7 Combustion The ideal cycles previously considered use fluids which remain unchanged chemically as they pass through the various processes of the cycle, In practical engines and power plants the source of heat is the chemical energy of substances called fuels. This energy is released during the chemical reaction of the fuel with oxygen, The fuel elements combine with oxygen in an oxidation process which is rapid and is accompanied by the evolution of heat. The combustion process takes place in a controlled manner in some form of combustion chamber after initiation of combustion by some means (e.g, in a petrol engine the combustion is started by an electric spark), The most convenient source of oxygen supply is that of the atmosphere which contains oxygen and nitrogen and traces of other gases. Normally no attempt is made to separate out the oxygen from the atmosphere, and the nitrogen, etc. accompanies the oxygen into the combustion chamber. Nitrogen does not oxidize easily and is inert as far as the combustion process is concerned, but it acts as a moderator in that it absorbs some of the heat of combustion and so limits the maximum temperature reached. As combustion proceeds the oxygen is progressively used up and the proportion of nitrogen plus products of combustion to the available oxygen increases. For a given amount of fuel there is a definite amount of oxygen, and therefore ait, which is required for the complete combustion of a given fuel. To ensure complete combustion it is usual to supply air in excess of the amount required for chemically correct combustion. The oxygen not consumed in the reaction passes into the exhaust with the products of combustion, Internal-combustion engines are run on liquid fuels which are grouped as, ‘petrols’ (known as gasoline in the USA), and diesel oils, or gaseous fuels, commonly used in combined heat and power plant; gas turbines are run mainly on kerosene although natural gas is now commonly used. Engines burning solid fuels have been built but are mainly experimental. In the many and diverse applications in industry, solid, liquid, and gaseous fuels are used, Generalization is not possible on the selection of fuels, since the fuel used and its necessary firing equipment depend on the particular application, the practical circumstances, and economic considerations.7A Table 7.1 Relative atomic and molecular ‘masses of some ‘common substances 7.1 Basic chomistry Basic chemistry It is necessary to understand the construction and use of chemical formulae, before combustion problems can be considered. This involves elementary concepts which have been met before by most students, but a brief explanation will be given here. Atoms. Chemical elements cannot be divided indefinitely and the smallest particle which can take part in a chemical change is called an atom. If an atom is split as in a nuclear reaction, the divided atom does not retain the original chemical properties. Molecules. Elements are seldom found to exist naturally as single atoms. Some elements have atoms which exist in pairs, each pair forming a molecule (c.g, oxygen), and the atoms of each molecule are held together by strong inter-atomic forces. The isolation of a molecule of oxygen would be tedious, but possible; the isolation of an atom of oxygen would be a different prospect. The molecules of some substances are formed by the mating up of atoms of different elements. For example, water (which is chemically the same as ice and steam) has a molecule which consists of two atoms of hydrogen and one atom of oxygen. ‘The atoms of different elements have different masses and these values are important when a quantitative analysis is required. The actual masses are infinitesimally small, and the ratios of the masses of atoms are used. These ratios are given by the relative atomic masses quoted on a scale which defines the atomic mass of isotope 12 of carbon as 12 (see Ch. 2, p. 40). The relative ‘atomic mass of a substance is the mass of a single entity of the substance relative to a single entity of carbon-12, Table 7.1 gives the relative atomic masses of some common elements rounded off to give values accurate enough for most, purposes. Element Oxygen Hydrogen Carbon Sulphur Nitrogen Atomic symbol ° H G s N Relative atomic mass 16 1 2 32 4 Molecular grouping H, c s Na Relative molecular ‘mass (rounded) 32 2 12 32 2B Accurate values 31999 2016 12 32.030 28013 Relative molecular masses are based on the relative masses of the atoms which constitute the molecule. In chemical formulae one atom of an element is represented by the symbol for the element, i, an atom of hydrogen is written as H, and other examples are given in Table 7.1. If a substance exists as a molecule containing, say, two atoms, as for hydrogen, it is written as Hy. ‘Two molecules of hydrogen is written as 2H,, etc. Table 7.1 includes relative molecular masses, rounded off, and, for comparison, the accurate values, 77Combustion Table 7.2. Compounds and their relative molecular masses 178 7.2 Relative Compound Formula molecular mass Water, stea H,0 (2x 1) +(1 x 16) Carbon monoxide CO (1x 12)+(1 x 16) Carbon dioxide CO, (1 x 12)+(2 x 16) Sulphur dioxide SO, (1 x 32) +(2 x 16) Methane cH, (1x 12) +4 x 1) Ethane Gig (2x 12)+(6x 1) Propane CHE (3x12 +8x1) n-Butane CiHyo (4x 12) + (10x 1) Ethylene GH (2x 12)+(4« 1) Propylene GH, Bx 12)+(6x 1) n-Pentane CsHi2 (5x 12) + (12x 1) Benzene cH (6 x 12) + (6x 1) Toluene Hy (7x 12) 48x 1) n-Octane CoHye (8x 12) +(18 x 1) Some of the other substances met in combustion work are given in ‘Table 7.2 to illustrate the calculations of the relative molecular mass from the relative atomic masses of the elements. Fuels ‘The most important fuel elements are carbon and hydrogen, and most fuels, consist of these and sometimes a small amount of sulphur. The fuel may contain some oxygen and a small quantity of incombustibles (e.g. water vapour, nitrogen, or ash), Coal is the most important solid fuel and the various types are divided into groups according to their chemical and physical properties. An accurate chemical analysis by mass of the important elements in the fuel is called the ultimate analysis, the elements usually included being carbon, hydrogen, nitrogen, and sulphur. The main groups are shown in Table 7.3, and their ultimate analyses are given, The analyses are typical but may vary from one sample to another within the group, and hence can be taken only as a guide, Another analysis of coal, also shown in Table 7.3 called the proximate analysis, gives the percentages of inherent moisture, volatile matter, and combustible solid (called fixed carbon). The fixed carbon is found as a remainder by deducting the percentages of the other quantities. The volatile matter includes the water derived from the chemical decomposition of the coal (not to be confused with free, or inherent moisture), the combustible gases (e.g. hydrogen, methane, ethane, etc.), and tar (ic. a complex mixture of hydrocarbons and other organic compounds). The procedures for both analyses are given in ref. 7.1; see also ref, 7.2 and a concise treatment in ref. 7.3. Most liquid fuels are hydrocarbons which exist in the liquid phase at atmospheric conditions. Petroleum oils are complex mixtures of sometimes hundreds of different fuels, but the necessary information to the engineer is the relative proportions of carbon, hydrogen, etc. as given by the ultimate analysis, Table 7.4 gives the ultimate analyses of some liquid fuels.Table 7.3 Analysis of solid fuels Table 74 Analyses of liquid fuels Table 75 Analysis by volume of a typical ‘natural gas 7.2 Fuels Ultimate analysis, Percentage by mass of dry fuel Mineral Fuel Rank CH ON S$ ___ matter Anthracite 101 82270 LT O52 Medium-rank coal 401 a8 4904481952 Low-rank coal 902 750 46 107 16 21 60 Coke 90 04 19 7 Proximate analysis on a mineral matter-free basis Percentage by mass of fuel Inherent Volatile Fixed Fuel moisture matter carbon Anrhtacite 2 6 92 Medium-rank coal 3 39 58 Low-rank coal 10 2 48 Fuel Carbon Hydrogen Sulphur Ash, ete 100-octane petrol 85.1 149 01 = Motor petrol 855 144 oul — Benzole 917 80 03 = Kerosene (paraflin) 863 136 o4 = Diesel oil 863 128, 09 - Light fuel oil 862 124 14 = Heavy fuel oil 86.1 118 24 = Residual fuel oil 883 95 12 10 Methane Ethane Propane = Butane trogen Carbon dioxide CHy CH Hy CiHyo Na CO, 92.6% 3.6% 08% 03% 2.6% 01% Gaseous fuels are chemically the simplest of the three groups. The main gaseous fuel in use occurs naturally but other gaseous fuels may be manufactured by the various treatments of coal. Carbon monoxide is an important gaseous fuel which is a constituent of other gas mixtures, and is also a product of the incomplete combustion of carbon. A typical analysis of a natural gas is given in Table 7.5. The table gives the analyses by volume, each constituent having been measured by volume at atmospheric pressure and temperature. The volumetric analysis is the same as the molar analysis (see equation (6.14)). Fuels are tested according to standardized procedures and for further information ref, 7.2 should be consulted. 179Combustion 180 7.3 Combustion equations Proportionate masses of air and fuel enter the combustion chamber where the chemical reaction takes place, and from which the products of combustion pass to the exhaust. By the conservation of mass the mass flow remains constant (ic. total mass of products equals total mass of reactants), but the reactants are chemically different from the products, and the products leave at a higher temperature. The total number of atoms of each element concerned in the combustion remains constant, but the atoms are rearranged into groups having different chemical properties. This information is expressed in the chemical equation which shows (i) the reactants and the products of combustion; (ii) the relative quantities of the reactants and products. The two sides of the equation must be consistent, each having the same number of atoms of each element involved. It should not be assumed that if an equation can be written, that the reaction it represents is inevitable or even possible. For possibility and direction the reaction has to be considered with reference to the Second Law of Thermodynamics. For the present the only concern is known combustion equations. ‘The equation shows the number of molecules of each reactant and product. ‘The amount of substance, introduced in section 2.3, is proportional to the number of molecules, hence the relative numbers of molecules of the reactants and the products give the molar, and therefore the volumetric, analysis of the gaseous constituents. As stated earlier the oxygen supplied for combustion is usually provided by atmospheric air, and it is necessary to use accurate and consistent analyses of air by mass and by volume. It is usual in combustion calculations to take air as 23.3% Oz, 76.7% Ny by mass, and 21% O;, 79% N; by volume. The small traces of other gases in dry air are included in the nitrogen, which is sometimes called ‘atmospheric nitrogen’. Consider the combustion equation for hydrogen: 2H, + 0, +2H,0 (14) This tells us that (i) hydrogen reacts with oxygen to form steam or water; (ii) two molecules of hydrogen react with one molecule of oxygen to give two molecules of steam or water, ie. 2 volumes H, +1 volume O, +2 volumes H,O The H,O may be a liquid or a vapour depending on whether the product has been cooled sufficiently to cause condensation. The proportions by mass are obtained by using relative atomic masses, 2H, + O, +2H,0 therefore 2x (2x 1) + (2 x 16) +2 x {(2 x 1) + 16}7.3. Combustion equations ie. 4kgH, + 32kg Oz > 36 kg H20 or LkgH, + 8kg 0, 9kgH,0 ‘The same proportions are obtained by writing equation (7.1) as Hy + $0, > Hy and this is sometimes done. It will be noted from equation (7.1) that the total volume of the reactants is 2 volumes H, + 1 volume ©, = 3 volumes. The total volume of the product is only 2 volumes. Theres therefore a volumetric contraction on combustion. Since oxygen is accompanied by nitrogen if air is supplied for the combustion, then this nitrogen should be included in the equation. As nitrogen is inert as far as the chemical reaction is concerned, it will appear on both sides of the equation. With 1 kmol of oxygen there are 79 /21 kmol of nitrogen, hence equation (7.1) becomes 19 19 2H, +O, + —N2 > 2H,0 + Nz 7.2} a+ O2 $5 20 +7 (7.2) Similar equations can be found for the combustion of carbon, There are two possibilities to consider: (i) The complete combustion of carbon to carbon dioxide C+0,5C0, (73) and including the nitrogen 79. 79 C+0,+—N, CO, +N 74) FOrt oy ana a Considering the volumes of reactants and products 0 volume C+ 1 volume O, + 2 volumes Ny 9 ~+1 volume CO, + 55 volumes Nz The volume of carbon is written as zero since the volume of a solid is negligible in comparison with that of a gas. By mass 79 12kgC +(2 x 16)kgO. + 2 x 14) kg Ny 9 {12+ (2 x 16)} kg CO, + 57-02 x 14) ke Na ie, 12g C+ 32kg Oz + 105.3 kg Nz +44 kg COz + 105.3 kg Nz 1053 44 1053 2 tke + ekg 0, + ken, +S kg co, + ken, a BO RRO tay ea Dn 181Combustion 182 7.4 (ii) The incomplete combustion of carbon. This occurs when there is an insufficient supply of oxygen to burn the carbon completely to carbon dioxide, ie 2C+.0,4200 5) 79. 79. IC +O, N,>2CO. N; 7.6) 20-40, +5Na +5)Na (76) By mass (2x 12)keC + (2x 16) kg 0, + 202 x 14) kg Ns 7 2(12 + 16)kgCO + Zea x 14) kgN, ie, 24g C + 32kgO, + 105.3 kg Np + 56 ke CO + 105.3 kg Ny 2 105.3 56 1053 kg C + << kg O, + "kg N, > kg CO + 2 ke N or BC+ keO, + ke 3g 800 +S ke, Ifa further supply of oxygen is available then the combustion can continue to completion p p 2CcO0 —N,> —N) VT, +02 + 57N2 +2003 + > 0) By mass, 56 kg CO + 32 kg O; + 105.3 kg Ny + 88 kg CO, + 105.3 kgN, 32 1053 88 105.3 1kg CO + — kg CO, + 2 kg ny, or CO + 56 kB O2 + Se kBNa > ke CO. + ke Ne Stoichiometric air-fuel ratio A stoichiometric mixture of air and fuel is one that contains just sufficient oxygen for the complete combustion of the fuel, A mixture which has an excess of air is termed a weak mixture, and one which has a deficiency of air is termed a rich mixture. The percentage of excess air is given by the following: Percentage excess air _ actual A/F ratio — stoichiometric A/F ratio aa stoichiometric A/F ratio (78) where A denotes air and F denotes fuel. For gaseous fuels the ratios are expressed by volume and for solid and liquid fuels the ratios are expressed by mass. Equation (7.8) gives a positive result when the mixture is weak, and a negative result when the mixture is rich. For boiler plant the mixture is usually greater than 20% weal; for gas turbines it can be as much as 300% weak. Petrol engines have to meet various conditions of load and speed, and operate over a wide range of mixture strengths. The7.5 Example 7.1 Solution 7.8 Exhaust and flue gas analysis, following definition is used: stoichiometric A/F ratio Mixture strength = : actual A/F ratio (79) The working values range between 80% (weak) and 120% (rich) (see section 13.6). ‘Where fuels contain some oxygen (e.g. ethyl alcohol C;H.0) this oxygen is, available for the combustion process, and so the fuel requires a smaller supply of air. Exhaust and flue gas analysis The products of combustion are mainly gaseous. When a sample is taken for analysis it is usually cooled down to a temperature which is below the saturation temperature of the steam present. The steam content is therefore not included in the analysis, which is then quoted as the analysis of the dry products. Since the products are gaseous, it is usual to quote the analysis by volume. An analysis, which includes the steam in the exhaust is called a wet analysis. The following, examples illustrate the principles covered in this chapter up to this point. ‘A sample of dry anthracite has the following composition by mass. C 90%; H 3%; O 2.5%; N 1%; S 0.5%; ash 3% Calculate: (i) the stoichiometric A/F ratio; (ii) the A/F ratio and the dry and wet analysis of the products of combustion by mass and by volume, when 20% excess air is supplied. (j) Each constituent is taken separately and the amount of oxygen required for complete combustion is found from the relevant chemical equation. Carbon: C40, CO, — 12kgC + 32kgO:—> 44kg CO, ie, Oxygen required = 0.9 x 2 = 24kg/kg coal where the carbon content is 0.9 kg per kilogram of coal Carbon dioxide produced = 0.9 x t =33kgCO, Hydrogen (H): Hy, +}0,+H,O 2kgHy + 16kgO, + 18kegH,0 or ikgH, + 8kgO,+ 9kgH,O ,03 x 8 = 0.24 ke/kg coal and Steam produced = 0.03 x 9 = 0.27 kg/kg coal Oxygen required = 183Combustion Table 76 Solution for Example 7.1(i) 184 Sulphur: S+0,+SO, 32kgS + 32kg 0, + 64 kg SO, or 1kgS + 1kgO,+ 2kgSO, ie, Oxygen required = 0.005 kg/kg coal Sulphur dioxide produced = 2 x 0.005 = 0.01 kg/kg coal These results are tabulated in Table 7.6; the oxygen in the fuel is shown as a negative quantity in the column ‘oxygen required’. ‘Oxygen required Product mass Constituent Mass fraction (kg/kg coal) (kgikg coal) Carbon (C) 0.900 2.400 3.30 (CO,) ‘Hydrogen (H) 0.030 0.240 0.27 (H,0) Sulphur (S) 0.005 0.005 0.01 (S03) Oxygen (0) 0.025 =0.025 — Nitrogen (N) 0.010 = 001 (Nz) Ash 0.030 ~ = 2.620 From Table 7.6 O, required per kilogram of coal = 2.62 kg therefore 2.62 Air required per kilogram of coal = quired per kilogram of coal = “= = 11.245 kg where air is assumed to contain 23.3% O, by mass, ie. stoichiometric air-fuel ratio = 11.245. (ii) For an air supply which is 20% in excess, using equation (7.8), Actual A/F ratio = 11.245 + ‘e x11 245) = 12x 11.245 = 13.494/1 Therefore N; supplied = 0.767 x 13.494 = 10.350 kg Also; supplied = 0.233 x 13.494 = 3.144 kg In the products, then, we have Nz = 10.350 + 0.01 = 10.360 kg and excess O2 = 3.144 — 2.620 = 0.524 kgTable 7.7. Solution for Example 7.1(ii) Example 7.2 Solution 7.5 Exhaust and flue gas analysis The products are entered in Table 7.7 and the analysis by volume is obtained. In column 3 the percentage by mass is given by 100 times the mass of each product divided by the total mass of 14.464 kg. In column 5 the amount of substance per kilogram of coal is given by equation (2.7), n; = m,/ri,. The total of 0.4764 in column 5 gives the total amount of substance of wet products per kilogram of coal, and by subtracting the amount of substance of HO from this total, the total amount of substance of the dry products is obtained as 0.4616. Column 6 gives the proportion of each constituent of column 5 expressed as a percentage of the total amount of substance of the wet products. Similarly column 7 gives the percentage by volume of the dry products. Wet Dry my m,/m wi, n/n n/n Product (kg/kg coal) (%) —(kg/kmol) m= mili, (%) ——_(%) 1 2 3 4 5 6 1 CO; 2282 44 0.0750 1574 16.25 H,0 18718 0.0150 315 sO, 007 64 0.0002 0.04 0; 362 32 0164 3.44 Na 163 28 0.3700 7163 100.01 0.4766 00.00 {weet (04616) (ary) The analysis of a supply of gas is as follows: H, 494%; CO 18%; CHy 20%; CyHy 2%; O 0.4% Nz 6.2%; CO, 4%. Calculate: (i) the stoichiometric A/F ratios (ii) the wet and dry analysis of the products of combustion if the actual mixture is 20% weak. (i) The example is solved by a similar tabular method to Example 7.1; a specimen calculation is shown more fully as follows. CH, + 20, > CO, + 2H,0 ie, 1 kmol CHy + 2kmol 0 +1 kmol CO, + 2 kmol HzO ‘There are 0.2 kmol of CH, for 1 kmol gas, hence 02 kmol CH, + (0.2 x 2)kmol 0, ~+0.2kmol CO, + (0.2 x 2)kmol H,0 Therefore the oxygen required for the CH, in the gas is 0.4 kmol/kmol gas. ‘The results are summarized in Table 7.8; the oxygen in the gas, (0.004 kmol/kmol of gas) is included in column 4 as a negative quantity. From Table 7.8, 53 Air required = 757 = 4.062 kmol/kmol gas 185Combustion Table 78 Results for Example 7.2 Table 7.9 Solution for Example 7.2 Example 7.3 186 Products kmol/kmol kmol/ fuel kmol 02 kmol/ fuel Combustion equation kmol fuel CO, HO 1 2 3 4 $ 6 Hy 0494 2H, + 0,+2H,0 0.247 0.494 CO 018-20 +.0, 260, 0.09 og = CH, 02, CH, +20, + CO, + 2H,0 04 02 04 CuHy 0.02 Cy + 60, -+ 4CO, + 4H0. 02 008 008 oO, 0.004 —0008 = = Na 0.062 — ~ = = cO, 004 = 0.04 Total = 085300 OH where air is assumed to contain 21% oxygen by volume, ic, Stoichiometric A/F ratio = 4,062 by volume (ii) For a mixture which is 20% weak, using equation (7.8), Actual A/F ratio = 4.062 + (0.2 x 4.062) = 12 x 4.062 = 4874 by volume Associated nitrogen = 0.79 x 4.874 = 3.851 kmol/kmol gas Excess oxygen = (0.21 x 4,874) — 0.853 = 0.1706 kmol/kmol gas Total amount of nitrogen in products .851 + 0.062 = 3.913 kmol/kmol gas ‘The analysis by volume of the wet and dry products is then as shown in Table 7.9 kmol/kmol Product fuel % by vol. (dry) % by vol. (wet) co, 0.50 10.90 HO 0974 - 0; oa7 Na 3.912 Total wet 5.557 -H;,0 0974 Total dry 4.583 Ethyl alcohol is burned in a petrol engine, Calculal (i) the stoichiometric A/F ratio; (ii) the A/F ratio and the wet and dry analyses by volume of the exhaust gas for a mixture strength of 90%;Solution 7.6 Exhaust and flue gas analysis (iii) the A/F ratio and the wet and dry analyses by volume of the exhaust gas for a mixture strength of 120%. (i) The equation for the combustion of ethyl alcohol is as follows: C,H,O + 30, + 2CO, + 3H,0 Since there are two atoms of carbon in each mole of C>H,O then there must be 2 mol of CO; in the products, giving two atoms of carbon on each side of the equation. Similarly, since there are six atoms of hydrogen in each mole of ethyl alcohol then there must be 3 mol of HO in the products, giving six atoms ‘of hydrogen on each side of the equation. Then balancing the atoms of oxygen, it is seen that there are {(2 x 2) + 3} = 7 atoms on the right-hand side of the equation, hence seven atoms must appear on the left-hand side of the equation. ‘There is one atom of oxygen in the ethyl alcohol, therefore a further six atoms of oxygen must be supplied, and hence 3 mol of oxygen are required as shown. Since the O, is supplied as air, the associated Ny must appear in the equation, 19 ie. C,H,O +30, + (3 x Bn, + 2CO, + 3H,0 + (@ x zs 1 kmol of fuel has a mass of (2 x 12) + (6 + 16) = 46 kg; 3 kmol of oxygen have a mass of (3 x 32) = 96 ke. Therefore O, required per kg of fuel -3 2.087 kg, Therefore Stoichiometric A/F ratio = (ii) Considering a mixture strength of 90% then, from equation (7.9), stoichiometric A/F ratio actual A/F ratio 09 ‘Therefore 8.957 ‘Actual A/F ratio 9952/1 ‘This means that 1/0.9 times as much air is supplied as is necessary for complete combustion. The exhaust will therefore contain (1/0.9) ~ 1 = 0.1/0.9 of the stoichiometric oxygen, 1 79 C,H.O + {2s + (3 x Ps} 0.1 1 po 3 2c =~ x 3}0,+(— No 2c0, +3H,0+(35 x ) 1+ (5 x32) 187‘Combustion 188 i.e. the products are 2kmol CO, + 3kmol H,O + 0,333 kmol O, + 12.540 kmol N, The total amount of substance = 2 + 3 + 0.333 + 12.540 = 17.873 kmol Hence wet analysis is 2 «100 = 11.19% CO; 3 x 100= 16.79% H,0 17873 17873 2233100 = 1.86% 03; 12540. 