inversely proportional to break thermal efficiency.

BSEC =

Fuel consumption inkg / hr

Break power inkw

Q.2(b) (i) Cylinder bore (D) : The internal diameter of the cylinder of an engine is

known as bore. (ii) Stroke (L) : The nominal distance through which the working piston moves

between two successive reversals of its direction of motion or the distance travelled by the piston from one of its dead centre positions to the other dead centre position is called ‘stroke’.

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artmospheric eric

power of engine. of engine.

efficiency to the cross pointing acy to the cross poi

ncyiency

output to the utput to the heat energy of the fueat energy o

fficiency y y based on the break d on the break power is

nd is given by en

btn =

yaHeat equivalent of B.P.eat equivalent of B.P.

Heat suppliedHeat supplied

ndicated thermal efficiency dicated thermal effEfficiency based on indicated powency based on indand is given by d is given

ith = = dyHeat equivaleHeat equivHeat suat s

(iv) BSEC (Brek Speci BSEC (Brek It is the height of It is the

1 hr. and it is i a

BSEC

(b) (i) Cy


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(i) Clearance Volume (VC) : The nominal volume of the space on the combustion side of the piston at top dead centre.

(ii) Cylinder volume: Cylinder volume is sum of piston swept volume and the numerical value of the combustion space or clearance volume.

V = VS + VC (iii) Dead centre : The position of working piston and the moving parts which are

mechanically connected to it at the moment when the direction of the piston motion is reversed. For vertical engine it is known as Top Dead Centre (TDC) and Bottom Dead Centre (BDC). For horizontal engine, these positions are known as Inner Dead Centre (IDC) and Outer Dead Centre (ODC).

(iv) Swept Volume (VS) : The nominal volume generated by the working piston when travelled from one dead center to the other, is calculated as the product of piston area and stroke. VS = A × L, where A is piston area and Listength of stroke.

Q.2(c) (i) I.P = PmLAN

= Pm × 2d LN4

= 6.65 × 105 × 4

(0.24)2 × 0.40 × 11560

= 23064.317 N.m/s = 23.064 kw (ii) B.P = (w s) r 2 N

= 85 × 0.75 × 2 25060

= 1668.97 9.81 N.m/s = 16372.60 Watts B.P = 16.372 kw

(iii) btn = B.P

100mf cv

= 6.372 kJ / sec

1000.21

1863260

btn = 25.10% Q.3(a) Catalytic Converter

Fig. 1 : Catalytic converter

Catalytic converters are widely used in car all over the world.

which af the piston ston

Centre (TDC) (TDCse positions are ns are

e (ODC). C). by the working piston working piston

er, is calculated as the culated as the

h of stroke. of stroke.

× 0.40 ×

an115115600

N

0.75 × 2 2 al25026060

1668.97 9.81 N.m/s 9.81 N.m/s = 16372.60 Watts 2.60 Watts

B.P = 16.372 kw 2 kw

) btnbtn = =

dyB.PB.P

100100mf cvmf cv

= dydy6.372 kJ / sec6.372 kJ / sec0.210.21

18632326060

btn = 25.10% = 25

Q.3(a) Catalytic ConvertQ.3(a) Catalytic C


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It is cylindrical canister placed between exhaust manifold and silencer. It contains plastic pallets coated with the catalyst. Catalytic converter is designed for the oxidation of pollutant gases escaping after primary combustion in the engine, within the exhaust system. The simplest type of catalytic converter is as shown in figure 1. The temperature required for bulk gas oxidation and reduction of hydrocarbon gases (HC), carbon monoxide (CO) and oxides of nitrogen (NOx) is about 600 to 700 C. The temperature of exhaust gases in exhaust system is lower. The catalytic converter oxidizes and reduces all the three pollutant gases at lower temperature because of catalytic chemical reaction. A catalytic converter becomes effective in the temperature between 250 to 300 C. The two basic types of catalytic converters used in car are

i) Two way catalytic converter A converter is filled with monolithic substrate coated with small amount of

platinum and palladium.

Through catalytic action, a chemical change converts carbon mono oxide (CO) and hydrocarbons (CH) into carbon dioxide and water. Such a converter is called ‘two way catalytic converter’.

ii) Three way catalytic converter Three way catalytic converter is installed on cars to check pollution. Such a

converter uses thin coating of platinum, palladium and rhodium over a support metal (generally alumina) and acts on all three major constituents of exhaust gas pollution i.e. hydrocarbons, carbon monoxide and oxides of nitrogen, oxidizing these to water, carbon dioxide and free hydrogen and nitrogen respectively.

