ENGINEERING DRAWING

अर्धबार्षिक परीक्षा प्रश्न पत्र  2013 

इंजीनियरिंग ड्राइंग 

प्रथम वर्ष  यांत्रिक इंजीनियरिंग 

___________________________________________________________________

[ समय : 2 घंटा ]                                                                                   पूर्णांक : 30 

               नोट : किन्ही तीन प्रश्नो के उत्तर दीजिये। प्रश्न संख्या 4 अनिवार्य है। 

                         प्रत्येक प्रश्न के अंक प्रश्न के सामने अंकित है। 

               1 ) 10 mm ऊंचाई के एकल प्रयास उर्ध्वाधर कैपिटल अक्षरों मैं निम्नलिखित वाक्य को        

                    स्वच्छता पूर्वक लिखिए।                                                                        (7. 5 )

                         "CIVIL ENGINEERING IS THE BACKBONE OF CIVILIZATION."

               2 ) एक रेखा AB , HP से 60^o का कोण बनाती है, तथा VP से 25 mm पीछे समानांतर है।  
                    यदि रेखा के सिरा 'A' H.P. से 25 mm निचे है , तथा रेखा की लम्बाई 50 mm है , तब रेखा 
                   के Front  View एबं Top View बनाइये।                                                    (7. 5 )

               3) (1 /50) निरूपक भिन्न (R. F.) वाली एक निकर्ण मापनी बनाइये जिस पर मीटर , 
                   डेसीमीटर तथा सेन्टीमीटर का मापन किया जा सके तथा 9 मीटर लम्बाई तक पढ़ने के 
                   लिए पर्याप्त हो। मापनी पर 6.47 मीटर की दूरी चिह्नित कीजिये।            (7. 5 )


               4) चित्रों मई एक समपरिमन दृश्य (Isometric View) दर्शाया गया है , दी गई दिशाओं से 
                   देखकर इनका अनुविक्षेप (Plan/ Top View) एबं  उत्सेध (Front View) तथा पार्श्व दृश्य 
                   ( Side View ) को III^rd Angle मैं बनाइये।                                                ( 15 )

ME-301: THERMODYNAMICS FOR THIRD SEMESTER; UPTU

SECTION A: Each question carry 2 marks

01) What is point function and path function?

02) Define Enthalpy.

03) What is SFEE?

04) What is internal energy of a system

05) What is vapour dome and dryness factor?

06) What is saturated liquid line and saturated vapour line?

07) What is triple point of water?

08) Define specific heats of ideal gases.

09) Write the reduced form of Vander Waals equation for real gases.

10) Define a thermodynamic system.

11) State with reasoning whether the following systems are closed, open or isolated
i) Refrigerator; ii) Pressure Cooker

12) Distinguish between isolated system and adiabatic system.

13) Explain the concept of flow work

14) What is control volume and control surface?

15) When does a real gas behave like an ideal gas?

16) What are extensive and intensive properties?

17) What is enthalpy of evaporation of steam?

18) Define degree of under-cooling and degree of superheat.

19) What is COP of a heat pump?

20) Define throttling process.

21) Define entropy.

22) Two moles of an ideal gas occupy a volume of 4.24 m³ at 400 K temperature. Find the pressure exerted by the gas.

23) Distinguish between refrigerator and heat pump.

24) What is Free Expansion?

25) Explain Zeroth law of thermodynamics.

26) Explain the concept of compressibility factor?

27) What is PMM-1.

28) What is a superheated steam?

29) Distinguish between universal gas constant and characteristic gas constant.

30) What is Exergy?

31) Distinguish between quasi-static process and reversible process.

32) What is a diathermal system boundary?

33) What is a steady flow open system?

34) What is the difference between latent heat and sensible heat?

35) What is a thermodynamic cycle?

36) Distinguish between restraint and unrestraint process.

37) What is a thermodynamic definition of work?

38) What is work of evaporation?

39) What is a pure substance?

40) What is the concept of continuum?

41) Define thermodynamic state, process and path.

42) Distinguish between thermal equilibrium and thermodynamic equilibrium.

43) What are the conditions for reversible process?

44) Distinguish between heat and work.

45) What are the differences between gas and vapour?

SECTION B: Attempt any three of the following questions. Each question contains two parts of 5 marks each. Total marks In this section is 3x10 = 30

01) a) State Zeroth law of thermodynamics and explain how it leads to the concept of temperature.

b) Explain different types of temperature scale and the relations among them.

02) a) Explain the corollaries of first law of thermodynamics.

b) 2 kg of air is confined in a rigid container of 0.42 m³ at 4 bar pressure. When heat energy of 164 kJ is added, its temperature becomes 127°C.
Find :

i) Work done by the system.
ii) Change in internal energy.
iii) Specific heat at constant volume.

03) a) Derive an expression for heat transfer and work done in a polytropic process.

b) 1.5 kg of oxygen contained in a cylinder at 4 bar pressure and 300 K expands three times its original volume in a constant pressure process. Determine
i) Initial volume, ii) Final temperature, iii) Work done by the gas, iv) Heat added and v) Change in internal energy.
; Assume C_p = 1.005 kJ/kg-K and R = 260 J/kg-K

04) a) Make steady flow energy analysis on a turbine.

b) Find the velocity and diameter at exit of a nozzle if 5 kg/s air at 9 bar and 200°C expands through the nozzle up to pressure at 1.1 bar. Approach velocity is 50 m/s.

05) a) Differentiate between absolute pressure and gauge pressure. What is a manometer?

b) An ideal gas of molecular weight 42.4 has a pressure of 10 bar and occupies a volume of 0.3 m³ at 27°C. Determine the characteristic gas constant for the ideal gas, its mass and number of moles.

06) a) Write the first law of thermodynamics for a flow process. Derive an expression for flow work.

b) Find the total work done and efficiency for a reversible Carnot cycle.

07) a) What is continuity equation in flow process?

b) 3 kg air at 2 bar pressure and 27℃ temperature has been compressed isothermally till the pressure reaches 6 bar. Next it has been heated at constant pressure and thereafter reaches the initial state by expanding adiabatically. Find the maximum Temperature reached in the cycle and total work done by the system.

08) a) Explain the Joule's experiment.

b) Prove that internal energy is a point function.

09) a) What is thermodynamic temperature scale?

b) i) A heat engine running between 300 K and 800 K generates 2000 kJ of energy. Find the total heat extracted from the source.
ii) Determine the power required to run a refrigerator that transfers 2000 kJ/min of heat from a cooled space at 0°C to the surroundings atmosphere at 27°C.

10) a) What is PMM - 2? State Kelvin-Planck statement of second law of thermodynics.

b) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

11) a) Explain the Vander Waal's gas equation.

b) 4 kg of steam at 16 bar occupies a volume of 0.28 m³. The steam expands at constant volume to a pressure of 8 bar. Determine the final dryness fraction, final internal energy and change in entropy.

12) a) Explain and derive Clausius Inequality.

b) 3 kg of air is heated reversibly at constant pressure of 2.5 bar from 23°C to 227°C. If the lowest available temperature is 20°C determine the increase in the available energy of air due to heating. Take C_p = 1.005 kJ/kg-K.

13) a) What is thermodynamic definitions of work? Distinguish between ∫pdV work and other types of work.

b) 3 kg of air at 1.5 bar pressure and 350 K is compressed isothermally to a pressure of 6 bar. Then heat of 350 kJ is added at constant volume. What will be the maximum temperature of air during the process? Find the total work done in the processes. Also find the change in internal energy of air.

14) a) Write the limitations of second law of thermodynamics. Prove that C_p - C_v = R

b) 10 kg of air at 300 K is stored in a totally insulated cylinder of volume 0.3 m³/kg. If 1 kg air has been taken out of the system, then what will be the value of new pressure?

15) a) Steam at 1.2 bar and a dryness fraction of 0.5 is heated at constant pressure until it becomes saturated vapour. Calculate the heat transferred per kg of steam.

b) Steam at 8 MPa and 500°C passes through a throttling process such that the pressure is suddenly dropped to 0.3 MPa. Find the expected temperature after throttling.