100 = 70.16% Ny 17873 17873 The total dry amount ofsubstance = 2 + 0.333 + 12.540 = 14.873 kmol Hence the dry analysis is 0.333 100 = 13.45% CO,: 100 = 2.24% 0,5 aera * 1° 2 Taga * 00 =? 2 12.540 BSB 5 100 = 84.31% N. 14873 * 2 (iii) Considering a mixture strength of 120%, then from equation (7.9), stoichiometric A/F ratio 1 actual A/F ratio Therefore 957 a = 7147/1 12 l Actual A/F ratio ‘This means that 1/1.2 of the stoichiometric air is supplied. The combustion cannot be complete, as the necessary oxygen is not available. It is usual to assume that all the hydrogen is burned to HO, since hydrogen atoms have a greater affinity for oxygen than carbon atoms. The carbon in the fuel will burn to CO and CO,, but the relative proportions have to be determined. Let there beakmol CO, and 6 kmol CO in the products. Then the combustion equation is as follows: 1 p 1,H.O + —430, +(3 x —)N, & Oris +( “F) } L co, +800 +28,04(,4 x3 «By, To find a and b a balance of the carbon and the oxygen atoms can be made,7.5. Exhaust and flue gas analysis ie. Carbon balance: 2 = a + b 1 Oxygen balance: 14(2 x a)=204643 Subtracting the equations gives =1 andthen 6=2-1=1 i, the products are (1 kmolCO,) + (1 kmol CO) + (3 kmol H,) + (9.405 kmol N,) =14143-4 9.405 = 14.405 kmol Hence wet analysis is 1 x 100 = 694% CO,; ——— x 100 = 94% CO 14.405 14.405 405 3 100=2083% Hy; 245 « 100 = 65.29% Ny 14.405 14405 The total dry amount of substance = 1 + 1 + 9.405 = 11.405 kmol Hence dry analysis is 1 x 100 = 8.77% CO; 11.405 11.405 405. 100 = 82.46% Nz 405 x 100 = 8.77% CO Example 7.4 For the stoichiometric mixture of Example 7.3, calculate: (i) the volume of the mixture per kilogram of fuel at a temperature of 65°C and a pressure of 1.013 bar; (ii) the volume of the products of combustion per kilogram of fuel after cooling to a temperature of 120°C at a pressure of 1 bar. Solution (i) As before 79 719 CaHeO + 30, +(3 x 57 JNa > 202 + 3H0 + (3 «57 No Therefore 79 ‘Total amount of substance reactants = 1 + 3 + (3 x z) = 15.286 kmol 189Combustion Example 7.5 Solution 190 From equation (2.8), pV = nT, therefore 86 x 10° x 8.3145 x 338 10° x 1.013 where T = 65 + 273 = 338 K. In I kmol of fuel there are (2 x 12) + (6 + 16) = 46 kg, therefore Volume of reactants per kilogram of fuel = aes = 9.219 m? Vv = 424.06 m?/kmol of fuel (ii) When the products are cooled to 120°C the HO exists as steam, since the temperature is well above the saturation temperature corresponding to the partial pressure of the HO. (This must be so since the saturation temperature corresponding to the total pressure is 99.6°C, and saturation temperature decreases with pressure.) The total amount of substance of the products is 2434 (3 x z) = 16.286 kmol From equation (2.8), p¥ 286 x 10° x 10° x 1 where T = 120 + 273 = 393K. aRT, therefore 145 x 39: v = 532.15 m3/kmol of fuel ie. Volume of products per kg of fuel = ae = 11.57 m? If the products in Example 7.4 are cooled to 15°C at constant pressure, calculate the amount of water which will condense per kilogram of fuel. At 15°C, since some condensation has taken place, the steam remaining is dry saturated, being in contact with its liquid. The saturation pressure at 15°C is 0.01704 bar, and this is the partial pressure of the dry saturated steam. Then using equation (6.14) % Pi Von p For the steam n, _ 0.01704 = 0.01704 n From Example 7.4 the total amount of substance of dry products is (16.286 — 3) = 13.286 kmol, therefore nm 0.01704 x 13.286 nn, + 13.286 = 0.01704 therefore n, = ( 1— 001704 ) = 0.2303 i, amount of substance of dry saturated steam remaining at 15°C is 0,2303 kmol,Example 7.6 Solution 7.8 Exhaust and flue gas analysis therefore amount of substance of water condensed is (3 — 0.2303) = 2.77 kmol. Also 1kmol of H,O contains (2+ 16) = 18kg, therefore mass of water condensed is (2.77 x 18) kg/kmol fuel ie, Mass of water condensed per kilogram of fuel = ——7=— = 1.084 kg, ‘Any problem in combustion can be solved using the amount of substance. The following examples illustrate the method for a solid and for a gaseous fuel; these examples should be compared with the method used in Examples 7.1 and 7.2, The gravimetric analysis of a sample of coal is given as 80% C, 12% H, and 8% ash. Calculate the stoichiometric A/F ratio and the analysis of the products by volume. 1 kg of coal contains 0.8 kg C and 0.12 kg H 1 kg of coal contains oS kmol Cand 0.12 kmol H Let the oxygen required for complete combustion be x kmol, the nitrogen supplied with the oxygen is then x x 79/21 = 3.76x kmol. For 1 kg of coal the combustion equation is therefore as follows: Se + O.12H + xOz + 3.76xN, -+ CO, + BHO + 3.76xNo, ‘Then Carbon balance: - a or a = 0.067 kmol Hydrogen balance: 0.12 = 2b or b = 0.060 kmol Oxygen balance: 2x =2a +b ie. x =a + b/2 = 0.067 + 0.030 = 0.097 kmol ‘The mass of 1 kmol of oxygen is 32 kg, therefore the mass of ©, supplied per Kilogram of coal is 32 x 0.097 kg, 32 x 0.097 shiometric A/F ratio = = 13.3/1 Stoichiometric A/F ratio = >> / ‘Total amount ofsubstance of products = a + b + 3.76x = 0.067 + 0.06 + (3.76 x 0.097) = 0492 kmol Hence wet analysis is 2067 100 = 13.6% COs; 2% x 100 = 12.2% Hy 0.492 0.492 0.365 o=° x 100 = 74.2% Nz 0.492 191Combustion Example 7.7 192 Solution 7.6 A gas engine is supplied with natural gas of the following composition: CH, 93%; CzH, 3%; Nz 3%; CO 1%, If the A/F ratio is 30 by volume, calculate the analysis of the dry products of combustion. It can be assumed that the stoichiometric A/F ratio is less than 30. Since we are told that the actual air—fuel ratio is greater than the stoichiometric it follows that excess air has been supplied. The products will therefore consist of CO;, H,0, O2, and Nz. The combustion equation can be written as follows: 0.93CH, + 0.03C,H, + 0.01CO + 0.03N; + (0.21 x 30)0, + (0.79 x 30)N, + aCO, + bH,O +O, +N, Then Carbon balance: 0.93 + (2 x 0.03) +0.01=a ora Hydrogen balance: (4 x 0.93) + (6 x 0.03) = 2b or b= 195 Oxygen balance: 0,01 + (0.21 x 30) = 2a + b + 2c Therefore = {6.31 —2~(2 x 195)}/2 = 0.205 Nitrogen balance: 0.03 + (0.79 x 30)=d ord = 23.73 + 0.205 + 23.73 24,935 kmol ie, Total amount of substance of dry products ‘Then analysis by volume is spare X 100 = 4.01 CO,; 24,935 24.935 23.73 24.935 x 100 = 0.82% O, x 100 = 95.17% Nz Practical analysis of combustion products The experimental investigation of a combustion process requires the analysis of the products of combustion. The most basic method is to take a sample of the gas and analyse it chemically as in an Orsat apparatus for example (see ref. 7.2), Modern instrumentation now provides a quicker, more accurate, and continuous means of analysis. Some of the methods used are summarized below, but manufacturers’ catalogues should be consulted for details of actual instruments since improvements are continually being made to the measuring systems. Non-dispersive infra-red (NDIR) In this method the concentration of the constituent gas is measured by recording its ‘optical’ absorption in the infra-red spectrum. The mixture is continuouslyFig. 7.1 Non-dispersive infra-red gas analyser 7.6 Practical analysis of combustion products examined by being drawn through a tube, the inside of which is subject to radiation from an infra-red source, each gas absorbing on a particular waveband of the radiation. Full descriptions of infra-red analysis and other methods are given in ref. 7.4 One particular instrument is shown diagrammatically in Fig. 7.1. The gas being analysed is passed through tube A and dry air is passed through tube B. The chambers C and D contain pure samples of the constituents to be detected. Radiation from nichrome elements is passed through tubes A and B and thence to chambers C and D where it is absorbed. The subsequent heating of the gases in C and D causes an increase in pressure in the two chambers; C and D are separated by a thin metal diaphragm which, together with an insulated, perforated plate, forms a capacitor. Radiator 1 | ey) ) a : > shutter ee YD Motor Sample Chart A | tubes >} B Disphragm eambly Amplifier Servo unit [Null balance recorder which Infrared analyser may replace the indicator If the constituent being sought is not present in tube A then equal amounts of radiation will be absorbed in C and D, which will be heated equally and there will be no pressure difference across the diaphragm. If the constituent is present in A, it will absorb some of the radiation admitted, and the rest will pass on to chamber C. The radiation absorbed in C will then be less than in D so that a greater pressure will be reached in D than in C, and'the diaphragm will be displaced. This displacement causes a change in capacitance of the capacitor and a current is produced which is amplified to give a reading on a microammeter. The microammeter scale is calibrated to give the corresponding concentration of the constituent in the gas being analysed. To avoid zero errors the radiation is cut off from both tubes simultaneously and allowed to fall on them simultaneously, by means of a vane which rotates 193Combustion 194 at a low frequency. The pressure changes are then related to the temperature changes produced by the differential absorption in C and D. To provide a continuously recording instrument as shown in the right-hand part of Fig. 7.1, a ‘null’ balance recorder is employed. The principle is that the pressure difference created should be nullified by cutting off from the vessel B a sufficient amount of radiation to balance that absorbed in A. This is done by means of a shutter driven by a servo-motor which receives a signal from the detector unit. A recording pen is linked to the balancing shutter mechanism and records on a circular chart. Another NDIR instrument uses photoacoustic detection: a sample is drawn into a closed cell where infra-red light of the correct frequency is absorbed by the gas to be measured; the pressure increases and decreases because the light is pulsing and the pressure wave is detected by a microphone in the cell wall; the acoustic information is processed by a computer and can be printed or plotted as required; an optical filter carousel is then turned so that infra-red light of a low frequency enters the cell so that the next constituent can be measured. The instrument is manufactured by Briiel and Kjaer of Denmark. The NDIR instruments are calibrated against accurately prepared samples of gas mixtures. This method is suitable for carbon monoxide, carbon dioxide, sulphur and nitrogen compounds, methane and other hydrocarbons, and organic vapours. Oxygen, hydrogen, nitrogen, argon, chlorine, and helium, do not absorb infra-red radiation and so will not be detected by this type of instrument. CO, and O, recorders Boiler-house engineers require a continuous indication of the quality of the combustion process in the plant under their control. This enables comparisons to be made and a falling-off in efficiency becomes immediately apparent. For continuous firing a continuous record is required, and digital instruments are available with in-built microchips which enable boiler efficiency to be obtained directly from individual readings of temperature, and percentage by volume of CO, and O;. The variations and applications of these instruments are many, and are made to suit particular requirements. The general principles only will be dealt with here, CO2 measurement by thermal conductivity variations The CO, content of a flue gas is an important criterion of efficient and economic combustion, and is important in observing the regulations governing smoke emission, When a heated wire is placed in a gaseous atmosphere it loses heat by radiation, convection and conduction. If the losses by radiation and convection are kept constant, the total heat loss is dependent on the heat loss by conduction, which varies with the constituents of the gas since each has a different and characteristic thermal conductivity. If a constant heat input is supplied to the wire there is an equilibrium temperature for each mixture, and if the CO, content alone is varied, then its concentration will be indicated by a measurementFig. 7.2 Paramagnetic analyser 7.6 Practical ai lysis of combustion products of the temperature of the wire. In an actual instrument the heat loss from the wire is mainly by conduction, the other means (e.g. convection, radiation, end cooling, diffusion) account for about 1% each of the total loss. The convection loss is reduced by mounting the wire vertically. The instrument is calibrated against mixtures of known composition. The sample of gas is passed over an electrically heated platinum wire in a cell which forms one arm of a Wheatstone bridge. In another arm of the bridge is a similar cell containing air. A difference in CO, content between the two cells causes a difference in temperature between the two wires and hence a difference in resistance. The out-of-balance potential of the bridge is measured by a recording potentiometer, calibrated to give the CO, content directly. Oxygen measurement by magnetic means Gases may be classified in two groups: (i) diamagnetic gases which seek the weakest part of a magnetic field; (ii) paramagnetic gases which seck the strongest part of a magnetic field. Most gases are diamagnetic, but oxygen is paramagnetic, and this property of oxygen can be utilized in measuring the oxygen content of gas mixtures. Referring to Fig, 7.2 the gas sample is introduced into the analysis cell and passes through the annulus as shown. The horizontal cross-tube carries two identical platinum windings, coils 1 and 2, which are connected in adjacent arms ofa Wheatstone bridge, and are heated by the applied voltage. The winding at A is traversed by a magnetic field of high intensity from a large permanent magnet. When the gas sample passes the end of the tube the oxygen is drawn into the cross-tube. It is heated and its paramagnetic property decreases due Gas out "AC supply 195Combustion Fig. 7.3 Flame ionization detector 196 to the increase in temperature as it passes through the tube. The induction of fresh cool oxygen continues, hence a continuous flow is established. The result is that the two windings are cooled by different amounts, the resistance changes, and the bridge goes out of balance. The resulting electromotive force (emf) is measured by a potentiometer, and since this is proportional to the oxygen content, the reading gives the oxygen content of the mixture, Zirconia cell An absolute method of measurement of oxygen may be obtained using a zirconia cell; the output of the cell is directly related to the oxygen concentration of the gas sample (see ref. 7.5). Flame ionization detector (FID) ‘This is a method used for detecting the hydrocarbon (HC) content of the exhaust from internal combustion (IC) engines; Fig, 7.3 shows a diagram of a typical system. The sample to be analysed is mixed with a special burner fuel which may be hydrogen, hydrogen plus helium, or hydrogen plus nitrogen. A polarizing voltage exists between the burner and the collector which causes a migration of ions in the flame and so a current to flow in the collector circuit. The current is proportional to the rate of ion formation which depends on the HC concentration in the gases and is detected by a suitable electrometer and displayed as an analogue output. The FID gives a rapid, accurate, and continuous reading of total HC concentration for levels as low as one part in 10°, Exhaust = Collector [ Flame Battery (100300 v) OO (I Electrometer x oe Air —7 Chemiluminescent analyser This method is preferred to NDIR for the analysis of the oxides of nitrogen found in the exhaust of IC engines; a diagram of a chemiluminescent analyser is shown in Fig. 7.4. The nitrous oxides, NO,, are converted to NO before analysis by passing the gas sample over a heated catalyst such as stainless steel,Fig. 14 Chemiluminescent analyser 7.6 Practical analysis of combustion products Recorder Electro — meter }o Photomltiplie Reaction |_Exhaust Chamber igh sla _| Sample a converter graphite, or molybdenum at about 600°C. The sample is now mixed with ozone where the reaction gives NO, and O,; some of the NO, contains an excess of energy within its atoms and this NO, then changes to the ground state with the emission of light (chemiluminescence) which is amplified then measured by a photomultiplier tube and the signal displayed as an analogue reading of the NO, content of the original sample. Results of the analysis ‘An analysis will show whether or not the combustion is complete, For instance the presence of CO will indicate that the combustion is not complete, and if an oxygen reading is obtained this will mean that excess air has been supplied. Both CO and O, may appear in the analysis as result of incomplete combustion, and dissociation (see section 7.7). Quantitatively the dry product analysis can be used to calculate the A/F ratio. This method of obtaining the A/F ratio is not so reliable as direct measurement of the ait consumption and fuel consumption of the engine. More caution is required when analysing the products of combustion of a solid fuel since some of the products do not appear in the flue gases (eg. ash and unburnt carbon), The residual solid must be analysed as well in order to determine the carbon content, if any. With an engine using petrol or diesel fuel the exhaust may include unburnt particles of carbon and this quantity will not appear in the analysis. The exhaust from internal combustion engines may contain also some CH, and H, due to incomplete combustion. The CH, content is approximately 0.22% of the dry products and the H, content is of the order of haif the CO content. 197‘Combustion 198 Example 7.8 Solution Considerable care is required in combustion calculations as inaccuracies are caused due to taking small differences between quantities and then using these as ratios. The use which can be made of the dry analysis is illustrated by the following problems. An analysis of the dry exhaust from an engine running on benzole shows a CO, content of 15%, but no CO. Assuming that the remainder of the exhaust contains only oxygen and nitrogen, calculate the A/F ratio of the engine and the dry and wet volumetric analyses of the products of combustion. The ultimate analysis of benzole is 90% C and 10% H. ‘There are many different methods used in combustion analysis. The following method is recommended; its algebraic approach makes it less confusing than some other methods; it also lends itself more easily to a computer solution, Consider 1 kg of fuel: . 0.233. 0.767 Sor oin} + 4(23o, + a) 12 32 28 = B{0.1SCO, + aO, + (1 — 0,15 ~ a)N,} + bHZO where A is the A/F ratio on a mass basis; B the amount of substance of dry exhaust gas per kilogram of fuel, kmol/kg; a the fraction by volume of oxygen in the dry exhaust gas; b the amount of substance of steam per kilogram of fuel, kmol/kg. A brief explanation of the above equation is given below. Lefi-hand side of the equation. There are (0.9/12) kmol C and 0.1 kmol H in 1 kg of fuel, A kg of air per kg of fuel contains 0.233 kg of oxygen and hence contains (0.23/32) kmol of oxygen; similarly there are (0.767/28) kmol of nitrogen in A kg of air for every kilogram of fuel. Right-hand side of the equation. There are 0.15 kmol CO, for { kmol of dry exhaust gas(given in the example), hence there are 0.15B kmol CO, per kilogram of fuel since there are B kmol dry exhaust gas per kilogram of fuel. Similarly, there are Ba kmol oxygen in the products per kilogram of fuel; the nitrogen in the dry exhaust gasis obtained by difference, ie.(1 — 0.15 — a). There are four unknowns in the above equation and four equations may be obtained from the mass balances on carbon, oxygen, nitrogen and hydrogen, Carbon balance: 09/12 =0.15B therefore B= 0.5 Hydrogen balance: 0.1 = 2b therefore b = 0.05 Oxygen balance: 0.2334/32 = 0.15B + Ba + b/2 ie, a = 0.014564 — 0.2 uy Nitrogen balance: 0.7674/28 = B(0.85 — a) ie. 0.85 — 0.054794 02] Solving equations [1] and [2] for A we have 0.014564 — 0.2 = 0.85 — 0.054794 therefore A=15.14 ie. A/F ratio = 15.14Example 7.9 Solution 7.6 Practical analysis of combustion products ‘The value of a can then be found from either equation [1] or equation [2] as 0.02. Hence the complete dry exhaust gas volumetric analysis is given by 15% CO, 20% O,, 65% Nz. The wet volumetric analysis can be found by totalling the amount of substance of wet products and taking the appropriate ratios, ie. Total amount of substance of wet products = (05 x 0.15)CO, + (05 x 0.2)0, + (0.5 x 0.65)N; +0.05 Hy = 055 kmol/kg fuel ‘Then the wet volumetric analysis is as follows: 5 x 0. 