Fuels And Combustion

Q.No.1. a) Give the classification of fuels with suitable examples. b) What is metallurgical coke? Describe any one method of manufacturing metallurgical coke.

Answer: a) Fuels have been classified according to their: 1. Occurrence (and preparation), and 2. The state of aggregation. According to the first classification, we have: A. natural or primary fuels, which are found in nature as such e.g., wood, peat, coal, petroleum, natural gas etc. B. artificial or secondary fuels are those which are prepared from the primary fuels. For example, charcoal, coke, kerosene, diesel, petrol, coal gas, oil gas, producer gas, blast furnace gas, etc. The second classification is based upon their state of aggregation like: a) Solid fuels; b) Liquid fuels, and c) Gaseous fuels. Type of fuel Solid Liquid Gaseous

Natural or primary Artificial or secondary Wood, peat, lignite, dung, Char coal, coke etc. bituminous coal and anthracite coal Crude oil Petrol, diesel and various other fractions of petroleum Natural gas Coal gas, oil gas, bio gas, water gas etc.

b) Good coke for metallurgy should possess the following characteristics: i. Purity: It should have moisture, ash, sulphur and phosphorous contents as low as possible. ii. Porosity: Coke should be porous so that oxygen easily come in contact with carbon of coke, there by helping in complete combustion at a high rate. iii. Strength: The coke should be quite compact, hard, strong to with stand dropping abrasion as well as the pressure of the over burden in the furnace. iv. Size: the size of metallurgical coke should be neither too big nor too small. v. Coke should be cheap and easily available. vi. Coke should burn easily. vii. The calorific value of coke should high. viii. The coke used for water gas manufacturing must be reactive to steam. Manufacture of coke by Beehive oven: The oven is dome-shaped structure of bricks. There are two openings, one at the top for the charging of coal and the other is one side for air supply as well as for coke discharge. These openings can be opened or clured as desired. Coal is charged through the top opening and leveled evenly to produce a layer about 0.6m deep. Some air is supplied in an the coal ignited. The volatile matter escapes or burns inside the partially closed side door. Combustion is allowed to proceed in a gradually diminishing supply of air so that slow carbonization from the top layer to the bottom, takes place completion of which takes about 3-4 days. When the carbonization is complete, the hot coke is quenched with water raked out through the side door leaving the oven hot enough to start the carbonization of next change batch. They yield of coke is about 80% of the coal charged and averages about 5-6 tons of coke per firing. To make the process economical many such ovens are worked in series so that the waste heat of the outgoing gases is utilized for heating purposes.

a) What is coal? How it is formed? Discuss the qualities of a good coal.

b) Give the classification of coal and explain each class with example.

Ans. a) Coal is highly carbonaceous matter formed by the alternation of vegetable matter. Its formation can be explained by the following theories. i) Insitu theory: as per this theory coal formed in place of vegetation. ii) Drift theory: according to this theory, trees when uprooted due climating conditions, in the deeper parts of the soil under transformation to coal due to high temperature, pressure absence of oxygen and presence of bacteria. Ans b) Coal is classified based on the carbon content. The following is sequence of conversion. Peat  Lignite  Bituminous  Anthracite  Moisture content, H, O, S, N, Volatile matter Carbon content, calorific value, hardness  Peat: Brown, fibrous, jelly like mass. It is the first stage of coalification. Uneconomical fuel. Contains 80-90% of H2O. Composition C = 57%, H= 6%, O = 35%, ash 2.5 to 6%. Calorific value = 5400 kcal/kg. Lignite: (Brown coal) soft, brown, colored lowest rank coal moisture content is 20 to 60%. Composition : C = 60%, O = 20%, Calorific value = 6,500 to 7,100 k.cal/kg Bituminous coal (common coal) Black pr coney colored. It has laminated structure it is sub classified based on carbon content. Bituminous coal: Bonded structure, % of C = 78 to 90%, VM = 20 to 45% , CV = 8000 to 8500 kcal/kg. Anthracite: Highest rank of coal. % of C = 98 % has lowest volatile matter hardest, dense, lustrous. CV = 8650 to 8700 kg. 6. a) With a neat diagram describe the orsat gas analysis method. What are the special precautions to be taken in the measurement.

b) Define calorific values of a fuel. Distinguish gross and net calorific value.

