1

T.Y. Diploma : Sem. V [ME/MH/MI]

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

Q.1(a) Attempt any THREE of the following : [12]Q.1(a) (i) Draw PV & TS diagram for Duel cycle. State name of the process. [4](A) The Dual Cycle

Fig.: Dual Cycle

In the Otto cycle, combustion is assumed at constant volume while in Diesel cycle combustion is at constant pressure. In practice they are far from real. Since, sometime interval is required for the chemical reactions during combustion process; the combustion cannot take place at constant volume. Similarly, due to rapid uncontrolled combustion in diesel engines, combustion does not occur at constant pressure. The Dual cycle, also called a mixed cycle or limited pressure cycle, is a compromise between Otto and Diesel cycles. Figures (a) and (b) show the Dual cycle on PV and T-S diagrams respectively. In a Dual cycle a part of the heat is first supplied to the system at constant volume and then the remaining part at constant pressure. Thermal Efficiency The efficiency of the cycle may be written as

Dual = S R

S

Q QQ

Q.1(a) (ii) Define the following terms related to air compressor.

(1) Volumetric efficiency (2) Free air Delivery [4]

(A) (1) Volumetric efficiency (v): It is defined as the ratio of actual volume of charge admitted during suction stroke

at N.T.P. condition to the swept volume of the piston. Volumetric efficiency (v)

= Actual volume of the inducedmeasured at NTP conditionSwept volume of the piston

= a

s

VV

The volumetric efficiency of a supercharged engine may be more than 100% if the air at about atmospheric temperature is forced into the cylinder at the pressure greater than that of the air surrounding the engine.

(2) FAD (Free Air Delivery) “Free Air Delivery is the actual volume of air delivered by the compressor, when

reduced to normal temperature and pressure condition”. The capacity of compressor is generally expressed in terms of free air delivery.

P

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Q.1(a) (iii) Give the detail classification of Air compressors. [4](A) Classification of Air compressors

Air compressors are classified in various ways according to : (a) Working (i) Reciprocating compressors (ii) Rotary compressors

(b) Action of Piston (i) Single acting : In this, suction, compression and delivery of air take place on single

side of piston. (ii) Double acting : In this, suction, compression and delivery of air take place on both

sides of piston.

(c) Number of Stages (i) Single stage : Here, compression of air is carried out in one cylinder only. (ii) Multistage : Here, compression of air is carried out in more than one cylinder.

(d) Pressure Developed (i) Low discharge pressure (upto 10 bar) (ii) Medium discharge pressure (between 10 bar to 80 bar) (iii) High discharge pressure (more than 80 bar) (e) Compressor Capacity (i) Low capacity (upto 0.15 m3/sec) (ii) Medium capacity (between 0.15 m3/sec to 5 m3/sec) (iii) High capacity (more than 5 m3/sec) (f) Method of Cooling (i) Air cooled (ii) Water cooled Q.1(a) (iv) Draw actual and theoretical indicator diagram for 4-stroke diesel engine. [4](A) Actual and Theoretical Indicator diagram of 4-stroke Diesel Engine

Figure shows the theoretical and actual P-V diagram. Actual P-V diagram deviates from theoretical P-V diagram.

Compressors

Positive displacement Dynamic compressors

Reciprocatin Rotary Centrifugal Axial

Roots blower Screw type Vane type

Fig. 1 : Indicator diagram for 4-stroke diesel

Prelim Paper Solution

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In actual P-V diagram the corners are rounded off because inlet and exhaust valves do not open and close instantaneously but they take some time to do so.

During suction stroke pressure inside cylinder is slightly less than the atmospheric pressure and during exhaust stroke the pressure inside the cylinder is higher than atmosphere.

Thus, the shaded loop is considered as negative loop known as pumping loss. This shaded area is subtracted from the area of larger loop to get network.

Q.1(b) Attempt any ONE of the following : [6]Q.1(b) (i) State the purpose of Morse test in petrol engine testing. Write stepwise

procedure for conducting Morse test. [6]

(A) Morse Test The Morse test consists of obtaining indicated power of the engine without any elaborate

equipment. The test consists of making inoperative, in turn, each cylinder of the engine and noting the reduction in brake power developed. With a gasoline engine each cylinder is rendered inoperative by shorting the spark plug of the cylinder; with a diesel engine by cutting off the supply of fuel to each cylinder. It is assumed that pumping and friction losses are the same when the cylinder is inoperative as well as during firing. This test is applicable only to multi cylinder engines. Referring to the Figure, the unshaded area of the indicator diagram is a measure of the gross power, gp developed by the engine, the dotted area being the pumping power, pp. Net indicated power per cylinder = gp pp

In this test the engine is first run at the required speed by adjusting the throttle in SI engine or the pump rack in CI engine and the output is measured. The throttle rack is locked in this position. Then, one cylinder is cut out by short circuiting the spark plug in the SI engine or by disconnecting the injector in the CI engine. Under this condition all the other cylinders will motor the cut out cylinder and the speed and output

drop. The engine speed is brought to its original value by reducing the load. This will ensure that the frictional power is the same while the brake power of the engine will be with one cylinder less.