05 X01 199 — 13.64% CO, 99%? x 100 = 18.18% O,, 0.55 0.55 05 x 065 00s x 100 = 59.09% Nz, > x 100 = 9.09% H, 0.55 055 ‘The analysis of the dry exhaust gas from an internal combustion engine gave 12% CO,, 2% CO, 4% CHy, 1% Hz, 4.5% Oz, 76.5% Nz. Calculate the proportions by mass of carbon to hydrogen in the fuel, assuming it to be a pure hydrocarbon, and the A/F ratio used. Let the unknown mass fraction of carbon in the fuel be x kg per kilogram of fuel. Then we can write the combustion equation as follows: x 0.233, 0.767 Xe+a—yn+ a(2o, 4 Oty, not) (CS achitgg :) + B{0.12CO, + 0.02C0 + 0.04CH, + 0.01H, + 0.0450, + 0.765N,} + aH,0 As in Example 7.8 there are four unknowns and four possible equations. Carbon balance: x/12 = B(0.12 + 0.02 + 0.04) x .16B (t] Hydrogen balance: (1 — x) =(4 x 0.04)B + (2 x 0.01)B + 2a a=0.5 —0.5x — 0.09B [2] Oxygen balance: 07334 _ a(ara +22, 0085) +8 32 2 2 a= 0014564 — 0.358 031 Nitrogen balance: —0,1674/28 = 0.765B A=27.927B [4] Equating [2] and [3] we have 0.5 — 0.5x — 0.09B = 0.014564 — 0.35B A= 34341 — 34.341x + 17.857B 199Combustion 200 77 Then using equation [4] 34.341 — 34.341x + 17.857B = 27.9278 B=341—341x Substituting in [1] X = 7.366 — 7.366x ie. x = 0.8805 The composition of the fuel is therefore 88.05% C, 11.95% H. Also, B= 3.41 x 0.1195 = 0.4075 ‘Then in equation [4], A= 27927 x 04075 = ie A/F ratio = 11.38 1.38 Dissociation Itis found that during adiabatic combustion the maximum temperature reached is lower than that expected on the basis of elementary calculation, One important reason for this is that the exothermic combustion process can be reversed to some extent if the temperature is high enough. The reversed process is an endothermic one, ie. energy is absorbed. In a real process the reaction proceeds in both directions simultaneously and chemical equilibrium is reached when the rate of break-up of product molecules is equal to their rate of formation. This is represented for the combustion of carbon monoxide and hydrogen respectively by the equations 2CO+0;=2CO, and 2H, +0,=2H,0 Both of these reactions can take place simultaneously in the same combustion process. The proportions of the constituents adjust themselves to satisfy the equilibrium conditions and their actual values depend on the particular pressure and temperature. The presence of CO and H, means that there is further energy to be released on their reaction with O, so the maximum temperature reached cannot be as high as that expected on the basis of complete combustion. As the combustion proceeds and the temperature level falls, due to expansion and/or subsequent heat loss, the amount of dissociation decreases (it is significant at temperatures >1500K) and combustion proceeds to completion. However, since the energy is released at lower temperatures and, in a positive expansion cylinder, at a lower effective compression ratio the efficiency of the process is reduced. ‘The condition of equilibrium during a reversible combustion process can be studied by means of a conceptual device known as the ‘Van’ t Hoff equilibrium box’ as shown in Fig. 7.5. Consider the general reversible combustion process akmol A + 6 kmol B=*c kmol C + dkmol DFig. 75. Van’t Hoff equilibrium box i —We Surroundings at Ts akmol of A Pa Mixture of A, B, C and Po _ckmol of C ap.and T 2 ate se a at pend Tras? | me | _entnivo ie PSH st \ / Semispermeable membranes mH We ‘which occurs at a fixed temperature T and a pressure, p, in the equilibrium box. The reactants A and B are each initially at p, and T and the products C and D are each finally at p, and T: As the process is reversible some energy transfer will take place in the form of work and this is allowed for by the inclusion of isothermal compressors and expanders. The equilibrium box contains a mixture of gases A, B, C and D at total pressure p and temperature T and to allow reversible mass flow of the constituents the pressure of each constituent at entry to the box must be equal to its partial pressure in the box. The pressure adjustments are made by the isothermal expanders and compressors and each constituent enters or leaves through a semi-permeable membrane. Some substances permit one gas to pass through but prevent other gases, e.g. a glowing aluminium sheet allows hydrogen to pass through but not, other gases. It is assumed here that such substances are available for gases A, B, Cand D. The process may proceed equally well in either direction but itis illustrated here as going from left to right in the equation and in Fig. 7.5. With a reversal of the process the heat and work transfers would be reversed in direction. The work input during an isothermal expansion by a perfect gas between states 1 and 2 is given by We marin(?2) = nkerin(22) Py Pi, and this can be applied to each of the compressors and expanders in the system of Fig. 75 W, = work input on A = akrin(22) = Rrin( £4) Pr Pi Wp work input on B= oarin(*) Pi We = work input on C = 201Combustion Wy = work input on D = dRT un() Po, ‘Therefore the net work output of the system —W= — Wy Wy — We Mo w= Er} (2) [BY om) +H} sane} ~Rrin( 2) Ps ie. Suppose that in a second similar system in the same surroundings the pressure in the equilibrium box is p’ then it will havea net work output, — W’, given by -We art in (Ze (PA) (Pa)? where p' =P, + Pa + Pc + Pb It is proposed that —W # —W’ and this statement is to be investigated. Suppose — W > —W’, then the second system can be reversed, as shown in Fig. 7.6, and a single system formed by using the work output from the first system, —W, to provide the work input for the second system reversed. sped Fig. 76 Hypothetical T r r (impossible) combination of two systems Q fo Q -9 fo - -w jew Systems 1 and 2 System 1 System 2 Combined reversed system The result isa single system giving a net output of work (— M7) — (—W’) while exchanging heat with a single source at temperature T: This is a contradiction of the Second Law of Thermodynamics, thus the proposition that W # W’ is not true so that W = W’, therefore PEPH _ (Po)'(Po)" PAPE (Pa)*(Pay? 2027.7 Dissociation where K is the thermal equilibrium or dissociation constant and has been shown, to be independent of the pressure in the equilibrium box. A standard thermodynamic equilibrium constant, K®, can be defined in dimensionless form by referring each partial pressure to a pressure of 1 bar, ; « : A ie xe=in(£5) +122) ~mn(2) ~ (22) P P p P or in general In K° = in(p,/p** (7.10) where 7; is the stoichiometric coefficient, taken as positive for the products and negative for the reactants, The thermodynamic equilibrium constant K° is a function of temperature and values of In K° are tabulated against temperature for each reaction equation, see ref. 7.6. As the partial pressures of the constituents are proportional to the molar proportions then K° is an indication of the ratio of products to reactants and so is a measure of the amount of dissociation. If Kis large then the proportion of product is high and the amount of dissociation is small. The above general expression for K®, equation (7.10), will now be applied to particular important reactions. For example the combustion of carbon monoxide to carbon dioxide can be written as CO +40;=C0, with molar proportions for CO, 0, and CO, of 1, 0.5, and 1. Thus K can be written as Peo, v®)!* 0 ae xa a For the combustion of hydrogen the equation is H, +40,=H,0 with molar proportions 1, 0.5, and 1. The equilibrium constant then becomes InK°= ey Ke = anol? is Pa,(Po,) In the combustion of hydrocarbon fuels both of the above reactions may occur simultaneously and another equilibrium constant can be defined by dividing (7.12) PaolP°)” yy Peo,(®)'? Pu,(Po,) Pco(Po,)” giving ko = Puree (7.13) Poo, which is also tabulated, see ref. 7.6, and can be used to form another equation in the analysis, 203Combustion 204 Example 7.10 Solution Itis readily seen that dissociation as described introduces an added complexity into the analysis of the combustion process but the complication does not end there. The conditions of equilibrium must be satisfied at any particular temperature and also the energy balance for the process must be satisfied. The temperature reached will depend on the amount of fuel burned, the proportions of the constituents and their thermodynamic properties, all of which are also dependent on pressure or temperature. Thus several conditions have to be satisfied before any particular state in the process can be determined and it is necessary to establish a sufficient number of equations to complete the analysis. This is discussed more fully in section 7.8 and for the immediate purpose it will be assumed that the energy requirements have been met in the process and only the dissociation effects are required. The analysis of combustion may be extended to include the formation of nitric oxide, 4N, + $0,=2NO, which occurs at high temperature and the dissociation of HO vapour into hydrogen and hydroxyl, $H, + OH=H,0, as well as into its constituent gases. There may also be the dissociation of molecules of oxygen, hydrogen and nitrogen into atoms. These aspects of combustion are detailed and involve small proportions of the charge. It has been stated previously that recombination occurs as the temperature falls so that combustion is completed with a loss in efficiency of the cycle such that, less work output is obtained than expected. The completion of combustion requires the absence of low-temperature quenching conditions, which may arrest the process, and a sufficiency of time for the reaction to be completed. ‘The importance of the analysis of combustion increases with advancing engine technology as typified in the petrol engine. This engine includes the extremes of combustion conditions starting from a complex chemical fuel and a mixture of air, water vapour and residual exhaust gas. The pressure and temperature levels passed through are large and.the duration of a cycle is only a fraction of a second. The cylinders are water-cooled usually and sudden exhausting of gas is provided for. Under these conditions the exhaust gas can have a complex analysis and some of the constituents are responsible for the polluting of the atmosphere with undesirable and irritating results. A combustible mixture of carbon monoxide and air which is 10% rich is compressed to a pressure of 8.28 bar and a temperature of 282°C. The mixture is ignited and combustion occurs adiabatically at constant volume, When the maximum temperature is attained analysis shows 0.228 kmol of CO present for 1 kmol of CO supplied. Show that the maximum temperature reached is 2695 °C. For stoichiometric conditions CO + $0, + (3.76 x $N,)> CO, + (3.76 x $N,) CO +050, + 1.88N, > CO, + 1.88N, 100 Actual A/Fratio = stoichiometric A/F ratio x 7 > = stoichiometric A/F ratio/1.1Therefore the actual reactants are CO + (0.50, + 1.88N,)/L.1 With dissociation there will be some break up of CO, giving CO and O, in the products such that CO +(0.50, + 1.88N,)/1.1 = aCO, + CO + cO, + (1.88/1.1)No ‘The question states that b = 0.228, therefore CO + 0.4550, + 1.709N, > aCO, +.0.228CO + cO, + 1.709N, Carbon balance: 1 =a + 0.228 therefore a = 0.772 ‘Oxygen balance: 1+ (2 x 0.455) = 2a + 0.228 + 2c therefore ¢ = 0.069 For the reaction CO + $0,=CO, 1 — Peo,(P?)? Pco(Po,)"* a and Pco, = Pa m therefore a [fmgp? Koad [}% 1 iv Pr } a where p, = the total pressure at the required temperature and n, = total amount of substance of products ng = a+b +6 + 1.709 = 0.772 + 0.228 + 0,069 + 1.709 = 2.778 kmol At ignition py =828bar and T, = 273 +282 = 555K pyV=n,RT, and p,V=n,RT, and V =constant therefore mT Pa PL where n, = amount of substance of reactants = 1 + 0.455 + 1.709 = 3.164. Then, assuming that T, = 273 + 2695 = 2968 K, we have 2.778 2968 6CO, + 3H,0(vap) Tg=14+75=85, mp =6+3=9 From equation (7.31) AUp = AHo — (np — MRT. = ~ (3169540 x 1) — {o-s x 83145 x oe} where Tp = 273 + 25 = 298 K ie. AUy = —3 169540 — 1239 = —3170779 kT {note that AUs is negligibly different from AH) 1 kmol of CgHg = (6 x 12) + (6 x 1) = 78 kg therefore 3170779 Aulp = — —40 651 kI/kg, B Change in reference temperature Tt has already been mentioned that the internal energy and enthalpy of reaction depend on the temperature at which the reaction occurs. This is due to the change in enthalpy and internal energy of the reactants and products with temperature, Tt can be seen from the property diagram of Fig. 7.11 that the enthalpy of reaction at temperature 7; AH; can be obtained from AH at Ty by the relationship —AHr= AHg + (Hr, — He,) — (Hp, — Hp,) (7.32) where Hy, — Hx, = increase in enthalpy of the reactants from 7, to T and Hp, |, ~ Hp, = increase in enthalpy of the products from T to T 213Combustion Example 7.14 Table 7.10 Molar enthalpies of gases for Example 7.14 214 Solution For carbon monoxide at 60°C, Af is given as 282990 kJ/kmol. Calculate Ah at 2800 K given that the molar enthalpies of the gases concerned are as given in Table 7.10. Molar enthalpy/(kJ/kmol) Gas At6o°C At 2800K Carbon monoxide 1018 86115 Oxygen 1036 90 144 Carbon dioxide 1368 140440 The reaction equation is CO +40, 4CO, Also, taking values from Table 7.10, and using equations (7.25) and (7.28) Hp, = (1 x 1018) + (0.5 x 1036) = 1536 kJ He, = (1 x 86115) + (0.5 x 90 144) = 131 187 kT Hp, = 1 x 1368 = 1368 kJ Hp, = 1 x 140440 = 140440 kJ ‘Then using equation (7.32) ~AH;= —AHy + (Hp, — Hp.) — (He, — Hp,) = (1 x 282990) + (131 187 — 1536) — (140440 — 1368) = 273569 kJ ie, Ah = —273 569 k3/kmol CO The property diagram and real processes The property diagram of U against T has already been referred to and is shown in Fig. 7.12(a); a diagram of H against T is shown in Fig. 7.12(b). The diagrams can be used effectively to demonstrate real processes of interest. The solid lines indicate the property variations with Tif the constituents are gascous. Ifreactants or products contain a liquid component then the property lines will be modified as shown by the dotted lines. It can be seen by inspection that the effect of condensation of the HO in the products is to increase AU, and AH. In processes 1-A, T, = Tp and there is a maximum energy transfer to the surroundings AU, or AH.Fig. 7.12 U-T and H-T diagrams for combustion Fig. 7.13 Energy-absolute temperature diagram between stages I and 2 7.8 Internal enorgy and enthalpy of reaction u Reactants (gaseous) Reactants (gaseous) Products (gaseous) Constituents in Fiquid form ZF Constituent in liquid form (a) (6) In processes 1B the internal energy, or enthalpy, initially and finally is the same so that the increase in temperature is a maximum and the combustion process is adiabatic. In 1—C the processes are general with heat transfer and possibly work transfer. Process 1A’ in Fig. 7.12(a) corresponds to the constant volume bomb calorimeter test and in Fig. 7.12(b) 1-A' corresponds to the steady-flow combustion Boys’ calorimeter test, Both of these tests are discussed later in section 7.12. Tn a general non-flow or steady-flow process the initial state (1) and final state (2) will be different and neither will be at the reference temperature T). The quantities of interest are U, — U,,and H — H, for the respective processes. It can be seen readily from Fig. 7.13 that Uz = Uy = — (Uy = Ua) = — {-AUp — (Un, — Ur.) = Ur, — Ur,)} U, — U, = Up, — Up, = (Up, — Ur,) + AUo + (Un, — Un,) Reactant ‘This is equation (7.16) repeated. A corresponding expression is obtained for H, — Hy, The principles dealt with in this section will now be applied to typical problems. 215Combustion Example 7.15 A combustible mixture of carbon monoxide and air which is 10% rich is compressed to a pressure of 8.28 bar and a temperature of 282°C. The mixture is ignited and combustion occurs adiabatically at constant volume. Calculate the maximum temperature and pressure reached assuming that no dissociation takes place. At the reference temperature of 25°C, Ah’ for CO is 282990 kJ/kmol. Solution For stoichiometric conditions 216 CO + 40, + (3.76 x 3)Ny > COz + (3.76 x H)Ny CO + 0.50; + 1.88N, + CO, + 1.88N; and for a mixture strength of 110% Actual A/F ratio = stoichiometric A/F ratio x in = stoichiometric A/F ratio/1.1 Therefore, for the actual conditions, CO + (0.50, + 1.88N,)/1.1 + aCO, + BCO + 1.71N3 Then, Carbon balance: 1=a+b ty ie, (2] From equation [1] and [2] =0.909 and b= 0.091 therefore CO + 0.4550, + 1.709N, + 0,909CO, + 0.091CO + 1.709N, | Also for the reaction CO +402 +C0, m= 1405=15 and mp=1 therefore AU g = AHy — (np — my )RT. —282990 x 1) — {(1 — 1.5) x 8.3145 x 298} — 282990 + 1239 = —281 751 kJ ie. A@§ = —281751 kI/kmol The non-flow process is defined by Q+W=(U,—U,) Also Q = 0; W = 0 at constant volume; U, = Up, Uz = Up,, therefore (Up, — Ur.) =0Example 7.16 Solution 7.8. Internal energy and enthalpy of reaction (Up, — Up,) + (Up, — Un.) + (Ur, — Ur,) = 0 (Up, — Up,) + AUy + (Un, — Up,) =0 03] ‘The values of molar internal energies for the various gases can be read from the tables of ref. 7.6. Then At 298.15K Ux, = —2479 — (0.455 x 2479) — (1.709 x 2479) = —7844 kJ p ALSSSK Up, = 2984.6 + (0.455 x 3233.1) + (1.709 x 2944.2) = 9487 kJ AL298.15K Up, = —(0.909 x 2479) — (0.091 x 2479) — (1.709 x 2479) 6716 kI Substituting in [3], (Up, + 6716) — (0.909 x 281 751) + (—7844 — 9487) = 0 Note: there are 0.091 kmol CO in the products therefore only (1 — 0.091) = 0.909 kmol of CO releases energy of reaction, ie, Up, = 266727 kJ To find the temperature of products a trial-and-error method is now necessary. AtT, Up, = {0.909 x (i, for CO;)} + {0.091 x (fi, for CO)} + {1.709 x (iy for Nz)} Try T = 3200K Up, = (0.909 x 138720) + (0.091 x 74391) + (1.709 x 73555) = 258.572 kJ ‘This compares with the figure of 266 727 kJ calculated above, hence the actual temperature of the products is slightly greater than 3200 K; a second guess from tables gives a value of 3280 K. ‘The temperature calculated in Example 7.15 is that which would be obtained by adiabatic combustion of the mixture. This temperature would not be attained in practice due to the effect of dissociation; this was discussed in section 7.7 and the inclusion of this eflect is illustrated in Example 7.16. Calculate the final temperature for the mixture defined in Example 7.15 including the effect of dissociation. ‘The proportions of the constituents depend on the temperature so the combustion equation is written as CO + 0.4550, + 1.709Nq + aCO + bCO + 60, + 1.709Nz and b and c can be expressed in terms of a by an atomic balance. 217Combustion 28 Carbon balance: 1=a+b therefore b=1—a Oxygen balance: 1 + (0.455 x 2) = 2a +b + 2c 191 =2a +b +2c ie. 2c = 191 —2a—(1 ~a) therefore c= 0.455 —0.5a and CO + 0.4550, + 1.709N, + aCOz + (1 — a)CO + (0.455 — 0.5a)O, + 1.709N, ‘The energy equation can be written as previously (Up, — Up,) + AUg + (UR, — Ur,) =0 Using the values calculated in Example 7.15, Up, = —(a x 2479) — {(1 — a) x 2479} — {(0.455 — 05a) x 2479} — [1.709 x 2479] = — (3.164 — 0.5a) x 2479 ky AUy = ~a x 281751 kJ Uz, = —T844KT— Up, = 9487KI Substituting, Up, + {(3.164 — 0.5a) x 2479} — 281 751a— 17331 = 0 therefore Up, A second equation is derived using the equilibrium constant for the reaction 2 ofr} af |" Poo LPo,Ji2 (1 = a) {(0.455 — 0.5a)p2 where p» is the pressure of the mixture after combustion and n, the number of kmol of the products. Now piV=n,RT, and p,V=n,RT, my _ Tam _ 555, (1+0.455 + 1.709) _ 212.08 Pe TP, Th 8.28 T, Substituting in the expression for K gives Kou_4 { 21208 __ ye 121 (=) (0455 — 05a), ‘When the value of 7; is known then Up, can be found by reading off @ for each of the products and multiplying by the amount of substance of each; equating this to the value of Up, in [1] allows a value of a to be calculated. 487 + 2829910 C1] kmol/bar79 7.9 Enthalpy of formation, Aly Similarly, when 7; is known a value of In(K®) can be read from tables, and from [2] a value of a is then found. Therefore, one convenient trial-and-error method is to choose a value of T;, caloulate a value of a from [1], and then find a value of K® from [2]. Compare this value of K° with that found from tables at T,. When the two values of K° are the same then the correct value of T; has been found, For example, at T, = 3000 K, from tables: Up, = 127920a + 68 598(1 — a) + 73155(0.455 — 0.5a) + (67795 x 1.709) = 217745 + 2274450 ie. in [1], 217745 + 22744.5a = 9487 + 282991a therefore a=08 Substituting for a and 7, in [2] xe = 08 212.08 ye © 0.2 ((0.455 — 0.5a) x 3000. = 454 From tables at T, = 3000 K, In(K*) = 1.110 therefore K° = 3.03 (too low) By such a method the value of 7; is found to be 2949 K to the nearest degree. This value should be compared with the value of 3280 K from Example 7.15 when dissociation was ignored. Enthalpy of formation, Ah, ‘The combustion reaction is a particular kind of chemical reaction in which products are formed from reactants with the release or absorption of energy as heat is transferred to or from the surroundings. As some substances, for instance hydrocarbon fuels, may be many in number and complex in structure the enthalpy of reaction may be calculated on the basis of known values of the molar enthalpy of formation, Af? of the constituents of the reactants and products at the reference temperature Ty. ‘The molar enthalpy of formation Ah, is the increase in enthalpy when a compound is formed from its constituent elements in their natural form and in a standard state. Something needs to be said about the standard form. The normal forms of oxygen (O2) and hydrogen (H,) are gaseous, so Ali? for these can be put equal to zero. The normal form of carbon (C) is graphite, so Ai? for solid carbon is put to zero. Carbon in another form, e.g. diamond or gas, 219Combustion heats of formation Ah? of various species at 25°C (298 K) and 1 bar 220 Example 7.17 Solution is not ‘normal’ and Ai? is quoted. The standard state is 25°C, and 1 bar pressure, but it must be borne in mind that not all substances can exist in the natural form, e.g. HO cannot be a vapour at 1 bar and 25°C. For calculation purposes, for a particular reaction Al? = Yn, Ah? — Yn, Akg (7.33) F © Typical values of Ai? are quoted for different substances in Table 7.11 Ale Substance Formula State (kS/kmol) ° Gas 249170 Oxygen 0; Gas 0 Water 1,0 Liquid = 285820 Water H,0 Vapour —241 830 Carbon c Gas 714990 Carbon c Diamond 1900 Carbon c Graphite 0 ‘arbon monoxide CO Gas 11030 Carbon dioxide CO, Gas —393520 Methane CH, Gas 74870 Methyl alcohol CH,OH Vapour = 240532 Ethyl alcohol CH,OH Vapour —281 102 Bthane CoH. Gas 83870 Ethene GH, Gas 52470 Propane Hy Gas 102900 Butane CuHy Gas —125000 Octane CoH Liquid =247 600 Calculate the molar enthalpy of reaction at 25°C of ethyl alcohol, CH,OH, using the data of Table 7.11. ‘Using equation (7.33) Alig = J n,Ahi —Y n,Ahe ¥ ® and the combustion equation C,H,OH + 30, > 2CO, + 3H,0 Ln Ahe =1 x (—281 102) +3 x 0 = —281 102 Ln Ake = 2 x (—393 520) +3 x (—241 830) = — 1512530 P therefore 1512530 —(—281 102) = —1231 428 kJ/kmol7.10 711 7.11. Power plant thermal efficiency Calorific value of fuels The quantities Ah® and Au® are approximated to in fuel specifications by quantities called calorific values which are obtained by the combustion of the fuels in suitable apparatus. This may be of the constant volume type (e.g. bomb calorimeter) or constant pressure, steady-flow type (e.g. Boys’ calorimeter). Both of these are described in section 7.1. Definitions of calorific value are as follows: (1) The energy transferred as heat to the surroundings (cooling water) per unit quantity of fuel when burned at constant volume with the HO product ‘of combustion in the liquid phase is called the gross (or higher) calorific value (GCY) at constant volume Q,,.. This approximates to —Au° at a reference temperature of 25°C with the H,O in the liquid phase. If the H,O products are in the vapour phase the energy released per unit quantity is called the net (or lower) calorific value (NCV) at constant volume, Qyer- This approximates to — Au? at 25°C with the HO in the vapour phase. (2) The energy transferred as heat to the surroundings (cooling water) per unit quantity of fuel when burned at constant pressure with the H,O products of combustion in the liquid phase is called the gross (or higher) calorific value (GCV) at constant pressure, Q,,,». This approximates to —AA° at a reference temperature of 25°C with the HO in the liquid phase. If the HO products are in the vapour phase the energy released is called the net (or lower) calorific value (NCV) at constant pressure, Qn,» This approximates to —Ah® at 25°C with the H,O in the vapour phase. Contrary to the definition of Au® and Ah? it is usual to quote calorific values as positive quantities, If Qy,,, and the fuel composition is known, the other quantities can be calculated. The above quantities are related as follows. For the constant volume process Qero = Aner + Melty (7.34) For the constant pressure process erp = Anetsn + Moby (7.35) where m, is the mass of condensate per unit quantity of fuel. Ugg = try at 25°C for HO = 2304.4 kI/kg Iigg = hgg at 25°C for HO = 2441.8 kI/kg The calorific values differ from Ah® and Au® due to the departure of experimental conditions from ideal with regard to temperatures of products and reactants and also heat transfer conditions. This topic is discussed further in section 7.12 Power plant thermal efficiency ‘The purpose of any power plant is the power output which should be obtained as economically as possible consistent with capital cost and running conditions. 