16) a) What are the causes of irreversibility?

b) Distinguish between a quasi-static process and reversible process.

17) a) 3 kg of air at 400 K and 4 bar pressure adiabatically mixed with 4 kg of air at 500 K and 3 bar pressure. Find the change in entropy of the universe.

b) Explain the principle of increase of entropy.

SECTION C : marks 50, 5 questions of 10 mark each. Each question contains 3 parts. Attempt any two parts out of three from each question.

01) a) Steam at 20 bar pressure and 300°C expands isentropically in a turbine to a pressure of 2 bar. Find the final condition of the steam. Also Calculate the work delivered by the turbine.

b) What is isentropic efficiency of a turbine? Calculate internal energy of steam at 6 bar pressure and 300°C.

c) Explain the steam formation process at constant pressure.

02) a) What is adiabatic mixing of two ideal gases? Derive the expressions for final temperature and pressure.

b) 5 kg of steam at 8 bar pressure and 200°C mixed with 3 kg of steam at 5 bar and dryness fraction x = 0.8 adiabatically. Find the final condition of the steam.

c) 5 kg of air at 4 bar pressure is heated at constant pressure from 300 K to 500 K. Find the change in entropy of the system.

03) a) Prove that in an adiabatic process pV^γ = Constant.

b) Polytropic compression of air from state 1 to state 2 where p₁ = 100 kPa and T₁ = 300 K, p₂ = 300 kPa and n = 1.2 where as mass of air is 3 kg. If R = 0.287 kJ/kg-K. Then find

i) heat exchange during the process
ii) change in internal energy
iii) total work done by the air
iv) change in entropy


c) A non flow reversible process occurs for which pressure and volume are correlated by the relation p = (V² + 6V), where V is the volume in m³ and pressure p is in bar. Determine work done if volume changes from 3 to 6 m³.

04) a) A gas expands according to the equation pv = 100, where " p " is the pressure in kPa and " v " is the specific volume. Initial and final pressures are 1000 kPa and 500 kPa respectively. The gas is then heated at constant volume back to it'd original pressure of 1000 kPa. Determine the net work done. Also sketch the processes in p-v coordinates.

b) What is the definition of thermodynamic work?

c) What is the efficiency of a thermodynamic cycle?

05) a) What is paddle work? Distinguish between ∫pdV work and ∫-vdp work.

b) If pV = mRT, determine whether the expression (V/T).dp + (p/T).dV is a property of a system.

c) 2 kg of air at 1 bar pressure and 300 K is compressed adiabatically to a pressure of 6 bar. Then heat of 200 kJ is added at constant pressure. What will be the maximum temperature of air during the process? Find the total work done in the processes. Also find the change in internal energy of air.

06) a) Find the expression for heat transfer in terms of work done in a polytropic process.

b) What is the specific heat C_n for a polytropic process?

c) 2 kg of air at pressure 2 bar and 300 K is compressed reversibly to 4 bar and 650 K temperature in a polytropic process. Determine the polytropic index (n) of the process.

07) a) Find an expression for mechanical work in steady flow process.

b) What is the meaning of - vdp work?

c) Air flows through a gas turbine system at a rate of 5 kg/s. It enters with a velocity of 150 m/s and an enthalpy of 1000 kJ/kg. At exit the velocity is 120 m/s and enthalpy is 600 kJ/kg. If the air passing through the turbine looses 30 kJ/kg of heat to the surroundings, determine the power developed by the system.

08) a) Write the assumptions considered in Kinetic theory of gases? Prove that C_p - C_v = R

b) Explain the law of corresponding states.

c) 10 kg of air at 300 K is stored in a cylinder of volume 0.3 m³/kg. Find the pressure exerted by air using Vander Waals gas equation. Critical properties of air are: P_c = 37.7 bar, T_c = 132.5 K, v_c = 0.093 m³/kg_mole, R = 287 J/kg-K

09) a) What are the limitations of Vander Waals gas equation? Explain reduced properties of a real gas?

b) What is a undercooled liquid and degree of undercooling? Also define enthalpy of water.

c) What are the properties of steam at critical state? Explain sublimation process and triple point line.

10) a) What are the differences between dry saturated steam and superheated steam at a same pressure? Also, explain vapourdome, saturated liquid line, saturated vapour line and critical point.

b) What are the differences between work of evaporation and enthalpy of evaporation?

c) An inventor claims to have developed a refrigeration unit which maintains −10℃ in the refrigerator which is kept at a room where the surrounding temperature is 25℃ and which has COP of 8.5. Find the claim of the inventor is possible or not.

11) a) Prove that the absolute zero temperature is impossible to achieve according to second law of thermodynamics.

b) Two reversible heat engines A and B are arranged in series. A rejects heat directly to B. Engine A receives 200 kJ at a temperature of 421℃ from the hot source while engine B is in communication with a cold sink at a temperature of 5℃. If work output of A is twice that of B, find :
(i) Intermediate temperature between A and B.
(ii) Efficiency of each engine.
(iii) Heat rejected to the sink.


c) Prove that the reversible heat engines are the most efficient.

12) a) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vessel until it becomes saturated vapour. Calculate the heat transferred per kg of steam.

b) Steam at 9 MPa and 600°C passes through a throttling process such that the pressure is suddenly dropped to 0.4 MPa. Find the expected temperature after throttling.

c) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

13) a) Explain the change of entropy in a perfectly isolated system during a process in the system.

b) Explain the conditions those must be satisfied by a reversible process.

c) Two kg of water at 90℃ is mixed with three kg of water at 10℃ in a perfectly isolated system. Calculate the change in entropy of the system.

14) a) Explain the second law of thermodynamics and prove that no engine can have a 100% efficiency.

b) Explain the theoretical Carnot cycle and derive its efficiency with diagrams.

c) A reversible engine working in a cycle takes 4800 kJ of heat per minute from a source at 800 K and develops 35 kW power. The engine rejects heat to two reservoirs at 300 K and 360 K. Determine the heat rejected to each sink.

15) a) What are the causes of external irreversibility?

b) Write the first and second Tds equations and derive the expression for the change of entropy during a polytropic process.

c) Prove that reversible engines are most efficient.

16) a) Explain the second law of thermodynamics.

b) Derive the Clausius inequality.

c) Steam at 160 bar and 550℃ is supplied to a steam turbine. The expansion of steam is adiabatic with increase in entropy of 0.1 kJ/kg-K. If the exhaust pressure is 0.2 bar, calculate specific work of expansion.

17) a) 5 kg of water at 400 K is isobarically and adiabatically mixed with 3 kg of water at 500 K. Find the change in entropy of the universe.

b) Explain i) Second law efficiency, ii) Effectiveness of a system and iii) Availability of a closed system.

c) Explain the principle of increase of entropy.

18) a) Explain Helmholtz and Gibbs function.

b) Explain the concept of PMM-I and PMM-II.

c) Find an expression of exit velocity C₂ in terms of pressure ratio when air passes through a nozzle from a pressure of p₁ and temperature T₁ to a pressure p₂.

19) a) Distinguish between enthalpy and internal energy.

b) What is absolute or thermodynamic temperature? Explain briefly.

c) Two Carnot engines work in series between the source at temperature 500 K and sink at temperature 300 K. If both develop equal power, determine the intermediate temperature.

20) a) Show that two adiabatic curves on p-V diagram never intersects each other.

b) Define and classify thermodynamic systems.

c) In an isentropic flow through nozzle, air flows at the rate of 600 kg/hr. At inlet to the nozzle pressure is 2 MPa and temperature is 27℃. The exit pressure is 0.5 MPa. Initial air velocity is 300 m/s, determine
i) exit velocity of air
ii) inlet and exit area of the nozzle

THE END

AIR-FUEL MIXTURE AND STOICHIOMETRIC RATIO

AIR-FUEL MIXTURE AND STOICHIOMETRIC RATIO

CHEMICAL COMBUSTION OF FUEL

Subhankar Karmakar
Assistant Professor; SGIT
Jindal Nagar; Ghaziabad

---

Chemical Combustion is basically a rapid oxidation process of hydro-carbon fuel inside thekjm combustion chamber in the presence of air. The oxidation of fuel is basically a Exothermic or heat liberating chemical process.