Ans:: a) Orsat’s apparatus: It consists os a water – jacketed measuring burette, connected in series to a set of three absorption bulbs, through stop cocks. The other end is provided with a three way stop cock, the free end of which is further connected to a U – tube packed with glass wool (for avoiding the incoming of any smoke particles, etc.) The graduated burette is surrounded by a water jacket to keep the temperature constant of gas during the experiment. The lower end of the burette in connected to a water reservoir by means of along rubber tubing. The absorption bulbs are usually filled with glass tubes, so that the surface area of contact between the gas and the solution is increased. The absorption bulbs have solutions for the absorption of CO2, O2 and CO respectively. First bulb has potassium hydroxide solution (250 g KOH in 500ml of boiled distilled water), and it absorbs only CO2. The second bulb has solution of alkaline pyrogallic acid (25 g pyrogallic acid + 200g KOH in 500 ml of distilled water) and it can absorb CO2 and O2. The third bulb contains ammonium cuprous chloride (100g cuprous chloride + 125ml liquor ammonia + 375 ml of water) and it can absorb CO2, O2 and CO.

Orsat’s gas analysis apparatus Hence, it is necessary that the flue gas it passed first through potassium hydroxide bulb, where CO2 is absorbed, then through alkaline pyrogallic acid bulb, when only O2 will be absorbed (because CO2 has already been removed) and finally through ammonical cuprous chloride bulb, where only CO will be absorbed. Working: Step 1: To start with, the whole apparatus is thoroughly cleaned, stoppers greased and then tested for air tightness. The absorption bulbs are filled with their respective solutions to level just below their rubber connections. Their stop cocks are then closed. The jacket and leveling reservoir are filled with water. There three way stop cock is opened to the atmosphere and reservoir is raised, till the burette is completely filled with water and air is excluded from the burette. The three way stop cocks is now connected to the flue gas supply and the reservoir is lowered to draw in the gas, to be analyzed, in the burette. However, the sample gas mixed with some air present in the apparatus. So the three way stop cock is opened to the atmosphere, and the gas expelled out by raising the reservoir. This process of sucking and exhausting of gas is repeated 3-4 times, so as to expel the air from the capillary connecting tubes etc. Finally, gas is sucked in the burette and the volume of the flue gas is adjusted to 100ml at atmospheric pressure. For adjusting final volume, the three way stop cock is opened to atmosphere and the reservoir is carefully raised, till the level of water in it is the same as in the burette, which stands at 100ml mark. The three ways stop cock is then closed. Step 2: The stopper of the absorption bulb, containing caustic potash solution, is opened and all the gas is forced into the bulb by raising the water reservoir. The gas is again sent to the burette. This process is repeated several times to ensure complete absorption of CO2 [KOH solution]. The unabsorbed gas is finally taken back to the burette, till the level of solution in the CO2 absorption bulb stands at the fixed mark and then, its stop cock is closed. The levels of water in the burette and reservoir are equalized and the volume of residual gas is noted. The decrease in volume gives the volume of CO2 in 100ml of the gas sample. Step 3: The volumes of O2 and CO are similarly determined by passing the remaining gas through alkaline pyroggallic acid bulb and ammonical cuprous chloride bulb respectively. The gas remaining in burette after absorption of CO2, O2 and CO is taken as nitrogen. 6.b) Calorific value of w fuel may be defined as “the total quantity of heat liberated, when a unit mass (or volume) of a fuel is burnt completely”. Or

“Calorific value is the amount of heat liberated by the complete combustion of a unit weight of the fuel and in usually expressed as cal gm-1 or kcal gm-1 or Btu Differences between gross calorific value and net calorific value: Gross calorific value is the heat liberated when a unit quantity of fuel is completely burnt and the products of combustion are cooled to room temperature. This heat includes the latest heat of condensation of water. Because when a fuel containing hydrogen is burnt, the hydrogen present is converted to steam. As the products of combustion are cooled to room temperature, the steam gets condensed into water and the latent heat is evolved. Thus the latent heat of condensation of steam, so liberated, is included in the gross calorific value. Net calorific value is the gross calorific value excluding the latent heat of condensation of water (the weight of water formed is nine times the weight of hydrogen in the fuel). Therefore, Net calorific value = Gross calorific value – (Mass of hydrogen per weight of fuel burnt x 9 x latent heat of vaporization of water of water). Latent heat of steam is 587 kcal/g. H Net calorific value = Gross calorific value – 9 x ---- x 587 100 = Gross calorific value – 52.83 x %H Where % H = percentage of hydrogen. The gross and net calorific values of coal can be calculated by bomb calorimeter. A known mass (about 0.5 to 1.08) of the given coal is taken in a crucible. The crucibles is then supported over the ring. A fine magnesium wire touching the fuel sample is the stretched across the electrodes. The bomb lid is tightly screwed and bomb filled with oxygen to 25 atmospheres pressure. The bomb is than into copper calorimeter, containing a known mass of water is noted. The electrodes are than connected to 6 volt battery and circuit completed. The sample burns and heat is liberated. Uniform stirring of water is continued and the maximum temperature attained is recorded. Calculations: Let mass of fuel sample taken in crucible = x g Mass of water in the calorimeter =Wg Water equivalent in g of calorimeter + Stirrer + thermometer + bomb, etc. = w g Initial temperature of water in calorie meter = T10C Final temperature of water in calorie meter = T20C Higher calorific value of fuel = L cal/g Heat liberated by burning of fuel = xL and heat absorbed by water + apparatus = (W + w) (T2 – T1) But heat liberated by the fuel = Heat absorbed by water and apparatus. xL = (W + w) (T2 – T1) or HCV of fuel (Gross calorific value), (W+ w) (T2 – T1) (GCV)= L = ------------------------ cal/g …………..(1) x If ‘H’ is the percentage of hydrogen in fuel 9H ---- g = mass of H2O from 1 gm of fuel = 0.09 Hg. 100 Heat taken by water in forming steam = 0.09 H x 587 cal.

