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1113/TY/Pre_Pap/2013/CP/Vac/PE_Soln 1

Vidyalankar

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

Power Engineering Prelim Question Paper Solution

(i) Otto cycle (Constant volume cycle)

Otto cycle

Otto cycle is the idealized cycle for spark ignition engine (i.e. petrol engine)

where heat addition and rejection is taking place at constant volume. Actual engine cycle consists of four strokes : suction, compression, expansion and exhaust. But idealized cycle is free from suction and exhaust. Therefore we neglect suction and exhaust stroke.

Otto cycle consists of four processes: Process 1-2 : Isentropic compression of air raising pressure and

temperature. Process 2-3 : Addition of heat at constant volume increasing pressure and

temperature. Process 3-4 : Isentropic expansion of air from high pressure and

temperature to low pressure and temperature. Process 4-1 : Rejection of heat at constant volume reducing pressure and

temperature. Considering 1 kg of air. Heat supplied = qA = Cv (T3 T2) Heat rejected = qR = Cv (T4 T1) Work done = qA qR = Cv (T3 T2) Cv (T4 T1) Air standard efficiency of Otto cycle is

= Net work doneHeat addition

= A R

A

q q

q

1. (a)

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= v 3 2 v 4 1

v 3 2

C (T T ) C (T T )

C (T T )

= 1 4 1

3 2

T T

T T

Compression ratio, r = 1 4

2 3

V V

V V

Consider process 1-2 : PV = C

2

1

T

T =

11

2

V

V

T2 = T1

11

2

V

V

T2 = T1 (r)1

Consider process 3-4: PV = C

3

4

T

T =

14

3

V

V

T3 = T4 1

4

3

V

V

T3 = T4 (r)1

Substitute the value of T3 and T2 in equation (1.4)

= 1 4 11 1

4 1

T T

T (r) T (r)

= 1

4 11

4 1

T T1T T(r)

= 11

1

(r)

(ii) The different additives with their function

Additive Function 1) Detergent Control of high temperature deposits. It can also act as

acid neutralizer. 2) Dispersant Control of low temperature sludge and varnish deposits. 3) Anti-wear Reduce wear and prevent scoring, galling and seizure. 4) Anti-rust Reduce rusting by acid neutralization of formation by

protective film 5) Viscosity

Index Improver

Increases viscosity index reducing sensitivity of viscosity with temperature of oil.

6) Pour point depressant

Reduce pour point of oil by interfering with wax crystallization.

7) Anti-foam Reduce oil foaming by causing collapse of bubbles due to air entrainment.

1. (a)

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Prelim Question Paper Solution

1113/TY/Pre_Pap/2013/CP_Vac/PE_Soln 3

8) Anti-oxidant Reduce oil oxidation to protect alloy bearings against corrosive attack.

(iii) SCAVENGING The clearing or sweeping out of the exhaust gases from the combustion

chamber of the cylinder is known as scavenging. It is necessary that cylinder should not have any burnt gases because

they mixed with the fresh incoming charge and reduce its strength. Power will lost if the fresh charge is diluted by the exhaust gases.

The scavenging is necessary only in two stroke engine since piston does not help for clearing the burned gas from the cylinder.

The scavenging methods in two stroke cycle engine are: (a) Cross flow scavenging :

The admission ports are provided on the sides of the cylinder and exhaust ports are kept on the opposite cylinder wall.

The charge or air entering through the scavenge (admission) ports is directed upward which pushes out the exhaust gases through oppositely situated exhaust ports as shown in Figure (a).

(b) Full-loop or backflow scavenging :

The exhaust ports and scavenge ports are provided on the same side of the cylinder wall and the exhaust ports are situated just above the scavenge port as shown in Figure (b).

This method is particularly suitable for double-acting I.C. engine.

(c) Uniform flow scavenging :

The scavenging ports are provided on one side of the cylinder wall and exhaust port are kept in the cylinder head for the removed of the exhaust gases.