In SI engines generally we use volatile hydrocarbon as fuel. The intermixing of fuel with air takes place outside the engine and the device that prepares air-fuel mixture of required mixture strength is called CARBURETION and the device is known as CARBURETTOR.

Estimation of air quantity needed for complete combustion of a given fuel

We know that any chemical reaction can be represented by the corresponding chemical equation like
CH₄ + 2O₂ = CO₂ + 2H₂O
here molecular weight of CH₄
μ_CH4 = 12 + 4x1 = 16
μ_2O2 = 2x16x2 = 64
μ_CO2 = 12 + 2x16 = 44
μ_2H2O = 2x(2+16) = 36

For complete combustion of CH₄
16 kg CH₄ needs 64 kg O₂ therefore,
1 kg of CH₄ needs (64/16) = 4 kg of O₂
For 23 kg of O₂ air needed is 100 kg
hence, for 4 kg of O₂ air needed is (100/23)x4 = 17.39 kg of air.
Air-fuel ratio will be 17.39 : 1

The ratio of air fuel mixture, needed for the complete combustion of the fuel or the chemically correct ratio of air fuel mixture required for complete combustion of the fuel is called " Stoichiometric Air fuel mixture. "

If the amount of air in the air-fuel mixture is less than the chemically correct amount of air, then the mixture is called rich mixture, where as if the quantity of air is more than the chemically corrected amount of air it is called lean mixture.

The strength of air-fuel mixture has a profound influences on the process of combustion. The required mixture strength for different operation conditions are different.

Different Operating Conditions	Required air-fuel Mixture Strength
For Max. Efficiency	17 : 1, 16.4% weak
For Max. Power	12 : 1 17.8% rich
For Starting, Idling, & Low load running	11 : 1 ~ 16 : 1 very rich mixture
For accelerated motion	13 : 1 rich mixture
For Part Load running Cruising Range	17 : 1 Lean Mixture Strength

RECIPROCATING COMPRESSORS

Q.2) Classify the compressors
(i) On the basis of operations employed
(ii) On the basis of pressure achieved
(iii) On the basis of pressure ratio
(iv) On the basis of capacity of compressors.

A.2) Depending upon different parameters, compressors can be classified on the basis of operations employed, the delivery pressure achieved, pressure ratio and capacity of compressors as follows.

On the basis of operations employed, compressors are classified into two groups:
i) :Reciprocating compressors : It uses piston cylinder arrangement and due to positive displacement of air in the cylinder, the air is compressed and delivered to a vessel called Receiver. These are capable to produce high delivery pressure with low volume flow rate.
ii) Rotary Compressors : These compressors operate at high speeds, therefore, can handle large volume flow rates compared to reciprocating compressors.

In rotary compressors, the dynamic head is imparted to the gas with the help of very high speed impeller rotating at a confined space so that the air is compressed due to centrifugal action.

On the basis of delivery pressure, compressors are classified into three categories
i) Low Pressure Compressors : Delivery pressure upto 1.1 bar
ii) Medium Pressure Compressors : Delivery pressure upto 7 bar
iii) High Pressure Compressors : Delivery pressure between 7 to 10 bar.

On the basis of pressure ratio, we can classify the devices as follows,
Fans : Pressure ratio upto 1.1
Blower : Pressure ratio upto 1.1 to 4.0
Compressors : Pressure ratio above 4.0

On the basis of capacity, compressors can be classified as follows,
Low capacity compressors : Volume flow rate upto 10 m³/min, or less
Medium capacity compressors : Volume flow rate 10 m³/min to 300 m³/min
High capacity compressors : Volume flow rate above 300 m³/min

Q.3) Find an expression for required work done to drive a compressor, when compression is,
adiabatic in nature
isothermal compression
polytropic compression

A.3) During the analysis of the operations of a reciprocating air compressor, we consider some assumptions to simplify the analysis. They are as follows
i) There is no Clearence Volume
ii) Working substance air is an ideal gas
iii) There is no frictional loss.
iv) There is no wire drawing in the valve or pipe lines.

WORK REQUIRED TO DRIVE A COMPRESSOR

Suppose, we are running a single stage air compressor, which draws air at pressure P₁ and temperature T₁ during the suction or induction process. The air thus drawn inside the cylinder then compressed to achieve a delivery pressure, P₂ by adiabatic process. In the adjacent figure, process a - b is the suction process. Process b - c is the adiabatic compression of the air from pressure P₁ to pressure P₂. Process c - d is the delivery stroke, delivering the compressed air at a pressure P₂.

W_ad = P₂V₂ + {( P₂V₂ - P₁V₁)/(γ - 1)} - P₁V₁

=> (P₂V₂ - P₁V₁){1 + 1/(γ - 1)}

=> {γ /(γ - 1)}P₁V₁ {(P₂V₂)/(P₁V₁) - 1}

=> {γ /(γ - 1)}mRT₁{(P₂/P₁)^{{(γ - 1)/γ}} - 1}

POLYTROPIC WORKDONE IN RECIPROCATING COMPRESSORS

W_poly = P₂V₂ + {( P₂V₂ - P₁V₁)/(n - 1)} - P₁V₁

=> (P₂V₂ - P₁V₁){1 + 1/(n - 1)}

=> {n/(n - 1)}P₁V₁ {(P₂V₂)/(P₁V₁) - 1}

=> {n/(n - 1)}mRT₁{(P₂/P₁)^{{(n - 1)/n}} - 1}

ISOTHERMAL WORKDONE REQUIRED TO DRIVE RECIPROCATING COMPRESSOR

W_iso = P₂V₂ + {P₁V₁ln(V₁/V₂)} - P₁V₁

but as the process is isothermal, P₁V₁ = P₂V₂

W_iso = P₁V₁ln(V₁/V₂)

W_iso = mRT₁ln r_p

V₁/V₂ = P₂/P₁ = r_p (pressure ratio)

THERMODYNAMICS: 2nd MINOR TEST AND ITS SOLUTION.

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

SOLUTIONS

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

Ans: The boiling point of water is a function of the pressure, as pressure increases, boiling point is also elevated. For a certain pressure, water has a fixed boiling temperature known as saturation temperature and denoted by t_s. If the temperature of water at a given pressure is lower than the corresponding saturation temperature i.e. t < t_s, then the water is called sub-cooled or under-cooled water.

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

COMPRESSORS AND COMPRESSED AIR

COMPRESSORS:

Q.1) What is a Compressor? What is the difference between a Compressor and a Pump? What are the practical uses of Compressed Air?

A.1) A compressor is a device which is extensively used to raise the pressure of a compressible fluid like pure air. In a compressor, the pressure is increased at the expense of work done on the fluid, which is generally provided by an electric motor, IC engines or Gas Turbines. In a compressor, fluids are compressed by reducing the specific volumes of the working fluids. Due to compression, the temperature of the fluid is also increased.

If air is used as the working fluid in a compressor and air is compressed into a high pressure by the application of work on the fluid, then it is known as Air Compressor.

PRACTICAL USES OF COMPRESSED AIR

In industry, compressed air is so widely used that it is often regarded as the fourth utility, after electricity, natural gas and water. Compressed air, commonly called Industry's Fourth Utility, is air that is condensed and contained at a pressure that is greater than the atmosphere. The process takes a given mass of air, which occupies a given volume of space, and reduces it into a smaller space. In that space, greater air mass produces greater pressure. The pressure comes from this air trying to return to its original volume. It is used in many different manufacturing operations.