[latent heat of steam = 587 cal/g] LVC (or Net calorific value) = HVC – latent heat of water formed = (L – 0.09 H x 587 (cal/g)) ……..(2)

6. a) How calorific value of a gaseous fuel is determined by Junker’s gas calorimeter. Describe the experiment with a neat diagram.

Ans: a) Calorific value of a fuel may be defined as “the total quantity of heat liberated, when a unit mass (or volume) of a fuel is burnt completely”. Calorific value of gaseous fuel can be determined by using Junker’s calorimeter, it consists of a vertical cylindrical combustion chamber where combustion of gaseous fuel can be carried out with the help of Bunsen burner. The supply of gaseous fuel is regulated with the help of pressure governer. The volume of gas, flowing in a particular time, is measured with the help of gasometer. The combustion chamber is surrounded by an annular water space. Inside the outer flues, heat exchanger coil are also fitted. Radiative and convective heat loss from the calorimeter is prevented with the help of outer jacket which is chromium plated. Moreover, the outer jacket contains air which is very good heat insulator. Around the combustion chamber, there is an annular space where water is made to circulate. At the appropriate places there are the openings where thermometers are placed for measuring the temperatures of the inlet and outlet water. Junker’s gas calorimeter A known volume of gas is burned in excess of air at a constant rate in combustion chamber in such a manner that all the heat produced is absorbed in water. Water is flowing at a constant rate in annular space around the combustion chamber. The increase in the temperature of the water is measured and the heat evolved from the burning of the gas can be readily calculated. The weight of water flowing is also recorded for the calculation of calorific value of gaseous fuel. Let V = Volume of gas burnt in certain time t at S.T.P. T1 = Temperature of incoming water, T2 = Temperature of outgoing water, W = Weight of water collected in that time t. W(T2-T1) Then, Higher calorific value (HCV)= ------------- kcal/m3 V Now suppose, m = mass of steam condensed in certain time t in graduated cylinder from V m3 of gas. And as latent heat of steam = 587 kcal/kg. m Thus, lower calorific value (LCV) = [HCV - ---- x 587] kcal/m3 V 6.b) Mass of cooling water = 25, T2 = 30, T1 = 25, V = 0.09m3; W-(T2-T1) 25(30-25) GCV = L = ------------ = ------------ = 1388.88 V 0.09 M = 0.02 kg (water condensed) V = 0.09 m3 m 0.02 NCV = L - --- x 587 = 1388.88 - ----- x 587 = 1258.43 kcal/m3 V 0.09 a) What are Octane number? Explain how the molecular structure will affect the octane number.

b) What is leaded petrol? Discuss its advantages and disadvantages. c) Describe the fractional distillation of petroleum.

