BASIC THERMODYNAMICS

FOR

POWER PLANT OPERATORS

1

BASIC THERMODYNAMICS

FOR POWER PLANT OPERATORS This aim of this paper is to act as an introduction to thermodynamics that the operator will come into contact with during his day to day operational duties. The subject matter will be kept basic and simple with the use of diagrams and illustrations and the avoidance of more complicated mathematical formula. It is therefore hoped that the contents of this paper will give a practical understanding of thermodynamics as applied to personnel connected with power station operation, in particular the more junior grades who are still in the process of learning their profession. CONTENTS: Section 1: THE PROPERTIES OF STEAM. Section 2: THE STEAM CYCLE. Section 3: THE TOTAL HEAT—ENTROPY DIAGRAM AND TURBINE EFFICIENCY. Section 4: TYPES OF TURBINE. Section 5: LOSSESS WITHIN THE TURBINE. Section 6: FEED HEATERS AND HEAT EXCHANGERS. Section 7: COMBUSTION CHEMISTRY.

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Introduction. Applied Thermodynamics. Applied thermodynamics is the science of the relationship between heat, work, and systems that analyse energy processes. The energy processes that convert heat energy from available sources such as chemical fuels into mechanical work are the major concerns of this science. Thermodynamics consists of a number of analytical and theoretical methods which may be applied to machines for energy conversion. Related topics are: • Laws of thermodynamics.

1) The zeroth law of thermodynamics. 2) The first law of thermodynamics. 3) The second law of thermodynamics.

• Definition of:

Heat engines, Turbines, Steam turbine, Gas turbine, Compressor, Thermodynamic cycle, Working fluid, Ideal gas, System.

• Definition of processes: Isobaric, Isothermal, Isentropic, Isometric, Adiabatic, Adiabatic mixing, Throttling, Free expansion, Polytropic.

• Heat transfer: Heat transfer, Heat exchangers, Heat flow through a pipe, Heat flow through a wall. • Definition and / or units of:

Energy, Power, Enthalpy, Entropy, Temperature, Pressure, Specific volume, Density, Sensible heat, Latent heat, Thermal conductivity, Mass flow. The Zeroth Law of Thermodynamics. This law states that if any object “A” is in thermal equilibrium with another object “B” which in turn is in thermal equilibrium with an object “C”, then object “C” is also in thermal equilibrium with object “A”. This law allows us to build thermometers. For example the length of a mercury column “B” may be used as a measure to compare the temperatures of the other two objects. The First Law of Thermodymanics. ( Conservation of Energy ). The principle of conservation of energy states that energy can neither be created nor destroyed. If a system undergoes a process by heat and work transfer, then the net heat supplied plus the net work input, is equal to the change of intrinsic energy of the working fluid The Second Law of Thermodynamics. The second law of thermodynamics states that no heat engine can be more efficient than a reversible heat engine working between two fixed temperature limits.(Carnot cycle) i.e. the maximum thermal efficiency is equal to the thermal efficiency of the Carnot cycle.

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Definitions Energy: Energy is an inherent property of a system. Any system at a given set of conditions (eg. Pressure and Temperature) has a certain energy content. The concept of energy was invented to describe a number of processes such as conversion of work to heat. The SI unit of energy is the joule (J). It can also be expressed as 1 kcal = 4186.8 J Sensible Heat Sensible heat is defined as the heat energy stored in a substance as a result of an increase in its temperature. The SI units are expressed as kJ/kg. Latent Heat. Latent heat is defined as the heat which flows to or from a material without a change in temperature. The heat will only change the structure or phase of the material. e.g. melting or boiling of pure water. The SI units are expressed as kJ/kg. Internal Energy The internal energy of a system. is the energy content of the system due to its thermodynamic properties such as pressure and temperature. The change of internal energy of a system depends only on the initial and final stages of the system and not in any way by the path or manner of the change. This concept is used to define the first law of thermodynamics. Specific Volume. The specific volume, v, of a system is the volume occupied by the unit mass of the system. The relationship between the specific volume and density is v = 1/p The SI unit of specific volume is m3/kg. Entropy Entropy of a system is a measure of the availability of its energy. A system with high entropy can do less useful work. This concept was formally used to define the second law of thermodynamics. The SI unit for entropy is kJ/kg.K. Specific Entropy. Specific entropy of a system is the entropy of the unit mass of the system and has the dimension of energy/mass/ temperature. The SI unit of specific entropy is kJ/kg.K. Enthalpy Enthalpy of a system is defined as the mass of the system, m, multiplied by the specific enthalpy of the system, h i.e. H = m × h. The SI unit of enthalpy is kJ/kg. Specific Enthalpy Specific enthalpy of a working fluid, h, is a property of the fluid which is defined as: H = u + Pv. where, u = Specific internal energy P = Pressure v = Specific volume Specific enthalpy has the same dimension as [ energy/mass ] The SI unit of specific enthalpy is kJ/kg.

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Temperature Temperature is a measure of hotness and can be related to the kinetic energy of the molecules of a substance. A number of physical phenomena can be used for measuring the temperature of an object. An instrument used for measuring temperature is called a thermometer and is constructed by using one of the following principles: • The change of length, such as the length of a column of mercury, • The change of volume, such as the volume of a fixed mass of gas at constant pressure. • The change of pressure, such as the pressure of a fixed mass of gas at a constant temperature. • The change in electric resistance, such as in a thermister. • The flow of electricity due to the heating effect of the dissimilar materials in the thermocouple. • The radiation, as in a radiation pyrometer. All thermometers require a scale. For instance the Centigrade scale has been defined from the melting (0°C ) and boiling (100°C ) points of pure water at atmospheric pressure. Where the symbol C is for centigrade (or celsius), and K for kelvin. If the temperature on the Centigrade ( or celsius ) scale is zero, then the absolute temperature on the Kelvin scale will be: C + 273.15. The absolute temperature or thermodynamic temperature (degree kelvin, K) is a fundamental dimension. Pressure The pressure of a system is defined as the force exerted by the system on a unit area of its boundaries. This is the definition of absolute pressure. Often in pressure measurements a gauge is used to record the pressure difference between the system and the atmospheric pressure. This is called gauge pressure and can be stated by the following equation. P = Pg + Po, where, P = absolute pressure. Pg = gauge pressure. Po = atmospheric pressure. If the pressure of a system is below atmospheric, it is called a vacuum. The SI unit of pressure is the bar. The measurement of pressure in industrial application is normally the bar. 1 bar is approximately equal to atmospheric pressure. • Standard atmospheric pressure = 1.01325 bar = 101325 Pa.(pascals). State of Working Fluid The working fluid is the matter contained within the boundaries of a system. Matter can be in solid, liquid, vapour or gaseous phase. The working fluid in applied thermodynamic problems is either approximated by a perfect gas or a substance which exists as liquid and vapour. The state of the working fluid is defined by certain characteristics known as properties. Some of the properties which are important in thermodynamic problems are; • Pressure (P) • Temperature (T) • Specific enthalpy (h) • Specific entropy (s) • Specific volume (v) • Specific internal energy (u)

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The state of any pure working fluid can be defined completely by just knowing two independent properties of the fluid. This makes it possible to plot state changes on diagrams such as; • Pressure – volume (P-V) diagram, • Temperature − entropy (T-s) diagram, • Enthalpy – entropy ( h-s) diagram. Rankine Cycle The Rankine cycle is a heat engine with a vapour power cycle. The common working fluid is water. The cycle consists of four processes; • Isentropic expansion (Steam Turbine) • Isobaric heat rejection (Condenser) • Isentropic Compression ( Feed Pump) • Isobaric heat supply (Boiler). Turbines Turbines are devices that convert mechanical energy stored in a fluid into rotational mechanical energy. These machines are widely used for the generation of electricity. The most important types of turbines are: steam turbines, gas turbines, water turbines and wind turbines. Steam Turbines. Steam turbines are devices which convert the energy in steam into rotational mechanical energy. These machines are widely used for the generation of electrical power in a number of different cycles, such as; • Rankine cycle. • Reheat cycle. • Regenerative cycle. • Combined cycle. The steam turbine may consist of several stages. Each stage can be described by analysing the expansion of the steam from the higher pressure section of the turbine to the lower pressure section. The condition of the steam may be wet, dry saturated or superheated. Consider a steam turbine under steady state conditions, the output power would be represented by: P = m(h1− h2), where P is the output power, m represents the mass flow of steam through the turbine, and h1 and h2 are the respective specific enthalpy of the steam at the inlet and outlet of the turbine. The efficiency of steam turbines are often described by the isentropic efficiency for the expansion process. The presence of water droplets in the steam will reduce the efficiency of the turbine and can cause physical erosion of the blades. Therefor the dryness fraction of the steam at the outlet of the turbine should not be less than 0.9.

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Section 1

THE PROPERTIES OF STEAM Water boils at 100o Celsius at atmospheric pressure (1013.25 milibars absolute). If it is heated under a higher pressure the temperature at which it boils increases; and at a lower pressure, the temperature decreases. The three phases To achieve high temperature steam, heat must be applied through three specific phases: a water phase, a combination of water and steam phase and a steam only phase. The heat value at any particular point of pressure and temperature is well plotted and the engineer has a selection of tables at his disposal for calculations. Using these he is able to find the specific enthalpy (the amount of heat possessed by one kilogram of the substance), specific entropy (an abstract number a substance is said to possess which increases or diminishes as heat is received or rejected) and the specific volume. The units for the above three properties are kilojoules per kilogram (kJ/kg), kilojoules per kilogram degree Kelvin (kJ/kg K) and cubic decimetres per kilogram (dm3/kg) for enthalpy, entropy and volume respectively. An example of these tables is shown in Figure 1. Abs. Press

(bar) Temp (oC)

Specific Enthalpy (kJ/kg)

Specific Entropy (kJ/kg . K)

Specific Volume (dm3/kg)

Ps Ts Hf Hfg hg Sf Sfg Sg Vf Vfg Vg 100.0 310.961 1408.0 1319.7 2727.7 3.3605 2.2593 5.6198 1.4525 16.589 18.041 102.0 312.420 1416.7 1307.5 2724.2 3.3748 2.2328 5.6076 1.4596 16.146 17.605

Figure 1. An example of steam tables. If we apply heat to water, we term this heat "the specific enthalpy of the saturated liquid", better known to the "older" engineer as sensible heat - and used in some cases in this paper. As we continue to apply heat a temperature is reached (depending on the pressure exerted) when no further increase in temperature takes place and the water boils. The heat value at this point is tabulated in the steam tables under the column headed "hf". If we continue to apply heat, the water begins to evaporate into steam, until all the water has been evaporated. This heat value is termed "the increment of enthalpy for evaporation", better known to most engineers as latent heat. The value of this heat is tabulated in the column headed "hfg". At this point we have the total amount of heat added in the first two phases; this quantity is termed "the specific enthalpy of the saturated vapour" and is tabulated in the column headed "hg. We now have hf + hfg = hg in the kJ/kg column. We can continue to apply heat to the vapour, but the temperature will begin to rise. This is called superheat and termed "the increment of enthalpy for superheat". At any point of superheating steam the specific enthalpy is equal to the increments of enthalpy of the saturated liquid plus the increments for evaporation and for superheating the steam to that point. This "total" superheat enthalpy is in the steam tables under the column headed h, but not shown here. The temperature - entropy diagram To understand the process more easily we can plot the boiling water point and dry saturation points and show them on a temperature/entropy (T-S) diagram, Figure 2. Entropy, mentioned earlier, is a very difficult property to define. Steam having a lower energy value is said to have lower entropy. If the absolute temperature at which heat is added is multiplied by the change in entropy, the product is equal to the amount of heat added in the process. Conversely if the absolute temperature at which heat is rejected is multiplied by the difference between the entropy at the beginning and completion of the process the product is equal to the amount of heat rejected.