Compressed air is extensively used in industrial applications like pneumatic machines, as well as in the refrigeration and air-conditioning systems or supercharging CI engines to boost the output of the engine.
01) Compressed air is extensively used to operate pneumatic tools like drills, hammers, rivetting machines etc.
02) It is used to drive Compressed Air Engine.
03) Compressed air is used to spray painting.
04) Compressor is a vital component of Air-conditioning and Refrigeration industry.
05) Very often, compressed air can be used as a means of energy storage.
06) It is used in Gas Turbine power plants.
07) It is used to Super-charging an IC engines.
08) It is used to convey or pump to flow the materials like sand or concrete slurries along a pipe line.
09) It can be used as a means to pump water through the pipe lines.
10) It is used to drive minning machineries in a fire risky zone.
11) It is also used in blast furnaces.

Q.2) Classify the compressors
(i) On the basis of operations employed
(ii) On the basis of pressure achieved
(iii) On the basis of pressure ratio
(iv) On the basis of capacity of compressors.

A.2) Depending upon different parameters, compressors can be classified on the basis of operations employed, the delivery pressure achieved, pressure ratio and capacity of compressors as follows.

On the basis of operations employed, compressors are classified into two groups
i) Reciprocating compressors : It uses piston cylinder arrangement and due to positive displacement of air in the cylinder, the air is compressed and delivered to a vessel called Receiver. These are capable to produce high delivery pressure with low volume flow rate.
ii) Rotary Compressors : These compressors operate at high speeds, therefore, can handle large volume flow rates compared to reciprocating compressors.

In rotary compressors, the dynamic head is imparted to the gas with the help of very high speed impeller rotating at a confined space so that the air is compressed due to centrifugal action.

On the basis of delivery pressure, compressors are classified into three categories
i) Low Pressure Compressors : Delivery pressure upto 1.1 bar
ii) Medium Pressure Compressors : Delivery pressure upto 7 bar
iii) High Pressure Compressors : Delivery pressure between 7 to 10 bar.

On the basis of pressure ratio, we can classify the devices as follows,
Fans : Pressure ratio upto 1.1
Blower : Pressure ratio upto 1.1 to 4.0
Compressors : Pressure ratio above 4.0

On the basis of capacity, compressors can be classified as follows,
Low capacity compressors : Volume flow rate upto 10 m³/min, or less
Medium capacity compressors : Volume flow rate 10 m³/min to 300 m³/min
High capacity compressors : Volume flow rate above 300 m³/min

QUESTION PAPER: EME-505, IC ENGINES & COMPRESSORS

THERMODYNAMICS: BASIC CONCEPTS AND DEFINITIONS

UNIT- I:

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Fundamental Concepts and Definitions; Terminology, Definition and Scope; Microscopic and Macroscopic Approaches; Engineering Thermodynamics; it's definition and practical applications; Systems and Control volumes; Characteristics of System boundary and Control Surfaces; Surroundings and fixed, moving and imaginary boundaries; Thermodynamic States, state point; identification of a state through properties; definitions and units; extensive, intensive and specific properties, Thermodynamic planes and coordinate systems using properties; Change of state, path and processes; Quasi-static processes; Reversible processes, Restrained and unrestrained processes; Thermodynamic Equilibrium; diathermic wall, Zeroth Law of thermodynamics, Temperature as an important properties.

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Q.1) What is the meaning of Thermodynamics?

Ans:) The branch of science that deals with energy and its movements in the space is generally known as Thermodynamics. The study of this science is based upon experimental values and common experiences and the laws are empirical in thermodynamics.

¤ Introduction:

The most of general sense of thermodynamics is the study of energy and its relationship to the properties of matter. All activities in nature involve some interaction between energy and matter. Thermodynamics is a science that governs the following:

• (i) Energy and its transformation
• (ii) Feasibility of a process involving transformation of energy
• (iii) Feasibility of a process involving transfer of energy
• (iv) Equilibrium processes

More specifically, thermodynamics deals with energy conversion, energy exchange and the direction of exchange.

¤ Areas of Application of Thermodynamics:

All natural processes are governed by the principles of thermodynamics. However, the following engineering devices are typically designed based on the principles of thermodynamics.

Automotive engines, Turbines, Compressors, Pumps, Fossil and Nuclear Power Plants, Propulsion systems for the Aircrafts, Separation and Liquefaction Plant, Refrigeration, Air-conditioning and Heating Devices.

The principles of thermodynamics are summarized in the form of a set of axioms. These axioms are known as four thermodynamic laws:

• Zeroth law of thermodynamics,
• First law of thermodynamics,
• Second law of thermodynamics, and
• Third law of thermodynamics.

The Zeroth Law deals with thermal equilibrium and provides a means for measuring temperatures.

The First Law deals with the conservation of energy and introduces the concept of internal energy.

The Second Law of thermodynamics provides with the guidelines on the conversion of internal energy of matter into work. It also introduces the concept of entropy.

The Third Law of thermodynamics defines the absolute zero of entropy. The entropy of a pure crystalline substance at absolute zero temperature is zero.


¤ Different Approaches in the Study of Thermodynamics:

There are two ways through which the subject of thermodynamics can be studied

• Macroscopic Approach
• Microscopic Approach

¤ Macroscopic Approach:

Consider a certain amount of gas in a cylindrical container. The volume (V) can be measured by measuring the diameter and the height of the cylinder. The pressure (P) of the gas can be measured by a pressure gauge. The temperature (T) of the gas can be measured using a thermometer. The state of the gas can be specified by the measured P, V and T . The values of these variables are space averaged characteristics of the properties of the gas under consideration. In classical thermodynamics, we often use this macroscopic approach. The macroscopic approach has the following features.

• The structure of the matter is not considered.
• A few variables are used to describe the state of the matter under consideration. The values of these variables are measurable following the available techniques of experimental physics.

¤ Microscopic Approach:

On the other hand, the gas can be considered as assemblage of a large number of particles each of which moves randomly with independent velocity. The state of each particle can be specified in terms of position coordinates ( x_i , y_i , z_i ) and the momentum components ( p_xi , p_yi , p_zi ). If we consider a gas occupying a volume of 1 cm³ at ambient temperature and pressure, the number of particles present in it is of the order of 10²⁰. The same number of position coordinates and momentum components are needed to specify the state of the gas. The microscopic approach can be summarized as:

• A knowledge of the molecular structure of matter under consideration is essential.
• A large number of variables are needed for a complete specification of the state of the matter.

¤ Zeroth Law of Thermodynamics: 

This is one of the most fundamental laws of thermodynamics. It is the basis of temperature and heat transfer between two systems. Suppose we take three thermodynamic system named System A, System B and System C. Now let that system A is in thermal equilibrium with system B. By thermal equilibrium we mean that there is no heat transfer between system A and system B when they are brought in contact with each other. Now, suppose system A is in thermal equilibrium with system C too and there is no contact between system B and system C. It implies that although system B and C are isolated from each other, they will remain at thermal equilibrium to each other. It means that there will be no heat transfer between system B and C, when they are brought in contact with each other. This is called the Zeroth Law of thermodynamics.


¤ Basis of Temperature: 

When two bodies are kept at contact with each other and if there is no heat transfer between them we say that their body temperatures are same. It means that temperature is the property of a system which decides whether there will be any heat transfer between two different bodies. Heat transfer always occur from a higher temperature body to a lower temperature body. Further whenever there is any heat inflow to a body, it raises its temperature and conversely, if heat outflow occurs from a system it lowers its temperature.

Suppose we take two bodies one of which is at higher temperature than the other. Now when we bring the bodies at contact, heat will be transformed from a higher temperature body to that of lower temperature. Then what will be its effect, we may ask as a result of this heat transfer? Is this heat transfer a perpetual process? Our common life experiences tell us that it will not be the case. Although, at first heat transfer will take place, but its amount will be gradually decreased and after some time, a situation will come when there will be no heat transfer between the bodies or the bodies will come to a state of thermal equilibrium with each other. So, what is the reason for that? Can we justify the situation?