Three way catalytic converter operates in two stages. The first converter stage uses rhodium to reduce the NOx in the exhaust into

nitrogen and oxygen. In the second stage converter platinum or palladium acts as oxidation

catalyst to change HC and CO into harmless water and CO2. For supplying the oxygen required in the second stage air is fed into the

exhaust after the first stage. The catalyst allows the oxidation of the exhaust gas at a much lower temperature than in the combustion chamber.

Reactions within catalyst produce additional heat that reaches temperature of

900 C, which is required for the catalytic converter to operate at complete efficiency. To safeguard form this high temperature, the catalytic converter is made of stainless steel and special heat shields are also used.

No regular maintenance is required for catalytic converter.

Catalytic converter may be replaced at about 80,000 km or more.

The engine emissions HC, CO and NOx are removed from the exhaust gas by oxidation.

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ar hydrocarbon arbon

is about 600 to 600 tem is lower. The ower. The

utant gases at lower ases at lower A catalytic converter lytic convert

to 300 C. The two basic C. The two basic

rate coated with small amount of ed with small am

cal change converts carbon monoange converts carbo into carbon dioxide and water. bon dioxide and

atalytic converter’. converter’.

nverter nveronverter is installed on cars to cheerter is installed on cars

n coating of platinum, palladium g of platinum, paenerally alumina) and acts on all thy alumina) and acts o

pollution i.e. hydrocarbons, carboon i.e. hydrocarboxidizing these to water, carbon di these to water, ca

respectively. tively.

ee way catalytic converter operates ee way catalytic converter ophe first converter stage uses rhodiue first converter stage uses

nitrogen and oxygen. gen and oxygen.

In the second stage conve the second stage ccatalyst to change HC and atalyst to change HC an

For supplying the oxy For supplying exhaust after the firexhaust aftegas at a much lowas at a

Reactions w 900 C, wefficienmad


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Limitations of Catalytic Converter i) Since lead destroys catalytic activity, the engine cannot use leaded petrol. ii) Exhaust systems are hotter than normal as a result of exothermic

reaction in catalyst bed. iii) The emission of SO2 increases if fuel contains sulphur. iv) The use of equipment adds to the cost. v) Periodic replacement of air filter of induction system is required.

Q.3(b) BP = 60 kw, PM = 7.5 bar, No. of explosion per min = 50

1d

= 21

L = 2d m = 0.85

m = BPIP

0.85 = 60IP

IP = 70.85 kw IP = Pm LAN

70.58 × 103 = 7.5 × 105 × 2 2 90d 2d

4 60

d3 = 0.0399 d = 0.3418 m d = 34.18 cm L = 2d = 2 × 34.17 = 68.36 cm Piston speed = 2LN

= 2 × 0.68.36 × 9060

= 2 × 1.025 m/min = 2.0508 m/min

Q.3(c) Purpose of I.C. Engine Testing i) To get information, that is not possible to be determined by calculations. ii) To confirm the validity of data used while designing the engine. iii) To satisfy the customer as to rated power with guaranteed fuel consumption.

The tests carried out on I.C. Engine can be classified in two classes : i) Commercial tests ii) Thermodynamic tests Commercial Tests These tests are carried out on I.C. Engine in order to check the following : i) Rated power (output) with a guaranteed fuel consumption in kg/kW-hr. ii) The quantity of lubricating oil consumed on brake power basis per kW-hr. iii) The quantity of cooling water circulated on brake power basis in kg/kW-hr. iv) The steadiness of engine under varied load conditions. v) The overload carrying capacity of the engine.

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ard.