Ans: a) The resistance offered by gasoline to knocking cannot be defined in absolute terms. It is generally expressed on an arbitrary scale known as Octane rating. In an internal combustion engine, a mixture of gasoline vapor and air is used as a fuel. After the initiation of the combustion reaction, by spark in the cylinder, the flame should spread rapidly and smoothly through the gaseous mixture, thereby the expanding gas drives the piston down the cylinder. The ratio of the gaseous volume in the cylinder at the end of the suction-stroke to the volume at the end of compression ratio. The efficiency of an internal combustion engine increases with the compression ratio, which is dependent on the nature of the constituents present in the gasoline used. In certain circumstances (due to the presence of some constituents in the gasoline used), the rate of oxidation becomes so great that the last portion of the fuel air mixture gets ignited instantaneously, producing an explosive violence, known as knocking. The knocking results in loss of efficiency. Chemical structure and knocking: The tendency of fuel constituents to knock in the following order. Straight – chain paraffins > Branched- chain paraffins (i.e.,iso paraffins)> Olefines> Cycloparaffins (i.e.,naphthalenes)> aromatics. Thus, olefins of the same carbon chain length possess better anti knock properties than the corresponding paraffins and so on. Octane rating: It has been found that n-heptane, H H H H H H H H–C–C–C–C–C–C–C-H H H H H H H H Knocks very badly and hence, its anti-knock value has arbitrarily been given zero. On the other hand, isooctane (2: 2: 4 – trimethyl pentane) H CH3 H CH3 H H– C – C – C – C – C - H H CH3 H H H Gives very little knocking, so its anti-knock value has been given as ‘100’. Thus, octane number (or rating) of a gasoline (or any other internal combustion engine fuel) is the percentage of isooctane in a mixture of isooctane and n-heptane, which matches the fuel under test in knocking characteristics. In this way, an “80-octane” fuel is one which has the same combustion characteristics as a 80:20 mixture of isooctane and n-heptane. Improvement of anti-knock characteristics of a fuel: The octane number of many fuels can be raised by the addition of sum extremely poisonous materials as tetra ethyl lead, (C2H5)4 Pb or TEL and diethyl telluride, (C2H5)2 Te. In motor spirit (or motor fuel), about 0.5 ml and in aviation fuels, about 1.0 to 1.5 ml of TEL is added per lit of petrol. According to the most accepted theory, TEL is converted into a cloud of finely divided lead oxide particles in the cylinder and these particles react with any hydrocarbon peroxide molecules formed, thereby slowing down the chain oxidation reaction and thus, decreasing the changes of any early detonation. However, deposit of lead oxide is harmful to the engine life. Consequently, in order to help the simultaneous elimination of lead oxide. H H–C-H H–C-H H H H

H

H – C - C - Pb – C – C -H H H H H H–C-H H–C-H H Tetra ethyl lead (TEL) Formed, from the engine, a small amount of ethylene dibromide is also added to petrol. Ethylene dibromide removes lead oxide as volatile lead bromide along with the exhaust gases. The presence of sulphur compounds in petrol reduces the effectiveness of the TEL. 5.b) The variety of petrol in which tetra ethyl lead is added, it is a leaded petrol. C2H5 C2H5 – Pb – C2H5 Tetra ethyl lead (TEL) C2H5 Advantages: Usually petrol with low octane number is not a good quality petrol. It often knocks (i.e., produces huge noise due to improper combustion). As a result of knocking, petrol is wasted, the energy produced cannot be used in a proper way. When tetra ethyl lead is added, it prevents knocking, there by saves money and energy. Usually 1 to 1.5 ml of TEL is added per 1lit of petrol. The mechanism of action is as follows: First TEL will be transformed into finely divided particles of PbO which looks like a cloud. This takes place in the cylinder. Then the PbO particles react with hydrocarbon peroxide molecules formed, thus slowing down the oxidation process and prevents early detonation. Thus either knocking may be stopped or greatly reduced. Disadvantages: Deposits of PbO are harmful to engine. So PbO must be eliminated from the engine. For this purpose, little amount of ethylene dibromide is added to petrol. It converts the harmful PbO to volatile PbBr2 and eliminated through exhaust. Presence of any sulphur compounds reduces the efficiency of TEL. 5.c) Refining of petroleum: It involves three steps. Step1: Separation of water (Cotree’s process) crude oil forms an emulsion with oil. So, when crude oil is allowed to flow between two highly charged electrodes, colloidal water droplets coalesie to form large droplets is separate out. Step2: Removal of sulphur compounds: crude oil containing sulphur compounds when treated with copper oxide, the former are removed as copper sulphide (by filtration). Step3: Fractional distillation: Heating of crude oil around 4000C in an iron retort, produces hot vapor which is allowed to pass through fractionating column. It is a tall cylindrical tower containing a number of horizontal stainless trays at short distances and are provided with small chimney covered with loose cap. As the vapors go up they get cooled gradually and fractional condensation takes place. Higher boiling fraction condenses first later the lower boiling fractions. 6.

a) The analysis of flue gases was found to contain the following percentage by volume CO2 = 13%, O2 = 6%.

Find the air per Kg of coal if the carbon content of coal is 85%.

