FUEL USED IN IC ENGINES AND REFINERY PROCCESSES; EME-505

FUEL USED IN IC ENGINES

An article on fossil fuels

Internal Combustion Engines are the generators of the energy mainly used for transportation. Almost more than 90% of the total IC Engines run on fossil fuels or different derivatives of petroleum.

IC Engines are a kind of open cycle heat engine where heat is supplied to the engine by the combustion of working fluids thus releasing huge amount of energy due to the combustion processes of the working fluids. Combustible working fluids are called fuels.

The natural petroleum oil is the largest single source of internal combustion engine fuels. Petrol and Diesel are the most used among them. The boiling point of petrol is 30°C to 200°C and that of diesel oil is from 200°C to 375°C.

Fuels of most of the IC Engines are the derivatives of Petroleum like gasoline, diesel oil, kerosene, jet fuel etc. All of these fuels are produced during the fractional distillation of Petroleum Oil obtained from crude from oil wells.

The fuels used in the IC Engines are designed to satisfy the performance requirements of the engine system in which they are used. As a result the fuels must have certain

• (i) physical,
• (ii) chemical and
• (iii) combustion properties.

Following are the some characteristics a fuel must have in order to produce the desirable output to the engine performance.

1. A fuel must have a large energy density to be capable to release huge amount of energy during its combustion in side the combustion chamber.
2. A fuel must posses a good combustion quality to produce large amount of energy in smooth way.
3. A fuel must have high thermal stability or pre-ignition may occur.
4. A fuel must show a low deposit forming tendency else gum formation and other deposit forming processes will hamper the combustion process.
5. A fuel must be non-toxic, easy to handle and storage.

CRUDE PETROLEUM OIL:

Petroleum or often referred as "Crude Oil" is a naturally occurring inflammable mixtures of liquid and mud and it contains a complex mixture of different hydrocarbons of various molecular weights. It is mainly recovered through a process called "Oil Drilling".

Oil Wells and Gas Wells:

An oil well produces mainly crude oil with some natural gas dissolved in it. In contrast a gas well produces natural gases although it may contain heavier hydrocarbons like pentane, hexane or hepthane in gaseous state due to the extreme pressure and temperature inside the well, but at surface conditions condensation starts and forms "Natural Gas Condensate" or simply known as Condensate.



COMPOSITIONS OF CRUDE WELL:

Basically, crude well is the muddy mixtures of different hydrocarbons of different molecular weights. Alkanes, Cyclo-alkanes or napthenes, aromatics. It contains nitrogen, oxygen, sulfur and phosphorous. It may also contains metallic compounds too.

Four different types of hydrocarbon molecules appear in crude oil. The relative percentages are widely varied from oil to oil. They are:

• i) Paraffins (alkanes,  C_nH_{2n + 2} )
• ii) Olefins (alkenes, C_nH_2n),
• iii) Napthenes (cyclo-alkanes, C_nH_2n ),
• iv) Aromatics (having benzene ring, C_nH_{2n - 6}).

It is then refined by fractional distillation in oil refinery to obtain a large number of consumer products, from petrol or gasoline, diesel to kerosene, heavy oil, fuel oil, asphalt, chemical reagents, plastics etc.

Most of the derivatives of the petroleum have been used as fuel or heating purpose. The major products of a petroleum refinery are:

• (i) Gasoline,
• (ii) Kerosene,
• (iii) Diesel Oil,
• (iv) Fuel oil,
• (v) Heavy Oil,
• (vi) Lubricating Oil,
• (vii) Asphalts

INTRODUCTION: 

As the demands for gasoline, kerosene/ jet fuel and diesel oil are maximum, refineries around the world have started to convert heavy fuels and other higher hydrocarbons into gasoline, kerosene and diesel oil. To perform this, refineries have adopted several thermo-chemical processes those can convert high molecular weight hydrocarbons into lighter ones by breaking them.

GENERAL REFINERY PROCESSES:

Petroleum refining has evolved continuously in response to changing consumer demand for better and different products. The original requirement was to produce kerosene as a cheaper and better source of light than whale oil. The development of the internal combustion engine led to the production of gasoline and diesel fuels. The evolution of the airplane created an initial need for high-octane aviation gasoline and then for jet fuel, a sophisticated form of the original product, kerosene. Present-day refineries produce a variety of products including many required as feedstock for the petrochemical industry.

a) Distillation Processes:

The first refinery, opened in 1861, produced kerosene by simple atmospheric distillation. Its by-products included tar and naphtha. It was soon discovered that distilling petroleum under vacuum could produce high-quality lubricating oils. However, for the next 30 years kerosene was the product consumer wanted. Two significant events changed this situation. The invention of the electric light decreased the demand for kerosene and the invention of the internal combustion engine created a demand for diesel fuel and gasoline (naphtha). 