Yes, we can justify it as the hotter body releases heat to the colder body, the temperature of the hotter body decreases where as the temperature of the colder body increases and after sufficient time both the bodies will have equal temperature and a state of thermal equilibrium will be achieved.


¤ Temperature Measurement: 

We know the temperature of a body can be measured with a thermometer. How can we actually calculate the temperature of a body with the help of thermodynamics?


¤ Thermometer:

A thermometer is a temperature measuring instrument. It is made of a thin capillary glass tube, one end is closed and the other end is fitted with metallic bulb full of mercury. The mercury is in thermal equilibrium with the metallic bulb. Therefore, the temperature of the mercury is equal to the temperature of the metallic bulb. 
Mercury has a good coefficient of volume expansion and it means that as the temperature of the mercury increases, its volume increases too and as a result mercury column inside the capillary rises up. 

The capillary tube has been graduated with the help of calibrating with standard temperature sources. Therefore, the temperature of the mercury can be measured from the height of mercury column as the tube is finely graduated. 

Whenever we want to measure the temperature of a body, we kept the body in contact with the metallic bulb of the thermometer. When thermal equilibrium is established between the body and the metallic bulb of the thermometer, the temperature of both the body will be equal again the metallic bulb is in thermal equilibrium with mercury then the temperature of the mercury will be equal to the temperature of the metallic bulb and the temperature of the object.


As we can measure the temperature of the mercury from the column height, hence we can also determine the temperature of the object as they are equal to each other.

DISCUSSION:
Microscopic basis of temperature and pressure:
Here we shall try to discuss the basis of temperature and pressure only qualitatively, without any mathematical expression. 






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Thermodynamic Systems: 


If we want to analyze movement of energy over space, then we must define the space that would be used for the observation, we would call it as a System, separated from the adjoining space that is known as "Surroundings", by a boundary that may be real or may be virtual depending upon the nature of the observation. The boundary is called as System Boundary. So, we shall now define a system properly.


A thermodynamics system refers to a three dimensional space occupied by a certain amount of matter known as ''Working Substance'', and it is the space under consideration. It must be bounded by an arbitrary surface which may be real or imaginary, may be at rest or in motion as well as it may change its size and shape. All thermodynamic systems contain three basic elements:


• System boundary: The imaginary surface that bounds the system.
• System volume: The volume within the imaginary surface.
• The surroundings: The surroundings are everything external to the system.


So we get a space of certain volume where Energy Transfer (movement of energy) is going on, what may or may not be real, and distinct, it may be virtual (in case of flow system ), again if real boundary exists, then it may be fixed (rigid boundary like constant volume system) or may be flexible (like cylinder-piston assembly). For a certain experiment the system and surroundings together is called Universe.

The interface between the system and surroundings is called as "System boundary", which may be real and distinct in some cases where as some of them are virtual, but it may be real, solid and distinct. If the air in this room is the system, the floor, ceiling and walls constitutes real boundaries. The plane at the open doorway constitutes an imaginary boundary.



Classification of Thermodynamic Systems:

Systems can be classified as being (i) closed, (ii) open, or (iii) isolated.

(i) Closed System:

A thermodynamic system may exchange mass and energy with its surroundings. There are systems which allow only energy transfer with surroundings in the form of either heat transfer or work transfer or both heat and work transfer between a system and its surroundings. In these types of system, any sorts of mass transfer between the system and its surroundings are prohibited. These types of systems are classified as closed system. Examples of closed thermodynamic systems include a fluid being compressed by a piston inside a cylinder, a bomb calorimeter. In a closed system although energy content may vary over a period of time, but the system will always contain the same amount of matter.

(ii) Open System or Control Volume: 

An open system is a region in space defined by a boundary across which matter may flow in addition to work and heat exchange between the system and the surroundings. So, in an open system, the boundaries must have one or more opening through which mass transfer may take place in addition to work and heat transfer. Most of the engineering devices are examples of open system. Some examples are (a) a gas expanding from a container through a nozzle, (b) steam flowing through a turbine, and (c) water entering a boiler and leaving as steam. The boundary of an open system may be real or imaginary and it is called as control surface. The space inside an open system is called as control volume.

(iii) Isolated System:  

In an isolated system, there is no interaction between a system and its surroundings. Hence, the quantities of mass and energy in these types of system doesn’t change with time or we can say mass and energy remain constant in an isolated system. If there is no change in energy of a system, it indicates that there is neither any kind of heat transfer nor any kind of work transfer.  Our universe as a whole can be regarded as an isolated system.

Property, Equilibrium and State: 

A property is any measurable characteristic of a system. The common properties include: 

• pressure (P)

• temperature (T)

• volume (V)

• velocity (v)

• mass (m)

• enthalpy (H)

• entropy (S)

Properties can be intensive or extensive. Intensive properties are those whose values are independent of the mass possessed by the system, such as pressure, temperature, and velocity. Extensive properties are those whose values are dependent of the mass possessed by the system, such as volume, enthalpy, and entropy. 

Extensive properties are denoted by uppercase letters, such as volume (V), enthalpy (H) and entropy (S). Per unit mass of extensive properties are called specific properties and denoted by lowercase letters. For example, specific volume v = V/m, specific enthalpy h = H/m and specific entropy s = S/m 

*Note that work and heat are not properties. They are dependent of the process from one state to another state.

When the properties of a system are assumed constant from point to point and there is no change over time, the system is in a thermodynamic equilibrium.

The state of a system is its condition as described by giving values to its properties at a particular instant. For example, gas is in a tank. At state 1, its mass is 2 kg, temperature is 160°C, and volume is 0.1 m3. At state 2, its mass is 1 kg, temperature is 80°C, and volume is 0.2  m³..

A system is said to be at steady state if none of its properties changes with time.

State:

It is the condition of a system as defined by the values of all its properties. It gives a complete description of the system. Any operation in which one or more properties of a system change is called a change of state.

Phase:

It is a quantity of mass that is homogeneous throughout in chemical composition and physical structure. Examples of phase are solid, liquid, vapour, gas. Phase consisting of more than one phase is known as heterogenous system, where as if it consists of only one phase, it is called as homogenous system.

Process, Path and Cycle: 

The changes that a system undergoes from one equilibrium state to another are called a process. The series of states through which a system passes during a process is called path.

In thermodynamics the concept of quasi-equilibrium processes is used. It is a sufficiently slow process that allows the system to adjust itself internally so that its properties in one part of the system do not change any faster than those at other parts.

When a system in a given initial state experiences a series of quasi-equilibrium processes and returns to the initial state, the system undergoes a cycle. For example, the piston of car engine undergoes Intake stroke, Compression stroke, Combustion stroke, Exhaust stroke and goes back to Intake again. It is a cycle.

Quasi-static Processes:

Although the processes can be restrained or unrestrained, in practical purpose we need restrained processes.

A quasi-static process is one in which,

• The deviation from thermodynamic equilibrium is infinitesimal.

• All states of the system passes through are equilibrium states.

In a cylinder-piston assembly, several small weights are placed on the piston as shown in the figure. If we remove a weight, the pressure on the enclosed gas will be reduced by an infinitesimal amount. If we remove these weights one by one very slowly, then the pressure on the gas will be reduced by very small amount very slowly. Every time we remove a weight, the equilibrium state will be changed to a new equilibrium state at a very slow rate, such that the system will be appeared at a static condition as the change is infinitesimally small and the rate of change is also very small. The path of the change will be a series of quasi-equilibrium states. These types of processes are known as quasi-static processes.  

Equilibrium States:

A system is said to be in an equilibrium state if its properties will not be changed without some perceivable effect in the surroundings.

• Equilibrium generally requires all properties to be uniform throughout the system.

• There are mechanical, thermal, phase, and chemical equilibrium.

Nature has a preferred way of directing changes. As examples, we can say,

• Water flows from a higher to a lower level

• Electricity flows from a higher potential to a lower one

• Heat flows from a body at higher temperature to the one at a lower temperature

• Momentum transfer occurs from a point of higher pressure to a lower one.