221Combustion Example 7.18 Solution 222 It is necessary to assess the overall performance of a plant for comparison purposes and an important criterion is the overall thermal efficiency no. This is defined as work output fuel energy supplied Itis necessary to decide on the denominator for this definition. It is desirable, if we consider a steady-flow process, to relate the plant conditions to those for the steady-flow calorimeter in which the products are cooled to atmospheric temperature giving Q,,,y- However, it is not possible, or desirable, to cool the products of combustion in a real plant to atmospheric temperature so there is, a substantial energy loss to atmosphere. Complete cooling of the products would require large heat transfer surfaces which are expensive and the condensate produced would form corrosive acids. As achieving these conditions is not even attempted it seems that the use of Q,,, is unsatisfactory and Qgeup is more appropriate and this is often preferred. It does not matter for comparison purposes except that both values are in use making the definition of y somewhat Ww or (7.36) Queer It is necessary to state with the definition whether Qy..p OF Qneu is being used. In a plant it may be required to assess the boiler or steam generator performance only, in which case the boiler efficiency, given similarly in equation (8.16), is defined as . . heat transferred to working fluid Boiler efficiency =~ —v—v fuel energy supplied therefore heat transferred to working fluid ty = Li Qurp OF Qneuy tis again necessary to state whether Qr,,» OF Qnsry i8 being used when boiler efficiency is quoted. (737) A medium-size steam boiler required to supply a generator of output 25 000 KW has a performance specification as follows: steam output 31.6 kg/s; steam pressure 60 bar; steam temperature 500°C; feed water temperature 100°C; fuel, natural gas (96.5% CH,, 0.5% C,H, remainder incombustible); gross calorific value 38 700 kJ/m? at 1.013 bar and 15°C; fuel consumption 285 m3/s. Calculate the boiler efficiency and the overall thermal efficiency based on the net calorific value of the fuel. 1 kmol of CH, burns to give 2 kmol HO, therefore 0.965 kmol of CH, burns to give 2 x 0,965 x 18 = 34.74 kg H,O.712 742. Practical determination of calorific values 1kmol of C,H, burns to give 3kmol HO, therefore 0.005 kmol of C;H, burns to give 3 x 0,005 x 18=0.27kg HO. Therefore 1kmol of gas produces 34.74 +0.27=3501kg HO, and 1kmol of gas occupies a volumeof (RT/p) = (8314.5 x 288/1.013 x 10°) = 23.64m* at 1.013 bar and 15°C, therefore 35.01 Steam formed per m? of gas = 55 = 1481 ke Using equation (7.35) rep = Dorp + Melty therefore Queup = 38700 — (1.481 x 2441.8) = 35084 kI/m? Heat to working fluid = Hyappty stam — Hreedwa = 3421 — 419.1 = 3001.9 kI/kg ‘Using equation (7.37), therefore __ 3001.9 x 31.6 © 285 x 35084 tm 95(95%) and using equation (7.36) _ work output _ 0.25(25%} Quen oo%) No Practical deter: ation of calorific values The methods for determining calorific value depend on the type of fuel; solid and liquid fuels are usually tested in a bomb calorimeter, and gaseous fuels in a continuous-flow apparatus such as Boys’ calorimeter. The apparatus in every case is required to meet with a standard specification and the procedure to be adopted is also laid down, Solid and liquid fuel In the bomb calorimeter combustion occurs at constant volume and is a non-flow process; in Boys’ calorimeter the gas is burnt continuously under steady-flow conditions. The bomb is a small stainless steel vessel in which a small mass of the fuel is held in a crucible (see Fig. 7.14). If the fuel is solid, itis usually crushed, passed through a sieve, and then pressed into the form of a pellet in a special press. The size of pellet is estimated from the expected heat release, and is such that the temperature rise to be measured does not exceed 2~3 K. The pellet is ignited 223Combustion Fig, 7.14 Bomb calorimeter 224 Connections to external firing circuit Drive sO tectrcal connections Release valve ~ |__ Oxygen inlet valve Terminal insulated ft wa from bomb t ee | Stirrer— Cotton seule Platinum] H ee Pellet wire 4 Fixed range thermometer (or Beckmann type) —~ Water ‘Nylon feet by fusing a piece of platinum or nichrome wire which is in contact with it. The wire forms part of an electric circuit which can be completed by a firing button, which is situated in a position remote from the bomb. With a special form of press the pellet can be formed with the fuse wire passing through it. This facilitates the firing, particularly with some of the more difficult fuels. The crucible carrying the pellet is located in the bomb, a small quantity of distilled water is put into the bomb to absorb the vapours formed by combustion and to ensure that the water vapour produced is condensed, and the top of the bomb is screwed down. Oxygen is then admitted slowly until the pressure is above 23 atm. The bomb is located in the calorimeter and a measured quantity of water is poured into the calorimeter. The calorimeter is closed, the external connections to the circuit are made, and an accurate thermometer of the fixed range or Beckman type is immersed to the proper depth in the water. The watet is stirred in a regular manner by a motor-driven stirrer, and temperature observations are taken every minute. At the end of the fifth minute the charge is fired and temperature readings are taken every 10 s during this period. When the temperature readings begin to fall the frequency of readings can be reduced to every minute.Example 7.19 Table 7.12 Results for a bomb calorimeter test Solution The measured temperature rise is corrected for various losses. The cooling loss is the largest, but corrections are also necessary for the heat released by the combustion of the wire itself, and for the formation of acids on combustion. ‘The cooling correction can be determined graphically or by use of a formula developed by Regnault and Pfaundier. The allowance for the combustion of the wire is determined from its weight and known calorific value. The allowance for acids present is determined by a chemical titration. For most purposes only the correction for cooling need be applied. If a liquid fuel is being tested, it is contained in a gelatine capsule and the firing may be assisted by including in the crucible a little paraffin of known calorific value. ‘The water equivalent of the calorimeter is determined by burning a fuel of known calorific value (e.g. benzoic acid) in the bomb. The calculation for the test is then as follows: Mass of fuel x calorific value = (mass of water + water equivalent of bomb) x corrected temperature rise x specific heat capacity of water From this equation the calorific value of the fuel tested can be found, Table 7.12 gives the results of a bomb calorimeter test on a sample of coal. The mass of coal burned was 0.825 g and the total water equivalent of the apparatus was determined as 2500 g. Calculate the calorific value of the coal in kilojoules per kilogram. The temperature rise is to be corrected according to the formula by Regnault and Pfaundler, but no correction need be made for the acids formed. Precfiring Cooling period period time Temp. Heating time Temp. time Temp. (min. (°C) (min.) co) (min) (oy 0 25.730 | 16 27340 |, 10 27.880 1 25732 | 7 27.880 in 27878 2 25734 | 8 27.883 12 21816 3 25.736 | 149 27885 13 27814 4 25.738 14 21872 tS 25.740 15 27870 The Regnault and Pfaundler cooling correction is as follows: b, 2H,0 therefore 4kg H+ 32kg 0, 36kg H,O or Lkg H gives 9 kg H,0, therefore 0.144 kg H gives 0.144 x 9 = 1.296 kg HO ‘Then using equation (7.34) Qneww = Qarw — Mele therefore rere = 46900 — (1.296 x 23044) 43910 kI/kg Table 7.13 gives some typical calorific values of solid, liquid, and gaseous fuels, Air and fuel-vapour mixtures The mixture supplied to an engine fitted with a carburettor is one of air and fuel vapour, and the quality of the mixture is controlled by the carburettor. If the mixture is saturated with fuel vapour then the relative proportions of fuel to air can be determined from a knowledge of the temperature~pressure relationship for the saturated fuel. Such values are given for ethyl alcohol in Table 7.14. In Example 7.3 the stoichiometric A/F ratio for ethyl alcohol, C,H,O, was found to be 8,957/1. If the NCV of ethyl alcohol is 27800 kJ/kg, calculate the calorific value of the combustion mixture per cubic metre at 1.013 bar and 15°C. The molar mass of ethyl alcohol is 46 kg/kmol, therefore we have amount of substance of fuel per kilogram of fuel = = 0.02174 kmol/kg The molar mass of air is 28.96 kg/kmol, therefore amount of substance of air per kg of fuel = au = 0.3093 kmol/kg, ‘Total amount of substance of mixture = 0.021 74 + 0.3093 = 0.3310 kmol/kgTable 7.13 Typical calorific values of some fuels Table 7.14 Approximate saturation ‘temperatures and pressures for ethyl aleohol (CHO) 7.13. Air and fuel-vapour mixtures Calorific value at 15°C (kJ/kg) Fuel Gross Net Solid Anthracite 34.600 33900 Bituminous coal 33.500 32450 Coke 30750 30500 Lignite 21.650 20400 Peat 15900 14500 Liquid 100-octane petrol 47300 44000 Motor petrol 46.800 43.700 Benzole 42000 40200 Kerosene 46250 43.250 Diesel oil 46000 43.250 Light fue! oil 44800 42100 Heavy fuel oil 44000 41300 Residual fuel oil 42100 40.000 Calorific value at 15°C and 1 bar (MJ/m*) Gas Gross Net Butane 122.00 113.00 Propane 96.00 86.00 Natural gas 38.20 35.20 Coal gas 20.00 1785 Hydrogen 11.85 10.00 Producer gas 6.04 6.00 Blast-furnace gas Al 337 Temp/(°C) 0 10 20 30 40 50 Ey Pressure/(bar) 0.0162 0.0314 0.0584 0.1049 0.1800 0.2960 0.4690 From equation (2.8) pV =nRT therefore 0.331 x 8314.5 x 288 Tas iar 7 7824:m* per kg of ful 013 x 229‘Combustion Example 7.22 Solution 7A 72 73 230 = 27.8 MJ/kg Now —NCV of fuel = Qye., therefore 278 NCV of mixture = =~ = 3.55 MJ/m? c of mixture = 77 /m? of mixture For a stoichiometric mixture of ethyl alcohol and air calculate the temperature above which there will be no liquid fuel in the mixture. The pressure of the mixture is 1.013 bar. Using the results of Example 7.21, we have amount of substance of ethyl alcohol = 0.02174 kmol/kg and Total amount of substance = 0.331 kmol Then using equation (6.14), p; = (n/n), we have 0.021 74 Partial pressure ofethyl alcohol vapour x 1.013 0.331 0.0665 bar From Table 7.14, the saturation temperature corresponding to this pressure lies between 20 and 30°C. Therefore interpolating 0.0665 — 0.0584 204+ (ot * (cin — 0.0584 ) x (30 ~ 20) = 21.74°C Hence the minimum temperature of the mixture is 21.74°C for complete evaporation of the liquid fuel. Problems A sample of bituminous coal gave the following ultimate analysis by mass: C 81.9%; H 4.9%; O 6%; N 2.3%; ash 49%. Calculate: (i) the stoichiometric A/F ratio; (ii) the analysis by volume of the wet and dry products of combustion when the ait supplied is 25% in excess of that required for complete combustion, (10.8/1; CO 14.14%; HaO 5.07%; O; 408%; N, 76.71%; CO, 14.89%; 0, 4.30%; Np 80.81%) An analysis of natural gas gave the following values: CH, 93%; C,H, 4%; Nz 3%. Calculate the stoichiometric A/F ratio and the analysis of the wet and dry products of combustion when the A/F ratio is 12/1. (9524/1; CO, 7.76%; HzO 15.21%; O; 399%; N, 73.04%; CO; 9.15%; 0; 4.71%; Nz 86.14%) Calculate the stoichiometric A/F ratio for benzene (CgH,), and the wet and dry analysis of the combustion products. (13.2/1; CO, 16.13%; HO 8.06%; Nz 75.81%; CO, 17.54%; Nz 82.46%)7 75 16 a 74 79 7.40 7a 7212 Problems In the actual combustion of benzene in an engine the A/F ratio was 12/1. Calculate the analysis of the wet products of combustion. (CO, 13.38%; CO 3.94%; HO 8.66%; Nz 74.03%) ‘The ultimate analysis of a sample of petrol was 85.5% C and 14.5% H. Calculate: (i) the stoichiometric A/F ratio; (ii) the A/F ratio when the mixture strength is 90%; (iii) the A/F ratio when the mixture strength is 120%; (iv) the analyses of the dry products for (ii) and (iii); (v) the volume flow rate of the products through the engine exhaust per unit rate of fuel consumption for (iii) when the pressure is 1.013 bar and the temperature is 110°C. (14.76/15 16.4/1; 12.3/1; CO, 13.38%; Oz 2.24%; Na 84.38%; CO, 8.67%; CO 8.79%; Nz 82.54%; 15.11 m?/s per kg/s) ‘The ultimate analysis of a sample of petrol was C 85.5% and H 14.5%. The analysis of the dry products gave 14% CO, and some Oz. Calculate the A/F ratio supplied to the engine, and the mixture strength. (18.72/13 94%) Inan engine test the dry product analysis was CO, 15.5%; O 2.3% and the remainder ‘Nz. Assuming that the fuel burned was a pure hydrocarbon, calculate the ratio of carbon to hydrogen in the fuel, the A/F ratio used, and the mixture strength (15; 1484/1; 89.5%) ‘The ultimate analysis of a sample of petrol was 85% C and 15% H. The analysis of the dry products showed 13.5% CO,, some CO and the remainder N. Calculate: (i) the actual A/F ratio; (ii) the mixture strength; (ii) the mass of H, O vapour carried by the exhaust gas per kilogram of total exhaust gas; (iv) the temperature to which the gas must be cooled before condensation of the HzO vapour begins, if the pressure in the exhaust pipe is 1.013 bar. (1431/1; 104%; 0.088 kg/ke; 52.7°C) ‘A quantity of coal used in a boiler had the following analysis: 82% C; 5% H; 6% O} 2% N; 5% ash. The dry flue gas analysis showed 14% CO, and some oxygen. Calculate: (i) the oxygen content of the dry flue gas; (ii) the A/F ratio and the excess air supplied. (5.52%; 1429/1; 31.8%) For the mixture of Problem 7.4 calculate the calorific value per cubic metre of mixture at 1.013 bar and 38°C. The calorific value of benzene is 40700 kI/kg. (3.73 M3/m?) ‘The lower explosive limit of ethyl alcohol in air is 3.56% by volume at a pressure of 1013 mbar. If the pressure in a room is 1013 mbar calculate the lowest temperature at which the explosive mixture would be formed. What quantity of ethyl alcohol in litres would be needed in a room of volume 115m? to produce this mixture. The specific gravity of liquid ethyl alcohol is 0.794. Use the data of Table 7.14 for this problem. (11.73°C; 10.15/1) ‘The products of combustion of a hydrocarbon fuel, of carbon to hydrogen ratio 0:85:0.15, are found to be CO, 8%, CO 1%, O; 8.5%. Calculate the A/F ratio for the process by two methods and hence check the consistency of the data, (23.70, 23.53) 231Combustion 232 713 716 TAS TAG TAT 7.18 7.19 7.20 721 7.22 A stoichiometric mixture of CO and air was burned adiabatically at constant volume and at the peak pressure of 7.2 bar the temperature was 2469°C; analysis showed the volume of CO present in the products to be 0.192 of the volume of CO supplied. Show by calculating the equilibrium constant for CO + $0, =*CO, that the data are consistent. The value of In K° for this reaction at 2600 K is 2.800 and at 2800 K it is 1.893, The products ofa fuel when measured at a high temperature gave the following analysis: 9.21% CO3; 4.00% CO; 14.20% H0; 0.90% H; 71.03% Nz. Using appropriate values of equilibrium constants from tables such as those of ref. 7.3, estimate the temperature and pressure of the products. (2800 K; 2031 bar) A stoichiometric mixture of benzene (CoH) and air is induced into an engine of volumetric compression ratio 5 to 1, The pressure and temperature at the beginning of compression are 1 bar and 100°C. The estimated maximum temperature reached, allowing for dissociation, after adiabatic combustion at constant volume is 2727°C and at this temperature the standard equilibrium constants are Poo,(p?)\? ya 3034 PH,0Peo _ 7.214 Pu,Poos Pco(Po,) Show that about 74.4% of the carbon in the fuel is burned to CO, and calculate the maximum pressure reached, (41.7 bar) The enthalpy of combustion of propane gas, C,H, at 25°C withthe HO inthe products in the liquid phase is 50360 kJ/kg. Calculate the enthalpy of combustion with the H,0 in the vapour phase per unit mass of fuel and per unit amount of substance of ful. (46 364 kI/kg; —2040 030 kI/kmol) Calculate, for propane liquid, CsHg, at 25°C the enthalpy of combustion with the HO. in the products in the vapour phase. Use the data of Problem 7.16 and take lig at 25°C for propane as 372 kJ/kg. (45992 Kd/kg) Calculate the internal energy of combustion per unit mass of propane vapour, CsHy, at 25°C with the HO in the vapour phase from the corresponding value of Ah® = —46 364 KI /ke. (=46 420 kI/kg) Calculate the internal energy of combustion per unit mass for gaseous propane, CH, at 25°C with the HzO of combustion in the liquid phase from the corresponding value of Ah® = —50 360 kI/kg, (50191 kS/kg) Calculate the internal energy of combustion per unit mass for liquid propane, CsHg, at 25°C with the HO of combustion in the vapour phase from the corresponding value of ‘Ah® = —45992 kI/kg (46 105 kI/kg) Ah? for hydrogen at 60°C is given as —242400 kI/kmol. Calculate Af® at 1950°C given that the molar enthalpies of the gases concerned are as in Table 7.15. (~252087 kI/kmol) A stoichiometric mixture of hydrogen and air at 25°C is ignited and combustion takes, place adiabatically at a constant pressure of 1 bar. Af® for hydrogen at 25°C with theTable 7.15 Molar enthalpies of gases for problem 7.21 7.23 7.26 7.25 References Molar enthalpy/(ki/kmol) Gas °c 1950°C Hy 9492 69.250 O: 9697 76500 H,0 11147 94620 HO in the liquid phase is ~286000 kJ/kmol. Calculate, neglecting changes in kinetic energy, and using the tables of ref. 7.6: (i) the temperature reached if the process is assumed to be adiabatic and dissociation is neglected; (ii) the temperature reached after adiabatic combustion if the constituents are H, 02. HO, and Nz AL 25°C fig for H3O is 2441.8 KT/kg (2284°C; 200°C) Octane vapour, CsHyg, is to be burned in air in a steady-flow process. Both the fuel and air are supplied at 25°C and the product temperature is to be 760°C. Dissociation is negligible and the heat loss is to be taken as 10% of the increase in enthalpy of the products above the reference temperature. Ah for octane vapour is —5 510294 kJ/kmol with the products in the liquid phase. Calculate the A/F ratio by mass required; hy, at 25°C for HzO is 2441.8 kd /kg. Use the molar enthalpies in ref. 7.6. (49.2) Calculate the molar enthalpy of reaction of methane CH,, at 25°C and 1 bar pressure, Use Table 7.11 (p. 220) and assume the HO of combustion to be in the liquid phase. (890290 kJ/kmol) Calculate the molar enthalpy of combustion at 25°C and 1 bar pressure of a gas of volumetric analysis 12% Hz, 29% CO, 2.6% CH, 0.4% CHg, 4% CO, and 52% Nz for the H,O in the vapour phase and in the liquid phase. (—137240 kJ/kmol; — 145 158 kJ/kmol) References 7A BS 1016: 1989 Methods for Analysis and Testing of Coal and Coke 712 WaRKER S11 and pAcKHURST JR 1981 Fuel and Energy Academic Press 713. easror TD and croFr D x 1990 Energy Efficiency Longman A. sko0G DA and LEARY JL 1992 Principles of Instrumental Anal College Publishing 7S WILLARD HH, MERRITT (jr) LL, DEAN J A and SETTLE (jr) F A 1988 Instrumental Methods of Analysis 7th ed Wadsworth 7.6 noGERs GF C and MAYHEW Y R 1987 Thermodynamic Properties of Fluids and Other Data 4th edn Blackwell 7.7 BS 3804: Part 1 1964 Methods for the Determination of the Calorific Value of Fuel Gases is 4th edn Saunders 233Fig. 81 Carnot cycle for a wet vapour on a T-s diagram Bo Steam Cycles The heat engine cycle is discussed in Chapter 5, and it is shown that the most efficient cycle is the Carnot cycle for given temperatures of source and sink. This applies to both gases and vapours, and the cycle for a wet vapour is shown in Fig. 8.1. A brief summary of the essential features is as follows: 4 to 1: heat is supplied at constant temperature and pressure. 1 to 2: the vapour expands isentropically from the high pressure and temperature to the low pressure. In doing so it does work on the surroundings, which is the purpose of the cycle. 