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OCTOBER – 2019

32041/HEAT POWER ENGINEERING – ANSWER KEY

Part – A

Q.No. Description of answer Marks

1 NTP and STP Conditions:

NTP Condition: The conditions of temperature and pressure at 0°C (273 K) temperature and 760 mm of Hg pressure are termed as normal temperature and pressure. STP Condition: The temperature and pressure of any gas, under standard atmospheric conditions, is taken as 15°C (288 K) and 760 mm of Hg respectively.

2

2 Zeroth law of thermodynamics:

This law states that “when two bodies are separately in thermal equilibrium with a third body, then they are in thermal equilibrium with each other.

2

3 Detonation:

A very sudden rise of pressure during combustion accompanied by metallic hammer like sound is called Detonation. Pre-ignition:

it is the ignition of the homogeneous mixture of charge before the application of spark as it comes in contact with hot surfaces.

1+1

4 Excess air:

The theoretical air supplied may not be sufficient for the complete combustion of the fuel. The amount of air supplied above the theoretical air required is known as excess air.

2

5 Uses of compressed air:

The compressed air is used for many purposes such as for operating pneumatic drills, riveters, road drills, paint spraying,

In starting and supercharging of internal combustion engines.

In gas turbine plants, jet engines and air motors.

It is also utilized in the operation of lifts, ramps pumps and a variety of other devices.

In industry, compressed air is used for producing blast of air in blast furnaces.

2

6 Jet propulsion:

The term “propulsion” means moving forward. The jet propulsion engine is a form of internal combustion engine in which hot gases are produced by combustion of fuel.

2

7 Enthalpy of wet steam: (hwet)

It is amount of heat required to convert 1 kg of water at 00C into wet steam at constant

pressure. ℎ𝑤𝑒𝑡 = ℎ𝑓 + (𝑥. ℎ𝑓𝑔) in kJ/kg.

Enthalpy of dry steam: (hg)

It is the amount of heat required to convert 1 kg of water at 00C into dry steam at constant pressure. ℎ𝑔 = ℎ𝑓 + ℎ𝑓𝑔 in kJ/kg

1+1

8 Internal treatment given to boiler feed water:

Sodium carbonate, phosphate, colloidal treatment and Blow down 2

PART - B

9 Intensive and extensive properties:

Intensive or Intrinsic properties: These are the properties which are independent of the mass of the system. Examples: Temperature, pressure,

1.5+1.5

velocity, density, specific heat etc... Extensive or Extrinsic properties: These are the properties which are dependent on the mass of the system. Examples: specific volume, Specific enthalpy

10 Assumptions made in the analysis of steady flow energy equation:

1. The mass flow rate through the control volume is constant. 2. Only potential, kinetic, flow and internal energies are taken into account

and other energies such as electrical, magnetic and chemical are not considered.

3. The rate at which work and heat cross the control surface is constant. 4. The state of fluid at any point remains same at all times

3

11 Difference between reversible and irreversible process:

S.No Reversible process Irreversible process

1 The system and surroundings are completely restored back to their initial state when the process reversed.

The system and surroundings are not completely restored reversing the process.

2 There should not be any loss of heat due to friction, radiation or conduction, etc.

There is a loss of heat due to friction, radiation or conduction.

3 A cycle will be reversible if all the processes constituting the cycle is reversible.

A cycle will be irreversible, if any of the processes, constituting the cycle, is irreversible.

4 The initial conditions are restored.

The initial conditions are not restored.

3

12 Comparison of OTTO cycle and Diesel Cycle:

S.No OTTO CYCLE DIESEL CYCLE

1 It consists of two adiabatic and two constant volume processes.

It consists of two adiabatic, one constant pressure process and one constant volume process.

2 Heat addition takes place at constant volume.

Heat addition takes place at constant pressure.

3 Compression ratio is equal to expansion ratio.

Compression ratio is not equal to expansion ratio.

4 Efficiency of this cycle depends on compression ratio (r) only.

Efficiency of this cycle depends on compression ratio (r) and cut-off ratio.

5 For the same compression ratio and same heat input, the efficiency of Otto cycle is more than that of a diesel cycle.

For the same compression ratio and same heat input, the efficiency of Diesel cycle is less than that of a Otto cycle.

6 Compression ratio is less. It varies from 6 to 8.

Compression ratio is more. It varies from 12 to 22.

3

13 Merits and demerits of gas turbine: Merits:

1. Its mechanical efficiency is higher. 2. The lubrication and ignition systems are simple. 3. The exhaust to a gas turbine is free from smoke and less polluting. 4. The operation pressure is very low.

1.5+1.5

5. They are suitable for air crafts. 6. No flywheel is required. 7. It has no reciprocating parts.

Demerits:

(i) They are not self-starting; (ii) Low efficiencies at part loads; (iii) Non-reversibility; (iv) Higher rotor speeds; and (v) Overall efficiency of the plant is low.

14 Advantages of high pressure boilers:

1. Scale formation is eliminated since water is circulated with high

velocity through the tubes.

2. High pressure and high temperature steam can be obtained. This

increases efficiency of the plant.

3. Efficiency of the plant is increased up to 42%

4. It has less number of or complete elimination of drums.

5. Less floor space is reduced.

6. Less foundation cost due to lighter weight of tubes.

3

15 Procedure for starting the boiler from cold conditions:

1. All the joints and fittings are checked

2. Boiler mountings are checked for their proper functioning

3. The stokers, controls, etc., are checked for their proper functioning.

4. The boiler is filled with water upto the specified level. The drain and blow

off cocks are opened and water is allowed to escape for cleaning purpose.

Then the valves are closed again.

5. The boiler is again refilled with the treated water slightly above the normal

operating level.