Solution: 100gms of coal contains 85gms of carbon So 1kg contains 850gm 850 O2 required = ----- x 32 12

850 100 Air required = ----- x 32 x ---- = 9855gm/kg 12 23 6.b) A producer gas has the following composition by volume. H2 = 10.4%, CH4 = 35%, CO = 25%, CO2 = 10.8%, N2 = 50.3%.

Calculate the quantity of air required per m3 of gas. If 20% excess air is supplied, find the percentage composition of the products of combustion.

Solution: Combustion reaction

Vol. of O2 required 0.104 x ½ = 0.052m3 0.35 x 2 = 0.7m3 0.25 x ½ = 0.125m3 Total = 0.877m3

H2 + ½ O2  H2O CH4 + 2O2  CO2 + 2H2O CO + ½ O2  CO2 100 Volume of air required = 0.877 x --- = 3.81 m3 23 120 Volume of air supplied = 3.81 x ---- = 4.5 100 4500 In moles = ----- = 200.9 moles 22.4 In gms = 200.9 x 28.94 = 5813.89 = 5.8kg. Percentage of products of combustion. Reaction

Products

Percentage

H2 + ½ O2  H2O

18 0.4 x --- = 3.78 kg 2

CH4 + 2O2  CO2 + 2H2O

80 0.35 x --- = 1.75 16 44 0.25 x --- = 0.39 28 0.108 kg

% age of H2 = 7.78 x 100 ------------ = 59% 6.4 CH4 = 27%

CO + ½ O2  CO2 CO2 N2

77 0.5kg x ---- = 0.385 100 Total = 6.4kg

CO = 6% CO2 = 1.68% N2 = 6%

a) What is knocking of petrol ? Name any one anti knocking agent added and explain how it works.

b) What is cracking? Why it is done? How gasoline is obtained from moving bed catalytic cracking? What are the

advantages of this method over fixed bed method.

Ans: a) The resistance offered by gasoline to knocking cannot be defined in absolute terms. It is generally expressed on an arbitrary scale known as Octane rating.

In an internal combustion engine, a mixture of gasoline vapor and air is used as a fuel. After the initiation of the combustion reaction, by spark in the cylinder, the flame should spread rapidly and smoothly through the gaseous mixture, thereby the expanding gas drives the piston down the cylinder. The ratio of the gaseous volume in the cylinder at the end of the suction-stroke to the volume at the end of compression ratio. The efficiency of an internal combustion engine increases with the compression ratio, which is dependent on the nature of the constituents present in the gasoline used. In certain circumstances (due to the presence of some constituents in the gasoline used), the rate of oxidation becomes so great that the last portion of the fuel air mixture gets ignited instantaneously, producing an explosive violence, known as knocking. The knocking results in loss of efficiency. Chemical structure and knocking: The tendency of fuel constituents to knock in the following order. Straight – chain paraffins > Branched- chain paraffins (i.e.,iso paraffins)> Olefines> Cycloparaffins (i.e.,naphthalenes)> aromatics. Thus, olefins of the same carbon chain length possess better anti knock properties than the corresponding paraffins and so on. Octane rating: It has been found that n-heptane, H H H H H H H H–C–C–C–C–C–C–C-H H H H H H H H Knocks very badly and hence, its anti-knock value has arbitrarily been given zero. On the other hand, isooctane (2: 2: 4 – trimethyl pentane) H CH3 H CH3 H H– C – C – C – C – C - H H CH3 H H H Gives very little knocking, so its anti-knock value has been given as ‘100’. Thus, octane number (or rating) of a gasoline (or any other internal combustion engine fuel) is the percentage of isooctane in a mixture of isooctane and n-heptane, which matches the fuel under test in knocking characteristics. In this way, an “80-octane” fuel is one which has the same combustion characteristics as a 80:20 mixture of isooctane and n-heptane. Improvement of anti-knock characteristics of a fuel: The octane number of many fuels can be raised by the addition of sum extremely poisonous materials as tetra ethyl lead, (C2H5)4 Pb or TEL and diethyl telluride, (C2H5)2 Te. In motor spirit (or motor fuel), about 0.5 ml and in aviation fuels, about 1.0 to 1.5 ml of TEL is added per lit of petrol. According to the most accepted theory, TEL is converted into a cloud of finely divided lead oxide particles in the cylinder and these particles react with any hydrocarbon peroxide molecules formed, thereby slowing down the chain oxidation reaction and thus, decreasing the changes of any early detonation. However, deposit of lead oxide is harmful to the engine life. Consequently, in order to help the simultaneous elimination of lead oxide. H H–C-H H–C-H H H H H H – C - C - Pb – C – C -H H H H H H–C-H H–C-H H

Tetra ethyl lead (TEL) Formed, from the engine, a small amount of ethylene dibromide is also added to petrol. Ethylene dibromide removes lead oxide as volatile lead bromide along with the exhaust gases. The presence of sulphur compounds in petrol reduces the effectiveness of the TEL. 5. b) Decomposition of larger hydrocarbon molecules to smaller molecules is cracking. Cracking

Ex.