If there are k cylinders, then

ip1 + ip2 + ip3 + ip4+ …+ ipk = k

k k1

bp fp (1)

where ip, bp and fp are respectively indicated, brake and frictional power and the suffix k stands for the cylinder number. If the first cylinder is cut-off, it will not produce any power but it will have friction, then

ip2 + ip3 + ip4 + … + ipk = k

k k2

bp fp (2)

Subtracting Eqn. 2 from Eqn. 1

ip1 = k k

k k1 2

bp bp

Similarly we can find the indicated power of the cylinders, viz., ip2, ip3, ip4, …, ipk. The total indicated power developed by the engine, ipk, is given by

ipk = k

k1

ip (3)

when all the k cylinders are working, it is possible to find the brake power, ipk, of the engine. The frictional power of the engine is given by fpk = ipk bpk (4)

Fig. : P-V diagram of an Otto Engine

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Q.1(b) (ii) Write any three pollutants in exhaust gases of petrol & diesel engine withtheir effects on environment.

[6]

(A) Pollutants in exhaust gas of Petrol Engine : 1) Carbon Monoxide : Carbon monoxide is colourless, toxic gas with acute effects. 0.3 %

carbon monoxide in the air can be lethal within 30 minutes, 300 ppm (parts per million) are dangerous (1 ppm CO = 1.25 mg CO/m3).

In proposed MIC (Mean Imission Concentration), the term imission denotes air pollutions affecting humans, animals and vegetation objects which is 50 and 10 mg CO/m3 for 30 minutes and 24 hours respectively.

2) Nitrogen Oxide (NO) : Nitrogen oxide (NO) is colourless and rapidly forms reddish

brown nitrogen dioxide (NO2) in the presence of atmospheric oxygen. This is also a toxic substance with cute effects, which in concentration of more than 1500 ppm causes sever irritation of the respiratory organs.

When 500 to 1500 ppm are inhaled over extended periods of time, the lung tissue would be damaged to such an extent that death may ensue.

3) Oxides of Nitrogen (NOx) : Oxides of nitrogen i.e. mixture of nitrogen oxide and

nitrogen dioxide is less dangerous. In that, the high NOx emission (high concentrations and large exhaust gas quantities) are produced in the upper power range of the engine i.e. at higher speeds in less dense traffic, so they mix with air more quickly.

1 ppm NO = 1.34 mg NO/m3 and 1 ppm NOx = 2.5 mg/m3. The remaining toxic substances in exhaust gas are not critical in such concentrations. Some of the hydro carbons and oxidants cause sever irritations of the mucous

membranes of the eyes, nose and throat. The polycyclic hydrocarbons are designated as potential carcinogenic substances. It is

still under extensive investigations for the long term effect of these substances. Presently hydrocarbon emission concentrations are taken as indicator of fuel efficiency of an engine.

Pollutants in exhaust gas of Diesel Engine Exhaust gases from diesel engine contains : 1) Nitrogen Oxides : The mechanism of NOx formation is same as in petrol engine. The

height of local peak temperature and sufficient oxygen gives highest NOx

concentration in diesel exhaust. A precombustion engine produces less NOx than direct injection engine due to lower peak temperature.

The NOx production is also affected by injection system and time. It is also affected by variation in fuel characteristic such as cetane number, viscosity etc. NOx concentration also varies linearly with HP of engine.

2) Hydrocarbons : Diesel engine contains hydrocarbon compounds with higher boiling

points and higher molecular weight. Fuel in diesel engine can escape unburned due to - (a) The air-fuel mixture becomes too lean to auto ignite at local point. (b) The air-fuel mixture becomes too rich to ignite or to support a flame. Thus, the hydrocarbon remain un-consumed due to incomplete combustion of fuel. 3) Scoot or Smoke : In diesel engines, air is compressed in cylinder and finally fuel is

injected to it. As complete mixing of air-fuel takes inside the cylinder, complete mixing is not possible

and due to rich or lean fuel incomplete combustion takes place.

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Some of the fuel molecules undergo thermal decomposition and dehydration resulting information of smoke or scoot. Different types of diesel smoke are :

(a) White smoke : White smoke arises due to Too low operating temperature. Too long delay between the start of fuel injection and beginning of combustion. White smoke lasts for short time and not required paying attention from pollution point

of view but it has strong odour. (b) Black smoke : Black smoke is suspension of scoot particle in exhaust gas and results

from incomplete combustion. It appears after engine has warm up and accelerating or pulling under loads.

(c) Blue smoke : Blue smoke is usually due to excessive lubricating oil consumption and its emission indicates very poor condition in engine.

Q.2 Attempt any TWO of the following : [16]Q.2(a) Explain the construction & working of Screw compressor with a neat label sketch. [8](A) It is a type of dynamic compressor. Here, the air is not trapped in specified boundaries, but

it flows continuously and steadily through the machine. The energy from the impeller is transferred to air, when the air flows through the machine and pressure increases due to dynamic effects.

Construction It consists of two helically moving rotors. The male rotor consists of lobes and is

normally the driving rotor. The female rotor has gullies and is the driven rotor. The four helical lobes of male rotor are engaged in corresponding gullies (flutes) of

female rotor. Screw compressors are directly coupled to the prime mover and require low starting torque. Working When the driving rotor rotates, an interlobe space between a pair of lobes of male-

female rotors and the housing nearest to suction end gets opened and is filled with air. Now, the air so trapped is moved both axially and radially with the rotation of rotors and

gets compressed, until it reaches the discharge end. As the number of lobes of male rotor and flutes of female rotor vary, therefore the male

rotor and female rotor rotate at different speeds. A four lobe male rotor will drive six gully (flutes) female rotor at two third of its speed.