0

a2 9002d

606022d

m = 2 × 34.17 = 68.36 cm 34.17 = 68.36 cm

= 2LN N

= 2 × 0.68.36 × × 0.68.36 ×ya9006060

= 2 × 1.025 m/min = 2 × 1.02 = 2.0508 m/min = 2.0508 m

c) Purpose of I.C. Engine Testingc) Purpose of I.C. Engine Testi) To get information, that is i) To get information, that isii) To confirm the validity oi) To confirm the validity oiii) To satisfy the custom To satisfy the

The tests carried out The tests cai) Commercial tei) Commii) Thermodynii)

CommercCoThese ti) Rii)


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Thermodynamic Tests Thermodynamic tests are quite different from the commercial tests. These tests are carried out for the purpose of comparing actual results with theoretical or ideal performance. While carrying out such tests it is necessary to measure losses in addition to the useful part of the energy and also to draw up a heat balance account. After such tests are carried out and the performance of heat engine is compared with historical trends (records) the demand of improvement and development in heat engines can be satisfied. The measurements necessary to determine the mechanical and thermal efficiencies of the engine and to draw up the heat balance account are : i) Indicated power ii) Brake power iii) Frictional power iv) Rate of fuel consumption and its Calorific value v) Rate of flow of cooling water and rise in temperature for calculating the heat

carried by cooling water. vi) Heat carried by exhaust gas.

Q.3(d) Necessity of multi-staging a compressor i) As index of compression ‘n’ increases, work required for compression increases. ii) As pressure ratio of P2 / P1 increases, work required for compression

increases and at the same time size of cylinder increases. iii) As temperature at the section of compressor increases the work required for

compression increases, therefore, when pressure ration P2 /P1 increases beyond some certain limit, then instead of on cylinder, compression is carried out in more than on cylinder. This is called as multistage compression.

No. of states = No. of cylinders Advantages of Multi-staging i) Volumetric efficiency increases. ii) As compression being approximated is isothermal, power required to drive

the compressor reduced. iii) We get, better mechanical balance and uniform target size of flywheel is also

reduced. iv) Maximum temperature in the cycle is reduced then less difficulty in lubrications.

Q.3(e) Factors affecting volumetric efficiency of air compressure i) Mainly volumetric efficiency depends upon clearance. As clearance volume

increases, volumetric efficiency decreases. ii) Restricted passage and leakage at limit values. iii) Piston ring leakage. iv) As pressure ratio increases, volumetric efficiency decreases. v) If the speed of rotation is high, charge of air taken is loss, which decreases

efficiency. vi) Fresh air comes in contact with hot wall and gets expanded which decreases

the charge taken in therefore volumetric efficiency.

on to t

ne is compared mparend development in elopment in

mechanical and thermal nical and thermal ance account are : ance account are :

ific value ue se in temperature for calculating th emperature for calcu

a compressor a compressn ‘n’ increases, work required for com n’ increases, work required

of P2 / P P11 increases, work requ increases, w the same time size of cylinder increme time size of cylinde

e at the section of coe section of compressor incrempn increases, therefore, when presses, therefore, wh

ome certain limit, then instead ofertain limit, then inste on more than on cylinder. This isn on cylinder. This i called

No. of states = No. of cylinders No. of states = No. of cylind

vantages of Multi-staging ages of Multi-stag) Volumetric effilumetric e ciency increasesency incii) As compression being appros compression being a

the compressor reduced. he compressor reduced. iii) We get, better mechaniii) We get, better mechani

reduced. reduced. iv) Maximum temperat Maxim

Q.3(e) Factors affectinQ.3(e) Fai) Mainly voli)

increaseii) Restriii) Pisiv)


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Q.4(a) (i) Method of energy saving in air compressor In practice, it is not possible to achieve isothermal compression because isothermal process is quasistatic process, which requires very slow movement of piston but speed of compressor is very high. Therefore, it is not possible to achieve isothermal compression process in actual practices.

Therefore, to reduce work required to compression following methods are

adopted : (a) Spraying cold water into cylinder during compression. (b) Providing cooling jackets. (c) Multistaging of compressor.

Q.4(a) (ii) Uses of compressed air

(a) Cleaning of automobiles and workshops. (b) Starting of I.C. Engines. (c) Spraying of fuel in high speed diesel engine. (d) Spraying pains in paint industries. (e) Construction of bridges, roads, dams, structural work, sewage and tunnels. (f) Cooling of large building’s. (g) Operation of Pneumatic drills, air motors, hammers, also for riveting and

tightening nuts etc. h) Supercharging I.C. Engines and in working of gas turbine plants. Free air delivered (FAD) It is the volume of air delivered under the conditions of temperature and pressure existing at compressor intake in the assumes of free air conditions, these are taken as S.T.P. conditions i.e. pressure 1.01325 bar and 15 temp.