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If we look at the adding of heat during evaporation we may understand entropy better. The process takes place at constant pressure and constant temperature. Using the steam tables in Figure 1 we can calculate the amount of heat required to evaporate 1 kg of boiling water at 100 bar absolute pressure into dry saturated steam. We see from the table that at a pressure of 100 bar abs: The boiling water temperature = 310.961 oC Therefore the absolute temperature is 310.961 + 273.15 = 584.11 K Entropy at boiling point (column Sf) = 3.3605 kJ/kg . K Entropy at saturation point (column Sg) = 5.6198 kJ/kg . K Therefore the change in entropy (Sfg) 5.6198 - 3.3605 = 2.2593 kJ/kg . K Therefore heat added in the process is 584.11 x 2.2593 = 1319.68 kJ/kg

Figure2. Temperature/Entropy diagram

The shape of the boiling water/dry saturation curve is such that as the pressure exerted on the water increases, the less enthalpy for evaporation is required. As we apply heat during the evaporation process,

973.15

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the vapour produced becomes gradually dryer until it reaches the dry saturation point. It is then said to be 100 per cent dry. We can illustrate this by adding the lines of constant wetness to the temperature/entropy diagram. It should be remembered that to carry out calculations where the volume of wet steam is required, say for example, designing a chamber where wet steam will be present, the specific volume of the wet steam will be different from that shown in the steam tables at the boiling pint Vf or Vg. For example if the steam is x dry only x kilograms has the specific volume of the dry saturated vapour. Therefore (1-x) kilograms possess the specific volume of the saturated liquid. So we say that the specific volume of the wet steam is equal to x times the specific volume of the dry saturated vapour plus (1-x) times the specific volume of the saturated liquid. Usually the [(1 - x) (spec.vol.sat.liquid)] is small by comparison to [(x) (spec.vol.sat.vapour)] and can often be ignored. The critical point As we increase pressure, we increase the temperature at which water boils, and so move higher up the temperature/entropy diagram. By doing this we shorten the length of the evaporation line until we reach a point around 221.2 bar abs where the boiling water line meets the dry saturation line and there is no evaporation phase at all. This is the critical point. The temperature is 374.15oC, and the critical volume 3.17 dm3/kg. Figure 2 shows the complete temperature/entropy diagram with lines of constant pressure and critical pressure. At pressures higher than 221.2 bar abs the condition is supercritical; this zone is shown on the diagram. If water at a supercritical pressure is heated, the temperature will rise until at the point relating to that pressure it "flashes: instantly to steam and superheating begins. The specific volume of the dry steam will be no different from that of the liquid. The flashing temperature of water at a supercritical pressure is not known precisely. A typical power station supercritical boiler would operate at a steam pressure of around 250 bar absolute. The transition from water to steam takes place around 385oC. Knowing a little about the properties of steam we can look at the "practical cycle" for the operation of a turbine.

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Section 2

THE STEAM CYCLE Steam turbines for power generation use a closed loop cycle, the working fluid to be used over again. Any steam present after passing through the turbine has to be condensed so that it can be pumped (at high pressure) back to the boiler to begin the cycle again. The heat rejection in condensing the steam takes place at a pressure well below atmospheric. The sensible heat in condensate is retained. The ideal (Rankine) cycle We assume in the ideal cycle heat as the steam expands through the turbine, it does so isentropically (constant entropy) and without friction losses. Rankine of Great Britain and Clausius of Germany proposed this method of showing the cycle efficiency at about the same time. Let us work through an example of a basic cycle to find its heat values for the three phases and its thermal efficiency. We will take the pressure used previously - a turbine being presented with superheated steam at 100 bar abs, 566oC and the steam expanding through the turbine until it is at a pressure of 30 milibar absolute. The saturation temperature corresponding to that pressure is 24.1oC. Using the composite diagram, Figure 3, we see the isentropic expansion through the turbine at 6.8043 kJ/kg.K. Condensing the steam after the last row of blades would be isothermal (constant temperature) at 24.1oC.

Figure 3. Ideal (Rankine) cycles

condensingIsothermal

Evaporation

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The heat value of the superheated steam is made up of the following components: Sensible heat (Area below A, B) = 1307 kJ/kg Latent heat (Area below B, C) = 1319.7 kJ/kg Superheat (Area below C, D) = 811.6 kJ/kg Therefore the total heat value = 3438.3 kJ/kg Area ABCDFA has done work in the turbine, while the area below AF has not. It represents the amount of heat that has been rejected. To calculate the reject from this particular cycle multiply absolute temperature by the difference in entropy: = (24.1 oC + 273.15) . 6.8043 - 0.3544) = 297.25 x 6.45 = 1917.23 kJ/kg Therefore thermal efficiency = (Total heat - Rejected heat) x 100 Total heat = (3438.2 - 1917.2) x 100 3438.2 = 1521 x 100 3438.2 = 44.23% This value is the highest theoretical efficiency using Rankine Cycle with steam at the pressures and temperatures quoted. It assumes no losses in system so in practice the efficiency is considerably less than calculated figure. The cycle is only basic but can be improved in two ways: 1. By reducing the heat rejected, in the part below AF. To achieve this we would need to lower the

condensing temperature, which is dependent upon the sea water temperature over which we have little control.

2. By increasing the useful work of the cycle covered by ABCDF. Although to raise the superheated

steam temperature would mean building superheater tubes from far superior metal than we can provide at the present time.

Reheating There is no fundamental difference between superheating and reheating as far as the temperature – entropy diagram is concerned. Reheating is really re–superheating of the steam after partial expansion in the turbine. The best way to illustrate the effects of reheating is by use of the temperature – entropy diagram, whereby the shape of the diagram extends the cycle further to the right, by “reheating” the superheated steam after it has expanded through the high pressure turbine. It need only be reheated to its original temperature, say, 566oC. The efficiency of the reheat cycle is obtained in a similar way to that of the non – reheat cycle, but the steam gets heated twice in the cycle. The initial heat condition is from the feed water via the feed heating system, and then in the economiser and boiler water/steam circuit, and finally to the superheater where the steam is heated to turbine stop valve conditions. After exhausting from the HP turbine the steam is reheated up to its initial stop valve temperature.

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Look again at the cycle using superheated steam to the turbine at 100 bar abs and 566oC, expanding through the turbine until it reaches a pressure of 30 milibar abs with a saturation temperature of 24.1oC. This time the steam is reheated after the HP turbine at a high pressure exhaust of around 20 bar abs and will be reheated in the boiler to its original 566oC. Figure 3 shows how this can be represented on the temperature/entropy diagram, points G, H, J, H are now included in the cycle. The total heat value now would be: = 3438.2 + the heat added by reheating = 3438.2 + 564.8 kJ/kg = 4003.0 kJ/kg The heat rejected would be the area under JA: = (24.1oC + 273.15) . (7.6134 - 0.3544) = 297.25 x 7.259 kJ/kg = 2157.74 kJ/kg The thermal efficiency = (Total heat - Heat rejected) x 100 Total heat = 4003.0 - 2157.7 x 100 4003.1 = 46.1% Therefore, reheating the steam has improved the efficiency of this particular cycle form 44.23 per cent to 46.1 per cent. Comparing this with the previous cycle we see that the isentropic expansion has moved further to the right, moving the steam in the final stages of the turbine into a dryer region. This brings an additional practical advantage because it reduces erosion on the LP blades caused by water droplets present in the wet steam. ie; drier steam at exhaust. Regenerative Feedheating This is a final way of improving the ideal cycle efficiency. Steam is bled from an appropriate stage of the turbine and used to heat the feedwater (condensate) going back to the boiler. The bled steam will give up its superheat, latent heat and possibly some of its sensible heat and relieve the boiler of this work. Most importantly, it reduces the amount of rejected heat in the cycle that would otherwise be lost to the condenser. Looking again at the original cycle we see that although the area of useful heat bounded by LKF has been reduced, the area of rejected heat (below LF) has been reduced by a far greater amount. The heat lost from LKF has been transferred via heat exchangers to the boiler feedwater, so is not lost from the cycle. The subject of regenerative feedheating is discussed in greater detail later in the text.