Fig.: Screw compressor.

Advantages of Screw Compressor (i) Continuous and uniform flow of air. (ii) High Volumetric efficiency. (iii) Less power consumption even at part load operation. (iv) Can be directly coupled to prime mover like electric motor. (v) Low maintenance cost due to less number of moving parts.

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Q.2(b) Explain working and construction of window air-conditioner. [8](A)

Fig. 1 : Air circuit of window air-conditioner

It is room air conditioner mounted in lower hung of window. Window air conditioner is small unit having capacity 3/4, 1, 1.5, 2 or 2.5 ton capacity and usually does not incorporate heating capacity. It is usually capable of : (a) Cleaning air (b) Cooling air (c) Distributing air (d) Dehumidifying air Ventilation is also provided in window air conditioner to provide fresh air. The basic components of window air conditioner are filter, accumulator, capillary, thermostat, evaporator, evaporator fan, condenser, condenser fan and fan motor with controls.

The window air conditioner works on vapour compression cycle. High pressure, high temperature refrigerant vapour, coming out of compressor are condensed in condenser using outdoor air as cooling medium. The liquid refrigerant from condenser is passed through the drier filter, having silica gel capsule where it absorbs moisture present in the refrigerant. The liquid refrigerant is further passed through capillary to reduce pressure, and then passed to evaporator where it absorbs heat from air producing cooling effect. Due to absorption of heat, liquid refrigerant converts into vapour and these vapours are drawn into compressor. The cycle is repeated again and again.

Fig. 2 : Components of window air conditioner

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If air is cooled below dew point, dehumidification of air takes place and extracted moistures are removed from backside of window air conditioner. Window air conditioner has three separate air circuits. (a) Room air over evaporator : Air from room is drawn into unit through filter to remove

dust and impurities. Then clean air is passed over thermostat and evaporator coil. Thermostat measures temperature and if temperature is above the set level, the compressor is switched 'ON'. If it finds temperature below the set level, the compressor does not start. Finally air is passed over evaporator coil where it is cooled and once again supplied to room.

(b) Outdoor air over condenser : Air is drawn into the unit from outdoor and is passed over air cool condenser and condenser motor compressor caring heat from it.

(c) Outdoor fresh air circuit : If occupant load increases, it requires fresh outdoor air to maintain purity of air. This can be done by providing fresh air damper and ventilation switch.

The compressor used in window air conditioner is hermetic type. This type of compressor has its motor and all moving parts operating within a sealed gas tight housing. The condenser is of air cooled type having continuous cooper tubing. In order to increase heat transfer surface area, aluminum fins are bounded. Generally 2 or 4 rows of condenser coils are provided. A propeller type condenser fan is located immediately in front of condenser coil. This fan cools the condenser coil. The evaporator is located at the front end of window air conditioner. It is made of cooper tubing. The evaporator fan used in window air conditioner is of centrifugal or propeller type and is surrounded by insulating material to reduce noise. The evaporator cools the fresh outside air through ventilation damper, mixed together with room air and passed over evaporator coil to cool it. Air is filtered as soon as it enters the air conditioner. The front panel is provided with louvers for adjusting the direction of air up or down or sideways. Window air conditioner is provided with following controls, (a) Master control : Master control starts and stops the fan and compressor. The

occupant uses this control to choose the desired operation like cool, ventilate and to select the fan speed. It is provided with low, medium and high speed for fan and low, high or medium cooling.

(b) Thermostat control : Thermostat element in window air conditioner is usually located in return stream of air near filter.

Q.2(c) An I.C. Engine uses 5 Kg of fuel per hour having calorific value of

42,500 KJ/Kg. The brake power developed is 21 KW. The temperature rise of cooling water is 23C, when the rate of flow is 11 Kg/min. The temperature rise of exhaust gases is 260C, when rate of flow of exhaust gases is 4.6 Kg/min. Specific heat of water and exhaust gases are 4.187 KJ/KgK and 1 KJ/KgK respectively. Prepare heat balance sheet on minute basis.

[8]

(A) mf = 5 Kg/hr = 560

= 0.0833 KJ/min

C.V. = 42,500 KJ/Kg B.P. = 21 KW Two Twi = 23C mw = 11 Lg/min Tag Ta = 26C

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meg = 4.6 Kg/min Cpw = 4.187 KJ/KgK CPeg = 1 KJ/Kg K

(i) Heat supplied by fuel = mf × CV = 0.083 × 42,500 = 3540.25 KJ/min

(ii) Heat equivalent to B.P. = B.P. × 60 = 21 × 60 = 126 KJ/min

(iii) Heat rejected to cooling water = mw CPw (Two Twi) = 11 × 4.187 × 23 = 1059.311 KJ/min

(iv) Heat carried by exhaust gas = mag × Cpeg × (Tag Ta) = 4.6 × 1 × 260 = 1196 KJ/min

(v) Unaccounted heat = 3540.25 (1260 + 1059.311 + 1196) = 24.939 KJ/min Heat balance sheet

Heat input KJ/min Heat output KJ/min % Heat supplied by fuel

3540.25 (100%)

a) Heat equivalent for B.P. 1260 35.59 b) Heat rejected to cooling water 1059.311 29.92 c) Heat carried by exhaust gases 1196 33.78 d) Unaccounted heat 24.939 0.70

Total 3540.25 (100%)