C10H12

Decane

 C5H12 + C5H10 Pentane

Pentene

Cracking of two types: a. Thermal cracking of pyrolysis: If the cracking take place at high temperature then it is thermal cracking. It may take place at liquid phase or at vapour phase. The liquid phase cracking takes place at 4750C to 5300C at a pressure 100kg/cm2. While the vapor phase cracking occurs at 600 to 6500C at a low pressure of 10 to 20 kg/cm2 b. Catalytic cracking: If the cracking takes place due to the presence of catalyst than it is named as catalytic cracking. Al2O3  Cr2O3

Ex. n-heptane





Cyclization n-heptane

M.W= 100

Methyl Cyclohexane Cracking n-heptane

Ethyl benzene M.W = 90

Catalytic cracking may be fixed bed type or moving bed type. Octane number: It is the percentage of isooctane in a mixture of isooctane and n-heptane which matches the fuel under test in knocking characteristics. CH3 CH3 – C – CH2 – CH2 – CH3 CH3 – (CH2)5- CH3 CH3 CH3 2,24- trimethyl pentane n-heptane (isooctane) octane number 100 (good fuel) octane number zero (bad fuel) With the help of octane number purity of petrol can be known cetane number, it indicates the purity of diesel oil. The percentage of cetane in a mixture of cetane and n-hexadecane which has same ignition characteristics as the diesel fuel under test is certain number. CH3 (CH2)14- CH3 2-methyl naphthalene Cetane number = 0 (bad fuel)

n-hexadecane Cetane number = 100 (good fuel)

In the moving bed cracking, very finely divided catalyst which behaves like a fluid is circulated in the gas stream. The vapors of raw material mix with catalyst and get cracked. On the top of the reactor, there will be a part called cyclone, which can separate the gases with catalyst. It allows only gas to pass on but not catalyst. The gases will be fractioned. Catalyst after accumulation becomes heavy and settles down. From here, it will be sent to regenerator maintained 6000C through air blast. Here the catalyst will

be reactivated and further catalyses the cracking process. This is the advantages of moving bed cracking over fixed bed cracking.

6. a) Describe the method of determination of calorific value of a gaseous fuel by Junker’s gas calorimeter.

b) On burning 0.85 gm of a solid fuel in a Bomb calorimeter, the temperature of 3500 gm of water increased from 25.60C to

28.20C. Water equivalent of calorimeter and Latent heat of steam are 385 gm and 587 Cal/gm respectively. If the fuel contains 0.7% of hydrogen, calculate the gross and net calorific value. Express your answer in kj/kg.

Ans: a) Calorific value of a fuel may be defined as “the total quantity of heat liberated, when a unit mass (or volume) of a fuel is burnt completely”. Calorific value of gaseous fuel can be determined by using Junker’s calorimeter, it consists of a vertical cylindrical combustion chamber where combustion of gaseous fuel can be carried out with the help of Bunsen burner. The supply of gaseous fuel is regulated with the help of pressure governer. The volume of gas, flowing in a particular time, is measured with the help of gasometer. The combustion chamber is surrounded by an annular water space. Inside the outer flues, heat exchanger coil are also fitted. Radioactive and convective heat loss from the calorimeter is prevented with the help of outer jacket which is chromium plated. Moreover, the outer jacket contains air which is very good heat insulator. Around the combustion chamber, there is an annular space where water is made to circulate. At the appropriate places there are the openings where thermometers are placed for measuring the temperatures of the inlet and outlet water. Junker’s gas calorimeter A known volume of gas is burned in excess of air at a constant rate in combustion chamber in such a manner that all the heat produced is absorbed in water. Water is flowing at a constant rate in annular space around the combustion chamber. The increase in the temperature of the water is measured and the heat evolved from the burning of the gas can be readily calculated. The weight of water flowing is also recorded for the calculation of calorific value of gaseous fuel. Let V = Volume of gas burnt in certain time t at S.T.P. T1 = Temperature of incoming water, T2 = Temperature of outgoing water, W = Weight of water collected in that time t. W(T2-T1) Then, Higher calorific value (HCV)= ------------- kcal/m3 V Now suppose, m = mass of steam condensed in certain time t in graduated cylinder from V m3 of gas. And as latent heat of steam = 587 kcal/kg. m Thus, lower calorific value (LCV) = [HCV - ---- x 587] kcal/m3 V 6.b) W = Mass of H2O w = Water equivalent t2 = final temperature t1 = Initial temperature X = mass of coal (W+ w) (t2- t1) (3500-385) (28.2-25.6) (3115) (2.6)