b) Thermal Cracking Processes:

With the advent of mass production and World War I, the number of gasoline-powered vehicles increased dramatically and the demand for gasoline grew accordingly. However, distillation processes produced only a certain amount of gasoline from crude oil. In 1913, the thermal cracking process was developed, which subjected heavy fuels to both pressure and intense heat, physically breaking the large molecules into smaller ones to produce additional gasoline and distillate fuels. Visbreaking, another form of thermal cracking, was developed in the late 1930's to produce more desirable and valuable products. 

c) Catalytic Processes:

Higher-compression gasoline engines required higher-octane gasoline with better antiknock characteristics. The introduction of catalytic cracking and polymerization processes in the mid- to late 1930's met the demand by providing improved gasoline yields and higher octane numbers.   Alkylation, another catalytic process developed in the early 1940's, produced more high-octane aviation gasoline and petrochemical feedstock for explosives and synthetic rubber. Subsequently, catalytic isomerization was developed to convert hydrocarbons to produce increased quantities of alkylation feedstock. Improved catalysts and process methods such as hydrocracking and reforming were developed throughout the 1960's to increase gasoline yields and improve antiknock characteristics. These catalytic processes also produced hydrocarbon molecules with a double bond (alkenes) and formed the basis of the modern petrochemical industry. 

d) Treatment Processes:

Throughout the history of refining, various treatment methods have been used to remove non-hydrocarbons, impurities, and other constituents that adversely affect the properties of finished products or reduce the efficiency of the conversion processes. Treating can involve chemical reaction and/or physical separation. Typical examples of treating are chemical sweetening, acid treating, clay contacting, caustic washing, hydrotreating, drying, solvent extraction, and solvent dewaxing. Sweetening compounds and acids desulfurize crude oil before processing and treat products during and after processing. 

Following the Second World War, various reforming processes improved gasoline quality and yield and produced higher-quality products. Some of these involved the use of catalysts and/or hydrogen to change molecules and remove sulfur. 

 Basics of Hydrocarbon Chemistry:

Crude oil is a mixture of hydrocarbon molecules, which are organic compounds of carbon and hydrogen atoms that may include from one to 60 carbon atoms. The properties of hydrocarbons depend on the number and arrangement of the carbon and hydrogen atoms in the molecules. The simplest hydrocarbon molecule is one carbon atom linked with four hydrogen atoms: methane. All other variations of petroleum hydrocarbons evolve from this molecule. 

 

Hydrocarbons containing up to four carbon atoms are usually gases, those with 5 to 19 carbon atoms are usually liquids and those with 20 or more are solids. The refining process uses chemicals, catalysts, heat, and pressure to separate and combine the basic types of hydrocarbon molecules naturally found in crude oil into groups of similar molecules. The refining process also rearranges their structures and bonding patterns into different hydrocarbon molecules and compounds. Therefore it is the type of hydrocarbon (paraffinic, naphthenic, or aromatic) rather than its specific chemical compounds that is significant in the refining process. 

Principal Groups of Hydrocarbon

• Paraffins - The paraffinic series of hydrocarbon compounds found in crude oil have the general formula C_nH_2n+2 and can be either straight chains (normal) or branched chains (isomers) of carbon atoms. The lighter, straight chain paraffin molecules are found in gases and paraffin waxes. Examples of straight-chain molecules are methane, ethane, propane, and butane (gases containing from one to four carbon atoms), and pentane and hexane (liquids with five to six carbon atoms). The branched-chain (isomer) paraffins are usually found in heavier fractions of crude oil and have higher octane numbers than normal paraffins. These compounds are saturated hydrocarbons, with all carbon bonds satisfied, that is, the hydrocarbon chain carries the full complement of hydrogen atoms.
  • Example of simplest hydrocarbon molecule: Methane (CH₄), Examples of straight chain paraffin molecule (Butane) and branched paraffin molecule (Isobutane) with same chemical formula (C₄H₁₀)
    
    
    
    
• Aromatics - Aromatics are unsaturated ring-type (cyclic) compounds which react readily because they have carbon atoms that are deficient in hydrogen. All aromatics have at least one benzene ring (a single-ring compound characterized by three double bonds alternating with three single bonds between six carbon atoms) as part of their molecular structure. Naphthalenes are fused double-ring aromatic compounds. The most complex aromatics, polynuclears (three or more fused aromatic rings), are found in heavier fractions of crude oil.
  • Example of simple aromatic compound: Benzene (C₆H₆), Examples of simple double-ring aromatic compound: Naphthalene (C₁₀H₈)
    