Total 3540.25 100%

Q.3 Attempt any FOUR of the following : [16]Q.3(a) Explain different types of Lubricant additives. [4](A) Lubricant Additives and their Advantages:

The refined oil obtained by conventional methods are not satisfactorily used for I.C. engine as a Lubricant. Therefore to improve the properties by addition of chemical of compounds called additives. In an endeavour to improve their properties, certain oil soluble organic compounds containing inorganic elements such as phosphorous, sulphur, amine-derivatives and metals are added to the mineral-based lubricating oil. The main additives includes the following: Detergent dispersant Antioxidants and anticorrosives Pour-point depressors Anti-foam agent Rust inhibitors Viscosity-index improvers Oilness and film-strength agents

Detergent-dispersant Advantage: This additive improve the detergent action of the lubricating oil by keeping

the deposit in suspension form as this additives are oil soluble. e.g. Metalic salts or organic acids.

Antioxidants and anticorrosives Advantage : At high temperature oxidation of lubricating oil increases. The oxidation also increases by presence of certain metals such as copper which acts

as a catalyst for oxidizing hydrocarbons of the lubricants. Oxidation is undersirable because of sludge and varnish are created. Acid also form

which are corrosive. This additives used to decrease the oxidation and corrosiveness characteristics of

lubricating oil. These additives are in the form of sulphur and phosphorous compound or amine and phenol derivatives.

Zinc dithiophosphate used as an antioxidant and anticorrosive additives. Pour-point depressors Advantage: Lubricant contain paraffin compound and form wax precipitates as they cooled.

Prelim Paper Solution

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Wax reduce fluidity of oil at low temperature pour depressants are add to lower the pour point of lubricating oil.

e.g. polymerized phenols, easter, alkylated naphthalene oir methacrylate polymers. Anti-form agent Advantage: This additive prevent the formation of foam by reducing surface tension,

which allow air bubble to separate from oil more rapidly. e.g. Silicon polymers. Rust inhibitors Advantage: These prevent rusting of ferrous engine parts during storage and from

acidic moisture accumulation during cold engine operation. e.g. Metal sulphonates, fatty acid and amines.

Viscosity-index improvers Advantage: The viscosity index of lubricating oil can be improved by adding high

molecules polymers. e.g. butylene polymers, polymerized olefins or iso-olefin metharcrylate polymers. Oilness and film-strength agents Advantage: Oilness and high film strength are important in partial-film lubrication. Many organic compounds of sulphur, phosphorous and chlorin are used as additives to

improve the film strength of lubricants. The air standard efficiency of the cycle depends only on

compression ratio. The efficiency can be increased by increasing the compression ratio. But the compression ratio is limited to detonation.

Q.3(b) Give four application of gas turbine. [4](A) (i) Electric power generation. (ii) Turbo jet and turbo propeller engines. (iii) Pumping and refining processes of crude oil. (iv) Nuclear power plant. (v) Propulsion of ships and locomotives. (vi) Supercharging of I.C. engines. (vii) Aviation. Q.3(c) Draw actual valve timing diagram for 4-stroke petrol engine. [4](A) Actual and Theoretical Valve Timing Diagram of 4-Stroke Petrol Engine Figure 1(a) and 1(b) shows the theoretical and actual valve timing diagram of 4-stroke petrol

engine.

Fig.

Fig. 1 : Valve timing diagram of 4-stroke petrol engine

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In theoretical valve timing diagram valves are opened and closed exactly at TDC and BDC. But in actual practice it is difficult to open or close the valve instantaneously; so as to get better performance of the engine, valve timings are modified.

In actual practice inlet valve is opened upto 10° to 30° in advance of TDC, which allows the

fresh charge to enter the cylinder and helps the products of combustion to expel out of cylinder.

The suction of mixture continues upto 30° to 45° after BDC position. The inlet valve closes and the compression of the trapped mixture starts. The spark is produced at 30° to 40° before the TDC in order to allow the ignition lag. The maximum pressure is obtained at nearly 10° past TDC and expansion starts. The exhaust valve is opened 35° earlier in order to allow more time to expel products of

combustion out of cylinder. The exhaust valve is closed when piston is 10° past TDC. Q.3(d) Draw explain simple vapour absorption refrigeration system [4](A) A simple vapour absorption system is as shown in Figure 1.

The essential components of vapour absorption system are evaporator, absorber, generator, condenser, expansion valve, pump and reducing valve. The refrigerant used is ammonia and solution used is acqua ammonia. Strong solution of acqua ammonia contains as much as ammonia as it can and weak solution contains less ammonia. The compressor of vapour compression system are replaced by an absorber, generator, reducing valve and pump. The heat flow in system at generator, and work is supplied to pump.

Ammonia vapours coming out of evaporator are drawn in absorber. The weak solution containing very little ammonia is sprayed in absorber. The weak solution absorbs ammonia and gets converted into strong solution. This strong solution from absorber is pumped into generator.

The addition of heat liberates ammonia vapour and solution gets converted into weak solution. The released vapour is passed to condenser and weak solution to absorber through a reducing valve. Thus, the function of compressor is done by absorber, generator, pump and reducing valve.

Q.3(e) Describe types of Sensors along with their application. [4](A) Modern types of I.C. engines are managed by electronic systems, that receive signals

from numerous sensors. For example, measuring engine speed, manifold pressure or oxygen content.