Here the scavenge air and the exhaust gases moves in the same upward direction as shown in Figure (c).

The charge or air requires to be compressed before it is admitted to cylinder so that it will help in scavenging the cylinder.

1. (a)

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(iv) Cut-off Ratio : It is defined as Ratio of volume at outlet of heat addition to the volume at inlet.

() = 3

2

V

V

Compression Ratio : It is defined as Ratio of volume at inlet of compression stroke to the volume at end of compression stroke.

r = 1

2

V

V

(i) Stages of Combustion in S.I. Engine:

In S.I. engine sufficiently homogeneous mixture of vaporized fuel, air and residual gases is ignited by a single spark produced between electrodes of spark plug in the form of thin thread of flame. From this thin thread of flame, combustion spreads to envelop mixture immediately and burn completely. There are three stages of combustion: Ignition lag Propagation of flame After burning Ignition lag: It is the time taken for growth and development of self-

propagating nucleus of flame, called as ‘ignition lag or preparation phase’. After production of spark, fuel takes some time to ignite and produce a self propagating flame. Spark is produced 20 before TDC and fuel start burning at 8 before TDC.

Stages of combustion in S.I. Engine

Propagation of flame: Point ‘A’ is the point of production of spark in

above figure. At point ‘B’ the first rise of pressure can be detected and at point ‘C’, the peak pressure is reached. Thus, BC is the second stage i.e. propagation of flame. At point ‘C’, the travel of flame completes.

1. (a)

1. (b)

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After burning: Though point ‘C’ is the completion of flame travel, it does not mean that at this point the whole of the heat of fuel has been liberated, but fuel continues to burn. This continuous burning of fuel in expansion stroke is called as ‘after burning’.

(ii) Otto cycle may be represented on P-V and T-S diagram as follows:

Otto cycle

Pressure at the beginning of compression = P1 = 1 bar Temperature at the beginning of compression = T1 = 18C + 273 = 291 K

Compression ratio = 1

2

V

V= r = 8

Heat added, QA = 250 kJ/kg Consider process 1-2: PV = C

2

1

T

T =

11

2

V

V

T2 = T1 × r1 = 291 × (8)1.41 = 668.54 K

Heat added is given by, QA = Cv (T3 T2) 250 = 0.71 × (T3 668.54) T3 = 1020.65 K

Air standard efficiency of Otto cycle is given by,

= 11

1

r, where r = compression ratio

= 11.4 1

1

(8)

= 0.5647 = 56.47%

1. (b)

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Efficiency is also given by

= Work outputHeat input

0.5647 = Work output

250

Work output = 250 × 0.5647 = 141.175 kJ/kg Consider process 3-4: PV = C

3

4

T

T =

14

3

V

V

3

4

T

T = (r)1

4

1020.65T

= (8)1.41

T4 = 444.26 K Heat added is given by, QR = Cv (T4 T1) = 0.71 (444.26 291) QR = 108.81 kJ

(i) 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.

(ii) Simple Carburettor Simple carburettor consists of a float chamber, nozzle with metering orifice,

venturi and throttle valve.

2. (a)

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

Simple carburettor

The quantity of fuel discharged is regulated by butterfly type throttle valve.

As the throttle is closed, less air flows through the venturi and less air-fuel mixture is delivered to the cylinder and less power is developed. As throttle is opened, more air flows through the venturi and the power developed increases.

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Limitations of Simple Carburettor: A simple carburettor gives the correct mixture at only one engine speed

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

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.

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

The working of simple carburettor is affected by changes in atmospheric temperature and season.