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Section 3

THE TOTAL HEAT - ENTROPY DIAGRAM AND TURBINE EFFICIENCY In this section we consider the more practical way of determining the efficiency of a turbine stage. This is achieved by a Total Heat versus Entropy diagram, often called the Mollier Chart because it was devised by Richard Mollier (1863 - 1935). In years gone by every turbine efficiency engineer would have a large Mollier diagram pinned to his office wall. Nowadays the chart has been mostly superseded by computer programs. But to understand the calculations more easily we will look at the part of a Mollier chart in Figure 4. The y axis represents enthalpy (kJ/kg) and the x axis entropy (kJ/kg. K). Within the chart are lines of constant pressure, constant temperature, constant dryness and the dry saturation line. Having all these parameters on the chart enables the engineer to pin-point the exact condition of the steam being presented to the turbine and wherever tapings are fitted for pressure and temperature. Figure 4. Mollier Chart Figure 5. Stage efficiency

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Although we have shown the expansion through the turbine as isentropic, there are inefficiencies that prevent this in practice. The main ones are turbulence and friction of the steam over the blades which causes the steam to heat up. The effect is to move the condition of the steam at the exit further to the right on the Mollier chart. This can be seen more clearly in Figure 5 where a represents the temperature and pressure of the superheated steam at the inlet to the turbines stop valves, and b represent the temperature and pressure at the inlet to the turbine. The path from a to b is horizontal because there has been no loss of heat, just a drop in pressure and temperature due to the throttling effect of the inlet valves. The vertical line bc represent the ideal isentropic expansion, "d" being the final condition of the steam at the exit from turbine stage. Stage efficiency = (Enthalpy @ b - Enthalpy @ "d") x 100 (Enthalpy @ "b" - Enthalpy @ "C") = h x 100 H

Figure 6. Cylinder efficiency Figure 6 shows the expansion of steam through a complete high pressure and low pressure turbine, from a stage inlet pressure of 50 bar absolute to final stage exit pressure of 200 milibar absolute. = (3200 - 2460) x 100 (3200 - 2200) = 740 x 100 1000 = 17% Although the inlet to final condition is represented by a straight line bd the first stage and final stage are less efficient so line bd should in reality be made up of many shorter lines representing the efficiency of each set of blades. Each stage has a leaving loss due to kinetic energy possessed by the steam. We will consider this later on.

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We have mentioned bleeding off steam to heat the boiler feedwater, we can also improve efficiency by bleeding steam after it has done some useful work in the high pressure turbine by using it to drive the main boiler feed pumps, thereby bypassing the condensing stage where the greatest cycle losses occur. From looking at these efficiencies it is obvious that there are losses within a turbine cylinder, before going on to discuss them we should look at the types of blading in use.

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Section 4

TYPES OF TURBINE Steam turbines for electrical power generation are built in many different sizes, but all can be placed into two basic types: Impulse and Reaction. However, no one method of classifying turbines is adequate, since all large turbines are built for a particular installation. The following methods are most common: IMPULSE BLADING: As previously mentioned, thermal energy may be converted into mechanical energy by the impulse of the jet or by the reaction due to the jet leaving the nozzle. One stage of an impulse turbine will consist of a stationary nozzle ring attached to the casing and one row of moving blades attached to the shaft. The number of nozzles will depend on the quantity of steam required by the turbine. If the nozzles occupy the entire arc of the ring, the turbine is said to have full arc admission. With less than this number of nozzles, the turbine would have partial admission. One nozzle block and one rotating row of blades forms one Rateau stage. With normal blade speeds, a Rateau stage can handle an enthalpy drop of about 116 kJ / kg of steam with the best efficiency. Small units may use greater energy drops when the efficiency is not so important. When higher efficiencies are desired with a higher differential of thermal energy, the stages must be compounded by one or both of two methods. Pressure compounding consists of several Rateau stages in series, each with its own nozzle block and rotating row of blades. Velocity compounding employs one nozzle block with two or more rotating rows. In order to keep the direction of rotation the same for each rotating row, stationary reversing blades are placed between consecutive rotating rows. This type of staging, known as Curtis staging, and is often used in small single stage turbines and as the first stage of larger units. REACTION BLADING: Reaction stages are composed of one stationary row of blades and one rotating row of blades with a pressure drop occurring in each stationary and rotating row. Blades in both rows are shaped so that the area between two adjacent blades of the same row will form a converging nozzle; hence, there is a pressure drop and therefore an increase in the relative velocity across each row. Enthalpy differentials for each stage of a reaction turbine are usually lower than those for impulse stages. Thus, the reaction turbine requires more stages. Some turbine manufactures combine the impulse and reaction principles by employing a Curtis or Rateau stage as the first stage of a large turbine and by using reaction stages for the remainder of the expansion through the turbine. The advantage of this arrangement over a straight reaction turbine are the ease of governing during low load periods and the increased efficiency of the unit under these conditions coupled with a simpler steam chest design. Impulse type As previously mentioned, the impulse stage uses nozzles in the fixed blades where the heat is converted into kinetic energy (velocity). The stage has no pressure drop across the moving blades. To illustrate this point, let us now consider the diagrams as shown in Figure 7(a) and velocity/pressure profile in Figure 7(b). Generally speaking the greater the pressure drop across a nozzle the greater the velocity issuing from it, up to a certain maximum level. If we have a converging nozzle with a fixed inlet pressure Pi, and a throat pressure of Pt which can be varied by reducing the throat size, when Pi = Pt there will be no flow through the nozzle. As the ratio of Pt to Pi decreases a flow begins until it reaches the maximum when the ratio of Pt to Pi = 0.547. Reducing this ratio further will not increase the velocity. For lower values of Pt a converging/diverging nozzle must be used. To calculate the velocity of the steam at the exit the formula is:

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V = 44.72 V (KH) Where v = velocity (m/sec) H = the isentropic heat drop (kJ/kg) K = the ratio of the heat drops The angle of the nozzle is kept as low as possible to the moving blades as increasing the angle reduces efficiency. Other types of impulse stages include velocity-compounding, pressure-compounding and velocity-pressure-compounding. Figure 7(a). Pure impulse turbine

Figure7(b). Impulse stage pressure/velocity diagram Theory of Impulse Blades: The thermal energy of the steam is converted in a turbine first into kinetic energy by the nozzles and then into mechanical by the action of the jet on the blades. One stage of an impulse turbine consists of one or more nozzles admitting steam to a moving row of curved blades. As previously explained, this is a Rateau stage. A Curtis stage differs only in that the steam jet acts on two or three rotating rows of blades before leaving the turbine or before entering another set of nozzles. The impulse momentum law provides one method for solving problems on the design of turbine blades. The equation F = wa / g, where F is the force, w/g the mass, and a the acceleration, may be rearranged to become the impulse momentum relation.

VgwtF ∆=∆

Which may be stated as: an external force F acting on a moving particle which has mass w/g for a period time ∆t will change the velocity of the mass by the amount ∆V. Note that ∆t and the mass are scalar quantities, while the force and the velocity are not vector quantities. If the direction of either is known, the direction of the other will be established.

Stationary diaphragmSTEAM

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In the event that certain components of the force are desired,

( )xx VVgwtF 21 →=∆

( )yy VVgwtF 21 →=∆

(1) ( )zz VVgwtF 21 →=∆

The sign indicates the vector difference between the velocities. Next, it is necessary to apply these equations to a steam jet striking a moving blade.

FIG.1. Theoretical velocity diagram for an impulse blade The steam approaches the blade, with the absolute velocity represented by vector V1, at an angle α1 to the plane of rotation of the blade, which is moving at a velocity represented by vector U. Note that vector V1 is on the centerline of the nozzle supplying the steam and is the velocity leaving the nozzle that we previously called Vm. Vector V1 may be resolved into two components, C1, which is the velocity of the steam relative to the blade, and U, which is the mean velocity of the blade If the blade is assumed to be frictionless, then the only effect on the jet of steam will be the change in direction since there will be no change in magnitude. Therefore, the vector representing the jet leaving the blade, C2, will be in a path along the outlet edge of the blade. Vector C2 will be the relative velocity of the steam leaving the blade. In order to obtain the absolute velocity of the steam leaving the blade, add vectors C2 and U to get V2 at an angle α2 with the plane of the moving blade. The angles β1 and β2 are, respectively, the angles of the relative steam velocity entering and leaving the blade. Obviously, the steam should enter the blade with as little obstruction as possible, and so these angles are also the angles of the two edges of the blade. The force causing rotation of the wheel will be that component in the x direction, Eq (1), sometimes known as the whirl component of the jet velocity. Taking account of the directions and making w/∆t equal to W, substituting into Eq (1) gives

(2) ( )2211 coscos ββ CCgwFx +=

18

From the geometry of the figure, adding and subtracting the wheel velocity to this equation will result in the formula

(3) ( )2211 coscos αα VVgwFx +=

But the work done on the blade will be the force times the distance, or the power (P) will be the force times the velocity:

(4) ( )2211 coscos αα VVg

WUP +=

(5) ( )2211 coscos ββ CCg

WUP +=

Note that the units of P for this equation will depend on the unit of time employed to express the flow rate W, i.e., P has the units of kJ/sec if the flow is in Kg/sec or kJ/hr if the flow rate is in Kg/hr, etc. Because of the friction of the steam passing over the blade, the numerical value of C2 will be less than that of C1. Using k* to indicate the velocity efficiency of the blade, Eq. (5) will become

(6) ( )2111 coscos ββ kCCg

WUP +=

(7) ( )2111 coscos ββ CCg

WUP +=

And UVC −= 1111 coscos αβ * sometimes k is called the blade velocity coefficient. When setting the velocity ratio of the blade speed to the steam jet leaving the nozzle (U/V1) equal to ρ,

(8) UC

−= 1coscos 1

11 ραβ

gWUP

2

=

−1cos 1

ρα

+

1

2

coscos1

ββk

Since the energy supplied to the blade is the kinetic energy WV1

2/2g, the efficiency of the blade will be Eg.(8) divided by the kinetic energy of the jet,

(9) ( )

+−=

1

21 cos

cos1cos2ββραρ kEb

It is important to note from this equation that efficiency depends on the velocity ratio of the blade and the jet and not on the absolute values of either. In order to determine the conditions for which the efficiency will be the maximum, differentiate Eq.(9) with respect to the velocity ratio ρ, and set it equal to zero. Thus,