GCV = -------------------- = --------------------------X 0.85

=

-------------- = 95.28 cal/gm 0.85

LCV = L – 0.09H x 587 = 9528 – 0.09 x 0.7 x 587 = 949 cal/gm = 9.49 kcal/kg. Cetane Number: The knocking characteristics of diesel oil are expressed in terms of cetane number. Cetane, C16H34, is a saturated hydrocarbon which has a very short ignition lag as compared to any commercial diesel fuel. Hence, its cetane number is taken as 100. On the contrary, Alpha-methyl naphthalene. C11H10 has a very long ignition lag as compared to any commercial diesel oil. Hence its cetane number is taken zero. Then the cetane number of diesel oil is defined as the percentage by volume of cetane in a mixture of cetane and Alpha-methyl naphthalene which exactly mathaches in its knocking characteristics with the oil under test. ANALYSIS Proximate analysis involves in the following determinations: 1. Moisture 2. Volatile matter 3. Ash 4. Fixed carbon 1. Moisture: About 1 gram of finely powdered air-dried coal sample is weighed in a crucible. The crucible is placed inside an electric hot air-oven, maintained at 105 to 1100C. The crucible is allowed to remain in oven for 1 hour and then taken out, cooled in a desiccator and weighed. Loss in weight is reported as moisture. Percentage of Moisture = __Loss in weight__ X 100 Weight of coal taken 2. Volatile Matter: The dried sample of coal left in the crucible in (1) is then covered with a lid and placed in an electric furnace or muffle furnace, maintained at 925 + 20C. The crucible is taken out of the oven after 7 minutes of heating. The crucible is cooled first in air, then inside desiccators and weighed again. Loss in weight is reported as volatile matter on percentage-basis. Percentage of volatile matter = Loss_in _weight_due_to_removal_of_volatile_matter X 100 Weight of coal sample taken 3. Ash: The residual coal in the crucible in (2) is then heated without lid in a muffle furnace at 700 + 50 C for ½ hour. The crucible is then taken out, cooled first in air, then in desiccators and weighed. Hearing, cooling and weighing is repeated, till a constant weight is obtained. The residue is reported as ash on percentage-basis. Thus, percentage of ash = __Weight of ash left__ X 100 Weight of coal taken 4. Fixed carbon: Percentage of fixed carbon = 100 - % of (Moisture + Volatile matter + ash)

Importance of proximate analysis: Proximate analysis provides following valuable information’s in assessing the quality of coal. 1. Moisture: Moisture is coal evaporates during the burning of coal and it takes some of the liberated heat in the form of latent heat of evaporation. Therefore, moisture lowers the effective calorific value of coal. Moverover, it quenches the fire in the furnace, hence, lesser, the moisture

content, better the quality of coal as a fuel. However, presence of moisture, up to 10%, produces a more uniform fuel-bed and less of “fly-ash”. 2. Volatile matter: a high volatile matter content means that a high proportion of fuel will distil over as gas or vapour, a large proportion of which escapes unburnt, So, higher volatile content in coal s undesirable. A high volatile matter containing coal burns with a long flame, high smoke and has low calorific value. Hence, lesser the volatile matter, better the rank of the coal. A high volatile matter content means that high-proportion of fuel will be distilled and burned as a gas or vapour. The volatile matter present in the coal ay be as high as 50%. The volatile matter present in the coal may be combustible gases or non-combustible gases. The presence of noncombustible gases is always undesirable, as they do not add to heat value, but increases the volume of the furnace required. Moverover, the volatile matter affects the furnace volume and arrangement of heating space. Thus, a furnace with small combustion volume or of short flame travel is not suitable for burning high volatile coals at high rates of combustion, since a large proportion of volatile matter will escape unburnt. On the other hand, burning of low volatile coals necessarily requires forced draught and the intensity of draught increases with the decrease in volatile matter percentage. Volatile matter content is of special significance in coal gas manufacture and in carbonization plants, particularly when by-product recovery is the main object. Thus, high-volatile matter containing coals do not cake well; whereas medium-volatile matter content coals are capable of yielding hard and strong coke on carbonization. On the other hand, low-volatile matter containing coals do not cake at all and consequently, they are totally unsuitable for coke making. 3. Ash: Ash is a useless, non-combustible matter, which reduces the calorific value of coal. Moverver, ash causes the hindrance to the flow of air and heat, thereby lowering the temperature. Also, it often causes trouble during firing by forming clinkers, which block the interspaces of the grate, on which coal is being burnt. This in-turn causes obstruction to air supply; thereby the burning of coal becomes irregular. Hence, lower the ash content, better the quality of coal. The presence of ash also increases transporting, handling and storage costs. It also involves additional cost in ash disposal. The presence of ash also causes early wear of furnace walls, burning of apparatus and feeding mechanism. 4. Fixed carbon: Higher the percentage of fixed carbon, greater is its calorific and better the quality coal. Greater the percentage of fixed carbon, smaller is the percentage of volatile matter. This also represents the quantity of carbon that can be burnt by a primary current of air drawn through the hot bed of a fuel. Hence, high percentage of fixed carbon is desirable. The percentage of fixed carbon helps in designing the furnace and the shape of the fire-box, because it is the fixed carbon that burns in the solid state.