• Mass transfer occurs from higher concentration to a lower one


Equilibrium state will be achieved when there will not be any change of the values of the properties of a system. Neither the system will exchange 
Heat Energy nor any Work exchange nor any kind of mass exchange with its surroundings. There are mainly three kind of Equilibrium and they are as follows.

* Thermal Equilibrium
* Mechanical Equilibrium
* Chemical Equilibrium


Thermal Equilibrium: 

When two bodies are in contact, there will be heat exchange between the bodies if and only there exists a temperature difference (ΔT) between the bodies.

Due to the temperature difference between the bodies, heat will flow from the high temperature body to the low temperature body. 

As a result of this heat transfer, the temperature of the hot body will be decreased and the temperature of the cold body will be increased.

When the temperature of both the bodies becomes equal to each others, the flow of heat stops. This equilibrium condition is known as the Thermal Equilibrium. 


Mechanical Eqiilibrium : 

If there exists a pressure gradient (ΔP) inside a system, between two systems or between a system and its surroundings, then the interface surface will experience a net force not equal to zero and due to which work transfer will happen where the system having higher pressure will do work against the lower pressure system. 

Due to this work transfer, pressure of the high pressure system will be decreased as energy has flown out of the system. On the other hand, the pressure in the low pressure system will be increased. When the pressure becomes equal in both sides, the work energy flow will be stopped and this state is known as the state of Mechanical Equilibrium.;

Chemical Equilibrium:

If there exists a chemical potential (Δμ) within the components of the system or between the system and surroundings, then there will be a spontaneous chemical reaction which will try to neutralize the chemical potential, after sometimes when the chemical potential becomes zero, the reaction stops and then there will not be any more changes in chemical properties of the system. This condition is called Chemical Equilibrium.

When a system attains thermal, mechanical and chemical equilibrium simultaneously, the state of the system is called in a "THERMODYNAMIC EQUILIBRIUM".

INTERNAL COMBUSTION ENGINE COOLING

COOLING:

In general, one can define cooling as the reduction of temperature of an object or system due to heat extraction from it.

NECESSITY OF COOLING IN IC ENGINES:

Where as the energy generated during combustion of fuel inside the cylinder is the source of heat input Q_f. Out of this energy, approximately one third part is converted into useful work, one third has been carried away by the hot exhaust gas. A part of the remaining one third fraction of the energy has been accounted for various losses including frictional power, where as the remaining portion of heat energy flows into cylinder, thus making it hotter. The cylinder itself radiates heat to air, but rate of heat leakage is very small compared to the heat energy the cylinder is receiving from hot gases. Therefore, the temperature of the cylinder will start to increase until it becomes so hot that

average temperature of the cylinder wall equals to the temperature of hot gases inside. As a result, the cylinder metal properties will suffer and it will soon have a mechanical failure.

COMPARISON OF OTTO AND DIESEL CYCLE EFFICIENCY ON THE BASIS OF SAME MAXIMUM TEMPERATURE AND PRESSURE

Comparison of Otto & Diesel cycles efficiency on the basis of maximum temperature and maximum pressure:

Both Otto cycle and Diesel cycle are idealised thermodynamic cycles which can convert heat energy into useful work done and hence, be the basis of spark ignition and compression ignition internal combustion engines.

Although, both the cycles produce useful work, they have different efficiencies and they can be compared under different constraints and parameters. The most important of them is the comparison on the basis of same maximum cycle temperature and pressure, which is obviously a design constraint. In both the cycles, maximum temperature occurs at the end of compression ie., the state after compression will be same for both the cycles.

In the figure processes 1-2'-3-4-1 is the p-v diagram for Otto cycle, where as processes 1-2-3-4-1 represents diesel cycle and in both the cycles maximum temperature and pressure occurs at state 3, hence both have same maximum temperature and pressure. From the diagrams it has been seen that heat rejection is same for both the cycles and equal to Q_4-1. Where as heat input in diesel cycle is Q_2-3 and in Otto cycle it is Q_2'-3.

Efficiency_Otto = 1 - (Q_4-1 /Q_2'-3)

Efficiency_diesel = 1 - (Q_4-1 /Q_2-3)

From the T-s diagram, we can say area A-2-3-B represents heat transfer during heating in diesel cycle or Q_2-3 and area A-2'-3-B represents heat transfer during heating in Otto cycle or Q_2'-3. As the area A-2-3-B is larger than area A-2'-3-B, we can conclude,

Q_2-3 > Q_2'-3

Therefore,
(Q_4-1 /Q_2'-3) > (Q_4-1 /Q_2-3)
or, 1 - (Q_4-1 /Q_2'-3) < 1 - (Q_4-1 /Q_2-3)
or Efficiency_otto < Efficiency_diesel

FIRST MINOR TEST: IC ENGINES IN SGIT

Shree Ganpati Institute of Technology; Ghaziabad
From 23rd September, 2013 to 26th September first minor test has been organised. This semester, I am teaching IC Engines and Compressors (EME-505) and Thermodynamics (ME-301).
Here is the Question paper of EME-505

  

snapshot of the question paper
ME-301; Thermodynamics
3rd Semester; Mechanical Engg

THE CONCEPT OF VAPOUR LOCK IN IC ENGINES

VAPOUR LOCK

Vapour lock is a problem that mostly affects " Gasoline-fuelled internal combustion engine. " It occurs when liquid fuel changes state from liquid to gas while still in the fuel delivery system. This disrupts the operation of the fuel pump, causing loss of feed pressure to the carburettor or fuel injection system resulting in transient loss of power or even complete stalling.

REASONS OF VAPOUR LOCK

The fuel can vapourise due to being heated by the hot engine or by the local hot climate or due to a low boiling point at high altitude.

In regions where higher volatility fuels are used during winter to improve the cold starting, the use of winter fuels during summer can cause vapour lock more easily.

Vapour lock occurs in older type gasoline fuel systems where a low pressure mechanical fuel pump driven by the engine is located in the engine compartment and feeding a carburettor. These pumps are typically located higher than the fuel tank, are directly heated by the engine, and feed fuel. directly to the float bowl or float chamber of carburettor. As in these pumps fuel is drawn from the feedline and fed into the fuel pump under negative pressure, it lowers the boiling temperature of the liquid fuel. As a result fuel gets evaporated fast and totally invades the fuel pump system and carburettor. As the carburettor becomes devoids of liquid fuel, the mixture it prepares will have less amount of fuel as the volume of vapour of fuel is larger than the equal amount of liquid fuel.

The automotive fuel pump is designed to handle a mixture of liquid and vapour phases of fuel, hence it should handle both the phases of fuel. But, if the amount of fuel evaporated in the fuel system is critically high, the fuel pump stops functioning as per the design and started to pump more vapours than liquid fuel and hence, less amount of liquid fuel will go to the engine. The vapours of fuel will invade the fuel pump delivery system which stops the flow of liquid fuel into the engine.

Most carburettors are designed to run at a fixed level of fuel in the flat bowl of carburettor and reducing the level will reduce the fuel to air mixture and hence, will deliver a lean mixture to the combustion chamber which translates into uneven running of the engine or even stalling while idling or sometimes momentary stalling when running.

VAPOUR LOCK AND (V/L) RATIO

The vapour liquid ratio or (V/L) ratio of a gasoline, defined as the amount of vapour released from a gasoline to the amount of liquid remaining at a given temperature directly correlates with the degree of vapour lock likely to be experienced with this gasoline in the fuel system of a car. At V/L ratio = 24, vapour lock may start and at V/L ratio = 36, vapour lock may be severe. Therefore, the volatility of the gasoline should be maintained as low as practical to prevent vapour lock.