2 to 3: the vapour, which is wet at 2, has to be cooled to state point 3 such that isentropic compression from 3 will return the vapour to its original state at 4, From 4 the cycle is repeated, The cycle described shows the different types of processes involved in the complete cycle and the changes in the thermodynamic properties of the vapour as it passes through the cycle. The four processes are physically very different from each other and thus they each require particular equipment. The heat supply, 4-1, can be made in a boiler. The work output, 1-2, can be obtained by expanding the vapour through a turbine, The vapour is condensed, 2~3, in a condenser, and to raise the pressure of the wet vapour, 3~4, requires a pump or compressor. Thus the components of the plant are defined, but before these are discussed81 Fig. 82 Rankine cycle using wet steam on a T-s diagram 8.1 The Rankine cycle further, the deficiencies of the Carnot cycle as the ideal cycle for a vapour must be considered. The Rani e cycle It is stated in section 5.3 that, although the Carnot cycle is the most efficient cycle, its work ratio is low. Further, there are practical difficulties in following it. Consider the Carnot cycle for steam as shown in Fig, 8.1: at state 3 the steam is wet at T, but itis difficult to stop condensation at the point 3 and then compress it just to state 4. It is more convenient to allow the condensation process to proceed to completion, as in Fig. 82. The working fluid is water at the new state point 3 in Fig. 82, and this can be conveniently pumped to boiler pressure as shown at state point 4, The pump has much smaller dimensions than it would have if it had to pump a wet vapour, the compression process is carried out more efficiently, and the equipment required is simpler and less expensive. One of the features of the Carnot cycle has thus been departed from by the modification to the condensation process. At state 4 the water is not at the saturation temperature corresponding to the boiler pressure. Thus heat must be supplied to change the state from water at 4 to saturated water at 5; this is a constant pressure process, but is not at constant temperature. Hence the efficiency of this modified cycle is not as high as that of the Carnot cycle. This ideal cycle, which is more suitable as a criterion for actual steam cycles than the Carnot cycle, is called the Rankine cycle, The plant required for the Rankine cycle is shown in Fig. 8.3, and the numbers refer to the state points of Fig. 8.2. The steam at inlet to the turbine may be wet, dry saturated, or superheated, but only the dry saturated condition is shown in Fig. 8.2. The steam flows round the cycle and each process may be analysed using the steady-flow energy equation; changes in kinetic energy and potential energy may be neglected, then for unit mass flow rate Q+W=dh Each process in the cycle can be considered in turn as follows. Boiler: Qasi + W= hy he 235Steam Cycles Fig, 83. Basic steam plant 236 Boundary ett n “Be Condenser 1 a | water Ors Therefore, since W = 0, ass = hy hy (81) ‘Turbine: the expansion is adiabatic (ie. Q = 0), and isentropic (ie. s, = s2), and h, can be calculated using this latter fact. Then Qi2 + Wiz = hy — hy therefore Waa = ha — hy or Work output, —Wy2 = hy — hy (8.2) Condenser: Qr3 + W = hy — hy Therefore, since W = 0 Qa3 = hs — hy therefore Heat rejected in condenser, —Q,3 = hy — hy (8.3) Pump: Qs4 + Waa = hy — hy The compression is isentropic (ie. s, = s4), and adiabatic (ie. Q = 0), Therefore Waa = (hg — hs) ie. Work input to pump, Ws, 1g — hy (8.4) This is the feed-pump term, and as it is a small quantity in comparison with the turbine work output, — M2, it is usually neglected, especially when boiler pressures are low. ‘Net work input for the cycle BW = Wy2 + Wag8.1. The Rankine cycle ie, EW = (hz — hy) + (hg = hs) ot Net work output, —J W = (hy ~ ta) — (he hs) (85) Or, if the feed-pump work is neglected, Net work output, —EW=hy — hz (86) ‘The heat supplied in the boiler, Q4s1 = ty — hy. Then we have network output eee sa ear app lid in tie bole 67) ~ hy — hy ory = fa) = bh = bs) (88) (hy = ts) = (ig = hs) If the feed-pump term, hy — hs, is neglected equation (8.8) becomes hy = (8.9) “hy hy When the feed-pump term is to be included it is necessary to evaluate the quantity, Wes. From equation (8.4) Pump work = Wy, = hy — hy Tt can be shown that for a liquid, which is assumed to be incompr v= constant), the increase in enthalpy for isentropic compression is given by (Ita — ha) = (Pa ~ Ps) The proof is as follows. For a reversible adiabatic process, from equation (3.15), dQ =dh—vdp=0 therefore dh=vdp ie. f nf vdp For a liquid, since » is approximately constant, we have 4 ho =of dp = v(pa — Ps) 3 je. hg — hy = 0(74 — Ps) therefore Pump work input = hy — ty (P4 — Pa) (8.10) where v can be taken from tables for water at the pressure p.. 237Steam Cycles Fig. 84 Rankine cycle showing real processes on a T-s diagram 238 The efficiency ratio of a cycle is the ratio of the actual efficiency to the ideal efficiency, In vapour cycles the efficiency ratio compares the actual cycle efficiency to the Rankine cycle efficiency, cycle efficiency ie. Efficiency ratio = ———_____—_ z ankine efficiency (8:11) The actual expansion process is irreversible, as shown by line 1-2 in Fig. 8.4, Similarly the actual compression of the water is irreversible, as indicated by line 3-4, The isentropic efficiency of a process is defined by actual work Isentropic efficiency =“ “°"__ for an expansion process isentropic work and isentropic work ‘actual work Isentropic efficiency Hence Turbine isentropic efficiency (8.12) It has been stated that the efficiency of the Carnot cycle is the maximum possible, but that the cycle has a low work ratio. Both efficiency and work ratio are criteria of performance. By the definition of work ratio in section 5.3, net work output Work ratio = “Wor oupet gross work output (8.13) Another criterion of performance in steam plant is the specific steam consumption (ssc). It relates the power output to the steam flow necessary to produce it. The steam flow indicates the size of plant and its component parts, and the ssc is a means whereby the relative sizes of different plants can be compared. ‘The ssc is the steam flow required to develop unit power output. The powerExample 8.1 cycle output is — mW, therefore tt 1 siEW EW Neglecting the feed pump work we have -EW=h-h therefore ssc = (8.14) 1 (hy = ha) Note that when fy and hz are expressed in kilojoules per units of ssc are kg/kI or kg/kW h ssc jogram then the A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam consumption: (i) for a Carnot cycle using wet steam; ii) for a Rankine cycle with dry saturated steam at entry to the turbine; (iii) for the Rankine cycle of (ii), when the expansion process has an isentropic, efficiency of 80%. Solution (i) A Carnot cycle is shown in Fig, 8.5. Fig. 85 Carnot cycle for Example 8.1(a) 5262. 299.1 ‘Temperature/(K) T, = saturation temperature at 42 bar = 253.2 + 273 = 526.2 K T, = saturation temperature at 0.035 bar = 26.7 + 273 = 299.7 K Then from equation (5.1) 526.2 — 299.7 526.2 0.432 or 43.2% 239Stoam Cycles 240 Also. Heat supplied = hy ~ hy = hy at 42 bar = 1698 kJ/kg Net work output, — 5 W Gross heat supplied Therefore —P W = 0.432 x 1698, ie, Net work output, —, W = 734 kI/kg Then Nesmoe = 1432 To find the gross work of the expansion process it is necessary to calculate ha, using the fact that s; = 5). From tables hy = 2800 kJ/kg ands, = sz = 6049 kI/kg K Using equation (4.10) 5 = 6.049 = 5, + X25, = 0.391 + 28.13 therefore __ 6.049 — 0.391 “813 Then using equation (2.2) Ay = hy, + Xphyg, = 112 + (0.696 x 2438) = 1808 kJ/kg = 0.696 2 Hence, from equation (8.2) = Wy2 = hy — hy = 2800 — 1808 = 992 kI/kg Therefore we have, using equation (8.13), network output _ 734 _ 739 gross work output 992 Work ratio = Using equation (8.14) ey 734 ie, ssc = 0.001 36 kg/kW s = 491 kg/kWh (ii) The Rankine cycle is shown in Fig. 8.6. As in part (i) hy = 2800 kJ/kg and hy = 1808 kI/kg Also, hy = hy at 0.035 bar = 112 kI/kg, v, at 0.035 bar ssc Using equation (8.10), with » Pump work = (pq — p3) = 0.001 x (42 ~ 0.035) x z = 4.2 ki /kgFig 86 T-s diagram for Example 8.1(b) Fig. 87 T-s diagram for Example 8.1(c) 8.1 The Rankine cycle Using equation (8.2) Wy = hy — hz = 2800 — 1808 992 kd /kg, Then using equation (8.8) (hy = ha) = (h, hy hy) — (hh ie, ng = 368% ~hy)_ 992-42 hh) (2800 — 112) — 4.2 0.368 Using equation (8.13) network output _ 992-42 grosswork output 992 Work ratio 0.996 Using equation (8.14) 1 —yw se =! _ 9.00101 kg/kWs (992. = 4.2) ssc ie. 3.64 kg/kW h (iii) The cycle with an irreversible expansion process is shown in Fig. 8.7. Using equation (8.12) hy = hy Wa Tsentropic efficiency i. iy Tg Wane 2a1Steam Cycles Fig. 88 Steam cycle efficiency and specific, steam consumption against boiler pressure 242 therefore og = 2 992 ie — Wz = 0.8 x 992 = 793.6 KI /kg, Then the cycle efficiency is given by Cycleeficieney = = ha) — (ta = hs) ‘gross heat supplied a 793.6 — 4.2 © (2800 = 112) = 4.2 ie. Cyole efficiency = 29.4% = 0.294 = pump work _ 793.6 4.2 Wa 793.6 0.995 Also ———— 7936 — 42 = 0.001 267 kg/kW s = 4.56 kg/kW h The feed-pump term has been included in the above calculations, but an inspection of the comparative values shows that it could have been neglected without having a noticeable effect on the results. It is instructive to carry out these calculations for different boiler pressures and to represent the results graphically against boiler pressure, as in Fig. 88. As the boiler pressure increases the specific enthalpy of vaporization decreases, thus less heat is transferred at the maximum cycle temperature. Although the efficiency increases with boiler pressure over the first part of the range, due to the maximum cycle temperature being raised, it is affected by the lowering of the mean temperature at which heat is transferred, Therefore the graph for this, efficiency rises, reaches a maximum, and then falls. 30. —— —s0 Eicieney 40 pe 0 |__| {36 2» 24 L ° 70 40210 Boiler pressure/t(bar) Cyele eficieney/(%) Specific steam consumption /(kg/kW8.2 Fig, 89 Steam plant with a superheater (a) and the cycle on a T-s diagram (b) Drum [F A ==| (a) Example 8.2 Solution 8.2 Rankine cycle with superheat Rankine cycle ih superheat The average temperature at which heat is supplied in the boiler can be increased by superheating the steam. Usually the dry saturated steam from the boiler drum is passed through a second bank of smaller bore tubes within the boiler. This bank is situated such that it is heated by the hot gases from the furnace until the steam reaches the required temperature. The Rankine cycle with superheat is shown in Fig. 8.9(a) and 89(b), Figure 8.9(a) includes a steam receiver which can receive steam from other boilers. In modern plant a receiver is used with one boiler and is placed between the boiler and the turbine. Since the quantity of feedwater varies with the different demands on the boiler, itis necessary to provide a storage of condensate between the condensate and boiler feed pumps. This storage may be either a surge tank or hot well. A hot well is shown dotted in Fig. 8.9(a). Receiver Turbine ‘ae 2 Condenser (b) Compare the Rankine cycle performance of Example 8.1 with that obtained when the steam is superheated to 500°C. Neglect the feed-pump work. From tables, by interpolation, at 42 bar: hy = 3442.6 kI/kg and 5, = 8 = 7.066 kI/ke K Using equation (4.10) 5 = 5, +25, therefore 0.391 + x,8.13 = 7.066 ie, x, = 0.821 Using equation (2.2) iy = hr, + Xahg, = 112 + (0.821 x 2438) = 2113 kd /kg From tables: hay = 112 kI/kg Then, using equation (8.2) Wyy = hy — hy = 3442.6 — 2113 = 1329.6 kI/kg, 243Steam Cycles ig. 810 Steam cycle efficiency and specific steam consumption against steam temperature at turbine entry 244 Neglecting the feed-pump term, we have Heat supplied = hy — hy = 3442.6 — 112 = 330.6 kI kg Using equation (8.9) I= ha _ 132966 _ 0399 or 39.9% Cycle effcieney = FF? = oe 5 MA Also, using equation (8.14) 1 hy—h, 13296 ssc = 0.000 752 kg/kW s = 2.71 kg/kW h The cycle efficiency has increased due to superheating and the improvement in specific steam consumption is even more marked. This indicates that for a given power output the plant using superheated steam will be of smaller proportions than that using dry saturated steam. The condenser heat loads for different plants can be compared by calculating the rate of heat removal in the condenser, per unit power output. This is given by the product, ssc x (Itp — hy), where (ht, — ht) is the heat removed in the condenser by the cooling water, per unit mass of steam, Comparing the condenser heat loads for the Rankine cycles of Examples 8.1 and 8.2, we have with dry saturated steam at entry to turbine, using the results from Example 8.1 (ii) Condenser heat load = 0,001 01(1808 — 112) = 1.713 KW per kW power output With superheated steam at entry to the turbine, using the results from Example 82: Condenser heat load = 0.000 752(2113 — 112) = 1,505 kW per kW power output, For given boiler and condenser pressures, as the superheat temperature increases, the Rankine cycle efficiency increases, and the specific steam consumption decreases, as shown in Fig. 8.10. so 60 ficiency oC —Sh48 So | 368 3 Poh se mics 2 "7 0 S12 ° | 260 500 70 Steam temperature/(°C)Fig. 811. 7-s diagram showing the effect of a higher boiler pressure on the steam condition in the turbine 8.3. The enthalpy-entropy chart ‘There is also a practical advantage in using superheated steam. For the data of the Rankine cycle of Examples 8.1 and 8.2, the steam leaves the turbine with dryness fractions of 0,696 and 0.821 respectively. The presence of water during the expansion is undesirable, since the droplets are denser than the remainder of the working fluid and therefore have different flow characteristics. The result is the physical erosion of the turbine blades, and a reduction in isentropic efficiency. The modern tendency is to use higher boiler pressures, and a comparison of cycles on the T-s diagram shows that for a given steam temperature at turbine inlet, the higher pressure plant will have the wetter steam at turbine exhaust (see Fig. 8.11 in which p, > p2). It is usual to design for a dryness fraction of not less than 0.9 at the turbine exhaust. The enthalpy-entropy chart In this chapter, and in later ones, we are concerned with changes in enthalpy. It is convenient to have a chart on which enthalpy is plotted against entropy. The h-s chart recommended is that of ref. 8.1, which covers a pressure range of 0.01-1000 bar, and temperatures up to 800°C. Lines of constant dryness fraction are drawn in the wet region to values less than 0.5, and lines of constant temperature are drawn in the superheat region. In general h-s charts do not show values of specific volume, nor do they show the enthalpies of saturated water at pressures which are of the order of those experienced in steam condensers. Hence the chart is useful only for the enthalpy change in the expansion process of the steam cycle; the methods used in Examples 8.1 and 8.2 are recommended for problems on the Rankine cycle ‘A sketch for the h-s chart is shown in Fig. 8.12, Lines of constant pressure are indicated by p;, P2, etc.; lines of constant temperature by T;, 7, etc. Any two independent properties which appear on the chart are sufficient to define the state (e.g. p, and x, define state A, and hg can be read off the vertical axis). In the superheat region, pressure and temperature can define the state (e.g. ps and T, define the state B, and hy can be read off), A line of constant entropy between two state points B and C defines the properties at all points during an isentropic process between the two states. 245Steam Cycles Fig. 812 Sketch of an enthalpy-entropy chart for steam 246 8.4 Lines of constant temperature 8 Enthalpy (kJ/kg) Critical point Entropy/(ke}/kg K) The reheat cycle Its desirable to increase the average temperature at which heat is supplied to the steam, and also to keep the steam as dry as possible in the lower pressure stages of the turbine. The wetness at exhaust should be no greater than 10%. The considerations of section 8.2 show that high boiler pressures are required for high efficiency, but that expansion in one stage can result in exhaust steam which is wet, This is a condition which is improved by superheating the steam. The exhaust steam condition can be improved most effectively by reheating the steam, the expansions being carried out in two stages. Referring to Fig. 8.13, 1-2 represents isentropic expansion in the high-pressure turbine, and 6~7 represents isentropic expansion in the low-pressure turbine. The steam is reheated at constant pressure in process 2-6, The reheat can be carried out by returning the steam to the boiler, and passing it through a special bank of tubes, the reheat bank of tubes being situated in the proximity of the superheat tubes. Alternatively, the reheat may take place in a separate reheater situated near the turbine; this arrangement reduces the amount of pipe work required. The use of reheat cycles has encouraged the development of higher pressure, forced circulation boilers, since the specific steam consumption is improved, and the dryness fraction of the exhaust steam is increased. ‘The analysis is as follows: Heat supplied = Qys1 + Qe413 T-s diagram showing a reheat steam cycle Example 8.3 Solution Fig. 814. T-s diagram for Example 8.3 8.4 The reheat cycle ‘Neglecting the feed-pump work Qasr = hy — hy ‘Also, for the reheat process Qr5 = Nhe — he Work output = — Wy — Wor and —Wi2=h,—h, and —Woy = he — hy Cycle efficiency Waa — Wer Qas1 + O26 (hy = ha) + (Be = In) (hy = hy) + (Ae = hea) Calculate the new cycle efficiency and specific steam consumption if reheat is included in the plant of Example 8.2. The steam conditions at inlet to the turbine are 42 bar and 500°C, and the condenser pressure is 0.035 bar as before, Assume that the steam is just dry saturated on leaving the first turbine, and ig reheated to its initial temperature. Neglect the feed-pump term, The cy is shown on a T-s diagram in Fig, 8.14. 247‘Steam Cycles 248 8.5 It is convenient to read off the values of enthalpy from the h-s chart, ie. hy = 3442.6 kJ /Iegs hy = 2713 kJ/kg (at 2.3 bar); hg = 3487 kJ/kg (at 2.3 bar and 500°C); hy = 2535 kI/kg. From tables hy = 112 kJ/kg Then Turbine work = (hy — hy) + (he — hy) = (3443 — 2713) + (3487 — 2535) ie. Turbine work = 1682 kJ/kg Heat supplied = (h, — hs) + (ig — hy) = (3443 — 112) + (3487 — 2713) ic, Heat supplied = 4105 kJ/kg therefore Cycle efficiency = im OAl or 41% Also sso = = 0.000595 kg/k 168: ie, sso = 2.14 kg/kWh Comparing these answers with the results of Example 8.2 it can be seen that the specific steam consumption has been improved considerably by reheating (ie, reduced from 2.71 kg/KW h to 2.14 kg/kW h), The efficiency is greater (ie. increased from 39.9 to 41%). The cycle efficiency will be increased by reheating only if the mean temperature of the heat supply is increased; this will not be the case if the reheat pressure is too low. The regenerative cycle In order to achieve the Carnot efficiency it is necessary to supply and reject heat at single fixed temperatures, One method of doing this, and at the same time having a work ratio comparable to the Rankine cycle, is by raising the feedwater to the saturation temperature corresponding to the boiler pressure before it enters the boiler. This method is not a practical proposition, but is of academic interest, The feedwater is passed from the pump through the turbine in counter-flow to the steam, as shown in Fig. 8.15(a). The feedwater enters the turbine at ¢, and is heated to the steam temperature at inlet to the turbine. If at all points the temperature difference between the steam and the water is negligibly small, then the heat transfer takes place in an ideal reversible manner. Assuming dry saturated steam at turbine inlet, the expansion process is represented by line 1-2-2’ in Fig, 8.15(b). The heat rejected by the steam, area 12561, is equal to the heat supplied to the water, area 34783. The heat supplied in the boiler is given by area 41674, and the heat rejected in the condenser isBoiler Fig. 815 Steam plant operating (a) on a regenerative cycle and (b) the cycle on a T-s diagram Boller Fig. 8.16 Steam plant with (a) one open feed heater and (b) the cycle on a T-s diagram 8.5 The regenerative cycle Condenser a C) Turbine 4] ie rm 3 O Feed pump (a) (b) given by area 3'2'83'. This regenerative cycle has an efficiency equal to the Carnot cycle, since the heat supplied and rejected externally is done at constant temperature. This cycle is clearly not a practical proposition, and in addition it can be seen that the turbine operates with wet steam which is to be avoided if possible. However, the Rankine efficiency can be improved upon in practice by bleeding off some of the steam at an intermediate pressure during the expansion, and mixing this steam with feedwater which has been pumped to the same pressure. ‘The mixing process is carried out in a feed heater, and the arrangement is represented in Figs 8.16(a) and (b). Only one feed heater is shown but several could be used. i (=v) ke 2 Condenser ‘The steam expands from condition 1 through the turbine. At the pressure corresponding to point 6, a quantity of steam, say y kg per kilogram of steam supplied from the boiler, is bled off for feed heating purposes. The rest of the steam, (1 — y)kg, completes the expansion and is exhausted at state 2. This amount of steam is then condensed and pumped to the same pressure as the bleed steam. The bleed steam and the feedwater are mixed in the feed heater, and the quantity of bleed steam, y kg, is such that, after mixing and being 249Steam Cycles 250 Example 8.4 Solution pumped in a second feed pump, the condition is as defined by state 8. The heat to be supplied in the boiler is then given by (/1, — hg) kJ/kg of steam; this heat is supplied between the temperatures 7, and T,. If this procedure could be repeated an infinite number of times, then the ideal regenerative cycle would be approached. It is necessary to determine the bleed pressure when one or more feed heaters are used, and this can be based on the assumption that the bleed temperature to obtain maximum efficiency for such a cycle is approximately the arithmetic mean of the temperatures at 5 and 2 (see Fig. 8.16(b)), ts tty 2 je, thteea = (8.15) If the Rankine cycle of Example 8.1 is modified to include one feed heater, calculate the cycle efficiency and the specific steam consumption. The steam enters the turbine at 42 bar, dry saturated, and the condenser pressure is 0.035 bar. At 42 bar, t, = 253.2°C; and at 0.035 bar, t, Therefore by equation (8.15) _ 253.2 +.26.7 rt.) Sclecting the nearest saturation pressure from the tables gives the bleed pressure as 3.5 bar (ie. ty = 138.9°C). To determine the fraction y, consider the adiabatic mixing process at the feed heater, in which y kg of steam of enthalpy his, mix with (1 — y) kg of water of enthalpy hs, to give 1 kg of water of enthalpy ty. The feed pump can be neglected (ie. hg = hy). Therefore hg + (1 — y)hg = hy fy = hg _ hy — hy hig — hy he hs Now, hy = 584 IJ /kg; hy = 112 kJ/kg; ands, = 55 = 5 = 6.049 kJ/kg K. ‘Therefore 26.7°C. ts = 140°C ie ys 6.049 — 1.727 — = 0.829 ee S214 and x, = 5042 — 0.391 _ 4 696 8.130 Hence ig = hy, + Xehig, = 584 + (0.829 x 2148) = 2364 kJ/kg and hy = hy, + xahyg, = 112 + (0.696 x 2438) = 1808 kI/kgFig. 