19

(10) 0coscos1)(cos2

1

21 =

+−

ββραρ

ρk

dd

2cos 1αρ =

This shows that the best efficiency for the normal nozzle angle of 15 deg will be obtained when the velocity ratio is about 0.48; also that the angle α2 will be 90 deg. Thus, if the blade velocity were 213.36m/s, the maximum jet velocity for best efficiency should only about 426.721m/s, which represents a drop of 93kJ/kg of steam in the nozzle. Mean blade speeds of 107 to 229m/s are common, while the maximum for the largest low-pressure blading is about 345m/s. Only two ways remain to increase the outlet areas: use a large exit angle or increase the blade height. The latter method has limited possibilities, but the exit angle is usually found to be 5 or 10 deg larger than the nozzle angle. In the event that the outlet area of the blade is made smaller than the inlet area, the blade is said to have some reaction effect. It will be noted in the blade-efficiency curves given later that reaction blades are more efficient than impulse blades, and so a small pressure drop through the impulse blades, which is used to give some reaction, is desirable. The inlet angle must be chosen so that shockless entrance is assured, or else the steam jet will strike either the back of the blade or the concave front portion. If the jet strikes the front of the blade, the impulse will be increased but not efficiently because of the flow disturbances that will be set up. If the jet strikes the back of the blade, it will have a retarding effect that will be still worse. The inlet angle will about 10 to 16 deg larger than the nozzle angle, as can be seen from the velocity diagram (fig.1) Values of k decreases with the increased relative steam velocity, increase with increased blade widths, and decreases with increased total change in steam direction as the steam passes through the blade (called the angle of deflection); the values range from about 60 to 95%. When long blades are used in turbines of higher ratings, there will be some difference between the velocity ratio at the root of the blade and at the tip. In design, the mean velocity is used in the velocity ratio. As a compensation for the variation in blade velocity, the blades are warped or tapered so that the angles change throughout the length of the blade and the blade section decreases toward the tip, Fig. 9-10. Combined velocity diagrams using different methods of presentation are shown in Fig. 2. Part (a) and (b) are both drawn for an impulse blade with no allowance for friction. The type of diagram used in (a) indicates the velocities by the vector length, but the directions are incorrect. The methods of parts (b) and (d) have the advantage of showing both the magnitude and direction of the vectors. Diagrams such as these are used in practice in preference to the method used in Fig.1. Fig.2(c) is drawn the same as Fig.2(a) but indicates that friction has been taken into consideration The velocity diagrams indicate that the steam travels in the direction of the rotating blades. When the nozzle diaphragm has only a part of the arc occupied by nozzles, the second diaphragm for a multistage turbine must be displaced in the direction of rotation to receive the steam with the least turbulence. This lead placed on succeeding diaphragms permits efficient use of some of the residual velocity, V2. Obviously, lead has no meaning when there is full peripheral admission in the first-stage diaphragm.

20

FIG.2. Combined velocity diagrams for no friction loss at (a), (b), and (d); with friction loss at (c). The energy represented by the velocity V2 is known as the leaving loss and is expressed in kJ/kg. The energy gained by taking advantage of it in the design is known as carry-over and will amount to 50 to 75% of the leaving loss. Changing Eq.(9-1) to take care of the carry-over,

(11) Vm = 223.8 2

01 8.223)(

+− c

conhVEhhE

Where Enb = nozzle efficiency Eco = carry-over efficiency Vc = carry-over velocity, fps And other symbols are as for Eq.(1) and Eq.(5) From Eq. (9), the blade efficiency is

( ) %6.828251.0

8251.088.01439.09397.0439.02 =

×

+−×=bE

Leaving loss = 9.2kJ/kg. Velocity and Pressure Compounding. The Rateau stage of an impulse turbine has to absorb efficiently more kinetic energy than that represented by an enthalpy drop of about 93kJ/kg. When it is necessary to transform more than this amount, the turbine must be arranged in stages, each of which will transform a part of the total energy. The stages may be designed on the basis of velocity compounding so that the total energy can be used and at the same time wheel speeds kept relatively low. In velocity-compounding, the energy from the jet of steam issuing from a single nozzle, or group of nozzles, in one diaphragm is imparted to several rows of moving blades. Such a stage, known as a Curtis stage after its inventor, is shown in Fig.3

21

FIG.3. Nozzle block, stationary, and rotating blades of a Curtis stage.

In pressure compounding, the steam passes through a nozzle group, the imparts a portion of its energy to a single rotating row of blades, and then passes through the other successive stages, each stage being composed of a group of nozzles and one rotating row of blades. Only a part of the total energy is converted into kinetic energy in each Rateau stage. Under some conditions, velocity compounding is used for the first stage of a turbine, while the remaining stages are pressure-compounded. Curtis stages have been built with as many as four rotating rows of blades. However, it can be shown that the ratio of work obtained from each row is in the proportion of 7:5:3:1. In other words, the last row does only 1/16 of the total work of the stage. Two-row Curtis stages are very common. If the total pressure drop through the turbine is divided among several stages by pressure-compounding to reduce the velocity at each wheel, then the total enthalpy differential for the turbine may be divided equally among the several stages. The velocity for each stage will be, theoretically,

( )n

hhV 218.223 −=

In which V = steam jet velocity for each stage fps

h1 = initial enthalpy, kJ/kg h2 = final enthalpy, kJ/kg n = number of stages

Or the original jet velocity for the entire unit will be reduced to n for each stage if n is the number of stages. This means that for a two-stage turbine with the blade speed at 2133.6m/s and the steam-jet velocity twice the blade speed, the maximum allowable drop in enthalpy for efficient operation will be about 186kJ/kg. The effect of velocity-compounding in reducing the peripheral velocity of a two-row Curtis stage may be seen by examination of the velocity diagrams, Fig.4. Steam leaves the nozzle with the velocity V1 and enters the first moving blade with the relative velocity C1. The jet leaves the first moving blade with the relative velocity C2, numerically equal to C1 because friction is neglected, and enters the fixed reversing

22

blade with the absolute velocity V2. The absolute velocity V3 of the jet leaving the fixed blade is the same, numerically, as the entering velocity when friction is neglected. Velocities for the second moving blade are determined by the same means as for the first moving blade. The two rows of moving blades may be on separate wheels or on the same wheel. The values of the angles are customarily different from those for a Rateau stage. Note that the horizontal component of the velocity V1 in fig.4(b) is four times the blade velocity, whereas in a Rateau stage the relationship is two to one. In general, if there are n rows of moving blades in a Curtis stage, the velocity of the moving blades will be 1/n times the velocity of the single Rateau stage.

FIG.4. Velocity diagrams for a two-row Curtis stage without friction.

Applying this to the same blade speed used for one of the Rateau stages, the allowable enthalpy drop would be almost 372kJ/kg for a two-row Curtis stage. If friction were taken into account, the result would be slightly different. The variations in pressure and velocity for flow through a two-stage Curtis turbine and for a two-stage Rateau turbine are shown in Figs.5 and 6, respectively.

23

FIG.5. Pressure-velocity history of a two-stage Curtis turbine FIG.6. Pressure-velocity history of a

two-stage Rateau turbine Impulse-blade Efficiencies. Eq.(9) expressed the blade efficiency for an impulse stage as the ratio of the energy output of the blade and the kinetic energy of the steam jet entering the blade. Combining this efficiency with the efficiency of the nozzle, Fig. 7 results in a value frequently termed the diagram efficiency. This term is understood to include the effects of all losses due to friction in the nozzle and blade, blade angles, nozzle angles, velocity ratio, and leaving velocity.

FIG.7. Diagram efficiency Stage efficiency includes all the losses listed above, and in addition, the losses due to steam leakage through the diaphragm packing, windage or rotational losses, moisture losses due to wet steam, nozzle-end losses for partial peripheral admission, configuration losses, etc. thus, stage efficiency represents the ratio of the actual enthalpy difference across the stage to isentropic enthalpy difference across the stage for the same inlet and outlet pressure, Fig..3. Note that friction has the effect of reheating the steam at constant pressure.

24

The efficiencies used by turbine designers are determined by test, and a detailed discussion of them would be beyond the scope of this text. Briefly, the factors that must be considered are: (a) Nozzle Height. A high nozzle is usually more efficient than a low one. (b) Over-and Under expansion (c) Steam Displacement. This factor must be considered when there is partial admission. i.e., when

nozzles are used around only part of the nozzle ring. As the blades travel through the arc where there are no nozzles, they carry stagnant steam with them. Therefore, when a jet of steam enters the blades, it must displace the stagnant steam before imparting energy to the blades.

(d) Mean Diameter. It has been determined by tests that a large diameter wheel will be slightly more

efficient than a small one. (e) Moisture. Moisture in the steam will decrease the efficiency because (1) the moisture moves

slower than the steam and will strike the back of the blade and retard its forward motion, and (2) the moisture reduces the quantity of steam actually doing work. Also, since moisture in the steam has an erosive effect on the blade edges, it is kept to a maximum of about 13%. The moisture is estimated empirically to reduce the efficiency 1/2% for each 1% moisture present in the steam.

(f) Windage. This is a frictional loss due to the blades, shrouds, and discs rotating in a dense

atmosphere of steam.

FIG.8. Stage efficiency shown on Mollier diagram An interesting comparison of the efficiencies of reaction, Rateau, and Curtis stages is obtained by plotting the efficiencies against the enthalpy drop instead of the velocity ratio, Fig.9, for an assumed wheel velocity. The reaction stage must operate on a small enthalpy drop, and for this reason reaction turbines have more stages than impulse turbines for the same steam conditions. The Curtis stage, while having a low peak efficiency, maintain better efficiency with large pressure drops than either of the others-an important consideration for the first stage of a multi-stage turbine.

25

FIG.9. Nozzle and diagram efficiencies

Other losses that must be included in the turbine performance are those due to mechanical losses which include bearing friction, oil pumps, glands, and the governor system. All these losses are determined by test. The turbine is revolved at the desired speeds by some external source, and only sufficient steam is admitted to the casing to maintain the correct density inside. The power required to rotate the turbine will be the sum of the windage, friction, and mechanical losses, which should be deducted form the power generated by the blades to determine unit output. The blade windage losses may be reduced where there are no nozzles by using shields or baffles, Fig.10. An expression for diagram efficiency would be a ratio of the power output of the blades and the energy supplied to the nozzle.

(12)

( )

+−

=2

01 8.223778 c

dVhhW

PE

where Ed = diagram efficiency and other symbols are as in Eqs.(11)-1 and (4) through (8)

FIG.10. Shields to reduce windage loss of a Curtis stage where there are no nozzles

26

Reaction type The pure reaction stage is shown diagrammatically in Figure 7(c). A reaction stage is usually 50 per cent impulse and 50 per cent reaction. This type has a pressure drop across both sets of blading. Various degrees of reaction can be used. Modern turbines incorporate the best features of all types. For example, a turbine may have a velocity-compounded first stage followed by a pressure-compounded impulse stage, with reaction blading at the low pressure stages. It is common to find turbines fitted with an impulse type first stage in both the high pressure and intermediate pressure turbines. The advantages of this type of arrangement are: 1. the heat drop is high 2. the length of the cylinder is reduced 3. the cost is reduced 4. the temperature and pressure after the stage is lower, avoiding expensive materials 5. with lower pressure comes lower gland leakage Figure 7(c). Pure reaction turbine Theory of Reaction Blading. Both the underlying principle and the construction of reaction turbines differ materially from those of the impulse unit. One row of stationary and one row of rotating blades constitute a reaction stage. The blades are unsymmetrical, with the outlet angle much larger than the inlet angle, so that the available area for the steam jet is in the shape of a converging nozzle. There is a drop in pressure in both the stationary and rotating rows with a corresponding increase in specific volume. The absolute velocity of the steam increases in the stationary blades and the relative velocity increases in the moving blades, but the absolute velocity of the steam leaving the moving blades is less than the absolute entrance velocity. Since the turbine runs full of steam, the volume flow at any section is the product of the free area and the velocity; but since the volume increases as the steam progresses through the turbine, either the area or the velocity, or both, must also increase. In order to keep the blade lengths within reasonable limits, the area and the velocity are increased toward the exhaust of the turbine. All turbine blades may be classified according to their degree of reaction, which is defined as the ratio of the enthalpy drop in the moving blades to the enthalpy drop in the stage consisting of one stationary row of blades and one rotating row of blades. The degree of reaction may vary from 0 to 100%, zero reaction being pure impulse blades. It is not at all uncommon for impulse blades to have a small amount of reaction (say, 5%) to improve their efficiency.