Q.4(a) (iii)

(I. P) P2 = 1 3P .P = 1 16 = 4 bar

IP =

n 12n3

11

P2nmRT 1

n 1 P

=

1.3 12 1.32 1.3 6 16

0.287 273 15 11.3 1 60 1

= 27 kW

P1 = 1 bar

P2

P3 = 16 bar 3 3 3

Delivery

Saving in work

Pv1.3 = C

21

Pv = C 2 L.P.

suction

H.P. intercooler Pr

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arethods are are

ngine.

ms, structural work, sewage and tunnctural work, sewa

s, air motors, hammers, also for rivors, hammers, al

ines and in working of gas turbine pnd in working of gas

AD) ) air delivered undervered unde the conditionse cr

at compressor intake in the assumpressor intake in the n as S.T.P. conditioT.P. conditions i.e. pressure ns i.e

(I. P) P

dydddyyayaayayaa

PP11 = 1 bar 1 ba

PP22

= 16 bar33 3 3 3

Delivery Delivery

L.P. P. suct

H.P. H.Pintercooler Pintercool r


11

Isothermal power = mRT1 logePdPs

= 6 16

0.287 273 15 loge60 1

= 22.917 kW

niso =

Isothermalpower

Actualpower

= 22.917

10027

= 84.87%

FAD at intake conditions 1 bar and 15 C can be obtained by

1 × 105 H/m2 × V1 = 6

287 273 1560

V1 = 0.0826 m3/sec

Q.4(a) (iv) Comparison Rotary and reciprocating compressor

Rotary Compressor Reciprocating compressor 1) Compression of air takes

place due to rotary motion of blades.

Compression of air takes place with the help of piston and cylinder arrangement with reciprocating motion of piston.

2) Delivery of air is continuous. Delivery of air is intermediate. 3) Delivery of pressure is low

i.e. pressure ratio is low. Delivery of pressure is high i.e pressure ratio is high.

4) Flow rate of air is high. Flow rate of air is low. 5) Speed of compressor is high

because of perfect balancing. Speed of compressor is low because of unabalanced forces.

6) It required less maintenances. Reciprocating air compressor has more number of moving ports. It need proper lubrication and more maintenance.

7) Are used where large quantity of air at lower pressure is required.

Are used when small quantity of air at high pressure is required.

Q.4(b) (i) Non-positive Displacement Rotary or Steady Flow Compressor

Centrifugal Rotary compressor It mainly consists of impeller and diffuser. The impeller consist of an

impeller disc and impeller vanes, attached on the impeller disc radially forming radial diverging passage as shown in Figure 1.

The impeller rotates with the shaft at high speed and air is drawn into the impeller eye in an axial direction. The air is flown radially outwards through the impeller passages due to centrifugal force, and kinetic energy is imported to the air with some static pressure rise shown in Figure 1(b).

The remaining pressure rise is obtained in the diffuser. The diffuser which is stationary, consist of a number of fixed diverging passages. The air leaves the impeller tip with high velocity and enters the diffuser.

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artained by by

ing compressor mpressor

Reciprocating compressor procating comprnaotion of Compression of air takes plaCompression of air tahelp of piston and cylindehelp of piston anwith reciprocating motioth reciprocatinanla continuous.tinuou Delivery of air is inteDelivery of ailalaessure is low low

e ratio is low. is lowDelivery of pressuDelivery oi.e pressure rate presslalae of air is high. is high. Flow rate of Flalalad of compressor is high pressor is high

cause of perfect balancing. of perfect balancing. Speed of Spunabalallt required less maintenances. less maintenanc Recnyadd

7) Are used where large quant Are used where laof air at lower pressure is of air at lower pressrequired. red. dyididididyQ.4(b) (i) Non-positive Displac.4(b) (i) Non-positive

Centrifugal Rotary Centrifugal It mainly con It m

impeller dimpforming

The im


12

Fig. 1

The diffuser in which reduces the high velocity thus by diffusion process or deceleration of air in the diffuser, kinetic energy is converted into pressure energy.

The flow from the diffuser is collected in a spiral passage from which it is discharged from the compressor.

Single stage compressor can develop a pressure ratio 4:1 pressure ratio as high as 10:1 can be developed with the help of multi-stage centrifugal compressor.

The impeller may be single or double sided.