    
    
    
• Naphthenes - Naphthenes are saturated hydrocarbon groupings with the general formula C_nH_2n, arranged in the form of closed rings (cyclic) and found in all fractions of crude oil except the very lightest. Single-ring naphthenes (monocycloparaffins) with five and six carbon atoms predominate, with two-ring naphthenes (dicycloparaffins) found in the heavier ends of naphtha.
  • Example of typical single-ring naphthene: Cyclohexane (C₆H₁₂), Examples of naphthene with same chemical formula (C₆H₁₂) but different molecular structure: Methyl cyclopentane (C₆H₁₂)

Other Hydrocarbons

• Alkenes - Alkenes are mono-olefins with the general formula C_nH_2n and contain only one carbon-carbon double bond in the chain. The simplest alkene is ethylene, with two carbon atoms joined by a double bond and four hydrogen atoms. Olefins are usually formed by thermal and catalytic cracking and rarely occur naturally in unprocessed crude oil.
  • Example of simples Alkene: Ethylene (C₂H₄), Typical Alkenes with the same chemical formula (C₄H₈) but different molecular structures: 1-Butene and Isobutene
    
    
    
    
• Dienes and Alkynes - Dienes, also known as diolefins, have two carbon-carbon double bonds. The alkynes, another class of unsaturated hydrocarbons, have a carbon-carbon triple bond within the molecule. Both these series of hydrocarbons have the general formula CnH2n-2. Diolefins such as 1,2-butadiene and 1,3-butadiene, and alkynes such as acetylene,occur in C5 and lighter fractions from cracking. The olefins, diolefins, and alkynes are said to be unsaturated because they contain less than the amount of hydrogen necessary to saturate all the valences of the carbon atoms. These compounds are more reactive than paraffins or naphthenes and readily combine with other elements such as hydrogen, chlorine, and bromine.
  • Example of simplest Alkyne: Acetylene (C₂H₂), Typical Diolefins with the same chemical formula (C₄H₆) but different molecular structures: 1,2-Butadiene and 1,3-Butadiene

Non-hydrocarbons

• Sulfur Compounds -  Sulfur may be present in crude oil as hydrogen sulfide (H₂S), as sulfur compounds such as mercaptans, sulfides, disulfides, thiophenes, etc. or as elemental sulfur. Each crude oil has different amounts and types of sulfur compounds, but as a rule the proportion, stability, and complexity of the compounds are greater in heavier crude-oil fractions. Hydrogen sulfide is a primary contributor to corrosion in refinery processing units. Other corrosive substances are elemental sulfur and mercaptans. Moreover, the corrosive sulfur compounds have an obnoxious odor.  Pyrophoric iron sulfide results from the corrosive action of sulfur compounds on the iron and steel used in refinery process equipment, piping, and tanks. The combustion of petroleum products containing sulfur compounds produces undesirables such as sulfuric acid and sulfur dioxide. Catalytic hydrotreating processes such as hydrodesulfurization remove sulfur compounds from refinery product streams. Sweetening processes either remove the obnoxious sulfur compounds or convert them to odorless disulfides, as in the case of mercaptans.
  
  
• Oxygen Compounds -  Oxygen compounds such as phenols, ketones, and carboxylic acids occur in crude oils in varying amounts. 
  
  
• Nitrogen Compounds -  Nitrogen is found in lighter fractions of crude oil as basic compounds, and more often in heavier fractions of crude oil as nonbasic compounds that may also include trace metals such as copper, vanadium, and/or nickel. Nitrogen oxides can form in process furnaces. The decomposition of nitrogen compounds in catalytic cracking and hydrocracking processes forms ammonia and cyanides that can cause corrosion. 
  
  
• Trace Metals -  Metals, including nickel, iron, and vanadium are often found in crude oils in small quantities and are removed during the refining process. Burning heavy fuel oils in refinery furnaces and boilers can leave deposits of vanadium oxide and nickel oxide in furnace boxes, ducts, and tubes. It is also desirable to remove trace amounts of arsenic, vanadium, and nickel prior to processing as they can poison certain catalysts. 
  
  
• Salts -  Crude oils often contain inorganic salts such as sodium chloride, magnesium chloride, and calcium chloride in suspension or dissolved in entrained water (brine). These salts must be removed or neutralized before processing to prevent catalyst poisoning, equipment corrosion, and fouling. Salt corrosion is caused by the hydrolysis of some metal chlorides to hydrogen chloride (HCl) and the subsequent formation of hydrochloric acid when crude is heated. Hydrogen chloride may also combine with ammonia to form ammonium chloride (NH₄Cl), which causes fouling and corrosion. 
  