P2 P2

P1STEAM

27

Parsons referred to the blading he developed as impulse-reaction blading. Since his blades were 50% reaction, they were also 50% impulse. However, blades containing an appreciable percentage of reaction effect, such as 50% reaction, are known by the shortened term of reaction blades. Since there is an equal enthalpy drop in the stationary and moving rows of 50% reaction blades, the pressure drop across the moving and stationary rows is equal; also, the blade passages have the same shape. This represents a manufacturing advantage. It also follows that the velocity diagrams for both the stationary and rotating rows are identical, Fig.11. because of its importance, and because space does not permit a detailed discussion of all varieties of blade, the equations developed for reaction blades will be confined to 50% reaction. Therefore. V1 = C2, C1 = V2, α1 = β2, and β1 = α2

FIG.11. Velocity diagrams for reactions blades.

Then ( )2211 coscos ββ CCg

WFx −= (13)

( )2211 coscos αα VVg

W−= (14)

Then UCV −= 2222 coscos βα But 1122 coscos αβ VC = Therefore UVV −= 1122 coscos αα (15)

Then ( )UVg

WFx −= 11 cos2 α

Since the power, P, developed by the blade is UFx and 1V

U=ρ

( )21

21 cos2 ραρ −=gWVP (16)

28

If ∆h is the isentropic enthalpy drop per kg for a stage and en is the nozzle efficiency, then the amount

∆

2hWen is the enthalpy converted into kinetic energy per row, since the fixed and moving rows are

symmetrical. The exit velocity V2 will be available as carry-over to the next stage and may be assumed available for the stage under consideration from previous stage. Since the rows are symmetrical, this carry-over also exists between the stationary and moving rows.

If the carry-over efficiency is eco, the energy is

gWVeco

2

22 . The energy available per row is, using

7781

=A , Ag

WVeWVhWe con

−=

∆22

22

21

Or per stage Ag

WVeWVhWe con

−=∆

22

21

Age

VeVhn

co

−=∆

22

21

The diagram efficiency is ( ) ( )

22

21

21

21

22

21

21

21

cos2cos2

VeVVe

geVeV

gV

Eco

n

n

cod −

−=

−

−=

ραρραρ

( )

21

22

21

1

cos2

VVe

e

co

n

−

−=

ραρ

From the Law of Cosines,

2222

22

2 cos2 βUCUCV −+=

1122

1 cos2 αUVUV −+=

12

21

22 cos21 αρρ −+=

VV

( )( )1

2

21

cos211cos2

αρρραρ

−+−−

=co

n

eeEd (17)

Several interesting conclusions can be obtained from Eq.17. If the carry-over efficiency (eco) is 100%, then the diagram efficiency is the same as the nozzle efficiency and is independent of either the velocity ratio (ρ) or the blades angles. If the nozzle efficiency (en) were also 100%, then it would follow that the diagram efficiency would be 100%. With a carry-over efficiency of zero, the diagram efficiency would be

( )21cos2 ραρ −= nd eE ; and also with 100% nozzle efficiency, the diagram efficiency would be

29

21cos2 ραρ −=dE . In order to determine the velocity ratio that will give maximum efficiency, the last

equation may be differentiated with respect to ρ and equated to zero to obtain.

1cosαρ = (18) The carry-over efficiency of the blading is very difficult to determine by test and is included with the diagram efficiency as determined from testing. However, it is undoubtedly less than 100 %and may be in the neighbourhood of 60 to 80%. It can be seen that there would be no carry-over to the first stage of a turbine and that the carry-over from the last stage would be a complete loss. In the latter case it is called the leaving loss and is found in impulse as well as in reaction turbines. Exact values for leaving loss cannot be given because of the many factors involved, such as the design conditions versus the operating conditions, the possible variations in exhaust pressure, and the load. In general, the leaving loss is less for noncondensing turbines than for condensing turbines. It may be in order of 1% of the total isentropic enthalpy drop for the first case at full load and 4 or 6% in the later case. The difference is due to the tremendous volumes that must be handled at the exhaust of a condensing unit. The specific volume of the exhaust from a noncondensing unit is much smaller than that for a condensing unit. As in the case of impulse blading, the losses throughout the stage appear as reheat, and so the stage efficiency is also the ratio of the actual enthalpy difference for the stage divided by the isentropic enthalpy difference for the same initial and final pressures. The more recent designs of reaction blades are of the curved-foil impulse nozzle. The entrance is thick and well rounded, with the back of the blade as thin as possible to reduce flow disturbances at that point. Because of the rounded shape of the inlet edge of the blade, it is difficult to assign a definite angle for the inlet, although it would probably be some 80 or 90 deg. The exit angle is not often spoken of as such but is called per cent gaging, which is the sine of the exit angle. Gaging indicates the amount of the annular area (about 25 to 60%) that is available as actual blade area for the steam. It will be noticed that the diagram efficiency curve for reaction blading is flat at the peak so that the best efficiency may be obtained with a rather wide latitude in the velocity ratio, fig.7. Values of 0.8 to 0.85 for the velocity ratio seem to be customary for constant-speed drives, but at times it is best to favour low-speed operation when the drive is variable speed such as for marine units. Then the velocity ratio for design conditions may be chosen somewhat larger than cos α1. The factors involved in the efficiency of reaction stages are similar to those for impulse stages. They include, in addition to the diagram efficiency, (1) blade leakage, (2) blade height, (3) blade width, (4) trailing edge thickness, and (5) moisture in the steam. The moisture correction for reaction blades may be higher than that for impulse blades-perhaps 11/4% for each 1% moisture in the steam after isentropic expansion. Blade leakage is a factor that is peculiar to the reaction blades. The pressure drop across the rotating blades of a reaction stage will encourage steam to by-pass the blade by going through the clearance area; thus, the steam does not do work. Shrouds (peripheral metal strips that tie the blade tips together) are used for reaction blades. They also act as seals because of the small clearances between them and the stationary portion of the turbine.

30

The pressure-velocity history for a reaction turbine may be visualized by referring to Fig.12.

FIG.12. Pressure-velocity history of a group of reaction stages.

Since the reaction turbine runs full of steam at all times, the height of the blades is determined by the continuity equation and must increase as the steam progresses through the turbine. The height may be reduced to some extent by increasing the velocity of the steam in the quantity flowing. When the unit is to operate condensing, the annular area is increased by using a conical drum (shaft). For this arrangement the mean blade velocity will increase towards the low-pressure end of turbine, but the velocity ratio can be maintained the same throughout the turbine by increasing the steam velocity. As already mentioned, increased steam velocity will decrease the require area. Dummy Pistons. In addition to causing leakage across the fixed and moving blades, the pressure differential across the moving blades causes an end thrust on the shaft. If this thrust were to be absorbed entirely by the thrust bearing, it would require bearings that would be out of proportion to the turbine. A balance may be achieved, however, by using a dummy piston or balancing piston similar in principle to that used on a high-pressure pump. A small part of the steam received by the turbine is by-passed through a labyrinth sealing strip, Fig.13, and then to a low-pressure section of the turbine by a pipe shown schematically in fig.14. By properly proportioning the areas and pressures for the dummy piston to counteract the blading thrust, the load on the thrust bearing may be reduced considerably.

FIG.13. Reaction-turbine dummy piston labyrinth.

Considering the turbine shown schematically in Fig.14, we see that the pressure drop across the rotating rows of blades will cause a trust to the right which will be equal to the pressure on the blades times the area. In order to simplify the calculations, assume that this is only one rotating row of blades and that its area is mean for the first and last row of reaction blades. A hypothetical mean blade row would have a height of (2.54 +22.86) / 2 = 12.7 cm. with 50% reaction, the pressure drop across the rotating rows, and therefore the drop across the hypothetical mean blade row, would be (12.3 - 0.035) 0.50 = 6.13kg/cm2.

31

The area of the annulus represented by the hypothetical mean blade row may be approximated by multiplying its height by the circumference at the centre of the row: area = 12.7(43.18 + 12.7) = 2230 cm2. Then Blade thrust = 6.13 x 2232 = 13682kg. Thrust for area A = 0.035 × π ( 43.182 − 20.322 ) = 40kg 4 Thrust for area B = 12.3 × 8.9 ( 61 + 43 ) π = 17886kg. 2

FIG.14. Example of dummy-piston forces. From the diagram in Fig .14., conversions from Imperial to SI units used are as follows: 175psi = 12.3kg/cm2 30psi = 2.11kg /cm2 1” Hg abs = 0.035 kg/cm2 1 inch = 2.54 cm. 8 inch = 20.32 cm. 9 inch = 22.90 cm. 17 inch = 43.2 cm. 24 inch = 61 cm. The pressure of the steam acting on area C will be determined by the point at which the leak-off pipe reinducts the steam into the turbine because the flow through the blading is much greater than the flow through the dummy piston. Therefore, the flow through the blading will be the controlling factor and will determine the pressure at the outlet of the leak-off pipe. The pressure acting on area C must be the pressure at the outlet of the leak-off pipe plus a very small friction drop through the pipe. Using the pressure at C as 2.11kg/cm2. Thrust for area C = 2.11( 612 − 20.322 ) π = 5483 kg. 4 Total thrust towards the right = 13682 + 5483 = 19165 kg