Q.4(b) (ii) V1 = 0.028 m3 P1 = 1.25 bar T1 = 25 C = 25 + 273 = 298 K V2 = 0.0042m3

Process t2 Pvm = C

2

1

TT

= n 1

1

2

VV

2T298

= 1.3 10.028

0.5042

T2 = 526.48 K

2

1

TT

=

n 1n2

1

PP

526.48

298=

1.3 11.32

1

PP

2

1

PP

= 11.77

P2 = P1 × 11.77 P2 = 1.25 × 11.77 = 14.71 bar

Delivery 2

P2

P

3

4 1

P1

Suction

Compression

V

Pv1.3 = C

W

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ary thus by diffusion process or thus by diffusion process nergy is converted into pressure is converted into pres

d in a spiral passage from which i spiral passage

velop a pressure ratio 4:1 pressure pressure ratio 4:1ped with t he help of multi-stage help of mult

e or double sided. e or double sided.

+ 273 = 298 = 298 K Km3

ess t2 Pvm = C C

d22

11

TT22

TT11 = =

n 111VV1111

V11

2V2V

id2TT2

29829 =

1.3 13 10 0280 00.0280.00.50420.5040 50420 504

T T22 = 526.48 K = 526.

i221TT22

TT1 = dn 1

n2P22

P2

PPV526.48298

a

aaDelPP

3


13

W =

n 1n2

1 11

PnP V 1

n 1 P

= 1.3 11.3

1.31.25 105 0.028 11.77 1

1.3 1

= 11.62 kW v2 = 0.0042 m3 T3 = 298 K

C.V. process, 3

3

P

T = 2

2

PT

3P

298 =

14.71526.48

P3 = 8.32 bar

Q.5(a) Turbo Jet Engine Most common type of air breathing engine is turbo jet engine is shown in figure 1. This engine consists of diffuser, a mechanical compressor a combustion chamber, mechanical turbine and an exhaust nozzle. The function of diffuser is to convert the kinetic energy of engine air into a static pressure rise. After this, air enters to the mechanical compressor, either axial or centrifugal which further compresses the air, and delivers it to the combustion chamber. The fuel nozzle feeds fuel continuously and continuous combustion takes place at constant pressure. The high pressure and high temperature gases then enters the turbine, where they expand to provide power output of the turbine.

Fig. 1 : Turbo jet

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are is turbo jet engine is shown in figuo jet engine is sho

a mechanical compressor a cnical compress an exhaust nozzle. haust nozzle.

o convert the kinetic energy of engi convert the kinetic energy air enters to thnters to the mechanical compr mechanical

er compresses the air, and delivesses the air, and

e feeds fuel continuously and cont uel continuously an pressure. The high pressuree. The high pressure and hi

e, where they expand to provide po hey expand to pro


14

The turbine is directly connected to the compressor and all the power developed by the turbine is absorbed by the compressor and the auxiliary apparatus. The main function of the turbine is to provide power to drive compressor. After the gases leave the turbine, they expand further in the exhaust nozzle and leaves with a velocity greater than the flight velocity to produce a thrust for propulsion. The turbo jet engine is a continuous flow engine. Hence a compressor and a turbine is used to provide additional pressure rise, which could not be obtained in a Ram jet engine. However increase in flight velocity improve its performance because of the benefit of ram pressure achieved by the diffuser. Advantages of Jet Propulsion over Other Systems i) The specific weight of jet propulsion is 1/4 to 1/2 of the reciprocating engine. ii) There are no reciprocating parts. Therefore jet propulsion is free from

unbalanced forces. iii) The jet propulsion engine has small frontal area and better aerodynamic

shape. iv) The speed of jet propulsion is not limited by propeller hence high speeds can

be obtained. v) Since the thrust is applied directly there is no loss of power transmission. vi) The continuous combustion gives continuous thrust and unit has a smooth

running free from vibration. Disadvantages of Jet Propoulsion i) Particularly at low pressure, the thermal efficiency is lower. At low attitude

and speed upto 148 m/s the fuel consumption is 2 to 3 times that of reciprocating engine.

ii) The plant is very noisy, materials are costly and life short. iii) Compression ratio is not constant as in reciprocator but varies approximately

with the square of the rotational speed.