  
• Carbon Dioxide -  Carbon dioxide may result from the decomposition of bicarbonates present in or added to crude, or from steam used in the distillation process. 
• Naphthenic Acids -  Some crude oils contain naphthenic (organic) acids, which may become corrosive at temperatures above 450° F when the acid value of the crude is above a certain level.

 Major Refinery Products

• Gasoline. The most important refinery product is motor gasoline, a blend of hydrocarbons with boiling ranges from ambient temperatures to about 400 °F. The important qualities for gasoline are octane number (antiknock), volatility (starting and vapor lock), and vapor pressure (environmental control). Additives are often used to enhance performance and provide protection against oxidation and rust formation.
• Kerosene. Kerosene is a refined middle-distillate petroleum product that finds considerable use as a jet fuel and around the world in cooking and space heating. When used as a jet fuel, some of the critical qualities are freeze point, flash point, and smoke point. Commercial jet fuel has a boiling range of about 375°-525° F, and military jet fuel 130°-550° F. Kerosene, with less-critical specifications, is used for lighting, heating, solvents, and blending into diesel fuel.
• Liquified Petroleum Gas (LPG). LPG, which consists principally of propane and butane, is produced for use as fuel and is an intermediate material in the manufacture of petrochemicals. The important specifications for proper performance include vapor pressure and control of contaminants.
• Distillate Fuels. Diesel fuels and domestic heating oils have boiling ranges of about 400°-700° F. The desirable qualities required for distillate fuels include controlled flash and pour points, clean burning, no deposit formation in storage tanks, and a proper diesel fuel cetane rating for good starting and combustion.
• Residual Fuels. Many marine vessels, power plants, commercial buildings and industrial facilities use residual fuels or combinations of residual and distillate fuels for heating and processing. The two most critical specifications of residual fuels are viscosity and low sulfur content for environmental control.
• Coke and Asphalt. Coke is almost pure carbon with a variety of uses from electrodes to charcoal briquets. Asphalt, used for roads and roofing materials, must be inert to most chemicals and weather conditions.
• Solvents. A variety of products, whose boiling points and hydrocarbon composition are closely controlled, are produced for use as solvents. These include benzene, toluene, and xylene.
• Petrochemicals. Many products derived from crude oil refining, such as ethylene, propylene, butylene, and isobutylene, are primarily intended for use as petrochemical feedstock in the production of plastics, synthetic fibers, synthetic rubbers, and other products.
• Lubricants. Special refining processes produce lubricating oil base stocks. Additives such as demulsifiers, antioxidants, and viscosity improvers are blended into the base stocks to provide the characteristics required for motor oils, industrial greases, lubricants, and cutting oils. The most critical quality for lubricating-oil base stock is a high viscosity index, which provides for greater consistency under varying temperatures.

Common Refinery Chemicals

• Leaded Gasoline Additives: Tetraethyl lead (TEL) and tetramethyl lead (TML) are additives formerly used to improve gasoline octane ratings but are no longer in common use except in aviation gasoline.
• Oxygenates: Ethyl tertiary butyl ether (ETBE), methyl tertiary butyl ether (MTBE), tertiary amyl methyl ether (TAME), and other oxygenates improve gasoline octane ratings and reduce carbon monoxide emissions.
• Caustics: Caustics are added to desalting water to neutralize acids and reduce corrosion. They are also added to desalted crude in order to reduce the amount of corrosive chlorides in the tower overheads. They are used in some refinery treating processes to remove contaminants from hydrocarbon streams.
• Sulfuric Acid and Hydrofluoric Acid: Sulfuric acid and hydrofluoric acid are used primarily as catalysts in alkylation processes. Sulfuric acid is also used in some treatment processes.
  
  USEFUL LINKS:   
  refinery topics

MOCK TEST OF I.C. ENGINE; EME-505

1^st SESSIONAL EXAMINATION 2012-13

BRANCH – ME

SEMESTER- V

Course: B.Tech.                                                                                                                Year: 3^rd

Sub: IC Engines & Compressors                                                           Subject Code: EME- 505

Maximum Marks: 30                                                                                               Time: 1.30 hrs

Section A:                                                                                                       2x6 = 12

Q.1) Answer the following question in brief                                                                                  

(a)    Explain the difference between inlet valve and inlet port.

(b)   What is cracking?

(c)    Define volumetric efficiency of an IC engine with diagram.  

(d)   What are the assumptions of a air standard cycle?

(e)    Explain the concept of indicator thermal efficiency.

(f)    What are the general refinery processes?