IMPORTANT PROPERTIES OF SI ENGINE FUEL

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THE FUEL CHARACTERISTICS OF INTERNAL COMBUSTION ENGINE:

The fuel characteristics that are important for the performances of
Internal combustion engines are

• Volatility of the Fuel
• Detonation Characteristics
• Power and Efficiency of Engines
• Good thermal properties like heat of combustion and heat of evaporation
• Gum Content
• Sulphur Content
• Aromatic Content
• Cleanliness

Book For a Trial Class:

IMPORTANT CHARACTERISTICS OF SI ENGINE FUELS

SI (spark-ignition) engines, also known as gasoline engines, use a fuel-air mixture that is ignited by a spark from a spark plug to produce power. Some of the important properties of SI engine fuel include:

 

1. Octane rating: The octane rating of a fuel measures its resistance to knocking, which is an uncontrolled explosion in the engine cylinder that can damage the engine. The higher the octane rating, the more resistant the fuel is to knocking.
2. Energy content: The energy content of the fuel determines how much power can be produced from a given amount of fuel. Gasoline has a higher energy content per unit of volume than ethanol, for example.
3. Volatility: Volatility refers to the ease with which a fuel evaporates. High-volatility fuels can vaporize quickly, which is important for good cold-start performance. However, if a fuel is too volatile, it can also cause vapor lock in hot weather, which can disrupt fuel delivery to the engine.
4. Stability: Fuel stability refers to the ability of a fuel to resist oxidation and degradation over time. Stable fuels are less likely to form deposits or gum up fuel injectors, which can negatively impact engine performance and fuel efficiency.
5. Chemical composition: The chemical composition of the fuel can affect its combustion characteristics, including its flame speed and emissions. Gasoline typically contains hydrocarbons, oxygenates (such as ethanol), and various additives to improve performance and reduce emissions.
6. Cost: The cost of fuel is an important consideration for consumers and businesses alike. Gasoline is typically less expensive than alternative fuels like diesel or natural gas, but its price can fluctuate depending on supply and demand, as well as other market factors.

 

Every SI engines are designed for a particular fuel having some desired qualities. For a good performance of a SI engine the fuel used must have the proper characteristics.
The followings are requirements of a good SI engine fuels or Gasolines.

• It should readily mix with air to make a uniform mixture at inlet, ie. it must be volatile
• It must be knock resistant
• It should not pre-ignite easily
• It should not tend to decrease the volumetric efficiency of the engine.
• It should not form gum and varnish
• Its Sulphur content should be low as it is corrosive
• It must have a high calorific value

VOLATILITY OF THE FUEL

It is the most important characteristics of a SI engine fuel. Volatility is a physical concept that loosely defined as the tendency to evaporate at a temperature lower than their boiling temperature. It is the most dominant factor that controls the air-fuel ratio inside the combustion chamber.
One of the most important requirements for proper and smooth combustion is the availability of a highly combustible air-fuel mixture at the moment of the start of the ignition inside the combustion chamber.
A highly volatile (of low molecular weight) fuel generates a rich fuel air ratio at low starting temperature, to satisfy the criteria at the starting of the ignition. But, it will create another problem during running operation; it creates vapour bubble which choked the fuel pump delivery system. This phenomenon is known as vapour lock.

A vapour lock thus created restricts the fuel supply due to excessive rapid formation of vapour in the fuel supply system of the carburetor.
High volatility of fuel can also result in excessive evaporation during storage in a tank which will also pose a fire hazards.
Low volatile fuel like kerosene and distillates can be used for SI engines for tractors.

VOLATILITY AND ITS EFFECT ON ENGINE PERFORMANCES

Volatility greatly affects the engine performances and fuel economy characteristics. The most important of them are

1. ·         Cold and Hot starting
2. ·         Vapour Lock in fuel delivery system
3. ·         Short and Long trip economy
4. ·         Acceleration and Power
5. ·         Warm Up
6. ·         Hot Stalling
7. ·         Carburetor Icing
8. ·         Crankcase Dilution Deposit formation and Spark Plug Fouling
   

When the percentage evaporation of the fuel is 0% ~ 20%, it is called front end of volatility curves, and there are 3 major problems that we encounter in this region of volatility curves which is also known as Distillation curves. They are 
    • Cold Starting
    • Hot Starting
    • Vapour Lock

If front end volatility is very low of a SI engine fuel the engine may show the symptoms of "Cold Starting."  


THE CONCEPT OF COLD STARTING

In order to start an engine a highly combustible mixture rich in fuel is needed at starting temperature near the spark plug. 
As the ambient temperature is low during starting condition, hence the fuel-air mixture must be rich to ensure the start of combustion as sparking of spark plug is not able to start a chemical reaction of combustion near the spark plug.


The limit of air-fuel mixture at the start is
• for rich mixture it is 8:1
• for lean mixture it is 20:1


MECHANISMS OF COLD START:

At low ambient temperature, only a small fraction of total fuel fed to the combustion chamber is able to be effectively evaporated and it creates a insufficiently lean fuel-air mixture that is unable to combust and sustain the combustion process. As a result, the combustion never be able to provide a steady rate of heat supply and engine never starts in this condition. 
This phenomenon is known as cold starting of an IC engine.


To get rid of this problem, we generally apply Choking Process at the start of an engine at ambient temperature. When an Engine becomes hot enough to engineered a sufficiently rich fuel air mixture, the combustion becomes steady and it is known as Warming Up of an IC engine.

Choking is a process generally used to control or regulate air flow into the carburetor where fuel gets mixed with air homogeneously and been fed into combustion chamber. By decreasing air-flow rate into the carburetor, a rich mixture of fuel and air is prepared and fed into the cylinder or combustion chamber, one can increase the vapour content of fuel in the mixture as the reduced air makes the mixture fuel rich and the mixture becomes a combustible inside the combustion chamber.

DETONATION CHARACTERISTICS OF A SI ENGINE FUEL:

 

The detonation characteristics of a fuel refer to its tendency to detonate or explode prematurely in the engine cylinder, leading to engine knock or detonation. This is undesirable as it can cause damage to the engine and reduce its performance and efficiency.

 

In spark-ignition (SI) engines, the detonation characteristics of the fuel are influenced by several factors, including:

 

1. Octane rating: The octane rating of a fuel is a measure of its ability to resist knocking or detonation. Fuels with higher octane ratings are less prone to detonation and are therefore more suitable for use in high-performance engines.
2. Chemical characteristics: Fuels with higher percentages of aromatic hydrocarbons or olefins tend to have lower resistance to detonation.
3. Air-fuel ratio: The air-fuel ratio (AFR) is the ratio of air to fuel in the combustion mixture. AFRs that are too lean (i.e., too much air relative to fuel) can increase the risk of detonation.
4. Compression ratio: The compression ratio is the ratio of the volume in the engine cylinder when the piston is at the bottom of its stroke to the volume when it is at the top of its stroke. Higher compression ratios can increase the risk of detonation.
5. Engine operating conditions: The operating conditions of the engine, such as load, speed, and temperature, can affect the detonation characteristics of the fuel.

 

In general, fuels with higher octane ratings and lower percentages of aromatic hydrocarbons and olefins are more resistant to detonation and are therefore preferred for use in SI engines. Additionally, controlling the air-fuel ratio, compression ratio, and engine operating conditions can help to reduce the risk of detonation.

 

 

FACTORS OF DETONATION CHARACTERISTICS:

 

THE OCTANE RATING:

The octane rating is a measure of a fuel's ability to resist knocking or detonation in internal combustion engines. Knocking or detonation occurs when the air-fuel mixture in the engine's cylinder ignites prematurely or unevenly, leading to a rapid and uncontrolled burning of the remaining fuel. This can cause engine damage and reduce overall performance.

Fuels with higher octane ratings have better anti-knock properties and can withstand higher compression ratios and temperatures before auto-ignition occurs. High-performance engines, such as those found in sports cars or high-powered motorcycles, often operate at higher compression ratios and temperatures, which can lead to a greater tendency for knocking. Using a fuel with a higher octane rating helps prevent knocking and maintains engine performance.