Q.5(b) Open Cycle Gas Turbine With Intercooling The network of gas turbine cycle may be increased by saving some compression work. This is done by sung several stages of compression with inter cooling of air between stages. The air from the first stage compressor is cooled in the intercooler approximately to initial temperature before entering to the second stage compressor. The effect of inter cooling is to decrease the network and increase the efficiency as compared to the simple ideal cycle without inter cooling. The ideal open cycle gas turbine with inter cooling can be shown as 1-2-3-4-5-6. In first stage compressor atmospheric air is compressed from P1 to P2, it is then cooled from temperature T2 to T3 = T1 in the intercooler at constant intermediate pressure Px and finally compressed from Px to P2 in the second stage of compressor.

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arhrust

mpressor and a and a not be obtained in e obtained in

formance because of the ce because of the

ystems /4 to 1/2 of the reciprocating engine/2 of the reciprocating

Therefore jet propulsion is free ore jet propulsion

small frontal area and better aerntal area and be

n is not limited by propeller hence h limited by propeller

plied directly there is no loss of pow directly there is no loss mbustion gives continuous thrust a gives continuous

m vibration. tion.

s of Jet Propoulsion Propoulsion rly at low pressure, the thermal e ow pressure, the the

peed upto 148 m/s the fuel cono 148 m/s the procating engine. procating engine.

The plant is very noisy, materials aree plant is very noisy Compression ratio is not pression ratio is no constan

with the square of the rotationath the square of the ro

5(b) Open Cycle Gas Turbine Wi5(b) Open Cycle Gas Turbine WThe network of gas turbine The network of gaswork. This is done by suork. This is donebetween stages. een st

The air from the Thto initial tempeto iof inter coof icompare

The


15

Fig. 1 : Open cycle gas turbine with intercooling

Fig. 2 : T-S diagram for open cycle gas turbine with intercooling

Open Cycle Gas Turbine with Reheating By reheating or adding heat to the exhaust gases after they have passed through a part of the rows of turbine blading (or stages), a further increase in work done is obtained. In reheating, the gas temperature which has dropped due to expansion is brought back to approximately the initial temperature, for expansion in the next stage. Since the working fluid contains about 85% air, additional fuel can be burnt by injecting it into the gases without any additional air supply. The reheat cycle can be shown as 1-2-3-4-5-6. The combustion gases from combustion chamber CC1 at temperature T3 are partially expanded in the HP turbine from pressure P2 to intermediate pressure Px. After this, it is then passed through combustion chamber CC2 where it is reheated at constant pressure Px so that the temperature of gas is raised from T4 to T5. After this, gas is expanded in second stage of turbine from pressure Px to P1.

The increase in work due to reheating is shown by shaded area.

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artercooling ng

T-S diagram for open cycle gas turbagram for open cy

Gas Turbine with Reheating urbine with Reheating or adding heat to the exhaust gaing heat to the exh

f the rows of turbine blading (or staf the rows of turbine bladingained. ined

n reheating, the gas temperatureeating, the gas tembrought back to approximately thht back to approximatestage. stage.

Since the working fluid cSince the working injecting it into the gasecting it into t

The reheat cycle cThe rehea

The combusThe partially expa

After at


16

Fig. 3 : Open cycle gas turbine with reheating

Fig. 4 : T-S diagram for open cycle gas turbine with reheating

Q.5(c) Heat Pump

Heat pump is defined as thermodynamic system operating in a cycle which removes heat from low temperature body and delivers it to high temperature body. It is also defined as thermodynamic system operating in a cycle, which pumps heat to high temperature body by consumption of work.

Heat source

Heat Engine

Heat Sink

QR

QA

Fig. 1 : Heat pump

Wnet = QR QA Vidy

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ar eheating ng

4 : T-S diagram for open cycle gas tagram for open cyc

mp ump is defined as thermodynamiump is defined as thermod

oves heat from low temperature bes heat from low tody.

yHeaso


17

Q.6(a) (i) Dry Bulb Temperature (TDB) Dry bulb temperature of air is the temperature recorded by ordinary

thermometer and it is not affected by the moisture present in air. It is denoted by to tdb.

(ii) Wet bulb temperature (TWB) Wet bulb temperature is the temperature recorded by thermometer when its

bulb is covered with wet cloth known as ‘wick’ and it exposed to air, is known as ‘Wet bulb temperature’ it is denoted by twb.