Section B:

Q.2) Answer any three of the following questions                                         6 x 3 = 18

(a) Compare between four stroke and two stroke IC engines.

(b)  What is compression ratio? What is its range for (i) the SI engine (ii) CI engine? What factors limit the compression ratio in each type of engine?

(c) An ideal Otto cycle has compression ratio of 8 and initial conditions are 1 bar and 15°C. Heat added during constant volume process is 1045 kJ/kg. Find:

(i)                 Maximum cycle temperature

(ii)               Air standard efficiency

(iii)             Work done per kg of air

(iv)             Heat rejected

Take: C_v = 0.7175 kJ/kg-K and γ = 1.4

(d)  A diesel engine develops 5 kW. It’s indicated thermal efficiency is 30% and mechanical efficiency 57%. Estimate the fuel consumption in (i) kg/hr, (ii) litres/hr,

(iii) Indicated specific fuel consumption, and (iv) brake specific fuel consumption.

L.C.V. of diesel oil = 42000 kJ/kg.

(e) Compare the actual cycle and its deviation from theoretical cycles in IC engines.

MOCK CLASS TEST: THERMODYNAMICS Sub: Code: EME-303; Mahamaya Technical University

Time: 2 hrs                                                                                                   Maximum Marks: 50 

Attempt all the questions: 

SECTION A: 

1) Attempt the following questions:                                                                        (5 x 2 = 10) 

a) Define system, surroundings and universe. 

b) Distinguish between Heat pump and Refrigerator. 

c) What is Exergy and Anergy? 

d) Explain the law of degradation of energy. 

e) What is triple point of water? 

SECTION B: 

2) Attempt any three questions:                                                                               (3 x 5 = 15) 

a) Distinguish between macroscopic and microscopic approaches of thermodynamics. 

b) Discuss the neccessity of 2nd law of thermodynamics. 

c) 2 kg of a gas at 10 bar expands adiabatically and reversibly till its pressure drops to 5 bar. During the process 120 kJ of non-flow work is done by the system and the temperature drops from 377°C centigrade to 257°C. Calculate the value of the index of expansion and the characteristics gas constants. 

d) Steam at a pressure of 4 bar absolute and having dryness fraction 0.8, is heated at constant volume to a pressure of 8 bar absolute. Find the final temperature of the steam. Also, find the total heat absorbed by 1 kg of steam. 

e) 2 kg of air at NTP is heated at constant volume untill the pressure becomes 6 bar. Find the change of entropy of the system. 

SECTION C: 

Attempt part (a) or part (b) of the following questions                                                 (5 x 5 = 25) 

3) (a) Explain the thermodynamic equilibrium and quasi-static process. 

(b) Prove the equivalence of Kelvin-Planck statement and Clausius statement. 

4) (a) A steam turbine developing 110 kW is supplied steam at 17.5 bar with an internal energy of 2600 kJ/min, specific volume = 15.5 m³/kg and velocity of 275 m/s. Heat loss from the steam turbine  37.6 kJ/kg. Neglecting the changes in potential energy, determine the steam flow rate in kg/hr. 

(b) A reversible engine takes 2400 kJ/min from a reservoir at 750 K develops 400 kJ/min of work during cycle. The engine rejects heat at two reservoir at 650 K and 550 K. Find the heat rejected to each sink. 

5) (a) Explain the causes of internal and external irreversibility. 

(b) Explain the importance of Gibb's function and Gibb's free energy. 

6) (a) 5 kg steam at pressure 8 bar and temperature 300°C is adiabatically mixed with 4 kg steam at 6 bar and 250°C. Find the final condition of the mixture. Also find the change in entropy. 

(b) Hot steam is flowing through a perfectly adiabatic pipe. At point A the temperature of the steam is 250°C and pressure is 4 bar, while at the point B, its temperature is 275°C and pressure is 3.5 bar. Find the direction of the flow. 

7) (a) 5 kg of Oxygen is enclosed within a vessel of 0.05 m³ at a temperature 200°C, is being supplied 120 kJ of energy through heating. Find the final pressure and temperature. 

(b) One kg of an ideal gas is heated from 18.3°C to 93.4°C. Assuming R = 287 J/kg-K and  γ  = 1.18 for the gas. Find out (i) specific heats, (ii) change in internal energy, and (iii) change in enthalpy and entropy.

SMALL ENGINEERING COLLEGES OF GHAZIABAD: A BLEAK FUTURE

Economics says "When Supply is more than the Demand of a product, the Price falls." This is particularly true in the case of Technical Education in U.P. and perhaps upto some extent in the country itself.