These measurements are converted by transducers into electric signals.

Fig. 1: Simple vapour absorption system

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(a) Mass Air Flow Sensors These sensors measure the mass of air in the intake manifold of a combustion engine.

The engine can be S.I. or C.I. engine. In case of S.I. engines, it is important to measure the load of the engine demanded by

driver and to provide a signal to the engine controller to determine the ignition timing and fuel required for each individual combustion cycle.

In case of C.I. engines, the mass air flow sensor is used to find exact amount of recirculated exhaust gas by means of engine mapping and measurement of fresh air. The engine can then be adjusted to maintain stable combustion without high NOX emissions.

(b) Oil Pressure Sensors They measure the pressure of engine oil in combustion engines. For S.I. engines, oil pressure is an important indication for operation of systems, such

as, cylinder deactivation or variable valve timing. For C.I. engines, which can run very hot due to regeneration to modern exhaust gas

after treatment systems and can therefore have a need to assure that, the quality of oil is not deteriorated.

(c) Common Rail Pressure Sensors To inject the exact amount of fuel for optimum combustion at lowest emissions, the fuel

pressure inside the rails to be monitored closely. A slight variation in fuel quantity can cause exhaust emissions exceeding the legal limits. (d) Electronic Fuel Supply Pump Sensors Modern injection systems use a supply pump to provide fuel at moderate pressure. This is achieved by regulating the pump pressure in line with the demand to operate the

engine. (e) Gasoline Direct Injection (GDI) GDI was introduced in mid-nineties to achieve lower economy and higher power as

compared to conventional port injection systems. For GDI systems, a pressure sensor is needed as vital input to the engine controller so

as to control the fuel quality. Gasoline quality can vary widely around the world. It can contain various degrees of

sulfur, ethanol content and other additives. It is therefore, important to have a sensor able to work well with all fuels at high pressures and over typical wide temperature range.

Q.4(a) Attempt any THREE of the following : [12]Q.4(a) (i) Explain MPFI with neat diagram. [4](A) Multi-Point Fuel Injection (MPFI) System The main purpose of the Multi-Point Fuel Injection (MPFI) system is to supply a proper ratio

of gasoline and air to the cylinders. These systems function under two basic arrangements, namely.

Port injection Throttle body injection Port Injection In the port injection arrangement, the injector is placed on the side of the intake manifold

near the intake port (Figure 1). The injector sprays gasoline into the air, inside the intake manifold. The gasoline mixes with the air in a reasonably uniform manner. This mixture of gasoline and air then passes through the intake valve and enters into the cylinder.

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Every cylinder is provided with an injector in its intake manifold. If there are six cylinders, there will be six injectors. Figure 2 shows a simplified view of a port or multi point fuel injection (MPFI) system.

Fig. 1 : Port Injection Fig. 2 : Multi-point Fuel Injection (MPFI) near Port Throttle Body Injection System Figure 3 illustrates the simplified sketch of throttle body injection system (Single point

injection). This throttle body is similar to the carburettor throttle body, with the throttle valve controlling the amount of air entering the intake manifold.

Fig. 3 : Throttle Body Injection (Single Point)

An injector is placed slightly above the throat of the throttle body. The injector sprays

gasoline into the air in the intake manifold where the gasoline mixes with air. This mixture then passes through the throttle valve and enters into the intake manifold.

As already mentioned, fuel-injection systems can be either timed or continuous. In the

timed injection system, gasoline is sprayed from the injectors in pulses. In the continuous injection system, gasoline is sprayed continuously from the injectors. The port injection system and the throttle body injection system may be either pulsed systems or continuous systems. In both systems, the amount of gasoline injected depends upon the engine speed and power demands. In some literature MPFI systems are classified into two types: D-MPFI and L-MPFI.

D-MPFI System The D-MPFI system is the manifold fuel injection system. In this type, the vacuum in the

intake manifold is first sensed. In addition, it senses the volume of air by its density. Figure 4 gives the block diagram regarding the functioning of the D-MPFI system. As air enters into the intake manifold, the manifold pressure sensor detects the intake manifold vacuum and sends the information to the ECU. The speed sensor also sends information about the rpm of the engine to the ECU. The ECU in turn sends commands to the injector

Air fuel mixture ECU

Air in

Intake manifold

Engine

Injection volume control

Fuel in

Injector

Injection

Fig. 4 Engine RPM Sensor

Prelim Paper Solution

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to regulate the amount of gasoline supply for injection. When the injector sprays fuel in the intake manifold the gasoline mixes with the air and the mixture enters the cylinder.

L-MPFI System The L-MPFI system is a port fuel-injection system. In this type the fuel metering is

regulated by the engine speed and the amount of air that actually enters the engine. This is called air-mass metering or air-flow metering. The block diagram of an L-MPFI system is shown in Figure 5. As air enters into the intake manifold, the air flow sensor measures the amount of air and sends information to the ECU. Similarly, the speed sensor sends information about the speed of the engine to the ECU. The ECU processes the information received and sends appropriate commands to the injector, in order to regulate the amount of gasoline supply for injection. When injection takes place, the gasoline mixes with the air and the mixture enters the cylinder.

Q.4(a) (ii) State the norms of Bharat stage III & IV. [4](A) Emission standards are the requirements that, set specific limits to the amount of

pollutants that can be released into the environment. Many emission standards focus on regulating pollutants released by automobiles (motor cars) and other powered vehicles, but they can also regulate emission from industry, power plant, small equipment such as lawn movers and diesel generators. Frequently, policy alternatives to emission standards are technology standards.