(iii) Dual point temperature (TDP)

if the mixture of air and water vapour is cooled at constant pressure, the ability of air to hold water vapour reduces and saturated condition will be reached further lowering of temperature will result in condensation of water vapour or formation of dual.

(iv) Specific humidity ( ) It is defined as the ratio of mass of vapour to the mass of dry air in a given

sample of moist air.

( ) Specific humidity =

Mass of water vapour inmixtureMass of dry air inmixture

Q.6(b) Components of Vapour Compression Cycle

The main components of vapour compression cycle are : i) Compressor ii) Condenser iii) Expansion device iv) Evaporator i) Compressor Compressor is most important component of vapour compression

refrigeration system and is considered being the heart of the system.

Function Function of compressor is to compress the low pressure refrigerant from

evaporator to condenser pressure at a temperature more than saturation temperature corresponding to condenser pressure.

Location It is located between evaporator and condenser. Evaporator exit is

connected to compressor and compressor exit is connected to condenser.

Compressor used in Vapour compression cycle may be (a) Reciprocating compressor (b) Rotary compressor

It may be open type or hermetically sealed. (ii) Condenser Condenser is heat rejection component in vapour compression system.

Function Function of condenser is refrigeration system is to de-superheat and

condense the compressor discharged vapours and frequently to sub-cool the liquid with minimum pressure drop.

It is the heat rejection component in refrigeration cycle.

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arer when its n its

o air, is known know

constant pressure, the t pressure, the saturated condition will be ed condition will be

sult in consult in condensation of water densation of wa

vapour to the mass of dry air in a g to the mass of dry

anMass of water vapour inmixtureer vapour inmixtu

Mass of dry air inmixtureMass of dry air inm

Compression Cycle mpression Cycle f vapour compression cycle are : r compression cycl

) Condenser iii) Expansion denser iii) Expansio devic

r is most important componenmost important coon system and is considered being tem and is considere

ion nction of compressor is to comprenction of compressor is to

evaporator to condenser pressure aporator to condentemperature corresponding to conerature correspond

Location ocation It is located between e It is located between

connected to compressoonnected to compresso Compressor used in Compressor u (a) Reciprocating (a) R (b) Rotary com (b) Rot

It may be op

(ii) Conde(ii) Cond

F


18

Location Condenser is located between compressor and expansion device.

Compressor discharge is connected to condenser inlet and condenser exit is connected to expansion device.

Types of Condenser Condenser can be classified into three groups. (i) Air cooled condenser (ii) Water cooled condenser (iii) Evaporative condenser.

Water Cooled Condenser Air Cooled Condenser Advantages (i) Lower power requirement per ton

of refrigeration. (ii) Longer compressor life. (iii) Delivers the same amount of

cooling on the very hot day as it does on cooler day.

Disadvantages (i) Higher power requirement per

ton of refrigeration. (ii) Shorter Compressor life. (iii) On a day when more cooling is

required, least cooling is available.

Disadvantages (i) More trouble to install and more

cost. (ii) Problem of water cost and water

treatment.

Advantages (i) Less trouble to install and less

initial cost. (ii) Very few maintenance

problems.

(iii) Expansion Device Expansion device is the pressure reducing component in vapour

compression system.

Function (a) To reduce pressure of refrigerant from condenser pressure to evaporator

pressure by throttling. (b) To control mass flow rate of refrigerant entering in evaporator as per load

on evaporator.

Location Expansion device is located between condenser and evaporator of vapour compression cycle. Condenser exit is connected to expansion device and exit of expansion device is connected to evaporator.

Types of expansion device used (a) Capillary tube (b) Expansion valve (i) Thermostatic expansion valve (ii) Automatic expansion valve. (c) Solenoid valve (d) Float valve (i) Low side float valve (ii) High side float valve

(iv) Evaporator Evaporator is a component in which refrigerating effect is obtained.

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ared Condenser ndenser arrrgeser power requirement per er power requirement per

n of refrigeration. of refrigeraShorter Compressor life. er Compressor life.

) On a day when more cooling is y when more corequired, least cooling is equired, least cooling available. available. ka

rr more

and water and water

Advantagesdvant(i) Less trouble to install Less trouble to

initial cost. itial c(ii) Very few mainten(ii) Very few

problems. problnklalala vice is the pressure reducing s the pressure red

system. .

reduce pressure of refrigerant pressure of refrigera frompressure by throttling. throttling.