Over the past few years the supply is outstripping the demand for Engineering and Management seats in the country. Just take the example of Uttar Pradesh, the most populous state of India is home to about 333 Engineering colleges which cumulatively offer a total seats of 1,15,379 in Engineering Education where as according to University datas the total number of students took admission in various engineering colleges after qualifying SEE amounts to mere 25,903. So, what happens to the vacant seats? And this year the figures are not going to be improved it seems.

This year a total 1,60,561 candidates had registered for the State Entrance Examination, out of whom, 1,29,924 have qualified. But, there are approximately 1.33 lakh B.Tech seats in the Engineering colleges affiliated to GBTU and MTU.

There are clearly a huge gap between the supply and demand of Engineering seats. In this situation, it has been believed that many small colleges will be bankrupt due to the lack of students. Many colleges have defered the salaries of the teachers and other employees due to the revenue crunch. It seems a grim scenario ahead for those colleges.

Due to the revenue crunch, promoters of small colleges are taking the refuge of cost cutting, and as a part of that they are trying to trim their faculty strength. Surely, this will affect the quality of the education as the each teacher will be over burdened and perhaps have to take five classes per day, where as the AICTE limits the load at best 18 classes per week. Also, most of the colleges don't follow the exact teacher students ratio of 1:20 prescribed by the apex body.

Many promoters are planning to opt out by selling their stakes in the colleges. The causes of their exits are the facts that running colleges in western U.P. is no longer a profitable business. They have cited that due to lack of students in taking admission, the colleges are no longer the chickens that lay gold eggs, which were in fact so just three years ago. So, why these colleges suddenly loss their values? What are the reasons behind these failures?

There are several reasons for the fall in numbers of students opting B.Tech courses. The most vital reason is the very high tution fees in colleges under MTU and GBTU compared to colleges in other states like Karnataka, Punjab and Rajasthan. Most of the colleges here charge more than 90,000.00 in the first year B.Tech where as colleges in other states charges below 60,000.00, even colleges in Punjab and West Bengal charge below 50 thousand and this is going to be a major factor.

The 2nd factor is the placement after the completion of the degree. Although many colleges claim tall, citing a long list of companies taking interest to place their students in very good packages, but reality always bites hard. The negative publicity by the ex students are also eating the pie here and there is no solution other than boosting the placement record by making a good relation with the HR of these companies by the respective college authorities.

The third most important factor is the sagging quality of the available faculty members. Many teachers although possess M.Tech degrees are not competent to impart quality education due to lack of depth of required knowledge as well as the essential communicating power required to be a good teacher. In some cases, due to over burdened schedule, a good teacher becomes unable to teach in the class. Just imagine the mental fatigue a teacher experienced while taking 5th or 6th class in a day when each class is of 55 min. duration.

A college has to show money to run the next three years during the visits from AICTE. So all the colleges have to show enough balance to pay the salaries of the employees for atleast three years, otherwise they won't get the permission to run the colleges, still some of them couldn't pay the salaries of the teachers and staffs. Why? Becouse they must have showed enough balances to acquire the clearences during the AICTE visits. Where does the money go? Vanished! Or siphoned off? There are several "skips" of the rules and regulations these college authorities used to practise.

QUESTIONS BANK 5 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 4 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 3 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 2: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

1)      Explain the principle of Super-position.

Ans: The principle of superposition states that “The effect of a force on a body does not change and remains same if we add or subtract any system which is in equilibrium.”

In the fig 4 a, a force P is applied at point A in a beam, where as in the fig 4 b, force P is applied at point A and a force system in equilibrium which is added at point B. Principle of super position says that both will produce the same effect.

2)      What is “Force-Couple system?”

Ans: When a force is required to transfer from a point A to point B, we can transfer the force directly without changing its magnitude and direction but along with the moment of force about point B.

As a result of parallel transfer a system is obtained which is always a combination of a force and a moment or couple. This system consists of a force and a couple at a point is known as Force-Couple system.

      In fig 5 a, a force P acts on a bar at point A, now at point B we introduce a system of forces  in equilibrium (fig 5 b), hence according to principle of superposition there is no change in effect of the original system. Now we can reduce the downward force P at point A and upward force P at point B as a couple of magnitude Pxd at point B (fig 5 c).

3) What do you understand by Equivalent force systems?

Ans: Two different force systems will be equivalent if they can be reduced to the same force-couple system at a given point. So, we can say that two force systems acting on the same rigid body will be equivalent if the sums of forces or resultant and sums of the moments about a point are equal.

4)      What is orthogonal or perpendicular resolution of a force?

Ans: The resolution of a force into two components which are mutually perpendicular to each other along X-axis and Y-axis is called orthogonal resolution of a force.