Otto cycle

Volume at beginning of compression = V1 = 0.5 m3 Pressure at beginning of compression = P1 = 1 bar Temperature at beginning of compression = T1 = 30C + 273 = 303 K Pressure at the end of compression = P2 = 13.8 bar Heat added = QA = 210 kJ Consider process 1-2: PV = C

2

1

T

T =

1

2

1

P

P

2T

303 =

1.4 11.413.8

1

T2 = 641.39 K Similarly,

1

2

V

V =

1

2

1

P

P

2. (b)

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2

0.5V

=

11.413.8

1

V2 = 0.07669 m3 210 = 0.71 (T3 641.39) T3 = 937.16 K Consider process 2-3: V = C, V2 = V3 = 0.07669 m3

2

2

P

T = 3

3

P

T

13.8

641.39= 3P

937.16

P3 = 20.16 bar Consider process 3-4: PV = C

3

4

T

T =

14

3

V

V

3

4

T

T =

11

2

V

V

4

937.16T

= 1.4 1

0.50.07669

T4 = 442.72 K

Similarly, 4

3

V

V =

1

3

4

P

P

1

2

V

V =

1

3

4

P

P

0.5

0.07669 =

11.4

4

20.16P

P4 = 1.46 bar QR = Cv (T4 T1) = 0.71 × (442.71 303) = 99.19 kJ/kg Net work done = QA QR = 210 99.19 = 110.80 kJ/kg

Pressure, temperature and volume at all points, P2 = 13.8 bar V2 = 0.07669 m3 T2 = 641.39 K P3 = 20.16 bar V3 = 0.07669 m3 T3 = 937.16 K P4 = 1.46 bar V4 = 0.5 m3 T4 = 442.72 K

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(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’.

(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.

Actual and Theoretical Valve Timing Diagram of 4-Stroke Diesel Engine

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

Figure (a) and (b) shows theoretical and actual valve timing diagram of 4-stroke diesel engine. Theoretical valve timing diagram is similar to that of 4-stroke petrol engine. In actual valve timing diagram, inlet valve opens 10 to 25 in advance of TDC position and closes 25 to 50 after the BDC position and closes 10 to 15 after the TDC position. The fuel injection takes place 5 to 10 before TDC and continues up to 15 past TDC. Mechanical Efficiency (m) : It is a measure of mechanical perfection of the engine or its ability to transmit power developed in the engine cylinder to the crank shaft.

3. (a)

3. (b)

2. (c)

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

Mechanical efficiency = Break power

Indicatedpower

m = Break power

Brake power Frictionalpower

Mechanical efficiency of engine varies with the load. When load on engine is decreased efficiency falls down. Also efficiency decreases with increase in speed. Thermal Efficiency: It is the ratio of power output (developed) to the heat energy of the fuel supplied during the same duration. (Brake Thermal Efficiency) : The efficiency based on the brake power is known as brake thermal efficiency, and is given as:

b.th = Heat equivalent of brake power

Heat supplied

= BP inkW

Mass of fuel inkg / sec CV inkJ / kg

(Indicated Thermal efficiency (i)) : Efficiency based on Indicated power is termed as indicated thermal efficiency and is given as:

i = Heat equivalent of indicatedpower

Heat supplied

= IP inkW

Mass of fuel inkg / sec CV inkJ / kg

CATALYTIC CONVERTER

Fig. : Catalytic converter

Catalytic converters are widely used in car all over the world. 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 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 700C. The temperature of exhaust gases in exhaust system is lower. The catalytic converter oxidizes and reduces all the three pollutant gases at lower

3. (c)

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temperature because of catalytic chemical reaction. A catalytic converters becomes effective in the temperature between 250 to 300C. Two way catalytic converter: A converter is filled with a 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’.