Bharat stage emission standards are the emission standards constituted by Government of India to regulate the output of air pollutants from I.C. engine equipment’s, including motor vehicles.

The standards and timeline for implementation are set by central pollution control board under the ministry of environment and forests.

Details on emission norms proposed for petrol and diesel vehicles in India are as follows :

Table 1 : Emission Norms for Passenger Cars (Petrol)

Norms CO (g/km) HC + NOX (g/km)Bharat stage – II 2.2 0.5 Bharat stage – III 2.3 0.35 Bharat stage – IV 1.0 0.18

Table 2 : Emission Norms for Heavy Duty Vehicles

Norms CO (g/kW hr)

HC (g/kW hr)

NOX

(g/kW hr) PM

(g/kW hr) Bharat stage – II 4.0 1.1 7.0 0.15 Bharat stage – III 2.1 1.6 5.0 0.10 Bharat stage – IV 1.5 0.96 3.5 0.02

Intake manifold

Engine

Fig. 5

Air

Air flow meter

Throttle Valve

ECU

Injection line

Direction of air intake volume

Fuel

Injecto

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14

Q.4(a) (iii) Explain motoring test. [4](A) Motoring Test

In motoring test the engine is steadily operated at the rated speed by its own power and allowed to remain under the given speed and load conditions for sufficient time so that the temperature of the engine components, lubricating oil and cooling water reaches a steady state. A swinging field type electric dynamometer as used to absorb the power during this period which is most suitable for this test. The ignition is then cut-off and by suitable electric switching devices the dynamometer is converted to run is a motor so as to crank the engine at the same speed at which it was previously operating. The power supply from the above dynamometer is pressured which is a measure of the frictional power of the engine at that speed. The water supply is also cut-off during the motoring test so that the actual operating temperatures are maintained to the extent possible. This method though determines the fp at conditions very near to the actual operating temperatures at the test speed and load, it does not give the true losses occurring under firing conditions due to following reasons: The temperatures in the motored engine are different from those in a firing engine. The pressure on the bearings and piston rings is lower than in the firing engine. The clearance between piston and cylinder wall is more (due to cooling) and this reduces

the piston friction. The air is drawn at a temperature much lower than when the engine is firing because it

does not get heat from the cylinder (rather losses heat to the cylinder). Motoring method, however, gives reasonably good results and is very suitable for finding the losses imparted by various engine components. This insight on the losses caused by various components and other parameters is obtained by progressive stripping off of the engine. First the full engine is motored, then the test is conducted under progressive dismantling conditions keeping water and oil circulation intact. Then the cylinder head can he removed to evaluate by difference, the compression loss. In this manner, piston rings, pistons, etc. can be removed and evaluated for their effect on overall friction.

Q.4(a) (iv) Define the terms

(a) Indicated power (b) Mechanical efficiency (c) Brake power (d) BSFC

[4]

(A) (a) Indicated power The power developed inside the engine cylinder is called as Indicated power. (b) Mechanical efficiency The ratio of break power to the indicated power is called as Mechanical efficiency. (c) Brake power The power obtained at output of crank shaft is called as Breakdown Power.

(d) BSFC It is defined as the mass off fuel required to develop 1 Kw brake power for a period of one

hour and is inversely proportional to brake thermal efficiently. Q.4(b) Attempt any ONE of the following : [6]Q.4(b) (i) State different methods for improving thermal efficiency of gas turbine and

explain Regeneration method along with P-V & T-S diagram. [6]

(A) Different methods for improving thermal efficiency of gas turbine i) Reheating ii) Intercooling iii) Regeneration

The diagrammatic representation of simple cycle with regenerator or heat exchanger is shown in Fig. (a).

Prelim Paper Solution

15

Fig. (a) : Gas turbine plant with regenerator. Fig. (b) : TS diagram It is the method of preheating the air coming out from the compressor, before it

enters into the combustion chamber. This is carried out in the heat exchanger (known as regenerator). As a result of regeneration, compressed air is heated from T2 to T3 and exhaust gas is cooled from T5 to T6.

With the use of regenerator in the circuit, there is no change in the compressor work and turbine work.

But there is substantial reduction in the fuel consumption to bring the air to required temperature, so it results in increase in thermal efficiency.

Figure shows the thermal efficiency of a regenerator

cycle plotted against pressure ratio. Some points observed are as follows :

(i) Regeneration efficiency reduces as pressure ratio increases. This is due to the fact that as the pressure ratio increases, the delivery temperature from the compressor increases and ultimately exhaust gas temperature of turbine increases, then heat is lost from the air to the exhaust gas in regenerator.

(ii) Low pressure ratio and high temperature are favourable for regenerative cycle, since large heat recovery is possible.

Q.4(b) (ii) The following results were obtained during a Morse test on a four stroke

cycle petrol engine: (1) Brake power developed with all cylinder working = 16.2 kW (2) Brake power developed with cylinder No. 1 cutoff = 11.5 kW (3) Brake power developed with cylinder No. 2 cutoff = 11.6 kW (4) Brake power developed with cylinder No. 3 cutoff = 11.68 kW (5) Brake power developed with cylinder No. 4 cutoff = 11.57 kW Calculate mechanical efficiency of the engine.