On the other hand, some vehicles, especially those with lower compression ratios or engines designed for regular-grade fuel, do not require high-octane gasoline. In such cases, using fuel with a higher octane rating than what the engine needs might not provide any significant benefits and could be a waste of money.

It's essential to use the fuel recommended by the manufacturer for your specific vehicle, as using the wrong octane rating can lead to inefficient combustion and potentially harm the engine. Many modern vehicles have knock sensors and engine management systems that can adjust the engine's performance based on the octane level of the fuel being used, but it's still best to follow the manufacturer's guidelines.

 

THE CHEMICAL COMPOSITION OF A FUEL:

The chemical composition of a fuel can significantly influence its resistance to detonation or knocking. Fuels with higher percentages of aromatic hydrocarbons or olefins tend to have lower resistance to detonation compared to fuels with higher percentages of paraffins (saturated hydrocarbons). Let's explore this further:

1. Aromatic hydrocarbons: Aromatic hydrocarbons, such as benzene, toluene, and xylene, have a cyclic structure and are known for their high octane number, which indicates good resistance to knocking. However, when present in high concentrations in a fuel, they can contribute to pre-ignition issues and reduce the fuel's overall anti-knock properties. This is why modern gasoline formulations aim to limit the concentration of aromatic hydrocarbons to maintain optimal octane ratings.
2. Olefins: Olefins, also known as alkenes, are unsaturated hydrocarbons that contain at least one carbon-carbon double bond. Fuels with a higher content of olefins generally have lower octane ratings and are more prone to detonation. This is because the presence of double bonds in the molecular structure makes them more reactive, leading to premature ignition and knocking in high-compression engines.
3. Paraffins: Paraffins, also known as alkanes, are saturated hydrocarbons with single bonds between carbon atoms. Fuels with higher percentages of paraffins tend to have better anti-knock properties and higher octane ratings. They are less reactive compared to olefins, which makes them more resistant to detonation.

To improve the overall quality and anti-knock properties of gasoline, refiners often use various blending components and additives to achieve the desired octane rating while keeping the concentration of aromatic hydrocarbons and olefins within acceptable limits.

It's essential for fuel manufacturers to strike a balance in the chemical composition of gasoline to ensure optimal engine performance, fuel efficiency, and emissions control, while also meeting regulatory requirements and environmental standards.

 

THE AIR-FUEL RATIO:

The air-fuel ratio (AFR) refers to the ratio of the mass or volume of air to the mass or volume of fuel in the combustion mixture used by an internal combustion engine. It is a crucial parameter that significantly affects engine performance, fuel efficiency, and emissions.

In the context of detonation or knocking, an AFR that is too lean (meaning there is too much air relative to the amount of fuel) can indeed increase the risk of detonation. When the mixture is lean, there is an excess of oxygen compared to the available fuel molecules. This can lead to higher combustion temperatures and pressures, which can cause the air-fuel mixture to ignite prematurely or unevenly, resulting in knocking.

Detonation occurs because the rapid and uncontrolled burning of the lean mixture generates pressure waves that collide and produce a knocking sound. This can put excessive stress on the engine components and lead to engine damage over time.

On the other hand, an AFR that is too rich (meaning there is too much fuel relative to the amount of air) can also lead to knocking. A rich mixture tends to burn more slowly, and the unburned fuel can create hot spots in the combustion chamber, increasing the likelihood of pre-ignition and knocking.

To minimize the risk of knocking and achieve optimal engine performance, modern engines are equipped with sophisticated engine management systems and knock sensors that can adjust the air-fuel ratio in real-time based on various factors, such as engine load, speed, and temperature. These systems help maintain the AFR within the appropriate range to ensure efficient combustion and reduce the risk of detonation.

For high-performance engines or engines modified for increased power output, tuning the air-fuel ratio carefully is crucial to avoid knocking and maximize performance. It's important to follow the manufacturer's recommendations or consult with experienced tuners to ensure that the engine operates within safe and optimal parameters.

THE COMPRESSION RATIO:

The compression ratio is a crucial parameter in internal combustion engines, and it represents the ratio of the cylinder volume when the piston is at its bottom dead center (BDC) to the cylinder volume when the piston is at its top dead center (TDC). It is typically expressed as a numerical value, such as 10:1 or 12:1, representing the ratio of the larger volume (at BDC) to the smaller volume (at TDC).

Higher compression ratios indeed increase the risk of detonation, especially if the fuel used has a low octane rating or if other factors that promote knocking are present. Here's why:

1. Increased Temperature and Pressure: Higher compression ratios compress the air-fuel mixture more, resulting in increased temperature and pressure in the combustion chamber. This elevated pressure and temperature can cause the air-fuel mixture to autoignite prematurely, leading to knocking or detonation.
2. Reduced Time for Combustion: With higher compression ratios, the time available for the air-fuel mixture to burn completely is reduced. This can lead to incomplete combustion, which leaves unburned fuel and hot spots in the combustion chamber, increasing the likelihood of knocking.
3. Increased Sensitivity to Fuel Properties: Fuels with lower octane ratings are more likely to experience detonation under higher compression ratios. The lower the octane rating, the more susceptible the fuel is to pre-ignition, and the greater the risk of knocking in high-compression engines.

To mitigate the risk of detonation in high-compression engines, it is crucial to use fuels with higher octane ratings that can withstand the elevated pressures and temperatures without prematurely igniting. Additionally, modern engine management systems with knock sensors can detect knocking and adjust the engine's timing and air-fuel ratio to reduce the likelihood of detonation.

Engine designers and tuners carefully consider the compression ratio when developing or modifying engines to ensure optimal performance while avoiding harmful knocking or detonation. Following the manufacturer's recommendations regarding fuel type and engine specifications is essential to maintain the engine's longevity and performance.

 

THE ENGINE OPERATING CONDITION:

The operating conditions of an engine, including factors such as load, speed, and temperature, have a significant impact on the detonation characteristics of the fuel being used. Let's explore how these factors can influence the likelihood of detonation:

1. Engine Load: The engine load refers to the amount of power the engine is producing to meet the demands of driving or operating the vehicle. Higher engine loads, such as during acceleration or towing heavy loads, result in increased pressure and temperature in the combustion chamber. This elevated pressure and temperature can make the air-fuel mixture more prone to detonation, especially if the fuel used has a lower octane rating. As a result, engines under high load conditions are more susceptible to knocking.
2. Engine Speed: Engine speed, commonly measured in revolutions per minute (RPM), determines how frequently the combustion process occurs in the cylinders. Higher engine speeds mean that the air-fuel mixture is being compressed and ignited more frequently. If the engine is operating at high RPM, there is less time for the air-fuel mixture to burn completely, increasing the chances of knocking.
3. Engine Temperature: The temperature of the engine components, particularly the combustion chamber, plays a crucial role in the risk of detonation. Higher engine temperatures can cause hot spots in the combustion chamber, which can lead to premature ignition of the air-fuel mixture. This is especially true when the engine is running under heavy load or high RPM conditions.
4. Intake Air Temperature: The temperature of the intake air entering the engine also affects the likelihood of knocking. Cooler air is denser and can reduce the chances of knocking, as it allows for a higher air-to-fuel ratio without increasing the risk of detonation. Engines equipped with intercoolers or air intake temperature control systems can optimize the intake air temperature for improved performance and reduced knocking.
5. Ignition Timing: The ignition timing refers to the precise moment when the spark plug ignites the air-fuel mixture in the cylinder. Advanced ignition timing (igniting the mixture earlier) can increase the risk of knocking, especially under high load and high temperature conditions. Retarding the ignition timing (igniting the mixture later) can help reduce knocking in some cases.

To optimize engine performance and reduce the risk of detonation, modern engines use sophisticated engine management systems that continuously monitor various parameters and adjust ignition timing, air-fuel ratio, and other factors to maintain safe and efficient operation. Additionally, using high-quality fuels with appropriate octane ratings can also play a vital role in preventing knocking under varying operating conditions.