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 900C, which is required for the catalytic converter to operate at complete efficiency. To safeguard from 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. Limitations of Catalytic Converter: Since lead destroys catalytic activity, the engine cannot use leaded petrol. Exhaust systems are hotter than normal as a result of exothermic reaction in

catalyst bed. The emission of SO3 increases if fuel contains sulphur. The use of equipment adds to the cost. Periodic replacement of air filter of induction system is required. Piston diameter, d = 15 cm = 0.15 m Length of stroke, L = 40 cm = 0.40 m Mean effective pressure, Pm = 5 bar = 5 × 105 N/m2 120 explosions/min BP = 5 kW

3. (d) Vidyala

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Indicated power of engine is given by, IP = Pm LAN*

= Pm L4

d2 N*

= 5 × 105 N/m2 × 0.40m × 4

× (0.15)2 m2 × 12060

explosions/sec

= 7068.58 N.m/s = 7068.58 watts =7.068 kW Mechanical efficiency of engine is given by,

m = BPIP

= 5

7.068 100

m = 70.74% Important : N* is the number of explosions/sec or number of power strokes/sec. In case of engine running with N r.p.m.:

N* = N2

… [for four stroke engine]

= N … [for two stroke engine] Four cylinder : four stroke engine Mean effective pressure, Pm = 5 bar Speed, N = 1200 rpm Diameter of piston, d = 100 mm = 0.1 m Length of stroke, L = 150 mm = 0.15 m m = 80% Indicated power of one cylinder 4 stroke engine : IP = Pm LAN* where N* = No. of explosions

= Pm × L × 4

× d2 × N2

… [for four stroke engine]

= 5 × 105N/m2 × 0.15 m × 4

× (0.1)2 m2 × 120060 2

rev/s

= 5890.48 N.m/s = 5980.48 watts = 5.89 kW

indicatedpower of 4 cylinder

4 stroke engine

= 4 × 5.89 = 23.56 kW

Mechanical efficiency = BPIP

0.80 = BP

23.56

BP = 0.80 × 23.56 BP = 18.848 kW

3. (e)

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(i) Compressor Ratio : Compression ratio if the pressure ratio defined as delivery pressure to suction pressure.

Compressor capacity : It is to quantity of the free air actually delivered of by

compressor in m3/min.

Free air delivered (FAD) : It is volume of air delivered under the conditions of temperature and pressure existing at compressor intake. In the absence of free air conditions, these are taken as S.T.P. conditions i.e. pressure 1.01325 bar and 15C temperature.

Swept volume : It is actual volume of air taken during suction stroke. It is expressed in m3.

(ii) Screw Compressor

Screw type compressor is positive displacement compressor consists of two mutually engaged helical grooved rotors suitably housed in a casing. Out of two rotors generally male rotor is driver rotor having four helical lobes, which are engaged in corresponding flutes of female rotor. Female or driven rotor has six number of flutes. During the rotation of rotor, air is drawn through the suction port to fill the space between male and female rotor lobes. Since the air drawn between the lobes is isolated from the suction port, this air so trapped is moved both axially and radially with the rotation of rotors compressed due to volume reduction. When this compressed air is put in communication of discharge port, air leaves through discharge port. We can note that the male and female rotor rotates at different speeds due to different number of lobe and flutes.

Screw type compressor

Screw compressor is capable of handling 3.5 to 300 m3/min of air with maximum pressure ratio of 20. The rotors are usually cooled by internal circulation of lubricating oil and due to this these compressor requires elaborate lubricating oil system. These compressors are noisy in operation but due to its simplicity, durability and reliability it is used in refrigeration and air conditioning industry.

4. (a)

4. (a)

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(iii) Difference between Reciprocating Compressor and Rotary Compressor

Reciprocating Compressor Rotary Compressor 1) Compression of air takes place with

the help of piston and cylinder arrangement with reciprocating motion of piston.

Compression of air takes place due to rotary motion of blades.

2) Delivery of air is intermittent. Delivery of air is continuous. 3) Delivery pressure is high, i.e.

Pressure ratio is high. Delivery pressure is low, i.e. Pressure ratio is low.

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

of unbalanced forces. Speed of compressor is high because of perfect balancing.

6) Reciprocating air compressor has more number of moving parts. It needs proper lubrication and more maintenance.

Rotary air compressor has less number of moving parts, therefore less maintenance is required.