[6]

(A) Let B1, B2, B3, and B4 be the brake power of cylinder No. 1, 2, 3, 4 respectively. Indicated power of first cylinder I1 = (B. P.)all cylinder working (B. P)2,3,4

= 16.2 11.5 = 4.70 kW Indicated power of second cylinder I2 = (B. P.)engine (B. P.)1,3,4 = 16.2 11.6 = 4.60 kW

I3 = (B. P.)engine (B. P.)1,2,4 = 16.2 11.68 = 4.52 kW

I4 = (B. P)engine (B. P.)1,2,3 = 16.2 11.57 = 4.63 kW

Total I. P = I1 + I2 + I3 + I4 = 4.7 + 4.6 + 4.52 + 4.63 = 18.45 kW

Fig.: Efficiency Vs pressure ratio

Vidyalankar : T.Y. Diploma PE

16

Mechanical efficiency = B. P.I. P.

mech = 16.218.45

= 0.878 or 87.8%

Q.5 Attempt any TWO of the following : [16]Q.5(a) A vapour compression machine is used to maintain a temperature of –23°C in a

refrigerated space. The ambient temperature is 37°C. The compressor takes indry and saturated vapour of R-12. A minimum 10°C temperature difference isnecessary at the evaporator and condenser. There is no sub cooling of the liquid.If the refrigerant flow rate is 1 kg/min, find: (i) capacity of refrigeration (ii) power required (iii) C.O.P. of cycle (iv) carnot C.O.P. Properties of R-12 are given below :

Saturation Temp C

Enthalpy KJ/kg Entropy KJ/kg Khf hfg Sf Sg

23 159.250 172.682 0.8106 1.5701 33 170.146 166.630 08840 1.566837 241.710 128.361 1.1239 1.548747 245.715 123.765 1.1520 1.5386

[8]

(A) Since a minimum temperature of 10C is required in evaporator condenser, therefore evaporator temperature would be.

T1 = T4 = 23 10 = 33C = 33 + 273 = 240k and condenser T2 = T3 = 37 + 10 = 47 = 47 + 273 = 320k

1) Capacity of refrigeration per minute = mg (h1 h + 3) = 1 = 1 (336.630 245.715) = 90.915

Capacity of refrigeration = 90.915210

= 0.43 TR

2) Power required work done during compression of refrigeration = mg (h2 h1)

Enthalpy of super head vapour h2 = h 12

+ cp (T2 12T )

To find T2 entropy at point 2

S2 = 22S + 2.3 cp log 2T

320

1.5668 = 1.5386 + 2.3 0.64 log 2T320

log 2T320

= 1.5668 1.53862.3 0.64

= 0.02821.472

= 0.01916

2T320

= 1.01936

T2 = 326 k

h2 = h 12

+ Cp(T2 12T ) = 369.48 + 0.64 (326 320) = 369.48 + 3.84

h2 = 372.96 KJ/Kg

Power required = mg (h2 h1) = mg (372.96 336.776) = 1(36.184) = 36.20 kJ/min

Prelim Paper Solution

17

= 36.1860

Power required = 0.60 kJ/g

3) Cop of cycle = 1 fs

2 1

h hh h

= 336.776 245.715

372.96 336.776

= 91.06136.18

= 2.5

Cop of cycle = 2.5

4) Carnot cop = 1

2 1

TT T

= 240320 240

= 24080

= 3

Carnot cop = 3 Q.5(b) A single stage single acting air compressor delivers 0.6 kg of air per minute at 6.1

bar. The temperature and pressure at the end of suction stroke are 28C and 1.1 bar. The bore and stroke of the compressor are 100 mm and 150 mm respectively.The clearance is 3% of the swept volume. Assuming index of compression andexpansion as 1.25, find : (i) volumetric efficiency of compressor (ii) power required if mechanical efficiency is 85% (iii) speed of compressor in rpm

[8]

(A) m = 0.6 Kg/min = 0.660

Kg/min = 0.01 Kg/sec.

P1 = 1.1 bar = 1.1 105 N/m2 T1 = 28 + 273 = 301 K

1) Assume R = 0.287 kJ/KgK for air Indicated power

= nn 1

mRT1

n 1n

2

2

P1

P

= 1.251.25 1

0.01 0.287 301 1.25 11.256.1 1

1.1

= 5 0.01 0.287 301 [(5.5)0.2 1] = 5 0.01 0.287 301[1.41 1] = 5 0.01 0.287 301[0.41]

IP = 1.77 KJg

Kw 02 1770 Jg

02 w

If the mechanical efficiency is 85/0

Power required = 1.770.85

Power required = 2.08 kW. It is not affected by clearance volume.