817 Steam plant with two closed feed heaters 8.5. The regenerative cycle therefore 584 — 112 2364 — 112 ‘Neglecting the second feed-pump term (i.e. hy = hg), we have Heat supplied in boiler = (, — h) = 2800 — 584 = 2216 kI/kg Total work output, —W = —Wy — Wea = (iy — hg) + (1 — y)(hg — ha) 2800 — 2364) + (1 — 0.21)(2364 — 1808) = 876 kJ per kilogram of steam delivered by the boiler y 21 ke, ie. Work output ‘Therefore Cycle efficiency ie ss0= x 0.001 142 kg/kJ = 4.11 kg/kWh ‘Comparing these results with those of Example 8.1, it can be seen that the addition of one feed heater has increased the cycle efficiency from 36.8 to 39.6%, but the specific steam consumption has increased from 3.64 kg/kW h to 4.11 kg/kWh. The cycle efficiency continues to be increased with the addition of further heaters, but the capital expenditure is also increased considerably. Because of the number of feed pumps required, the heating of the feed water by mixing is dispensed with and closed heaters are used. The method is indicated in Fig, 8.17 for two feed heaters, but the number used could be as high as eight. Referring to Fig. 8.17, the feedwater is passed at boiler pressure through the feed heaters 2 and 1 in series. An amount of bleed steam, y, is passed to feed heater 1, and the feedwater receives heat from it by the transfer of heat through the separating tubes. The condensed steam is then throttled to the next feed heater which is also supplied with a second quantity of bleed steam, y>, and a lower temperature heating of the feedwater is carried out. ‘After passing through the various feed heaters the condensed steam is then fed to the condenser. The temperature differences between successive heaters [=| Feed. heater 1 yk On Fyn ke 251Steam Cycles are constant, and in the ideal case the heating process at each is considered to be complete (ic. the feedwater leaves the feed heater at the temperature of the bleed steam supplied to it), Example 8.5 __In a regenerative steam cycle employing two closed-feed heaters the steam is supplied to the turbine at 40 bar and 500°C and is exhausted to the condenser at 0.035 bar. The intermediate bleed pressures are obtained such that the saturation temperature intervals are approximately equal, giving pressures of 10 and 1.1 bar. Calculate the amount of steam bled at each stage, the work output of the plant per kilogram of boiler steam and the cycle efficiency of the plant. Assume ideal processes where required. Solution Referring to Fig. 8.17 and the T-s diagram of Fig. 8.18, from tabl hy =3445.8kI/kg ands; = 7.089 kJ/kg K =s, Fig. 818 Ts diagram r for Example 8.5 (ys) ke. (19 ya) kg At state 2, 0.391 + (x2 x 8.13) = 7.089 therefore x, = £8 _ ogg 8.13 ie, hy = 112 + (0.824 x 2348) = 2117 KI /kg Also hy = hy at 0.035 bar = 112 kJ/kg For the first stage of expansion, 1~7, s; = s; = 7.089 kJ/kg K, and from tables at 10 bar s, < 7.089, hence the steam is superheated at state 7. By interpolation between 250 and 300°C at 10 bar we have 7.089 8220) 09 2944) = 2944 + oe x 108 7124 — 6.926, ie, hy = 3032.9 KI/kg For the throttling process, 11-12, we have 11 = Aya = 763 kI/kg hy = 2944 + ( 2528.6 8.6 Further considerations of plant efficiency For the second stage of expansion, 7-8, Sy = sg = 5; = 7.089 kJ/kg K, and from tables at 1.1 bar s, > 7.089 kJ/kg K, hence the steam is wet at state 8, Therefore, 1.333 + (xg x 5.994) = 7.089 therefore xg = 0.961 ie hy = 429 + (0.961 x 2251) 2591 KI /kg For the throttling process, 9-10: hy = he = hyo = 429 k3/ke Applying an energy balance to the first feed heater, remembering that there is no work or heat transfer: Yala + hs = aha + he _hig—hs 163-429 iy — hy, 3032.9 — 763 Similarly for the second heater, taking hg = hs: je oy 0.147 Yalta + Yily2 + hy = hs + (v1 + Ya)ho ite. Ya(he — hg) + Ysa + hy = hs + Yilto ‘ya(2591 — 429) + (0.147 x 763) + 112 = 429 + (0.147 x 429) therefore yy = 2878 <2 2162 ‘The heat supplied to the boiler, Q,, per kilogram of boiler steam is given by Q, = hy — hg = 3445 — 763 = 2682 kI/ke, ‘The work output, neglecting pump work, is given by = W = (hy — hy) + (1 = ya Mtg hy) + (1 = y= Ya) Clty — fa) = (3445 — 3032.9) + (1 — 0.147)(3032.9 — 2591) + (1 = 0.147 — 0,124)(2591 — 2117) = 412.1 + 376.9 + 345.5 = 1134.5 kJ/kg Ww _ 11345 Q, 2682 Then Cycle efficiency = 0.423 or 42.3% Further considerations of plant efficiency Up to now the considerations of efficiency have been based on the heat which is actually supplied to the steam, and not the heat which has been produced 253Steam Cycles Fig, 819 Steam plant with economizer and ait pre-heater 254 by the combustion of fuel in the boiler. The heat is transferred to the steam from gases which are at a higher temperature than the steam, and the exhaust ‘gases pass to the atmosphere at a high temperature, To utilize some of the energy in the flue gas an economizer can be fitted. This consists of a coil situated in the flue gas stream. The cold feedwater enters at the top of the coil, and as it descends it is heated, and continues to meet higher temperature gas. For the Carnot, the ideal regenerative, and complete feed heating cycles, no use can be made of an economizer since the feedwater enters the boiler at the saturation temperature corresponding to the boiler pressure, To cool the flue gas even further and improve the plant efficiency, the air which is required for the combustion of the fuel can be pre-heated. For a given temperature of combustion gases, the higher the initial temperature of the air then the less will be the energy input required, and hence less fuel will be used, Plants which have both economizer and pre-heater coils in the boiler usually require a forced draught for the flue gas, and the power input to the fan, which is a comparatively small quantity, must be taken into account in the energy balance for the plant. Figure 8.19 represents diagrammatically a plant with economizer, pre-heater, and a re-heater. The boiler efficiency is the heat supplied to the steam in the boiler expressed as a percentage of the chemical energy of the fuel which is available on combustion, hy — (enthalpy of the feedwater) m, x (GCV or NCV) where h, is the enthalpy of the steam entering the turbine and m, the mass of fuel burned per kilogram of steam delivered from the boiler. The GCV and NCV are the higher and lower calorific. values of the fuel, and the determination of these quantities was considered in Chapter 7. ie. Boiler efficiency (8.16) | Fiue{ gas | Pre-heater SSS Hower oS -=-|— Economizer bank =| First superheater bank =| reteater bank Second superheater an [— 5 f z ater [Murine | orine FE tubes 7 ey oo Condens Pomp8.7 Turbine | ‘Steam to process ig. 820 Back-pressure turbine for process steam Steam from am Bos fotprocas | Turbine Condenser Fig, 821 Pass-out turbine for process steam 8.7 Steam for heating and process use The size of the boiler, or its capacity, is quoted as the rate in kilogram per hour at which the steam is generated, A comparison is sometimes made by an equivalent evaporation, which is defined as the quantity of steam produced per unit quantity of fuel burned when the evaporation process takes place from and at 100°C. Steam for heating and process use ‘When steam is required for heating or for a process it may be raised in a boiler and used directly, or passed to a calorifier to heat water which is then circulated. In factory complexes where power and process steam are both required it is usually more efficient to use a plant combining both requirements. In reaching the compromise between power and process demands two main possibilities are available as discussed below. Back pressure turbine The turbine works with an exhaust pressure which is appropriate to the process steam requirements; the steam leaving the turbine is not condensed but is passed to the process. A typical example is shown in Fig. 8.20. One of the disadvantages of the back pressure turbine is that if the demand for process steam falls off but the power requirement is unchanged then some steam must be blown off to waste at the process steam pressure. If the power requirement increases with the process demand unchanged then excess power requirements can be bought from the grid. If the power requirements falls off with the process demand unchanged the best solution is to arrange to sell excess power to the grid. A back pressure turbine used for process steam is suitable for high values, of the ratio of process energy to power, say 10 or above. Pass-out tur! ie Steam is bled from the turbine at some point or points between inlet and exhaust and is passed to process work. A typical example is shown in Fig. 8.21. In this system the boiler supply conditions and the condenser pressure can be fixed and the process steam load varied by varying the mass flow rate of process steam bled off the turbine. If the rate of process steam flow is much less than the design value then the excess power generated by the second stage expansion can be solid to the grid; alternatively process steam can be blown off or the boiler operated at part load, but both of these alternatives are wasteful of energy. Ifthe power requirement falls off with the process demand unchanged, the best solution is to sell the excess power to the grid; a more wasteful solution is to blow off excess steam at the bleed point to reduce the power from the second stage of expansion. If the rate of process steam demand increases above the design value the best solution is to use a stand-by boiler to raise the excess, 255Steam Cycles Example 8.6 Solution Fig. 822 Processes on the h-s chart for Example 8.6 256 steam direct. The pass-out turbine system for satisfying process and power requirements is most suitable for low process energy to power ratios, say in the range from 4 to 10. ‘When the process steam energy to power ratio falls below about 4 it becomes more efficient to generate power and steam for process use separately. Heat—power ratios using various prime movers are discussed in more detail in ref. 7.2. A pass-out two-stage turbine receives steam at 50 bar and 350°C. At 1.5 bar the high-pressure stage exhausts and 12000 kg of steam per hour are taken at this stage for process purposes. The remainder is reheated at 1.5 bar to 250°C and then expanded through the low-pressure turbine to a condenser pressure of 0.05 bar. The power output from the turbine unit is to be 3750 kW. ‘The relevant values should be taken from an h-s chart, Take the isentropic efficiency of the high-pressure stage as 0.84, and that of the low-pressure stage as 0.81. Calculate the boiler capacity. The processes are shown on an h—s chart in Fig, 8.22. High-pressure stage: Actual work output = Misestropic * (ty — has) ie, (hhy — ha) = 0.84 x (3070 — 2397) = 565.3 kI/kg 3070 2973 Specific enthalpy (KI /ke) 207 3302 Low-pressure stage: (hy — hha) = Msentropic * (ts ~ has) = 081 x (2973 — 2392) = 470.6 kI/kg 12000 Process steam flow = = 3.33 ke/s 3600 Steam flow through the boiler = ri kg/s Steam flow through low-pressure stage = (1 — 3.33) kg/s Turbine power output = 3750 kW.8a 8.2 84 85 Problems. therefore ri( hy — ha) + (9h — 3.33)(hy — hg) = 3750 ie, (vit x 565.3) + (th — 3.33) x 470.6 = 3750 therefore mm 5.14 kg/s ie, Boiler capacity 8 500 kg of steam per hour Problems (a) Steam is supplied, dry saturated at 40 bar to a turbine and the condenser pressure 0.035 bar. Ifthe plant operates on the Rankine cycle, calculate, per kilogram of steam: (j) the work output neglecting the feed-pump work; (ii) the work required for the feed pump; (iii) the heat transferred to the condenser cooting water, and the amount of cooling ‘water required through the condenser ifthe temperature rise of the water is assumed to be 55K; (iv) the heat supplied; (v) the Rankine efficiency; (vi) the specific steam consumption, (b) For the same steam conditions calculate the efficiency and the specific steam consumption for a Carnot cycle operating with wet steam. (986 KJ; 4 KI; 1703 kJ; 74 kg; 2685 kJ; 36.6%; 3.67 kg/KW h; 42.7%; 4.92 ke/kW h) Repeat Problem 8.1(a) for a steam supply condition of 40 bar and 350°C and the same condenser pressure of 0.035 bar (1125 KF; 4 kB; 1857 KI; 80.7 kg; 2978 kl; 37.6%; 3.21 ke/kW h) Steam is supplied to a two-stage turbine at 40 bar and 350°C. It expands in the first turbine until itis just dry saturated, then it is re-heated to 350°C and expanded through the second-stage turbine. The condenser pressure is 0,035 bar. Calculate the work output and the heat supplied per kilogram of steam for the plant, assuming ideal processes and neglecting the feed-pump term, Calculate also the specific steam consumption and the eyele efficiency. (1290 ki; 3362 kJ; 2.79 kg/kW h; 38.4%) 1 the expansion processes in the turbines of Problem 8.3 have isentropic efficiencies of 84% and 78% respectively, in the first and second stages, calculate the work output and the heat supplied per kilogram of steam, the cycle efficiency, and the specific steam consumption. ‘Compare the efficiencies and specific steam consumptions obtained from Problems 8.1, 82, 83, and 84. Compare also the wetness of the steam leaving the turbines in each case. (1028 KJ; 3311 KJ; 31.1%; 3.5 ke/kW h) (Dryness fractions at condenser in each case: 0.699, 0.762, 0.85, and 0.94.) ‘A generating station isto give a power output of 200 MW. The superheat outlet pressure of the boiler is to be 170 bar and the temperature 600°C. After expansion through the 257Steam Cycles 258 a7 8.10 first-stage turbine to a pressure of 40 bar, 15% of the steam is extracted for feed heating. ‘The remainder is reheated at 600°C and is then expanded through the second turbine stage to a condenser pressure of 0.035 bar. For preliminary calculations it is assumed that the actual cycle will have an efficiency ratio of 70% and that the generator mechanical and electrical efficiency is 95%. Calculate the maximum continuous rating of the boiler in kilograms per hour. (632000 kg/h) A steam turbine is to operate on a simple regenerative cycle. Steam is supplied dry saturated at 40 bar, and is exhausted to a condenser at 0.07 bar. The condensate is pumped to a pressure of 3.5 bar at which itis mixed with bleed steam from the turbine at 3.5 bar. The resulting water which is at saturation temperature is then pumped to the boiler. For the ideal cycle calculate, neglecting feed-pump work, (j) the amount of bleed steam required per kilogram of supply steam; (ii) the cycle efficiency of the plant; (iii) the specific steam consumption. (0.1906; 379%; 4.39 kg/kW h) Steam is supplied to a two-stage turbine at 40 bar and S00°C. In the first stage the steam expands isentropically to 3.0 bar at which pressure 2500 kg/h of steam is extracted for process work. The remainder is reheated to 500°C and then expanded isentropically 10 0106 bar. The by-product power from the plant is required to be 6000 kW. Calculate the amount of steam required from the boiler, and the heat supplied. Neglect feed-pump terms, and assume that the process condensate returns at the saturation temperature to mix adiabatically with the condensate from the condenser. (14950 kg/h; 15880 kW) For the plant of Problem 8.7 it is required to improve the efficiency by employing regenerative feed heating by taking off the necessary bleed steam at the same point as the process steam. The process steam is not returned to the boiler, but make-up water at 15°C is supplied. The bleed steam is mixed with the condensate and make-up water at 3.0 bar such that the resultant water is at the saturation temperature corresponding to 3.0 bar. Calculate: (i) the steam supply necessary to meet the same power and process requirements; }) the amount of bleed steam; (iii) the heat supplied in kW. Neglect feed-pump terms. (16480 kg/h; 2660 kg/h; 15460 kW) In a regenerative steam cycle employing three closed feed heaters the steam is supplied to the turbine at 42 bar and 500°C and is exhausted to the condenser at 0.035 bar. The bleed steam for feed heating is taken at pressures of 15, 4, and 0.5 bar. Assuming ideal processes and neglecting pump work, calculate (i) the fraction of the boiler steam bled at each stage; (ii) the power output of the plant per unit mass flow rate of boiler steam; (ii) the cycle efficiency. (0.0952, 0.0969, 0.0902; 1133.6 kW per kg/s; 43.6%) A boiler plant, see Fig. 8.19 (p. 254), incorporates an economizer and an air pre-heater, and generates steam at 40 bar and 300°C with fuel of calorific value 33 000 kI/kg burned at a rate of 500 kg/h, The temperature of the feedwater is raised from 40 to 125°C in the economizer, and the flue gases are cooled at the same time from 395 to 225°C. The flue gases then enter the air pre-heater in which the temperature of the combustion air is raised by 75K. A forced-draught fan delivers the air to the pre-heater at a pressure of 1.02 bar and a temperature of 16°C with a pressure rise across the fan of 180 mm ofReferences water. The power input to the fan is 5 kW and it has a mechanical efficiency of 78%. Neglecting heat losses, and taking c, as 1.01 kJ/kg K for the flue gases, calculate: (i) the mass flow rate of air; (ii) the temperature of the flue gases leaving the plant; (iii) the mass flow rate of steam; (iv) the efficiency of the boiler. The power required to drive the fan is given by pw tw. where h is the pressure rise across the fan expressed as a head of water, p, the density of water, g the acceleration due to gravity, V the volume flow rate of air, and my is the mechanical efficiency of the fan, (2.72. kg/s; 154°C; 1.37 kg/s; 83.6%) References 8.1 HICKSON Dc and TAYLOR F R 1980 Enthalpy-Entropy Diagram for Steam Basil Blackwell 8.2 EASTOP T D and WATSON W E 1992 Mechanical Services for Buildings Longman 2599.1 9 Gas Turbine Cycles ‘The simple constant pressure cycle and the open- and closed-cycle gas turbine units have been considered briefly in Chapter 5. In this chapter the various parts of the cycle will be considered in more detail and the practical limitations and modifications to the ideal cycle will be discussed. ‘The main use for the gas turbine at the present day is in the aircraft ficld, although gas turbine units for electric power generation are being used increasingly, usually using natural gas as fuel, Gas turbines are used in marine propulsion, but the oil engine and steam turbine are more frequently used, particularly for larger ships. The gas turbine is also used in conjunction with the oil engine, and as part of total energy schemes in combination with steam plant; this is discussed more fully in Chapter 17. The inefficiencies in the compression and expansion processes become greater for smaller stand-alone gas turbine units and a heat exchanger is frequently used in order to improve the cycle efficiency. A compact effective heat exchanger is necessary before the small gas turbine can compete for economy with the small oil engine or petrol engine. The use of constant pressure combustion with a rotary compressor driven by a rotary turbine, mounted on a common shaft, gives a combination which is ideal for conditions of steady mass flow over a wide operating range. The practical gas turbine cycle ‘The most basic gas turbine unit is one operating on the open cycle in which a rotary compressor and a turbine are mounted on a common shaft, as shown diagrammatically in Fig. 9.1. Air is drawn into the compressor, C, and after compression passes to a combustion chamber, CC. Energy is supplied in the combustion chamber by spraying fuel into the airstream, and the resulting hot gases expand through the turbine, T, to the atmosphere, In order to achieve net work output from the unit, the turbine must develop more gross work output than is required to drive the compressor and to overcome mechanical losses in the drive. The compressor used is either a centrifugal or an axial flow compressor andFig. 