7) Due to low speed of rotation it cannot be directly coupled to prime mover but it requires reduction of speed.

Rotary air compressor can be directly coupled to prime mover.

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

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

(iv) Methods 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 practice. Therefore, to reduce work required to compression, following methods are adopted: Spraying cold water into cylinder during compression: Drawback of this

method is air becomes wet and compressor is required to handle two phase mixture.

Providing cooling jackets: Increase in temperature due to compression is decreased by providing cooling jackets.

Multistaging of compressor: Instead of compressing air in single cylinder, compression is carried out in more than one cylinder.

(i) Single stage compression process can be shown on P-V diagram as follows:

Fig. : Single stage compression

4. (a)

4. (a)

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Indicated power with polytropic compression,

I.P. = n

n 1P1V1

n 1n

2

1

P1

P

=

1.2 11.251.2 3 5

1 10 11.2 1 60 1

= 9229.81 watts = 9.22981 kW

Isothermal power = P1V1loge2

1

P

P

= 1 × 105 × e3 5

log60 1

= 8047.18 watts = 8.04718 kW

Isothermal efficiency = Isothermalpower

100Indicatedpower

=

8.04718100

9.22981

= 87.19% (ii) RAM JET ENGINE

Ram jet is also called as ‘Athodyd or flying stovepipe’. It is a steady combustion of continuous flow engine and has the simplest construction of any propulsion engine. Ram jet is shown in figure. It consists of inlet diffuser, a combustion chamber and exit nozzle. Ram jet has no compressor as the entire compression depends upon ram compression. The function of supersonic and subsonic diffuser is to convert the kinetic energy or the entering air into a pressure rise, called as ‘ram pressure’.

Fig. : Ram jet

The air entering into the ram jet with supersonic speed is slowed down to sonic velocity in the supersonic diffuser, increasing air pressure. The air pressure is further increased in the subsonic diffuser increasing the temperature of the air. The diffuser section is designed to get the correct ram effect. Its job is to decrease the velocity and increase the pressure of the incoming air.

4. (b)

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The fuel injected into the combustion chamber is burned with the help of flame stabilizers. The high pressure and high temperature gases are passed through the nozzle converting the pressure energy into kinetic energy. The high velocity gases leaving the nozzle provide required forward thrust to the ram jet. The ram jet engines are used flight at supersonic speed (much number 1.5 to 2). Since the ram jet engine cannot operate under static conditions, as there will be no pressure rise in the diffuser, it is not self-operating at zero flight velocity. To initiate its operation the body propelled by the ram jet engine must either be launched from an airplane in flight or to be given an initial velocity by some auxiliary means, such as launching rockets. Basic Characteristics of Ram Jet Engine: No moving parts and lightweight as compared with turbo jet engine. Simple unit quite adaptable to mass production at relatively low cost. At low and moderate speeds, the fuel consumption is too large. The fuel

consumption however decreases flight speed. For steady combustion, certain devices in the form of flame holders or

pilot flames are required. Vapour Compression System

Fig. : Simple vapour compression system

5. (a)

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In air refrigeration system working medium is air, which absorbs heat in evaporator and rejects heat in heat exchanger. This absorption and rejection of heat involves only sensible heat therefore coefficient of performance of system is less and large quantity of air is required to get refrigerating effect. In vapour compression system working medium air is replaced by liquid, therefore it involves sensible as well as latent heat and coefficient of performance of vapour compression system is more than air refrigeration system. Most of the modern refrigerators work on vapour compression cycle with Freon 12, Freon 22 and eco-friendly refrigerant 143a or any other refrigerant as working fluid. Simple vapour compression cycle consists of four different processes. Compression Condensation Expansion Vaporisation Compression: Low-pressure vapour in dry state is drawn from the

evaporator during the suction stroke of compressor. During compression the pressure and temperature increases until the vapour temperature is greater than the temperature of condenser cooling medium. This process is known as ‘compression’.