32

Total thrust towards the left = 17886 + 40 = 17926 kg. Net thrust towards the right = 19165 - 17926 = 1239 kg. If the dummy piston were not used, the thrust bearing would have a load of 13682 kg. The thrust bearing area needed for the turbine designed without the dummy piston would be 968 sq cm versus 87.1sq cm. with the dummy piston. Of course it would be good engineering to use a trust bearing somewhat larger than 87.1 sq cm. Since there is practically no pressure drop across the wheels of an impulse turbine, dummy pistons are not required for those turbines. Performance. Steam entering a turbine must travel through the main steam valve (stop or throttle valve) which includes a strainer in the steam chest, and governor valves; all of which cause a pressure drop. Friction loss through this equipment would be about 4% of the initial pressure for a multivalve governor and 10% for a single valve turbine. This pressure drop is a throttling process represented by constant enthalpy from the entrance conditions p1, t1, h1, and s1 to point 2, Fig.15, which is the entrance to the first stage of the turbine. In the first stage, the steam theoretically would expand isentropically from point 2 to point 3'. Regardless of whether the stages are impulse or reaction, there will be losses occurring in the expansion through the blades and nozzles, as previously outlined, that will appear as thermal energy in the steam leaving the stage. This will result in "reheating" of the steam from enthalpy h'3 to h3 at the stage pressure p3. Upon entering the second stage, the steam will have the enthalpy h3. The process will continue in this manner throughout the turbine until the steam emerges from the last stage at enthalpy h7. Inasmuch as this is the last stage of the turbine, there is no possibility of obtaining useful work from the kinetic energy of the steam going into the condenser where the steam will come to rest. Thus, the kinetic energy known as leaving loss is converted into thermal energy. The leaving loss is represented by the enthalpy difference h8 - h7, while the energy transferred to the condenser circulating water is represented by the difference between h8 and the enthalpy of saturated liquid at pressure p7.

33

FIG.15. Turbine condition curve.

A curved line drawn through points 1, 2, 3, 4, 5, 6, and 8, indicating the condition of the steam throughout the turbine at points where it is possible to extract and measure the steam properties, is known as the condition curve. Note that the condition curve does not represent the path of the steam on the Mollier diagram during its flow through any particular stage of the turbine but indicates only the properties of the steam entering and leaving the stages of the turbine. Stage efficiency has been defined as the ratio of the actual enthalpy drop across the stage to the isentropic enthalpy drop for the same pressure differential. When this ratio is applied to a group of blades or to the entire turbine, it is referred to as the Rankine Cycle Ratio (RCR), or the internal efficiency of the turbine. This expression does not include such mechanical losses as bearing, oil pump power, etc.

34

Section 5 LOSSES WITHIN THE TURBINE These can be divided into two groups, external or internal. External losses Typical losses in this category are: 1. Shaft gland leakage, which is a small amount of steam bled off to provide a seal for the cylinders 2. Friction losses from the bearing journals, oil pump and the governor gear, for example. The external

losses speak for them selves Internal losses Most of the losses occur in this category, they include: 1. Nozzle friction 2. Blade friction 3. Disc friction 4. Diaphragm gland tip leakage 5. Partial admission 6. Wetness 7. Exhaust (leaving loss) The internal losses are considered in some detail. Nozzle friction. This reduces the heat drop through the nozzle and the steam velocity at the exit. The steam is therefore slower entering the next stage. Because the angle of the blades is designed to allow the steam to flow smoothly any deviation from the design velocity results in a "shock" as the slower steam strikes the next row of blades. Blade friction. Similar to the previous loss but blade friction causes a loss of kinetic energy over the sides of the blade; this causes the steam to leave the stage with a higher heat content than it should. This means a loss in stage efficiency because expansion has not been isentropic. The effect of this can be seen in Figure 6 Section 3. Disc friction. The disc of an impulse stage is in contact with steam and tends to drag a layer of steam with it causing slight turbulence, and so efficiency suffers. Diaphragm gland and tip leakage. There is a large pressure drop across impulse type stages; this occurs across the nozzles (the stationary row of blades called "diaphragms"). The small gap between the diaphragm and the rotor is a source of leakage. This is minimised by interstage glands between the rotor shaft and the diaphragm. Any wear on these glands increases steam leakage. This leaking steam also interferes with the normal steam path. There should be little or no pressure drop across the moving blades although in practice there is normally a slight drop. Holes are drilled across the disc to equalise the pressure (Figure 8(a)); with the impulse-reaction type of stage there is a pressure drop across each set of

35

blades whether fixed or moving. Consequently each set of blades has sealing arrangements, as shown in Figure 8(b). The seals prevent steam flowing around the tips of blades where it would do no useful work. This problem is sometimes overcome by fitting an impulse stage at the beginning of the turbine. Smaller turbines often use "end-tightening", moving the shaft axially to tighten any clearances, the machine start up from cold with large clearances, once warmed up and fully expanded the rotor can be adjusted to close the clearances to the minimum. (a) Impulse stage

(b) Reaction stage (with end tightening)

Figure 8. Sealing arrangements Partial admission. Some turbines control the quantity of steam by nozzle governing. Instead of one large inlet valve governing the flow, they have a series of smaller valves (nozzles) which allow an increasing amount of steam through as each opens in turn. At full load they are all open, at lower loads some are closed. There are areas (arcs) where steam flows to the first stage and areas where there is no steam flowing; this causes eddies and the steam flow is disturbed. Wetness. This loss usually occurs in the last few stages of the low pressure turbine, when expansion reaches the wet region. Tiny droplets of water begin to form and increase as the steam expands further into the wet region. The maximum wetness usually tolerated is around 12 per cent. Water droplets travel much slower than steam and in consequence impinge on the edge of the blade. To minimise the erosion a

Casing

Steam

Flow

Steam

Sealing arrangement(moving blade)

Balance hole(moving blade)

Rotor shaft

Sealing arrangement(stationary diaphragm)

Steam

Casing

Sealingarrangement

(moving blade)

Steam

Rotor shaft

Sealing arrangement(stationary blade)

Flow

36

special tip made from stellite, a cobalt based alloy with chromium, tungsten and carbon, is brazed to the tip in manufacture. New tips may be fitted if the originals wear out. Some water droplets can be removed by shaping the casing of the turbine so that when centrifugal force throws the droplets outwards it is entrained in cavities, where it can be collected and drained off to the condenser. It is a serious problem and causes considerable damage. We have already seen how wetness can be reduced on the final stages by reheating. Exhaust. This loss is generally known as "leaving loss". It occurs because as the steam leaving one set of blades has considerable kinetic energy which it can usually carry over to the next set of blades until it leaves the final row and the energy lost. It is absorbed in the condenser and given up to the cooling water.

Leaving loss = (Mass steam flows) . (absolute velocity after last blade) 2 Figure 9 shows the expansion through the low pressure turbine on the H-s diagram and indicates the carry over from stage to stage showing the last stage leaving loss. To keep the exhaust loss to the minimum the last row of blades must be as long as possible. Large modern turbines usually have two or three double flow low pressure turbines exhausting to the condenser so that each last row of blading only has to handle one quarter or one sixth of the total seam flow.

Figure 9. Expansion through a turbine cylinder showing the leaving loss.

FIRST STAGE INTERNAL WORK

FIRST STAGE CARRY - OVER

SECOND STAGECARRY - OVER

SECOND STAGEINTERNAL WORK

LAST STAGEINTERNAL WORK

LAST STAGELEAVING LOSS

STATE POINT OF STEAMLEAVING LAST STAGE

FINAL STATEPOINT STEAM INCONDENSER

Isen

trop

ic E

xpan

sion

ZY'1

2

3

4

5

6 AB

CURVE

CONDITION

dns1

dns2

dn1

dn2

dn3

dn4

dn9

dn8

Dn

= Σ

dn

Dn 9

37

SECTION 6: HEAT EXCHANGERS AND FEEDHEATERS. Heat Exchangers By far the most important heat exchanger in any power plant is the turbine condenser. With a large turbine using high pressure and high temperature steam conditions at the turbine stop valve, even a comparatively small increase in condenser back pressure will cause a much greater increase in steam consumption with a consequential reduction in thermal efficiency. The ideal condenser has been defined as one which will extract the whole of the remaining heat from the exhaust steam at constant temperature, the condensate and exhaust steam temperatures being equal. It would also use the minimum quantity of circulating water, of which the outlet would be at exhaust steam temperature. Such conditions cannot be achieved in practice, and in general, both condensate and circulating water outlet temperatures are lower than that of the exhaust steam. The vacuum that can be obtained from a given condenser in a clean condition will change with a number of factors, mainly the circulating water inlet temperature, the quantity of steam to be condensed, the quantity of circulating water, and the presence of any air leakage. To illustrate the important contribution made to the work done by operating at a vacuum, consider the diagram shown in figure 1.

Figure 1. Pressure vs specific volume for dry saturated steam.

38

Steam is admitted to a turbine at a pressure of 11bar absolute as shown by P1. The volume of steam is 0.177 m3/kg. If, after expansion in the turbine, it is rejected at a pressure P2 of 1 bar absolute the volume will have become 1.7m3/kg and the work done will be represented by the area under the curve between the limits shown by P1 and P2. If, now, the final pressure is reduced to 0.5 bar absolute the expansion will continue to P3 and the volume will be 3.3m3 /kg. Thus, the extra work obtained per kilogram of steam is represented by the shaded area. This represents a considerable amount of extra work, obtained by improving the back pressure by 0.5 bar. To achieve a comparable amount of extra work at the inlet to the turbine the steam pressure would have to be increased from P1 to P4, which would be from 11 to 17.5 bars as shown by the shaded area. Of course, in practice it is normal to operate with a considerably lower back pressures than illustrated in figure1. It is easy to see that even small changes in back pressure can cause considerable changes in the work done per kilogram of steam, and there are over a million kilograms of steam entering the condenser per hour on large steam turbine. From this we can see that the condenser is by far the most important heat exchanger in the power plant cycle there by making the turbine back pressure the most important terminal condition of all. Therefore, it is of the utmost importance to the efficient operation of the unit that its back pressure is always as near to optimum as possible. Effect of Varying The Back Pressure. From the foregoing, it follows that a large amount of extra work is done by the steam When the back pressure is reduced. If this was the complete picture, the lowering the back pressure would always result in an increased output from the unit. However, in practice as the back pressure improves certain losses increase. These mainly comprise of: 1) CW pumping power. 2) Leaving loss. 3) Reduced condensate temperature. 4) Wetness of the steam at exhaust. Considering each item in turn. Increased CW pumping power. Assuming that the CW inlet temperature is low enough the back pressure can be reduced by putting more and more CW through the condenser tubes. However, this in turn will require more and more CW pumping power and the gain from the improved vacuum must be offset against the extra power absorbed by the CW pumps. Therefore CW pumps should only be run when the cost of running them is less than the resulting benefit from increased unit output. In other words, the pumping operation should always be optimized.