9.1 Open-cycle gas turbine unit Fig. 92 Gas turbine cycle on a T-s diagram 9.1. The practical gas turbine cycle aE Ee | i : the compression process is therefore irreversible but approximately adiabatic. Similarly the expansion process in the turbine is irreversible but adiabatic. Due to these irreversibilities, more work is required in the compression processes for a given pressure ratio, and less work is developed in the expansion process. It is possible that the compressor and turbine may be so inefficient that the unit is not self-sustaining, and in fact it was the difficulties in improving the compressor and turbine design to cut down irreversibilities that retarded the development of the gas turbine unit. As stated in section 5.5 the open-cycle gas turbine cannot be compared directly with the ideal constant pressure cycle. The actual cycle involves a chemical reaction in the combustion chamber which results in high-temperature products which are chemically different from the reactants (see section 7.8). During combustion there is no energy exchange with the surroundings, the effect being a gradual decrease in chemical energy with a corresponding increase in enthalpy of the working fluid. The combustion reaction will not be considered in detail here, and a simplification will be made by assuming that the chemical energy released on a combustion is equivalent to a transfer of heat at constant pressure to a working fluid of constant mean specific heat. This simplified approach allows the actual process to be compared with the ideal and to be represented on a T-s diagram. ‘Neglecting the pressure loss in the combustion chamber the cycle may be drawn on a T-s diagram as shown in Fig. 9.2. Line 1-2 represents irreversible adiabatic compression; line 2-3 represents constant pressure heat supply in the 261Gas Turbine Cycles 262 combustion chamber; line 3-4 represents irreversible adiabatic expansion. The process 1-2s represents the ideal isentropic process between the same pressures >; and p2. Similarly the process 3~4s represents the ideal isentropic expansion process between the pressures p2 and p;. For the moment it will be assumed that the change in kinetic energy between the various points in the cycle is negligibly small compared with the enthalpy changes. Then applying the flow equation to each part of the cycle, we have the following for unit mass. For the compressor: Work input = ‘(Tz — Th) For the combustion chamber: Heat supplied = ¢,(T, ~ T;) For the turbine: Work output = ¢,(T, ~ T;) Then — Net work output = ¢,(Ts — Ty) — ¢p(T2 — T;) and Thermalefficiency = oT = Ts) ~ AB The value of the specific heat capacity of a real gas varies with temperature; also, in the open cycle, the specific heat capacity of the gases in the combustion chamber and in the turbine is different from that in the compressor because fuel has been added and a chemical change has taken place. Curves showing the variation of c, with temperature and air-fuel ratio can be used, and a suitable mean value of ¢, and hence y can be found. It is usual in the gas turbine practice to assume fixed mean values of c, and + for the expansion process, and fixed mean values of c, and 7 for the compression process. For the combustion process, curves as shown in Fig. 9.18 (p.282) are used; for simple calculations a mean value of c, can be assumed. In an open-cycle gas turbine unit the mass flow of gases in the turbine is greater than that in the compressor due to the mass of fuel burned, but it is possible to neglect the mass of fuel, since the air—fuel ratios used are large. Also, in many cases, airis bled from the compressor for cooling purposes, or in the case of aircraft at high altitude, bleed air is used for de-icing and cabin air-conditioning. This amount of ait bleed is approximately the same as the mass of fuel injected. The isentropic efficiency of the compressor is defined as the ratio of the work input required in isentropic compression between p, and p, to the actual work required, Neglecting changes in kinetic energy, we have Compressor isentropicefficiency, n_ = £2 7x — Ti) ep(T, — T;) Tas — Th T-T, (9.1)Example 9.1 Solution Fig. 93 Gas turbine unit (a) and Ts diagram (b) for Example 9.1 9.1. The practical gas turbine cycle Similarly the isentropic efficiency of the turbine is defined as the ratio of the actual work output to the isentropic work output between the same pressures. ‘Neglecting kinetic energy changes e(Ts— Te) (Ts = Ts) B-% “a=, “ Turbine isentropicefficiency, ny A gas turbine unit has a pressure ratio of 10/1 and a maximum cycle temperature of 700°C. The isentropic efficiencies of the compressor and turbine are 0.82 and 0.85 respectively. Calculate the power output of an clectrie generator geared to the turbine when the air enters the compressor at 15°C at the rate of 15kg/s. Take c, = 1.005 kJ/kg K and 7 =14 for the compression process, and take ¢,=1.11kJ/kgK and y = 1.333 for the expansion process. ‘A line diagram of the unit is shown in Fig. 9.3(a), and the cycle is shown on a T-s diagram in Fig. 9.3(b). In order to evaluate the net work output it is necessary to calculate the temperatures T, and T,. To calculate 7, we must first calculate 7;, and then use the isentropic efficiency. PO tusk ons} Generator T = g zB § i ; : AS 4273 288 (a) From equation (3.21) for an isentropic process t, (2 =n T ") therefore Ty, = 288 x (10)°4/44 = 288 x 1.931 = 556K Then using equation (9.1) 556 — 288 T, — 288 = 0.82 268 1, — 288) =“ = 3268 te (B88) = O95 263Gas Turbine Cycles therefore T, = 288 + 326.8 = 614.8 K Similarly for the turbine 1% [p\o T. \p therefore 973 973 Tes = Go ~ [qq 7 STAR Then from equation (9.2) h-T 93-T% nee = 085 Ty— Ty, 973 — 5874 ie, (973 — Ty) = 425.6 x 085 = 361.8 K therefore T, = 973 — 3618 = 611.2K Hence Compressor work inpui =o Ty — T,) = 1.005 x 326.8 = 328.4 kT /kg Turbine work output = ¢,(T3 — Ts) = 1-11 x 361.8 = 401.6 kI/kg therefore Net work output = (401.6 — 328.4) = 73.2 kI/kg ie, Power output = 73.2 x 15 = 1098 kW Example 9.2 Calculate the cycle efficiency and the work ratio of the plant in Example 9.1, assuming that c, for the combustion process is 1.11 kI/kg K Solution Heat supplied = ¢,(T, — T;) = 1.11(973 — 614.8) = L11 x 358.2 kJ/kg ie. Heat supplied = 397.6 kJ/kg, Therefore net work output _ 73.2 heat supplied 397.6 ic. Cycle efficiency = 0.184 or 18.4% Cycle efficiency = 2649.1. The practical gas turbine cycle From the definition of work ratio given in section 5.3, we have 1 networkoutput _ 732 _ 9 1g gross work output 401.6 Work ratio = Use of a power turbine In Examples 9.1 and 9.2 the turbine is arranged to drive the compressor and to develop net work. It is sometimes more convenient to have two separate turbines, one of which drives the compressor while the other provides the power output. The first, or high-pressure (HP) turbine, is then known as the compressor turbine, and the second, or low-pressure (LP) turbine, is called the power turbine. The arrangement is shown in Fig. 9.4(a). Assuming that each turbine has its own isentropic efficiency, the cycle is as shown on a T-s diagram in Fig, 9.4(b). The numbers on Fig. 9.4(b) correspond to those of Fig. 9.4(a). Neglecting kinetic energy changes, we have work from HP turbine = work input to compressor ie. ¢,(T3 — Ty) = (Tr — Ti) Fig. 94. Gas turbine unit with separate rey power turbine (a) and ah 3 the eyele on a T-s diagram (b) c up i 4 Air inet Genertor up LJ 7 SY Exhaust ®) ©) where ¢,, and cp, are the specific heat capacities at constant pressure of the gases in the turbine and the air in the compressor respectively. The net work output is then given by the LP turbine, ie, Net work output = ¢,,(T, — Ts) Example 9.3 A gas turbine unit takes in air at 17°C and 1.01 bar and the pressure ratio is 8/1. The compressor is driven by the HP turbine and the LP turbine drives a separate power shaft. The isentropic efficiencies of the compressor, and the HP and LP turbines are 0.8, 0.85, and 0.83 respectively, Calculate the pressure 265Gas Turbine Cycles 266 Solution and temperature of the gases entering the power turbine, the net power developed by the unit per kg/s mass flow rate, the work ratio and the cycle efficiency of the unit. The maximum cycle temperature is 650°C. For the compression process take c, = 1.005 kJ/kg K and y = 1.4; for the combustion process, and for the expansion process take c, = 1.15 kI/kg K and) = 1.333. Neglect the mass of fuel. The unit is as shown in Figs 9.4(a) and 9.4(b). From equation (3.21), for an isentropic process, ty (ey tT \p ie, Tay = 290 x 8°44 = 290 x 1.811 = 525K ‘Then, using equation (9.1) Tay — T, _ 525-290 08 "TT 290 therefore — 299 = 235 08 ie, T= 290 + 294 = 584K ‘Then Work input to the compressor = ¢,(T, — T,) = 1.005 x 294 = 295.5 kI/kg Now the work output from the HP turbine must be sufficient to drive the compressor, ie, Work output from HP turbit ¢,,(Ts — Ta) = 295.5 kJ/kg therefore By Ty = 285 957K 115 therefore Ty = Ty — 257 = 923 — 257 = 666K Then, using equation (9.2), Ty — 923 — 6 T= Ts $68 _ oss for HP turbine = = a e 923 —T, ie, 923 - Ty = = 3025K 085 therefore Ty, = 923 ~ 302.5 = 620.5 K9.1 The practical gas turbine cycle ‘Then from equation (3.21) for an isentropic process, , (yr Ts, \Pa P,\ =) 1.339/0.339 o = Ba zy " -(#) me = 49 pa \Tay 6205, Pa 8x 101 M49 49 Hence the pressure and temperature at entry to the LP turbine are 1.65 bar and 393°C, where f4 = 666 — 273 = 393°C. To find the power output it is now necessary to evaluate T;. The pressure ratio, pa/Ps, is given by (p4/Ps) x (P3/Ps) ie. = 1.65 bar ic, Pr (a, (since p2 = ps and Ps = Pi) Ps Ps Pt = 1.63 ps 49 (oy Then 7. (24) = 1,630-999/1-999 = 1.131 Ts, \Ps. therefore 7, = 856 1131 Then, using equation (9.2) T, ny for the LP turbine = — ie, Ty — Ty = 0.83(666 — 588) = 0.83 x 78 = 64.8 K Then Work output from LP turbine = ¢,,(T, — Ts) = LAS x 64,8 = 745 kI/kg ie. Net power output = 74,5 x 1 = 74.5 kW net work output 74.5 745 _ ooo gross work output 74.5 + 295. ” ~ 370 Heat supplied = ¢,,(T, ~ T,) = 1.15(923 — 584) ic, Heat supplied = 1.15 x 339 = 390 kI/kg network output _ 74.5 Then Cycleefficiency = BetworkoutPut _ 74.5 see eeney ~~ Feat supplied 390 191 oF 19.1% Work ratio 267Gas Turbine Cycles 9.5. Simple jet engine (a) with the cycle ona T-s diagram (b) 9.6 Turboprop engine (a) with the cycle ona T-s diagram (b) 268 Aircraft engines In a jet engine the propulsion nozzle takes the place of the LP stage turbine, as shown diagrammatically in Fig. 9.5(a). The cycle is shown on a T-s diagram in Fig. 9.5(b), and it can be seen to be identical with Fig. 9.4(b), The aircraft is powered by the reactive thrust of the jet of gases leaving the nozzle, and this, high-velocity jet is obtained at the expense of the enthalpy drop from 4 to 5. The turbine develops just enough work to drive the compressor and overcome mechanical losses. Aircraft velocity Ar se &, at 3 4 5 Exhaust (a) () Ina turbo-prop engine the turbine drives the compressor and also the airscrew, or propeller, as shown in Figs9.6(a) and 9.6(b). The net work output available to drive the propeller is given by ‘od Ts ~ Ta) ~ 5, Tz — T) (neglecting mechanical losses), Net work output Aircraft velocity Sone E> exhaust 3 a 2 HK] c T i Propeller @ (b) In practice there is also a smalll jet thrust developed in a turbo-prop aircraft. Jet engines and turbo-prop engines are considered again in Section 10.9.Fig. 9.7 Parallel-flow gas turbine unit 9.2 9.2 Modifications to the basic cycle Parallel flow units In some industrial and marine gas turbine units, the air flow is split into two streams after the compression process is completed. Some air is then passed to combustion chamber which supplies hot gases to the turbine driving the compressor, while the rest of the air is passed to a second combustion chamber and from thence to the power turbine, The system is shown diagrammatically in Fig. 9.7, and is called a parallel flow unit. In this system each turbine expands, the gases received by it through the full pressure ratio. The advantage of this system is that the net power output can be varied using the second combustion chamber, and the power turbine operates independently of the compressor turbine. Air inlet Exhaust ce Net power T output Exhaust Modifications to the basic cycle Tt can be seen from Examples 9.1, 9.2, and 9.3 that the work ratio and the cycle efficiency of the basic gas turbine cycle are low. These can be improved by increasing the isentropic efficiencies of the compressor and turbine, and this, is a matter of blade design and manufacture. In a practical cycle with irreversibilities in the compression and expansion processes the cycle efficiency depends on the maximum cycle temperatures as well as on the pressure ratio. For fixed values of the isentropic efficiencies of the compressor and turbine, the cycle efficiency can be plotted against pressure ratio for various values of maximum temperature. This is illustrated in Fig. 9.8, for a cycle in which the compressor isentropic efficiency is 0.89, the turbine isentropic efficiency is 0.92, and the air inlet temperature is 20°C. The ideal air standard cycle thermal efficiency is shown chain-dotted, In section 5.4 it is shown that the ideal constant pressure cycle efficiency is given by y Dir 269Gas Turbine Cycles Fig. 98 Gas turbine ‘oycle efficiency against pressure ratio for different maximum cycle temperatures Fig. 9.9 Specific power against pressure ratio for different maximum cycle temperatures 270 Air standard Cycle efficieney/(%) 1 3 10 1s 20 25 Pressure ratio where r, is the pressure ratio and is independent of the maximum cycle temperature. It can be seen from Fig. 9.8 that at any one fixed maximum cycle temperature there is a value of pressure ratio which will give maximum cycle efficiency. The net work output also depends on the pressure ratio and on the maximum cycle temperature, and curves of specific power output against pressure ratio for various maximum temperatures are shown in Fig. 9.9. The isentropic efficiencies of the compressor and turbine, and the air inlet temperature are the same as those used in deriving the curves of Fig. 9.8. It can be seen that the cycle efficiency reaches a maximum at a different value of pressure ratio than the work output. The choice of pressure ratio is therefore a compromise, 350 250 150 50 Specific power/(kW per kg/s) 1 5 10 15 20 25 Prossure ratio. The maximum cycle temperature is limited by metallurgical considerations. The blades of the turbine are under great mechanical stress and the temperature of the blade material must be kept to a safe working value. The temperature of the gases entering the turbine can be raised, provided a means of blade cooling is available, Various methods of blade cooling have been investigated and a discussion of these will be found in ref. 9.1. In aircraft practice where the life expectancy of the engine is shorter, the maximum temperatures used are usually higher than those used in industrial and marine gas turbine units; more expensive alloys and blade cooling allow maximum temperatures of above 1600 K.Fig. 9.10 Gas turbine unit with intercooting {a) and the eycle on the T-s diagram (b) 8.2 Modifications to the basic cycle It is important to have as high a work ratio as possible, and methods of increasing the work ratio, such as intercooling between compressor stages, and reheating between turbine stages, will be considered in this section. Intercooling and reheating, while increasing the work ratio, can cause a decrease in the cycle efficiency, but when they are used in conjunction with a heat exchanger then intercooling and reheating increase both the work ratio and the cycle efficiency. Intercooling When the compression is performed in two stages with an intercooler between the stages, then the work input for a given pressure ratio and mass flow is reduced, Consider a system as shown in Fig, 9.10(a); the T-s diagram for the unit is shown in Fig. 9.10(b). The actual cycle processes are 1-2 in the LP compressor, 2-3 in the intercooler, 3-4 in the HP compressor, 4~5 in the combustion chamber, and 56 in the turbine, The ideal cycle for this arrangement is 1-2s-3~4s—5—6s; the compression process without intercooling is shown as 1-A in the actual case, and 1-As in the ideal isentropic case. Intercooler Air ‘Finlet @) & The work input with intercooling is given by Work input (with intercooling) = ¢,(T,— T,) + ¢(T,— 73) (9.3) The work input with no intercooling is given by Work input (no intercooling) = ¢,(Ty — T;) (Ts — T;) + p(T, — Ts) Comparing this equation with equation (9.3), it can be seen that the work input with intercooling is less than the work input with no intercooling, when 6,(T, — T,) is less than c,(T, — T,). This is so ifit is assumed that the isentropic efficiencies of the two compressors, operating separately, are each equal to the isentropic efficiency of the single compressor which would be required if no intercooling were used. Then (T; — Ts) <(T,— Tp) since the pressure lines diverge from left to right on the T-s diagram. anGas Turbine Cycles 272 It can be shown that the best interstage pressure is the one which gives equal pressure ratios in each stage of compression; referring to Fig. 9.10(b) this means that p2/p; = p4/ps. The work input required is a minimum when the pressure ratio in each stage is the same, and when the temperature of the air is cooled in the intercooler, back to the value at inlet to the unit (ie. referring to Fig. 9.10(b), T; = T)). Now net work output ‘gross work output Work ratio work ofexpansion — work of compression work ofexpansion It follows, therefore, that when the compressor work input is reduced then the work ratio is increased, However, referring to Fig. 9.10(b), the heat supplied in the combustion chamber when intercooling is used in the cycle is given by Heat supplied (with intercooling) = c)(T — T,) whereas the heat supplied when intercooling is not used, with the same maximum cycle temperature Ts, is given by Heat supplied (no intercooling) = ¢,(T; — T,) Hence the heat supplied when intercooling is used is greater than with no intercooling. Although the net work output is increased by intercooling it is found in general that the increase in the heat to be supplied causes the cycle efficiency to decrease. It will be shown later that this disadvantage is offset, when a heat exchanger is also used. ‘When intercooling is used a supply of cooling water must be readily available, The additional bulk of the unit may offset the advantage to be gained by increasing the work ratio. Reheat As stated earlier, the expansion process is very frequently performed in two separate turbine stages, the HP turbine driving the compressor and the LP turbine providing the useful power output, The work output of the LP turbine can be increased by raising the temperature at inlet to this stage. This can be done by placing a second combustion chamber between the two turbine stages in order to heat the gases leaving the HP turbine. The system is shown diagrammatically in Fig. 9.11(a), and the cycle is represented on a T-s diagram in Fig. 9.11(b). The line 4A represents the expansion in the LP turbine if reheating is not used. As before, the work output of the HP turbine must be exactly equal to the work input required for the compressor (neglecting mechanical losses), ¢,(T2 — Ti) = (Ts — Ts)Fig. 9.11 Gas turbine unit with reheating (a) and the cycle on a T-: diagram (b) 9.2 Modifications to the basic cycle Net power output Air inlet Exhaust (by ‘The net work output, which is the work output of the LP turbine, is given by (Ts — Ts) Net work outpul Ir reheating is not used, then the work of the LP turbine is given by ATs = Ta) Since pressure lines diverge to the right on the T-s diagram, it can be seen that the temperature difference (7; — Tg) is always greater than (T, — T,), so that reheating increases the net work output. Also Net work output (no reheat) work ofexpansion — work of compression work ofexpansion Work ratio , work of compression ie. Work rati eS work ofexpansion ‘Therefore, when the work of expansion is increased and the work of compression is unchanged, then the work ratio is increased. Although the net work is increased by reheating, the heat to be supplied is also increased, and the net effect can be to reduce the thermal efficiency, ie, Heat supplied I) T,—T,) + 6, (Ts — However, the exhaust temperature of the gases leaving the LP turbine is much higher when reheating is used (ie. T; a8 compared with 7,), and a heat, exchanger can be used to enable some of the energy of the exhaust gases to be used. Heat exchanger The exhaust gases leaving the turbine at the end of expansion are still at a high temperature, and therefore a high enthalpy (e.g, in Example 9.3, t, = 328.2°C). If these gases are allowed to pass into the atmosphere, then this represents a Joss of available energy. Some of this energy can be recovered by passing the gases from the turbine through a heat exchanger, where the heat transferred from the gases is used to heat the air leaving the compressor. The simple unit 273Gas Turbine Cycles Fig. 9.12 Gas turbine unit with heat exchanger (a) and the cycle on a T-s diagram (b) Fig, 9.13 T-s diagram for a gas turbine unit with a heat exchanger showing temperature differences for heat transfer 274 Heat exchanger Exhaust rh - 6 tw Available 3 ‘temperature : Ss difference \ 3] Net power 5 output T a | 1 (a) ) with a heat exchanger added is shown diagrammatically in Fig, 9.12(a), and the cycle is represented on a T-s diagram in Fig. 9.12(b), In the ideal heat exchanger the air would be heated from 7; to T; = T; and the gases would be cooled from T; to Te = T;. This ideal case is shown in Fig. 9.12(b). In practice this is impossible, since a finite temperature difference is required at all points in the heat exchanger in order to overcome the resistance to the heat transfer. Referring to Fig. 9.13, the required temperature difference between the gases and the air entering the heat exchanger is(T, — T,), and the required temperature difference between the gases and the air leaving the heat exchanger is (Ts — 3). If no heat is lost from the heat exchanger to the atmosphere, then the heat given up by the gases must be exactly equal to the heat taken up by the air, Le. thCp(Ts — Tz) = tigep,(Ts — Ts) (9.4) The assumption that no heat is lost from the heat exchanger is sufficiently accurate in most practical cases. Equation (9.4) is therefore true whatever the temperatures T, and Tz may be. A heat exchanger effectiveness is defined to allow for the temperature difference necessary for the transfer of heat, heat received by the air ‘maximum possible heat which could be transferred from the gases in the heat exchanger ie. EffectivenessFig. 