Condensation: When high-pressure refrigerant vapour enters the condenser, cooling medium absorbs the heat and converts vapour into liquid. This process is known as ‘condensation

Expansion: After condensation liquid refrigerant is stored in the liquid receiver. The liquid from receiver is passed to evaporator through expansion valve. The expansion valve reduces the pressure by keeping enthalpy constant. This process is known as ‘throttling process’.

Vaporisation: The liquid low-pressure refrigerant after expansion enters in evaporator where a considerable amount of heat is absorbed by it and converts into vapour. This vapour at low pressure is sucked in suction line of compressor.

Fig. : Simple vapour compression cycle on P-h and T-S diagram

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Dry saturated vapour at point 1 enters into compressor. The compression process is isentropic. Which increases pressure from evaporator pressure to condenser pressure. At point 2, superheated vapours enters the condenser where heat is rejected at constant pressure. Due to rejection of heat, temperature of superheated vapour is reduced to saturation temperature. Further removal of heat results in condensation and vapour converts into liquid at point 3. At point 3 throttling of liquid takes place, which reduces pressure by keeping enthalpy constant. Due to reduction of pressure some part of liquid converts into vapour as shown by point 4. Liquid at low pressure absorbs latent heat and gets converted to saturated vapour, which enters in suction line of compressor. Refrigerating effect, RE = m (h1 h4) Compressor power = m (h2 h1)

COP of vapour compression cycle = 1 4

2 1

h hR.E.Compressor power h h

Fig. : Two stage compression.

Ideal intercooler pressure is given as: P2 = 1 3P P 1 7 = 2.64 bar

The indicated power of two stage compression with perfect intercooling is:

I.P. =

n 12n

31 1

1

P2nP V 1

n 1 P

=

1.25 12 1.255 2 32 1.25 7

1 10 N/m 0.3 m /s 11.25 1 1

= 64.443 × 103 N.m/s = 64.443 kW To find out volume of each cylinder when intercooling is perfect: P1V1 = P2V2

1×V1 = 2.64 × V2

1

2

V

V = 2.64

5. (b)

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Compression is taking place at a rate of 0.3 m3/s when speed of compression is 600 rpm.

0.3 m3/s = V1 m3 ×

N60

rev./s

0.3 = V1 × 60060

V1 = 0.03 m3 … [Volume of L.P. cylinder]

From equation (i), 1

2

V

V = 2.64

V2 = 0.032.64

= 0.01136 m3 … [Volume of H.P. cylinder] Assume freezing takes place at 0C.

Fig. : Freezing of 400 kg fruits

Amount of heat required to be removed to cool 1 kg of fruit from 19C to 5C. = 1 × 1.256 [19 0] + 105 + 1.256 [0 (5)]

= 135.144kJkg

For 400 kg of fruit, heat removed = 135.144 × 400 = 54057.6 kJ

The heat has to be removed in 10 hrs.

RE = 54057.6 kJ

10 60 60 s

= 1.5016 kW

Refrigeration capacity = RE 1.5016

3.517 3.517

= 0.4269 tons of refrigeration Difference between Heat Pump and Refrigeration

Heat pump Refrigerator 1) Heat pump pumps the heat to heat

source. Refrigerator removes the heat from heat sink.

5. (c)

6. (a)

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2) Heat pump is used to heat house in winter season.

Refrigerator is used to produce cooling effect.

3) Kelvin-Plank statement is applicable to heat pump.

Clausius statement is applicable to refrigerator.