Ultimate analysis involves in the following determinations: 1. Carbon and Hydrogen: About 1 to 2 gram of accurately weighed coal sample is burnt in a current of oxygen in a combustion apparatus. C and H of the coal are converted into CO2 and H2O respectively. The gaseous products of combustion are absorbed respectively in KOH and CaCl2 tubes of known weights. The increase in weights of these are then determined. C + 02  CO2 : H2 + ½ O2  H2O 2KOH + CO2  K2CO3 + H2O CaCl2 + 7 H2O  CaCl2.7H2O

Percentage of C = Increase in weight of KOH tube X 12 X 100 Weight of Coal sample taken X 44 And

Percentage of H = Increase in weight of CaCl2 tube X 2 X 100 Weight of Coal sample taken X 18 2. Nitrogen: About 1 gram of accurately weighed powdered coal is heated with concentrated H2SO4 along with K2SO4 (catalyst) in a long-necked Kjeldahl’s flask. After the solution becomes clear, it is treated with excess of KOH and the liberated ammonia is distilled over and absorbed in a known volume of standard acid solution. The unused acid is then determined by back titration with standard NaOH solution. From the volume of acid used by ammonia liberated, the percentage of N in coal is calculated as follows: Percentage of N = Volume of BaSo4 obtained X 32 X_100 Weight of coal taken 3. Sulphur: Sulphur is determined from the washings obtained from the known mass of coal, used in bomb calorimeter for determination of a calorific value. During this determination, S is converted in to Sulphate. The washings are treated with Barium chloride solution, when Barium sulphate is precipitated. This precipitate is filtered, washed and heated to constant weight. Percentage of Sulphur = __Weight of BaSO4 obtained X 32 X 100_ Weight of coal sample taken in bomb X 233 4.Ash: ash determination is carried out as in proximate analysis. 5.Oxygen: It is obtained by difference. Percentage of Oxygen = 100 – percentage of ( C + H + S + N + Ash)

Importance of ultimate analysis: a. Carbon and Hydrogen: Greater the percentage of carbon and hydrogen, better is the coal in quality and calorific value. However, hydrogen is mostly associated with the volatile mater and hence, it affects the use to which the coal is put. b. Nitrogen: Nitrogen has no calorific value and hence, its presence in coal is underirable. Thus, a good quality coal should have very little Nitrogen content. c. Sulphur: Sulphur, although contributes to the heating value of coal, yet on combustion produces acids like SO2, SO3, which have harmful effects of corroding the equipments and also cause atmospheric pollution. Sulphur is, usually, present to the extent of 0.5 to 0.3% and derived from ores like iron, pyrites, gypsum, etc., mines along with the coal. Presence of sulphur is highly undesirable in coal to be used for making coke for iron industry. Since it is transferred to the iron metal and badly affects the quality and properties of steel. Moverover, oxides of sulphur pollute the atmosphere and leads to corrosion. d. Oxygen; Oxygen content decreases the calorific value of coal. High oxygen-content coals are characterized by high inherent moisture, low calorific value, and low coking power. Moverover, oxygen is an combined form with hydrogen in coal and thus, hydrogen available for combustion is lesser than actual one. An increase in 1% oxygen

content decreases the calorific value by about 1.7% and hence, oxygen is undesirable. Thus, a good quality coal should have low percentage of oxygen.