6. The pressure gauge is connected by opening the steam valve

7. The steam is raised to the predetermined pressure by boiling the treated

water

8. The steam valves are opened to allow steam into super heater. Chemicals

are introduced into the boiler with the feed water.

9. Upto blow down, the water is heated till the half of the working pressure is

reached. Then the pressure is raised to the working pressure.

10. Maintain drum water level at normal

3

16 Given data: To find: Initial condition of steam

P1=11 bar P2=1.2 bar t=1200C CPS=2.3 kJ/kgK

3

PART – C

17. (a).

Given data: To Find: (i) Cv & Cp (ii) R

m=2kg P1=8 bar=800 kN/m2 P2=4 bar=400 kN/m2 W=120000 Nm=120 kNm T1=3770C+273=650K T2=2570C+273=530K

10

(OR)

17. (b).

Given data: To find: (i) T2 (ii) W (iii) Q

M=30 γ=1.38 pV1.25=C; n=1.25 P1=1 bar=100 kN/m2 T1=150C+273=288K P2=16 bar=1600 kN/m2 m=1 kg

10

18. (a).

Given data

T3=20000C+273=2273 K T4=8000C+273=1073 K γ=1.4To find: (i) r (ii) η

10

(OR)

18. (b).

(i). Comparison of detonation and diesel knock

S.No. DETONATION DIESEL KNOCK

1 It is caused by the auto ignition of the end gas towards the end of combustion.

Auto ignition of the first charge at the start of the combustion.

2 To avoid detonation auto ignition of the end gas has to be prevented.

To avoid diesel knock earlier possible auto ignition should occur.

3 In SI engines the air fuel mixture that self ignites is homogenous

In CI engines the air fuel mixture present at the time of self ignition is not homogeneous.

4 In SI engines compression ratio is limited beyond which detonation would occur.

In CI engines higher the compression ratio lesser the ignition delay and less the diesel knock.

5 Larger cylinder size promotes detonation.

Larger cylinder size reduces diesel knock.

(ii) Variables affecting the delay period: 1. Air-fuel ratio

2. Fuel properties

3. Intake temperature

4. Intake pressure

5. Compression ratio

6. Relative velocity between the fuel injection and air turbulence

7. Rate of fuel injection

5+5

8. Injection timing

9. Presence of residual gases

10. Engine speed

11. Engine load

19. (a).

Given data: To Find: (i) P (ii) m (iii) T2

d= 150mm=0.15 m l=200mm =0.2m P1=100 kN/m2 T1=200C+273=293 K pV1.3=C, n=1.3 P2=600 kN/m2 N=110 rpm R=0.287 kJ/kgK

4+3+3

(OR)

19. (b).

Closed cycle gas turbine: A closed cycle gas turbine is a turbine in which the air is circulated continuously within

the turbine. The components of this turbine are compressor, heating chamber, gas

turbine which drives the generator and compressor, and a cooling chamber.

The main components of a simplest form of a closed cycle gas turbine are:

1. Compressor: It is used to compress the gas.

2. Heating chamber: The heating of the compressed gas is takes place in the heating

chamber.

3. Gas turbine: it is used to produce the useful work which is used by the generator to

generate electricity.

4. Generator: It generates the electricity with the help of the gas turbine.

5. Cooling chamber: Cooling of the gas after passing from the turbine takes place in the

cooling chamber.

Process:

The cycle consists of the following processes.

1-2 Isentropic compression ( in a compressor)

2-3 Constant pressure heat addition ) in a Heat exchanger)

3-4 Isentropic expansion (in a Turbine)

4-1 Constant pressure heat rejection ( in Heat exchanger).

Working

The closed cycle gas turbine

works on the principle of

Joule’s or Brayton’s cycle

In this turbine, the gas is

compressed isentropically and

then passed into the heating

chamber. The compressor

generally used is of rotary type.

The compressed air is heated

with the help of some external

source and then made to flow

over the turbine blades. The turbine used here is of reaction type.

The gas while flowing over the blades of the turbine gets expanded. From the

turbine the gas is passed to the cooling chamber. Here the gas is cooled at

constant pressure with the help of circulating water to its original temperature.

Now the gas is again made to flow through the compressor to repeat the

process.

Here the same gas is circulated again and again in the working of a closed cycle

gas turbine.

Advantages: 1. Combustion of fuel is external. This permits the use of any kind of fuel.

2. Internal cleaning of the system is not necessary

3. The turbine blades are not fouled by the products of combustion.

4. Thermal efficiency is more

10

20.(a). Given data: To find: (i) Final condition of steam

P1=5 bar; Initial Condition of steam – Dry (ii) Power P2=0.2 bar; s1=s2; Mass flow m= 2kg/s

5+5

(OR)

20.(b) Given data: To find: Dryness fraction x

mc =0.99 kg mc+w =3.85 kg mc+w+s=4.04 kg Cpc=0.386 kJ/kgK Cpw=4.2 kJ/kgK t1=70C t2=44.50C ts=1550C

10

21.(a) Given data: To find: (i) ma (ii) F (iii) me (iv) ηb (v) BP

tw=500C P=5 bar x=0.95 mf=600 kg/hr CV=30400 kJ/kg mw=4800 kg/hr

2+2+2 +2+2

(OR)

21.(b). Given data: To find: ηb and Heat balance sheet

mw=690 kg/hr; tw=280C P=8 bar; x=0.97 CV=27200 kJ/kg; mf= 91 kg/hr CVu=2700 kJ/kg; mu=7.5 kg/hr mg=17.4 kg/kg of coal tg=3250C ta=170C Cpg=1.005 kJ/kgK Loss due to incomplete combustion = 1%

10

Prepared by:

M.SARAVANAKUMAR Lecturer

Department of Mechanical Engineering

PSG POLYTECHNIC COLLEGE, COIMBATORE – 641 004