4) If T1 = Temperature of heat source T2 = Temperature of heat sink

Then, (COP)pump = 1

1 2

T

T T

If T1 = Temperature of heat source T2 = Temperature of heat sink

Then, (COP)refrigerator = 2

1 2

T

T T

Tonnes of Refrigeration: The rate at which heat is absorbed from the body or space to be cooled is known as ‘refrigerating effect’. The capacity of mechanical equipment is generally given in H.P. and of electrical equipment in kW. Similarly the capacity of refrigeration unit is given in Tons of Refrigeration. Due to the fact that refrigeration was first produced by ice, the refrigerating effect of refrigeration machine is compared with refrigeration produced by ice. A Ton of refrigeration is defined as ‘the quantity of heat required to be removed to from one tone of ice at 0C within 24 hours when initial condition of water is 0C, because same cooling effect will be obtained by melting the same ice. In S.I. units, 1 ton of refrigeration = 12660 kJ/hr = 211 kJ/min = 3.517 kJ/s = 3.518 kW 1 ton = 3.517kW Thus ton of refrigeration is not a unit of mass but a measurement of the rate of heat transfer. Psychometric Properties of Air Dry air: Dry air is the mixture of nitrogen and oxygen neglecting the water

vapour and other gases. The volumetric composition of dry air is 77% nitrogen and 23% oxygen. Dry

air does not contain any moisture.

Moist air: Moist air is mixture of dry air and water vapour. The amount of water vapour present in the air depends on absolute pressure and temperature of mixture and partial pressure of water vapour.

Saturated air: The water vapour present in air is known as ‘moisture’ and its quantity in air is an important factor in all air-conditioning systems. The mixture of air and water vapour at a given temperature is said to be saturated when it contains maximum amount of water vapour that it can hold. If the temperature of mixture of air and water vapour is above saturation temperature of water vapour, the vapour is called as ‘superheated vapour’.

For saturated air, Relative humidity = 100% and Degree of saturation = 1

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 tDB.

6. (b)

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Web bulb temperature (tWB): Wet bulb temperature is the temperature recorded by thermometer when its bulb is covered with we cloth known as ‘wick’ and is exposed to air, is known as ‘wet bulb temperature’. It is denoted by tWB.

The difference between dry bulb temperature and wet bulb temperature is known as ‘wet bulb depression’.

Cooling with Dehumidification The removal of water vapour from air is known as ‘dehumidification of air’. Dehumidification of air is possible only if air is cooled below dew point temperature. Theoretically air-water vapour mixture is first cooled along constant moisture line where sensible heat is removed till saturation line is reached. This is shown by process 1-4 in Psychrometric chart.

Fig. : Cooling with dehumidification

Further removal of heat result in condensation of water vapour and moisture is started extracting from air along with reduction in dry bulb temperature. This is shown by 4-2 in psychrometric chart. But in actual practice 1-4-2 is not representation of actual process. It is probably the locus of a series of steps of sensible cooling and condensation. It is represented by straight line, joining point 1 and 3. Where point 1 is inlet condition of air and point 3 is condition of air leaving. During the process, moisture content is reduced from 1 to 3. Under the ideal condition, air coming out of the coil will be at condition 2, which is nothing but coil temperature known as ‘apparatus dew point’. But no cooling coil is 100% efficient, air coming out of coil will be at condition 3, somewhere inbetween point 1 and 2 depending on coil efficiency. Therefore the coil surface must be kept colder than expected air off the coil temperature.

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Heating with Humidification: In this process, heating of air takes place along with rise in moisture. It is achieved by spray of water heated at a temperature higher than the dry bulb temperature of entering air, the resulting process is that of heating and humidification as shown in Figure The leaving air has higher dry bulb temperature and higher ratio than those of the entering air.

Fig. : Heating with humidification

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

6. (d)

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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’. 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

Temperature of source, T1 = 27 + 273 = 300 K Temperature of sink, T2 = 6 + 273 = 267 K Maximum possible COP is Carnot refrigerator COP,

(COP)ref = 2

1 2

T 267T T 300 267

= 8.09 No machine can have COP more than COP of Carnot refrigerator operating between same temperature limits. Inventor’s claim is not correct as his COP is 8.5, which is more that Carnot

COP which is 8.09.

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