) To control mass flow rate of refrig To control mass flow rate on evaporator. on evaporator.

Location ocation Expansion device is locatexpansion device is locacompression cycle. Conompression cycle. Conexit of expansion devicexit of expansi

Types of expansTypes(a) Capillary tu Cap(b) Expansio (i) Th (ii) (c) S(d


19

Function Refrigerating effect is produced in evaporator. The liquid at low pressure

enters in evaporator, by absorbing heat it converts into vapours. The vapours are drawn in suction line of compressor.

Location Evaporator is located between expansion device and compressor. Exit of expansion device is connected to evaporator and exit of evaporator is

connected to suction line of compressor.

The different types of evaporators are (a) According to operating condition (i) Dry expansion type evaporator (ii) Flooded type evaporator (b) According to construction of surface (i) Bare tube evaporator (ii) Plate surface evaporator (iii) Finned evaporator

Q.6(c) Superheating

In simple vapour compression cycle, the compressor suction is dry saturated vapour. If some liquid refrigerant enters the suction line of compressor, due to wet compression, lubricating oil present in compressor will be washed off causing more wear and tear of compressor and effective life of compressor reduces. The compressor is the main important component in vapour compression cycle. Thus in order to increase the life of compressor, refrigerant vapour coming out of evaporator is allowed to stay for some more time in evaporator to ensure dry saturated or superheated vapour at evaporator exit. Thus, the increase in the temperature of refrigerant vapour more than saturation temperature is evaporator is known as ‘superheating’.

Superheating of refrigerant is as shown in Figure 1 by 1-1 .

Fig. 1: Superheating

Effect of Superheating Due to superheating suction temperature of compressor increases, increasing compressor power but it also increases the refrigerating effect therefore COP of system remain more or less constant. The superheating is not done to increase the refrigerating effect or COP but it is done to increase the life of compressor.

Subcooling The process of cooling refrigerant below condensing temperature for a given pressure is known as ‘subcooling or under cooling’.

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ar evaporator is ator is

e, the compressor suction is dry smpressor suction enters the suction line of compres the suction line of c

present in compressor will be washnt in compressor wressor and eessor and effective life of compresective life of co

portant component in vapour comprtant component in vapoue life of compressor, refrigerant vf compressor, refr

d to stay for some more time in eay for some more timrheated vapour at evaporator exit. d vapour at evap

efrigerant vapour more tnt vapour more than saturathan superheating’. eating’.

ing of refrigerant is as shown in Figrigerant is as show

Effect of SuEffeDue to suDucompresystethe


20

Effect of Subcooling Due to subcooling the refrigerating effect increases or for same refrigerating effect the circulation rate of refrigerant decreases and therefore COP of system increases. Thus, subcooling is desirable and is done to increase refrigerating effect and COP of system.

The process of subcooling is shown by 3-3 in P-h and T-S as shown in figure 2.

Fig. 2: Subcooling

Q.6(d) Sensible heating

Sensible Cooling

Outlet Air t3

Heating coil

Steam in

t2 Inlet Air t1

Steam out

Sensible heating

tDB

Air at t3

Cooling Coil

Refrigerant in

Refrigerant out

Air In t1

Sensible cooling

tDB

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ar figure 2.

cooling

Sensible Cooling ble Cooling

aaayaaaal

yayayayayayaalall

OutletOutletAir Air t3

yaHeating Heatcoil

Steam in Stea

t2

d Refrigeranrige

A


21

Q.6(e) Cermet cycle

(CoP)Ref = 2

1 2

TT T

T1 = 27 + 273 = 300 K T2 = 12 + 273 = 261 K

(CoP)Ref = 261

300 261= 6.6923

Amount of hart required to be removed to produce 20 tons of ice at 3 C from water at 10 C per 24 hrs if calculated as follows:

(0 ( 3))]RE = [4.2(10 0) + 336 + 2.1 (0 ( 3))]20000

24 3600

= 88.95 k

CoP = RE

Comprosser Power

6.6923 = 88.95

Comprosser power

Compressor power = 13.29 k

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ars of ice at 33 C C

k3600

9 k

t.y. diploma : sem. v [me/pg/pt] power engineering prelim ...vidyalankar.org/upload/pe_soln.pdf ·...

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