2) IP = n 1n

21 1

1

n NPPr 1n 1 60P

1770 = 1.251.25 1

1.1 105 0.00118 1.25 11.256.1 1

1.1

N60

= 5 1.1 105 0.0018[0.41] N/60 = 406 N/60

N = 1770 60406

N = 262 r.p.m.

r1 = 21d

4 =

4 (0.1)2 0.15 = 0.00118 m3

3) Clearance volume = 0.03 vs

Vidyalankar : T.Y. Diploma PE

18

Volumetric efficiency = 1 1

1.252

1

vc P1vs P

= 1 1

1.250.03VS 6.1 1VS 1.1

= 1 0.03 [(5.55)0.8 1] = 1 0.03 [3.93 1] = 1 0.088 Volumetric = 0.91 = 91% Q.5(c) Explain the construction and working of Ram jet with the help of neat labelled

schematic diagram. State its limitations (any two). [8]

(A)

Ramjet it consist of inlet difference, combustion chamber and tail pipe (exist nozzle). Ramjet has no compressor as the entire compression depends upon compression. Function of

supersonic and subsonic difference to convert the kinetic called the ram pressure. Working The air entering into ram jet with sup sonic speed is slowed down to sonic velocity in the air

pressure is further increase in the sup sonic different increasing also the temperature of air. The diffuser section is designed to get correct ram effect its into decrees the velocity and increase pressure of in cooling air. The duel injected into combustion chamber is burned with help of igniter the high tress engine temperature garb are passed through the nozzle converting into pressure energy into kind energy. The high velocity gas leaving the nozzle provide required toward thrust to ramjet.

Limitation : 1) Ramjet engine be launched from an air plane flight. 2) Fuel consumption is too large. The fuel consumption lower decrees flight need. Q.6 Attempt any FOUR of the following : [16]Q.6(a) Define : (i) Dew point temp. (ii) Relative humidity

(iii) WBT (iv) Degree of saturation. [4]

(A) (i) Dew 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 dew.

(ii) Relative Humidity () : Relative humidity is defined as ‘the ratio of the partial

pressure of water vapour in a given volume of mixture to the partial pressure of water vapour when same volume of mixture is saturated at the same temperature’.

(iii) 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 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'.

(iv) Degree of Saturation () : Degree of saturation is defined as 'the ratio of mass of water vapour associated with unit mass of dry air to mass of water vapour associated with saturated unit mass of dry air at same temperature'.

Prelim Paper Solution

19

Q.6(b) Write four uses of compressed air. [4](A) Industrial use of compressed air (i) For sand blasting. (ii) In blasting furnace. (iii) In pneumatic tools. (iv) In ground water tube well. (v) In manufacturing of acid and other chemical products.

Commercial Use of Compressed Air (i) Filling air in the tubes of automobiles. (ii) For spray painting. (iii) For cleaning and dusting of automobiles. (iv) For supercharging of I.C. Engines. Q.6(c) An engine working on Otto cycle has diameter of 150mm & stroke of 225mm.

Clearance volume is 1.25 10-3 m3. Find air standard efficiency. [4]

(A) d = 150 mm = 0.15 m L = 225 mm = 0.225 m VC = 1.25 × 103m3 J = 1.4

= 1 J 1

1r

r = T

C

VV

= S C

C

V VV

VS = 2d L4 = 20.15 0.225

4

´ = 3.976 × 103 m3

r = 3 3

33.976 10 1.25 10

1.25 10

´ ´´

r = 4.18

= 1 1.4 1

14.18

= 56.42% Q.6(d) Draw split air conditioner with a neat sketch & labelled it. [4](A)

Condenser coils

Expansion device Evaporator

coils

Cool air

Condensate oxidation pan

Low pressure

refrigerant line

Cooling coil / air handler

Compressor / Condenser

Compressor

Outside air

High pressure

refrigerant line

Vidyalankar : T.Y. Diploma PE

20

Q.6(e) Differentiate between open cycle & closed cycle gas turbine. (minimum 4 points) [4](A) Differentiate between open cycle & closed cycle gas turbine

Closed Cycle Gas Turbine Open Cycle Gas Turbine 1. Cycle of

operation It works on closed cycle. The working fluid is recirculated again and again. It is a clean cycle.

It works on open cycle. The fresh charge is supplied in every cycle and after combustion and expansion, it is exhausted to atmosphere.

2. Manner of heat input

The heat is transferred indirectly through a heat exchanger.

Direct heat supply. Heat is generated in the combustion chamber itself.

3. Method of heating

Compressed air is heated in air heater.

Compressed air is heated in combustion chamber.

4. Turbine exhaust

Air from turbine is passed into the precooler.

Gas from turbine is exhausted into the atmosphere.

5. Working medium

Working medium is circulated continuously.

Working medium is replaced continuously.

6. Blade life As the air do not come in direct contact with fuel, the turbine blades do not wear or corrode earlier. So blade life is more.

As the air gets mixed up with the combustible fuel, the turbine blades may get corroded. So blade life is less.

7. Type of fuel used

As the heat is supplied externally through a heat exchanger, solid, liquid or gaseous or a combination ofthese can be used for heat generation.

Since combustion is an integral part of system, it requires high quality liquid or gaseous fuel for burning in a combustion chamber.

8. Quality of heat input

The heat can be supplied from any source like waste heat from some processes, nuclear heat or solar heat.

It requires high grade heat energy for generation of power in a gas turbine.

9. Capital cost Costlier due to complex construction of plant.

Less costlier due to simple construction of plant.

10. Maintenance cost

High maintenance cost. Low maintenance cost.

11. Efficiency High thermal efficiency for given temperature limits.

Low thermal efficiency for same temperature limits.

12. Part load efficiency

Better. Comparatively less.

13. Control Better control on power production. Poor control on power production. 14. Suitability Since air from turbine is cooled by

the circulating water in precooler, it is suited for stationary installation or marine uses.

Since the gas from turbine is discharged into atmosphere, it is best suited for moving vehicles.

15. Working fluid (medium)

Other than air, the gases like helium or helium-C02 mixture can be used.

Air is used, but it leads to low thermal efficiency.