39

Increased leaving loss. Consider the last row of blades of a turbine. These present to the steam a fixed annulus through which the steam must pass to get to the condenser. Now, the steam leaves the last row of blading at a velocity which depends upon the condition existing at this point. As this velocity is not utilized usefully it represents a loss of possible work. This is known as the “ Leaving Loss.” There is always a leaving loss but as the back pressure is reduced its magnitude increases rapidly. For example, if the back pressure is 60 mb the loss would be represented by a certain value. If the back pressure is now reduced to 30 mb the specific volume of the steam will be approximately doubled, and so the velocity of the steam through the fixed exhaust annulus must also double. But since the leaving loss varies as the square of the velocity, it will consequently increase four times. Reduced condensate temperature. If the condensate in the condenser is at the saturation temperature corresponding to the back pressure it will be 36°C at 60 mb. Reducing the back pressure to 30 mb will cause the temperature to drop to 24°C. Therefore when it enters No1 LP heater it will be cooler than before. This will result in more steam being automatically bled to the heater because of the increased condensation rate of the steam. It therefore follows that the extra steam being bled to the heater is no longer available to do work in the turbine down stream of the tapping point, so the turbine will be deprived of some useful steam for doing work. Increased wetness of steam. The lower the back pressure the greater the wetness of the steam. This extra moisture could result in damage to the moving blades. In addition to this, the volume of steam is reduced. Thus at 30 mb back pressure the volume of the steam without wetness would be 45.7m3 / kg. If there were 10% wetness in the steam, the volume per kg would be reduced to 41.1m3. As a rough guide it can be assumed that every 1% of wetness will reduce the efficiency of the associated stage by 1%. The foregoing mentioned losses will eventually significantly effect the result. Continued reduction of the back pressure will result in the net improvement in heat consumption becoming progressively less until a point is reached at which the benefit due to improved back pressure is exactly neutralised by the losses, and this is the point of minimum heat consumption, as illustrated in figure 2. Further reduction of the back pressure will cause the heat consumption to increase, and so there is no point in operating at a lower value. It should be noted, though, that the back pressure for minimum heat consumption varies with load and so the operations staff should be provided with a curve such as illustrated in figure 3 to enable them to determine the minimum back pressure for any loading for their particular machines. Therefore the conclusion from the foregoing is obvious in so much that every effort should be made to operate the plant at the optimum back pressure.

40

Figure 2. Back pressure correction curve --- 120MW T/A at full load.

Figure 3. Minimum back pressure for various loads.---120 MW T/A.

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Graphical determination of losses contributing to departure from optimum back pressure. The usual reasons for departure of condenser conditions from optimum are: 1) CW inlet temperature different from design. 2) CW quantity flowing through the condenser incorrect. 3) Fouled tube plates. 4) Dirty tubes. 5) Air ingress into system under vacuum.

The contribution of some factors to a given departure of back pressure from optimum may be determined graphically. Consider a condenser with the following optimum and actual conditions at full load.

Condition Unit Optimum Actual 1 2 3 4 5 6 7 8

CW inlet temperature CW outlet temperature CW temperature rise (2) - (1) Saturated steam temp. corresponding to back pressure Terminal temp. difference (4) - (2) Air suction temperature Air suction depression (4) - (6) Back pressure

oC oC oC oC oC oC oC

mbar

16.5 25.0 8.5

30.5 5.5

26.0 4.5

43.65

18.2 28.2 10

35.0 6.8

27.0 8.0

56.22 Refer the above data to the “Condenser Condition Graph” illustrated in Figure 4. Deviation due to CW inlet temperature. Plot a line vertically from the actual CW inlet temperature of 18.2 °C to the intersection with the optimum CW rise of 8.5°C. Then plot horizontally to the intersection with the optimum “ terminal temperature difference” (TTD) line. The corresponding back pressure is 47.2mb. Therefore the loss due to high CW inlet temperature is the difference in the back pressure between 47.7mb and the optimum value of 46.65mb, i.e. 4.05mb. Deviation due to CW flow. Plot a line from the actual CW inlet temperature vertically to the intersection with the actual CW rise of 10°C. Then plot horizontally to the optimum TTD, then vertically downward to the saturation temperature of 33.7°C. The corresponding back pressure 52.0mb, therefore the loss due to incorrect CW flow is given by 52.0 − 47.7 = 4.30mb. Deviation due to air/dirty tubes. The effect of air and dirty tubes on heat transfer is to increase the TTD above optimum. As they both produce the same effect they are lumped together in this exercise. However, they can be segregated if required to do so. Plot from the actual CW inlet temperature to the actual CW rise and than across to the actual TTD line of 7°C. Plotting vertically downward the saturated steam temperature is 35.1°C, and the back pressure is 56.22 mb. So the deviation due to air/dirty tubes is given by 56.22 − 52.0mb =4.22mb. The effect of air ingress into the condenser is to increase the value of the “air suction depression” from optimum. In the case under consideration the optimum value, obtained from the acceptance tests, is 4.5°C, but it is actually 8.0°C, so there must be air present. Action should be taken to prevent the air

42

ingress. Then a new set of readings can be obtained and all the deviation due to “air /dirty tubes” can be attributed to dirty tubes. Incidentally, when condenser tubes are cleaned it is only the “dirty tube” which is eliminated.

43

Notes on the effect of air ingress on back pressure. When air mixes with steam it has very little effect on the absolute pressure in the condenser. For example, mixing saturated steam with 1/2000 of its weight of air the back pressure will increase will only increase by about one quarter of one per cent, (0.25%). If the back pressure is 34mb, when only Dalton’s law of partial pressures is considered. The real problem with air is that it is incondensable, and so, when steam condenses on the CW tubes, the air remains, and may form a film on the tube surface. With air being such an excellent insulator it only requires a film a few molecules thick to seriously interfere with the heat transfer, and as a result the back pressure suffers. Fortunately it is easy to determine whether air is present merely by measuring the temperature of the contents of the air suction pipe to the condenser evacuation plant. When there is only a little air present the temperature is very little below the saturated steam temperature (4.5°C or less). As more and more air becomes present the temperature falls, the more air present the greater the depression of the air suction temperature compared to the saturated steam temperature. Preferably, the thermometer should be in direct contact with the contents of the air suction pipe. Condenser Performance: The vacuum at which the turbine exhausts is determined by the condenser. In the condenser the large quantity of heat remaining in the turbine exhaust steam has to be transferred to the circulating water. To make heat flow across a heat transfer surface a temperature gradient must exist. In a condenser the temperature of the condensing steam must be higher than the temperature of the circulating water. To obtain the highest operating efficiency this temperature difference between the exhaust steam and the circulating water must be kept as small as possible, thereby ensuring the lowest temperature of condensation and hence the best vacuum for any particular circulating water temperature. On the outside of the condenser tubes the temperature of condensation remains almost the same over all of the steam space except for the air cooling portion of the condenser where the air and vapour mixture is flowing towards the air extraction takeoff pipes. Modern condensers have wide steam lanes arranged to give the steam easy access to all parts of the tube surfaces and the whole steam space is virtually at one pressure and temperature. Inside the tubes the circulating water temperature rises throughout its passage through the condenser, the rise depending on the amount of heat being rejected and the amount of circulating water being used. The all important temperature difference between steam and circulating water will therefore vary , being relatively large at the circulating water inlet and small at the circulating water outlet. The very simple diagram illustrated in figure 5 shows typical temperature gradients in a condenser, the vertical scale being temperature and the horizontal scale represents the length of the circulating water path through the condenser. The vacuum of the condenser represented in figure 5 has been assumed to be 32mb so that the temperature on the steam side of all the tubes would be 26.1°C. On these diagrams the vacuum temperature is usually given the symbol t.3. The circulating water enters the condenser at 12.7°C ( t 1) and leaves the condenser at 21.1°C ( t 3 ). The rise in temperature of the circulating water is not uniform throughout its traverse of the condenser because of the diminishing temperature gradient between the steam and water. At the circulating water inlet the temperature difference is 13.3°C, the difference between steam at 26.1°C and the CW inlet temperature at 21.7°C. The amount of heat transferred per square metre of tube surface is proportional to the temperature difference, so at the circulating water inlet where the temperature gradient is at its highest the heat transfer will also be at its highest and the rise in circulating water temperature per metre length of tube will be at its highest as well. As the circulating

44

water is approaching the condenser outlet at 21.1°C the temperature gradient is now only 5°C so that heat transferred per square metre of tube surface will be reduced correspondingly, in fact, it will only be a little more than one third of what it was at the circulating water inlet. As a consequence the circulating water temperature rise per metre run of the condenser tube gets smaller and smaller from inlet to outlet and results in the curve of circulating water temperature as illustrated in figure 5. This phenomenon makes it rather difficult to measure directly, or visualize, an average temperature gradient between steam and circulating water and this average gradient is the best measure of condenser performance.

Temperature of circulating water

12.77oCt1

Initialtemperaturedifference θ1

Terminaltemperaturedifference θ2

Temperature of condensing steam

21.11oC

t3

t2

t326.05oC (982.05 mbar)

Fig 5. Condenser temperature diagram. However, the three temperatures which are available on the condenser instrumentation, the circulating water inlet and outlet temperatures and the condensing temperature corresponding to the vacuum, can be combined to give the required average or log mean temperature difference and is normally given in the following form , Log mean temp. difference ( L.M.T.D.) = θ1 − θ2 2.3 log10 ( θ1/ θ2 ). Where θ1 ( theta one ) = steam temp. ( t3 ) − c.w. inlet temp. (t1 ). And θ2 ( theta two ) = steam temp. ( t3 ) − c.w. outlet temp.( t2 ) Writing down the numerator in full shows that it is simply the temperature rise of the circulating water. θ1 − θ2 = ( t3 − t1 ) − ( t3 − t2 )

= t3 − t1 − t3 ÷ t2.

= t2 − t1, or the rise in circulating water temperature. Mean c.w. temperature difference = θ1 + θ2 2

45

As previously discussed, condenser back pressure can be effected by the following conditions: 1) Variation of c.w. inlet temperature. 2) Variation of c.w. flow. 3) Variation of heat transfer to the cw. ie; dirty tubes or air ingress. 1) Effect of variation of CW inlet temperature on back pressure.