If a force F acts on an object at an angle θ with the positive X-axis, then its component along X-axis is F_x = Fcosθ, and that along Y-axis is F_y = Fsinθ

5) What is oblique or non-perpendicular resolution of a force?

Ans: When a force is required to be resolved in to two directions which are not perpendiculars to each other the resolution is called oblique or Non-perpendicular resolution of a force.

   
       F_OA = (P sin β)/ sin (α +β)

 F_OB = (P sin α)/ sin (α +β)

QUESTION BANK 1: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTION BANK: ENGINEERING MECHANICS

by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force and force system?

Ans: A force is a physical quantity having magnitude as well as direction. Therefore, it is a   vector quantity. It is defined as an "external agency" which produces or tends to produce or destroys or tends to destroy the motion when applied on a body.

Its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.

When two or more forces act on a body or particle, it is called force system. Therefore, a force system is a collection of two or more forces.

2)      What is static equilibrium? What are the different types of static equilibrium?

Ans: A body is said to be in static equilibrium when there is no change in position as well as no rotation exist on the body. So to be in equilibrium process, there must not be any kind of motions ie there must not be any kind of translational motion as well as rotational motion.

We also know that to have a linear translational motion we need a net force acting on the object towards the direction of motion, again to induce an any kind of rotational motion, a net moment must exists acting on the body. Further it can be said that any kind of complex motion can be resolved into a translational motion coupled with a rotating motion.

“Therefore a body subjected to a force system would be at rest if and only if the net force as well as the net moment on the body is zero.”

There are three types of Static Equilibrium

1.      Stable Equilibrium

2.      Unstable Equilibrium

3.      Neutral Equilibrium

3)      What are the characteristics of a force?

Ans: A force has four (4) basic characteristics.

·         Magnitude: It is the value of the force. It is represented by the length of the arrow that we use to represent a force.

·         Direction: A force always acts along a line, which is called as the “line of action”. The arrow head we used to represent a force is the direction of that force.

·         Nature or Sense: The arrow head also represent the nature of a force. A force may be a pull or a push. If a force acts towards a particle it will be a push and if the force acts away from a point it is pull.

·         Point of Application: It is the original location of a point on a body where the force is acting. 

4)      What are the effects of a force acting on a body?

Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.

·         A force may change the state or position of a body by inducing motion of the body. (External effect)

·         A force may change the size or shape of an object when applied on it. It may deform the body thus inducing internal effects on the body.

·         A force may induce rotational motion into a body when applied at a point other than its center of gravity.

·         A force can make a moving body into an equilibrium state at rest.

5)      What is composition and resolution of forces?

Ans: Composition of forces: Composition or compounding is the procedure to find out single resultant force of a force system

Resolution of forces: Resolution is the procedure of splitting up a single force into number of components without changing the effect of the same.

6)      What is Resultant and Equilibrant?

Ans: Resultant: The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

Equilibrant: Any concurrent set of forces, not in equilibrium, can be put into a state of equilibrium by a single force. This force is called the Equilibrant. It is equal in magnitude, opposite in sense and co-linear with the resultant. When this force is added to the force system, the sum of all of the forces is equal to zero.

7)      Explain the principle of Transmissibility?

Ans: The principle of transmissibility states “the point of application of a force can be transmitted anywhere along the line of action, but within the body.”

The fig 3 a shows a force F acting at a point of application A and fig 3 b, the same force F acts along the same line of action but at a different point of action at B and both are equivalent to each other.

Private Engineering Colleges in Ghaziabad: Will They Survive?

There are some very good Engineering colleges in and around Ghaziabad. These colleges not only topped the annual ranks of formerly UPTU or its later avatars GBTU and MTU, but during these periods they have curved a niche for themselves.

There are colleges like Ajay Kumar Garg Engineering College or AKGEC, ABES, KIET, RKGIT and IMSEC in Ghaziabad which are doing good in imparting Technical Education and already established a brand name in this arena. They draw fair numbers of students every year but there are other colleges which are practically starving due to the lack of students as well as quality students.

The second rung colleges in Ghaziabad:

All the engineering colleges in Western UP (including NCRs ie. Ghaziabad, Noida and Greater Noida) are affiliated to the Mahamaya Technical University, Noida. There are several good colleges in Ghaziabad like Ideal Institute of Technology in Govindpuram, VIET in Dadri, BBDIT in Meerat Road, Sunderdeep Engineering College in Dasna are as good as the private colleges of Karnataka. Then there are VITS, SGIT, LKEC near Jindal Nagar, SIET and RKGEC in Pilakhuwa.