9.14 Example of a cycle where a heat exchanger is not feasible 9.2 Modifications to the basic cycle therefore Effectiveness (9.5) A more convenient way of assessing the performance of the heat exchanger is to use a thermal ratio, defined as temperature rise of the air ‘maximum temperature difference available Ty- kh Th Comparing equations (9.5) and (9.6) it can be seen that the thermal ratio is equal to the effectiveness when the product, rig¢,,,is equal to the product, rg¢y,- ‘When a heat exchanger is used then the heat to be supplied in the combustion chamber is reduced, assuming that the maximum cycle temperature is unchanged. The net work output is unchanged and hence the cycle efficiency is increased. Referring to Fig. 9.13 ‘Thermal ratio = ie. Thermal ratio = (9.6) Heat supplied by the fuel (without heat exchanger) Heat supplied by the fuel (with heat exchanger) = ¢),(T, — Ts) A heat exchanger can be used only if there is a sufficiently large temperature difference between the gases leaving the turbine and the air leaving the compressor. For example, in the cycle shown in Fig. 9.14 a heat exchanger could not possibly be used because the temperature of the exhaust gases, T,, is lower than the temperature of the air leaving the compressor, 7. In practice, although the gas temperature may be higher than the temperature of the air leaving the compressor, the difference in temperature may not be sufficiently large to warrant the additional capital cost and subsequent maintenance required for a heat exchanger. Also, when the temperature difference is small in a heat exchanger, then the surface areas for the heat transfer must be made large in order to achieve a reasonably high value of the thermal ratio. For small gas turbine units (e.g. for pumping sets or for motor cars) a compact heat exchanger ‘must be designed before such units can hope to become competitive for economy 275Gas Turbine Cycles Example 9.4 Solution Fig. 9.15 Gas turbine plant (a) and T-s diagram (b) for Example 94 Air inlet 276 Intercooler with conventional internal combustion engines of equivalent power. In large gas turbine units for marine propulsion or industrial power, a heat exchanger may be used, although the trend now is towards combined cycles using the turbine exhaust to generate steam or heat water (see Chapter 17). A 5000 kW gas turbine generating set operates with two compressor stages with intercooling between stages; the overall pressure ratio is 9/1. A HP. turbincis used to drive the compressors, and a LP turbine drives the generator, The temperature of the gases at entry to the HP turbine is 650°C and the gases are reheated to 650°C after expansion in the first turbine. The exhaust gases leaving the LP turbine are passed through a heat exchanger to heat theair leaving the HP stage compressor. The compressors have equal pressure ratios and intercooling is complete between stages. The air inlet temperature to the unit is 15°C. The isentropic efficiency of each compressor stage is 0.8 and the isentropic efficiency of each turbine stage is 0.85; the heat exchanger thermal ratio is 0.75. A mechanical efficiency of 98% can be assumed for both the power shaft and the compressor turbine shaft. Neglecting all pressure losses and changes in kinetic energy, calculate: (i) the cycle efficiency; (ii) the work ratio; (iii) the mass flow rate, For air take c, = 1.005 kJ/kg K and y= 1.4, and for the gases in the combustion chamber and in the turbines and heat exchanger take Gp = 1.15 kI/kg K and = 1.333. Neglect the mass of fuel. (i) The plant is shown diagrammatically in Fig. 9.15(a), and the cycle is represented on a T-s diagram in Fig. 9.15(b). Since the pressure ratio and the isentropic efficiency of each compressor is the same, then the work input required for each compressor is the same since both compressors have the same air inlet temperature, ie. T; = T;and T, = Ty. HP Lo 4 Net power o output Exhaust fy (a) (b)9.2. Modifies jons to the basic cycle From equation (3.21) Tas (By a P2 aac and Ph = 9 T \py yD v therefore Tyg = 288 x 304114 = 394 K ‘Then from equation (9.1), ‘ic, LP compressor oT, therefore = 7, = 3942288 _ 106 395K 08 (08 ie, Ty = 288 + 132.5 = 420.5 K Also Work input per compressor stage = ¢,,(T, — T;) = 1.005 x 132.5 = 133.1 kI/kg The HP turbine is required to drive both compressors and to overcome mechanical friction, 2x 1334 ie. Work output of HP turbine = = 272 kI/kg therefore Op(Ts — Ty) = 272 ie, 1.15(923 — 5) = 272 therefore 22 923 — T, = = = 2365 K 7 115 ie, T, = 923 — 236.5 = 686.5 K From equation (9.2) ny HP turbine = @—™ = 085 fo — Tre therefore 2365 Tg — Ty = o> = 28K oe 085 ie. 923 — 278 = 645K 277Gas Turbine Cycles Then using equation (3.21) Do _ (Tq \"-) _ /93\133910.333 Et = =4.19 Py () 645, Ps_ 9 =—=2147 py 419 Using equation (3.21) Je (£2) " or Area per unit mass flow, (10.2) Then substituting for the velocity C, from equation (10.1) ‘Area per unit mass flow = ’ (103) V {2(hy — h) + CF} It can be seen from equation (10.3) that in order to find the way in which the area of the duct varies it is necessary to be able to evaluate the specific volume, v, and the enthalpy, h, at any section X-X, In order to do this, some information about the process undergone between section 1 and section X-X must be known. For the ideal frictionless case, since the flow is adiabatic and reversible, the process undergone is an isentropic process, and hence 5, = (entropy at any section X-X) = say Now using equation (10.2) and the fact that s, =, it is possible to plot the variation of the cross-sectional area of the duct against the pressure along the duct. For a vapour this can be done using tables; for a perfect gas the procedure is simpler, since we have po’ = constant, for an isentropic process. In either case, choosing fixed inlet conditions, then the variation in the area, A, the specific volume, v, and the velocity, C, can be plotted against the pressure along the duct. Typical curves are shown in Fig. 10.1. It can be seen that the area decreases initially, reaches a minimum, and then increases again. This can also be seen from equation (10.2), 5 . v ie, Area per unit mass flow = & Direction of Section 1 flow Section 2 ae Area Velocity Specific volume m P“Throat Inlet Outlet Fig. 10.2 Cross-section through a convergent-divergent nozzle 10.2 Fig. 103 Convergent-divergent Y Fig. 10.4 Inlet section of a nozzle 10.2. Critical pressure ratio ‘When » increases less rapidly than C, then the area decreases; when v increases more rapidly than C, then the area increases. A nozzle, the area of which varies as in Fig. 10.1, is called a convergent— divergent nozzle. A cross-section of a typical convergent—divergent nozzle is shown in Fig. 10.2. The section of minimum area is called the throat of the nozzle. It will be shown later in section 10.2 that the velocity at the throat of a nozzle operating at its designed pressure ratio is the velocity of sound at the throat conditions, The flow up to the throat is subsonic; the flow after the throat is supersonic. It should be noted that a sonic or a supersonic flow requires a diverging duct to accelerate it. ‘The specific volume of a liquid is constant over a wide pressure range, and therefore nozzles for liquids are always convergent, even at very high exit velocities (¢.g. a fire-hose uses a convergent nozzle). Cc ical pressure ratio It has been stated in section 10.1 that the velocity at the throat of a correctly designed nozzle is the velocity of sound. In the same way, for a nozzle that is convergent only, then the fluid will attain sonic velocity at exit if the pressure drop across the nozzle is large enough. The ratio of the pressure at the section where sonic velocity is attained to the inlet pressure of a nozzle is called the critical pressure ratio, Cases in which nozzles operate off the design conditions of pressure will be considered in section 10.4; in what follows it will be assumed that the nozzle always operates with its designed pressure ratio. Consider a convergentdivergent nozzle as shown in Fig. 10.3 and let the inlet conditions be pressure p,, enthalpy h,, and velocity C,. Let the conditions at any other section X-X be pressure p, enthalpy h, and velocity C. In most practical applications the velocity at the inlet to a nozzle is negligibly small in comparison with the exit velocity. Tt can be seen from equation (10.2), jth = v/C, that a negligibly small velocity implies a very large area, and most nozzles are in fact shaped at inlet in such a way that the nozzle converges rapidly over the first fraction of its length; this is illustrated in the diagram of a nozzle inlet shown in Fig. 10.4. ‘Now from equation (10.1) we have C= Jf {2(hy — h) + CF} and neglecting C, this gives C= (2h, — } (10.4) Since enthalpy is usually expressed in kilojoules per kilogram, then an additional constant of 10? will appear within the root sign if C is to be expressed in metres per second ‘Then, substituting from equation (10.4) in equation (10.2), we have A_o v Area per unit mass flow, mC J, =} (10.5) 289Nozzles and Jet Propulsion As stated in section 10.1 the area can be evaluated at any section where the pressure is p, by assuming that the process is isentropic (i.e. s, = s). When this is done for a series of pressures, the area can be plotted against pressure along. the duct, or against pressure ratio, and the critical pressure can thus be found graphically. For a perfect gas it is possible to simplify equation (10.5) by making use of the perfect gas laws. From equation (2.18), h cpT for a perfect gas, therefore Be Ve (T, » ‘Area per unit mass flow rat From equation (2.5), 0 = RT/p, therefore RT Ip Jeal-a)} Let the pressure ratio, p/p,, be x. Then using equation (3.21), for an isentropic process for a perfect gas z-(2)" ont T \m Substituting for p = xp,, for T = T,x-", and for T/T, = RT, x0" Pry (2c, T, (1 ~ x9 )} Area per unit mass flow rat 2h, we have Area per unit mass flow rate For fixed inlet conditions (ie. p, and ‘ fixed), we have xB ‘Area per unit mass flow rate = constant x = constant x POT constant therefore constant Area per unit mass flow rate = —[O"* ; R Yat — xo (10.6) To find the value of the pressure ratio, x, at which the area is a minimum it is necessary to differentiate equation (10.6) with respect to x and equate the result 29010.2. Critical pressure ratio to zero, ie. for minimum area d 1 2} Ng de Ge — x OMIA [Baan (: + 1) fora Me NYT : 2a — xt Hence the area is a minimum when 2 em 2 (@ + ttre ? ? xlotnin-1-amer 2 2 pel therefore ( 2 y » xe(— yt P 2 \na-0 ie, Critical pressure ratio, = = ( (10.7) PL y It can be seen from equation (10.7) that for a perfect gas the pressure ratio required to attain sonic velocity in a nozzle depends only on the value of y for the gas. For example, for air y = 1.4, therefore Dy 2 \ toe bo = 0.5283 Pi (a + i) Hence for air at 10 bar, say, a convergent nozzle requires a back pressure of 5.283 bar, in order that the flow should be sonic at exit and for a correctly designed convergent—divergent nozzle with inlet pressure 10 bar, the pressure at the throat is 5.283 bar. For carbon dioxide, 7 = 1.3, therefore _ 2 \1303 Pew = 05457 ys (sa) Hence for carbon dioxide at 10 bar, a convergent nozzle requires a back pressure of 5457bar for sonic flow at exit, and the pressure at the throat of a convergent-divergent nozzle with inlet pressure 10 bar is 5.457 bar. The ratio of the temperature at the section of the nozzle where the velocity is sonic to the inlet temperature is called the critical temperature ratio, Te _(p\ro?_ 2 Critical temperature ratio, = = () T, \p ytd 2 ie. z =— (10.8) T y+ Equations (10.7) and (10.8) apply to perfect gases only, and not to vapours. 201Nozzles and Jet Propulsion 292 However, it is found that a sufficiently close approximation is obtained for a steam nozzle if it is assumed that the expansion follows a law po‘ = constant. The process is assumed to be isentropic, and therefore the index k is an approximate isentropic index for steam. When the steam is initially dry saturated then k = 1.135; when the steam is initially superheated then k = 1.3. Note that equation (10.8) cannot be used for a wet vapour, since no simple relationship between p and Tis known for a wet vapour undergoing an isentropic process, More will be said of this in section 10.6. ‘The critical velocity at the throat of a nozzle can be found for a perfect gas by substituting in equation (10.1), ie. Cy = J {2(hy — h) + CF} Putting C, = 0, as before, and using equation (2.18) for a perfect gas, h = c, T; we have 26%, = 7)} = ffen(B- 1) From equation (10.8), 7/7, = 2/(y + 1), hence J) - th ]=Vterner- ») Also, from equation (2.22), C G; ea or ¢(y—1)=9R Hence substituting, C.= J (RT) ie. Critical velocity, C, = (RT, (10.9) The critical velocity given by equation (10.9) is the velocity at the throat of a correctly designed convergent-divergent nozzle, or the velocity at the exit of a convergent nozzle when the pressure ratio across the nozzle is the critical pressure ratio, It can be shown that the critical velocity is the velocity of sound at the critical conditions. The velocity of sound, a, is defined by the equation 2 _ oP a? = at constant entropy dp where p is pressure and p is the density. A proof of this expression can be found in ref. 10.2. Now p = 1/v, where » is the specific volume. dp = d(1/v) = — dvExample 10.1 Solution Fig. 10.5 Convergent-divergent nozzle for Example 10.1 I pressure ratio Hence a? = — 22y2 dv For a perfect gas undergoing an isentropic process, po’ = constant, : K ie where K is constant, therefore dp _ kK dot Therefore substituting v2yK ot ‘Also, K = po”, hence ypo?v? yaa PP 8 ie. Velocity of sound, a = /(ypv) = /(9RT) (10.10) It can be seen that the critical velocity of a perfect gas in a nozzle, as given by equation (109), is the velocity of sound in the gas at the critical temperature. Equations (10.9) and (10.10) cannot be applied to a vapour; however, if an approximate isentropic law, pot = constant, is assumed for a vapour, then the critical velocity can be taken as C, = /(kpo). (Note that for a vapour the critical velocity cannot be expressed in terms of the temperature.) It is usually more convenient to evaluate the critical velocity of a vapour using equation (10.4), C, = y/{2(0h, — h)}, where (/t; — h,) is the enthalpy drop from the inlet to the throat, which can be evaluated from tables or by using an h-s chart. Air at 86bar and 190°C expands at the rate of 4.5kg/s through a convergent-divergent nozzle into a space at 1.03 bar. Assuming that the inlet velocity is negligible, calculate the throat and the exit cross-sectional areas of the nozzle. The nozzle is shown diagrammatically in Fig. 10.5, From equ: critical pressure ratio is given by n (10.7) the pe (2. \tord ( 2 ye Pea (_ =(= = 05283 Ps G + i) 24, ic. pe = 0.5283 x 8.6 = 4.543 bar 293Nozzles and Jet Propulsion Also, from equation (10.8) 2 tT y+i 190 + 273 ~ = 385.8 K 12 2 From equation (2.5) », aRT. 287 x 385.8 Spe 108 x 4.543 Also, from equation (10.9) C. = JORT,) = (14 x 287 x 385.8) = 393.7 m/s [or from equation (10.4) Co= af {2(lhy — hed} = {2eg(T, — T)} ie. C, = y/{2 x 1.005 x 109(463 — 385.8)} = 393.8 m/s] To find the area of the throat, using equation (1.11), we have A, ees 0.002 79 m? C. 393.7 0.00279 x 10° = 2790 mm? = 0.244 m3/kg, ic, Area of throat Using equation (3.21) for a perfect gas =a ois Bix a) ie (*5) = 1834 Ty \ Da, 1.03, ie = = “3. 2 a525K 1,834 Then from equation (2.5) _ Ry _ 287 x 2825 = 0.7036 m3 x Pa 10° x 103 mike 2% Also, from equation (10.4) Cr = J {2h —ha)} = V (2ey(T, - T)} ive. Cy = y/{2 x 1.005 x 10°(463 — 252.5)} = 650.5 m/s Then to find the exit area, using equation (1.11) vy _ 45 x 0.7036 C 650.5 ie, Exit area = 0.00487 x 10° = 4870 mm? A = 0.004 87 m? 29410.3 Fig. 10.6 Convergent nozzle with back-pressure variation Fig. 10.7 Propagation of a small disturbance in a flowing fluid 10.3. Maximum mass flow Maximum mass flow Consider a convergent nozzle expanding into a space, the pressure of which can be varied, while the inlet pressure remains fixed. The nozzle is shown diagrammatically in Fig. 10.6. When the back pressure, pp, is equal to p,, then no fluid can flow through the nozzle, As p, is reduced the mass flow through the nozzle increases, since the enthalpy drops, and hence the velocity, increases. However, when the back pressure reaches the critical value, it is found that no further reduction in back pressure can affect the mass flow. When the back pressure is exactly equal to the critical pressure, p,, then the velocity at exit is sonic and the mass flow through the nozzle is at a maximum, If the back pressure is reduced below the critical value then the mass flow remains at the maximum value, the exit pressure remains at p., and the fluid expands violently outside the nozzle down to the back pressure. It can be seen that the maximum mass flow through a convergent nozzle is obtained when the pressure ratio across the nozzle is the critical pressure ratio. Also, for a convergent-divergent nozzle, with sonic velocity at the throat, the cross-sectional area of the throat fixes the mass flow through the nozzle for fixed inlet conditions. i RZ Back — QValve — a» — pressure Pe When a nozzle operates with the maximum mass flow it is said to be choked. A correctly designed convergent-divergent nozzle is always choked. ‘An attempt can be made to explain the phenomenon of choking, by considering the velocity of any small disturbance in the stream. Any small disturbance in the flow is propagated as small pressure waves travelling at the velocity of sound in the fluid in all directions from the centre of the disturbance. ‘This is illustrated in Fig, 10.7; the pressure waves emanate from point Q at the velocity of sound relative to the fluid, a, while the fluid moves with a velocity, C. The absolute velocity of the pressure waves travelling back upstream is therefore given by (a — C). Now when the fluid velocity is subsonic, then C < a, Absolute ‘veloc Absolute a 295Nozzles and Jet Propulsion 296 Example 10.2 Solution and the pressure waves can move back upstream; however, when the flow is sonic, or supersonic (ie. C = a or C > a), then the pressure waves cannot be transmitted back upstream. It follows from this reasoning that in a nozzle in which sonic velocity has been attained no alteration in the back pressure can be transmitted back upstream. For example, when air at 10 bar expands in a nozzle, the critical pressure can be shown to be 5.283 bar. When the back pressure of the nozzle is 4 bar, say, then the nozzle is choked and is passing the maximum mass flow. If the back pressure is reduced to 1 bar, say, the mass flow through the nozzle remains unchanged. Even if the air were allowed to expand into an evacuated space, the mass flow would be no greater than that through the nozzle when the back pressure is 5.283 bar. A fluid at 6.9 bar and 93 °Centers a convergent nozzle with negligible velocity, and expands isentropically into a space at 3.6 bar. Calculate the mass flow per square metre of exit area (i) when the fluid is helium (c, (ii) when the fluid is ethane (c, 19 kI/kg K); 88 kJ/kg K), Assume that both helium and ethane are perfect gases, and take the respective molar masses as 4 kg/kmol and 30 kg/kmol. (i) It is necessary first to calculate the critical pressure in order to discover whether the nozzle is choked. From equation (2.9), R = R/im, therefore for helium, 8314.5 R= E> = 2079 Nm/ke K Then from equation (2.22) ie. therefore = 1.667 °" T2048 Then using equation (10.7) P. 2 \Wo-n 2 \1667/0.667 Pea (_*_ = = 0.487 pela) = Gas5 ie. p, = 0.487 x 6.9 bar ie. Critical pressure p, = 3.36 bar The actual back pressure is 3.6 bar, hence in this case the fluid does not reach the critical conditions and the nozzle is not choked.2 A 3.6bar 69 bar Fig. 108 Convergent nozzle with helium for Example 10.2 1 Z 2. as — By 2 7 so3tar sober Fig. 10.9 Convergent nozzle with ethane for Example 10.2 10.3 Maximum mass flow The nozele is shown diagrammatically in Using equation (3.21) =o os fa(%) -(2) = 1297 Tt \ps 36 934273 _og99K 1.297 ig. 108. ie. ‘Then from equation (10.4) Cz = J {20h = ha)} = V 2ey(T ~ TY} ie Cz = {2 x 5.19 x 10°(366 — 282.2)} Also, from equation (2.5) 932.7 m/s 282. RT, _ 2079 x 2822 63 m3 kg, Om 108 x36 Hence from equation (1.11) AsCp_ 1x 932.7 oS = ——_ = 572.3 kg/s me 163 Bi ie. Mass flow per square metre of exit area = 572.3 kg/s (i) Using the same procedure for ethane, we have and eto 01a7 so) Linpan Then (2) = an) 057 pei 2172 51 x 69 bar ie, Critical pressure, p, = 3.93 bar ‘The actual back pressure is 3.6 bar, hence in this case the fluid reaches critical conditions at exit and the nozzle is choked, The expansion from the exit pressure of 3.93 bar down to the back pressure of 3.6 bar must take place outside the nozzle. The nozzle is shown diagrammatically in Fig. 10.9. Since the nozzle is choked, from equation (10.8) we have Tr 2 2 T, yi 2172 297Nozzles and Jet Propulsion 298 10.4 ice = Also, from equation (10.9) Cy = C, = J (PRT) = V (1.172 x 277.1 x 337) = 331 m/s From equation (2.5), _ RE _ 271.1 x 337 = 2 = 0.238 m3, 1 105 x 3.93 tke ‘Then using equation (1.11) A,C, 1x 331 = = 1391 kg/s v2 0.238 ie. Mass flow per square metre of exit area = 1391 kg/s Nozzles off the design pressure ratio When the back pressure of a nozzle is below the design value the nozzle is said to underexpand, In underexpansion the fiuid expands to the design pressure in the nozzle and then expands violently and irreversibly down to the back pressure on leaving the nozzle (e.g. the nozzle in Example 10,2(ii) shown in Fig. 10.9 is underexpanding). When the back pressure of a nozzle is above the design value the nozzle is said to overexpand. In overexpansion in a convergent nozzle the exit pressure is greater than the critical pressure and the effect is to reduce the mass flow through the nozzle. In overexpansion in a convergent~divergent nozzle there is always an expansion followed by a recompression, The two types of nozzle can be considered separately. Convergent nozzle The pressure variations of a fluid flowing through a convergent nozzle are shown in Fig. 10.10. Assuming that the design back pressure is the critical pressure, p,, then when the back pressure is above this value the nozzle is overexpanding as shown by line (a), and the mass flow is some value below the maximum, When the back pressure is equal to the critical pressure the expansion follows the line (b), the nozzle is choked, and the mass flow is a maximum. When the back pressure is below the critical pressure the expansion in the nozzle still follows the line (b), but there is an additional expansion from p. down to the back pressure, p,, outside the nozzle. It can be seen from Fig, 10.10 that in the expansion outside the nozzle the pressure oscillates violently, and in fact a shock wave is formed. In this latter case the nozzle is underexpanding.Fig. 10.10 Pressure variations for flow through a convergent nozzle Fig. 10.11 Pressure variations for flow through a convergent—divergent nozzle Pressure, ym Ps> Pe Pa= Pe De

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