The last rung colleges are the newly established colleges like Bhagwati Institute of Technology in Masuri, Aryan Institute of Technology, Jindal Nagar, Bhagwant Institute of Technology, MAIT in near Jindal Nagar, Satyam, ICE in Pilakhuwa. The problem they are suffering is the lack of students. Last year many seats remained vacant, even the concerned colleges offered more than 15% in commission, still number of students getting admission was very low.

Last year the scenario was very grim, many colleges were finding tough to pay the salaries to their employees. Moreover, as the number of quality students dwindled over the passage of time, pass rate also plunged dramatically.

Just imagine the predicament of the colleges here, in one side the students of the subsequent batches are coming more dull and blunt where as the syllabus has been being modified every third year and every new syllabus is tougher than its previous versions. So, can you guess the outcome? Yes, rapidly falling over all pass rate and the fall of the ranks of these poor colleges. The cascading effects of these events are the sharp fall of the revenue earned by these colleges which in turn makes them unable to pay good salary to its employees which again becomes the cause of mass exodus of the good teachers to the cash rich colleges of Greater Noida and as a result the survival of these colleges gradually becomes tougher. It's a vicious trap and none of the colleges know how to deal with the situation.

MOCK QUESTION PAPER: APPLIED THERMODYNAMICS (2 units only)

                                                                   Paper Code: EME-402

B.  Tech - ME

(SEM.IV) Sessional Examination, 2011 – 12

Applied Thermodynamics

Time:   3hrs                        Total Marks:  100

    Note:   (1)           Attempt all questions.

         (2)  Be precise in your answer.

SECTION-A:

Q.1: Answer the following questions as per the instructions.           

2X10=30

 (i) What is the importance of feed pump in steam engine?

(ii) What is reversible adiabatic process?

(iii) Explain the term isothermal compressibility?

(iv) What is missing quantity?

(v) What is Work Ratio in Carnot vapour cycle?

(vi) Explain the term “Specific steam consumption.”

(vii) What is thermal efficiency of a steam engine?

(viii) What is indicated power?

(ix) What is mean effective pressure of a steam engine?

(x) What is inversion temperature?

SECTION-B:

Q.2: Answer any three parts of the followings:     

                                                                                                               3X10=30

a) Derive the Tds equations.

b) Derive the expressions of mass discharge of steam through a Nozzle.

c) A single cylinder double acting steam engine is supplied with dry and saturated steam at 11.5 bar and exhaust occur at 1.1 bar. The cut-off occurs at 40% of the stroke. If the stroke equals 1.25 times the cylinder bore and engine develops 60 kW at 90 rev/min. Determine the bore and the stroke of the engine. (Assume hyperbolic expansion and diagram factor of 0.79.)

Also calculate the theoretical steam consumption

d) Dry saturated steam enters a steam nozzle at a pressure of 12 bar and is discharged at a pressure of 1.5 bar. If the dryness factor of the discharged steam is 0.95, what would be the final velocity of the steam? Neglect initial velocity of steam.

If 12% heat drop is lost in friction, find the % reduction in the final velocity.

SECTION C:
Q.3: Answer any two parts of the following: 

                                                                                         5X2=10

a) Explain the term “Joule-Thomson coefficient.”

b) With proper diagrams explain the term nozzle efficiency.

c) Explain the Clausius Clapeyron equation. Also write their field of application.

Q.4: Answer any one part of the following:   

                                                                                           1X10=10

a) Explain the effect of velocity and pressure in the flow of a nozzle. What is a choked flow? Also explain the concept of critical pressure in isentropic flow through nozzle.

b) Steam at a pressure of 20 bar, 250°C expands in a convergent-divergent nozzle up to the exit pressure of 2 bar. Assuming a nozzle efficiency of 0.94 for supersaturated flow up to the throat and nozzle efficiency as 90%, find (i) velocity at throat, (ii) mass flow rate if the throat diameter is 1 cm and (iii) velocity and diameter of the nozzle.

Q.5: Answer any three questions: 

                                                                         3X10=30

a) Derive the Maxwell’s Equations

b) Prove that C_p - C_V = -T(∂V/∂T)_p²(∂p/∂V)_T.

c) Steam at a pressure of 10 bar, dry saturated enters the nozzle when exit pressure is 0.3 bar. The nozzle efficiency for the convergent position is 96% and that of the divergent portion is 92%. The throat diameter for each nozzle is 6 mm. Find the mass flow rate of steam and the exit diameter required.

d) Air enters a nozzle at 5 bar, 350°C and comes out at 0.95 bar. The efficiency of expansion through the nozzle is 92%. If the mass flow rate of air is 1.5 kg/s, determine the exit diameter of the nozzle and velocity of air at exit.