There are two effects caused by a change in CW inlet temperature. The primary one is to alter the steam saturation temperature by the same amount as the CW change, assuming all the other factors remain constant. This, in turn, will change the corresponding back pressure. The secondary effect is caused by the fact that the heat transfer of the CW water film in contact with the condenser tubes changes with the temperature of the water. The primary and secondary changes are in opposite directions. The magnitude of the secondary effect is approximately equal to the fourth root of the mean CW temperature. For example, consider a condenser with the following conditions: Cw inlet temperature t1 = 25°C CW outlet temperature t2 = 33°C Steam saturation temperature t3 = 36.2°C Corresponding back pressure = 60mb. Mean CW temperature = 7.2°C Log mean temperature difference = 6.4°C If the CW inlet temperature is increased to 30°C the primary effect will be to increase the saturation temperature of the steam to 41.2°C. The corresponding back pressure is 80mb. The secondary effect depends upon the change of mean CW temperature from 7.2°C to 12.2°C; the new log mean temperature difference will be: New LMTD = 6.4 4√ [ 7.2 ] = 6.4 × 0.88 = 5.63°C. 12.2 But LMTD = θ1 − θ2 2.3 log 10 θ1 θ2 ∴ 5.63 = 8 2.3 log 10 t3 − t1 t3 − t2. log10 t3 − t1 = 11.2 = 0.86 t3 − t2 5.63 × 2.3

46

log10 t3 − t1 = antilog 0.86 = 7.24 t3 − t2 t3 − 30 = 7.24( t3 − 38.0 ). t3 − 30 = 7.24 t3 − 275.12. 6.24 t3 = 245.12 So the new t3 = 39.3°C. Equivalent back pressure = 71.1mb Therefore the final conditions are: CW inlet temperature t1 = 30°C CW outlet temperature t2 = 38°C Steam saturation temperature = 39.3°C Equivalent back pressure = 71.1mb. Equally as well, the effect of varying the CW flow on the back pressure and the effect of variation of heat transfer to the CW can be examined by use of the LMTD. Effect of Circulating Water Temperature and Velocity. The heat flowing from the condensing steam has to pass through the film of condensed steam on the tubes, then through any tube surface, through the tube wall itself, trough any dirt or scale on the inner tube surface, through the stagnant layer of circulating water and into the main body of circulating water. The thickness of the stagnant water film and hence its resistance to heat flow, is dependent on the temperature of the water and also its velocity. The following empirical formula for a clean condenser applies. Heat transfer K = 650 × √ Vt/5 × 4√ t/100. Where K = heat transmission rate in KJ/m2/hr/°C Vt = water velocity through the tubes in m ⁄ sec.( 2 m ⁄ sec). T = arithmetic mean circulating water temperature. ( 30°). ∴Heat transfer K = 650 × √2⁄5 × 4√30⁄100.

= 650× 0.632 × 0.74 = 304 KJ ⁄ m2 ⁄ hr ⁄ °C.

The operations engineer need not be to precise about the effect of warmer circulating water on heat transfer providing he understands that the condenser terminal temperature difference is only going to vary

47

a few degrees throughout the year. Very approximately the effect of warmer circulating water is to reduce the terminal temperature difference by 0.55°C for every 5.5°C increase in circulating water temperature. As already pointed out the other operating variable which complicates assessment of condenser performance is the quantity of the circulating water. Quantity has a two fold effect on the condenser performance, the velocity of the circulating water through the tubes being, of course, directly proportional to the quantity. Firstly the velocity effects heat transfer rate. The more vigorous the scrubbing action with higher velocities thinning the stagnant water film and reducing the heat transfer. Secondly, for a fixed amount of heat rejection the temperature rise of the circulating water will be inversely proportional to its quantity. That is doubling the c.w. quantity will half the c.w. temperature rise. This effect alters the mean temperature of the circulating water and this also alters the heat transfer rate. As previously stated, θ1 = ( t3 − t1 ) and is often referred to as the initial temperature difference. Also θ2 = ( t3 − t2 ) and this is known as the terminal temperature difference ( TTD ), and can be used as a guide to condenser performance, particularly if there is little variation in the circulating water quantity. However, the log mean temperature difference is considered as the real driving force behind the heat transfer and as previously stated this must be kept as low as possible in order to achieve the lowest heat rejection temperature, which is the same as saying that the heat transfer across the tube surfaces must be as high as possible for the best results. Since the heat transfer is effected by the cleanliness of the tube surfaces and by the amount of air present in the steam space, these are the items which should be carefully controlled to obtain the best condenser performance. In addition heat transfer is effected by the velocity of the circulating water through the tubes and by its temperature; these two effects rather complicate the checking of condenser performance because if they are taken into account it is difficult to assess the cleanliness of the condenser. This effects the normal variations in circulating water quantity and temperature will have on heat transfer rate, therefore, log mean temperature difference and vacuum will have to be considered. If these are understood it should be possible to assess the cleanliness of a condenser under various operating conditions. Causes of poor condenser performance Should the continuous monitoring of condenser performance indicate an increasing terminal temperature difference, then there must be some corresponding deterioration of the heat transfer rate. The two main causes of poor hear transfer are usually: 1) Contamination of the circulating water side of the tubes by any suspended foreign material contained

in the circulating water. 2) Excessive air leakage into the steam space. Every effort should be made to keep the tube surfaces as clean as possible and the methods used will depend on the nature of the circulating water. For a sea water cooled system, the ball cleaning method is often adopted. This involves the circulation of a large number sponge rubber balls that are carried through the condenser tubes by the cw flow. There action is to scour the condenser tube surface whereby any solid material that has dropped out of suspension will be removed and carried away in the cw outfall.

48

Another problem that can arise is the partial blocking of the condenser tube plate by any foreign material that may find its way into the system whereby restricting the cw flow. Some designs of condenser water box incorporate cross connection valves at the inlet and outlet sides. By controlled sequence of operation of these valves together with the cw inlet and outlet valves the direction of the cw flow can be reversed whereby any deposit that may have accumulated on the tube plate can be removed and carried away by the cw outlet flow. Another very effective method of dealing with condenser tube fouling is the use of direct chlorination into the cw system. The injection points chosen can either be at the condenser itself or at the circulating water inlet. Chlorination can control organic contamination and in the case of a sea water cooled station it can be particularly effective in preventing the growth of marine life within the cw system. Any marine growth particularly by way of mussels or barnacles that are allowed to take place within the condenser tubes will eventually lead to condenser tube leaks with the resultant contamination of the feed system by sea water. To be most effective, for cw temperatures in excess of 25°C, a target chlorine level of 0.5ppm should be maintained at the cw outfall with the dosing rate at the cw inlet adjusted to achieve this target. (For cw temperatures below 25°C, a target of 0.25ppm should be sufficient) Most sea water cooled power plants have for some years employed the use of electrochlorination plants since a supply of sea water is readily available at little or no cost. Effect of Air in the Condenser Steam Space. Air is one of the main causes of poor vacuum. As parts of the turbine itself together with the low pressure feed heaters operate below atmospheric pressure during normal operation there are innumerable places where a leak would allow air to be sucked into the steam space. As already outlined in this section, air in the steam space reduces the heat transfer rate, which causes the condensing temperature to rise in order to transfer the heat across the tubes, and so results in a worse vacuum. In this respect air has just the same effect as tube fouling. On some units, the air extraction plants have an air metering device installed at the air discharge line from the air pumps, in which case it is an easy task to monitor the air discharge level to see if it has increased above normal. Feed Heaters. The most important modification of the straight condensing steam cycle was the introduction of regenerative feed heating by means of steam bled from the lower stages of the turbine. Here the object is not to increase the heat drop, but to decrease added heat by saving some of the latent heat that would otherwise be lost to the condenser and passing it to the feed water. Clearly the ideal point is to extract heat from the turbine is when the steam has done all the work that it can do on the shaft; i.e. at the exhaust flange. In practice this is impossible since there is no heat head or temperature difference to cause heat to flow from the exhaust steam to the condensate. Therefore a point of higher temperature must be chosen, when the steam has accomplished some 80% to 85% of possible work in the turbine, yet still has sufficient temperature to create a heat flow to the condenser. The theoretical limit to which bled steam heating can proceed from any point is when the condensate temperature has risen to that of the steam extracted. The design of modern high pressure feed heaters incorporates sections of desuperheating, condensing and drain cooling and as such, is capable of achieving a temperature of less then 1°C between the bled steam temperature and the condensate leaving the feed heater.

49

For low pressure feed heaters this difference is usually less the 3°C. These temperatures are often referred to as the heater terminal temperature difference. (TTD). In a large modern high pressure high temperature steam cycle, there may be as many as eight stages of feed heating leading up to a final feed temperature of around 285°C. The limits of this are fairly plain, since the final feed temperature must not be so high as to approach the saturation value at the boiler pressure, nor must it so restrict economizer performance to the extent that the boiler gas loss exceeds the bled steam saving. Low Pressure Heaters The current generation of surface type U- tube low pressure (LP) feed heaters has been developed from the traditional surface LP heater that has been used in power plant for many years. The design has been updated to take advantage of modern manufacturing techniques and materials. The latest design of heaters are constructed to take full advantage of a horizontal attitude. Other designs of LP heaters that have been used is the direct contact type. They are similar in design and function in the same way as the deaerating section of the standard type deaerator. Steam extraction points for surface type LP heaters are normally located on the LP turbine cylinders. The degree of superheat even on the highest pressure LP heater does not justify the provision of a desuperheating section within the heater. Drain cooling sections can be provided but the complication and cost of a drains level control system is seldom justified. It is usual practice to have LP heaters constructed as a condensing only heater and to provide a drain cooler upstream of the low pressure heater to recover some of the heat from the drains. The drain cooler adopted is usually of the flash type. A flash type drain cooler is a separate heater placed in the feed train up steam of the lowest pressure LP heater. It is a small heater with no bled steam supply but has a chamber at one end where the combined LP drains are flashed into steam and water. The steam is condensed onto the heater tubes and the drains are discharged to the condenser via some form of loop seal system. In the first generation of the larger units, to reduce the wetness burden of the steam flowing to the last row of LP turbine blades, a specially designed heater called the LP turbine moisture extraction condenser (TMEC) is provided. The TMEC is designed to extract the steam water mixture from the water extraction belt of the LP cylinders, which are located immediately before the last row of turbine blades. The more modern design of large generating units employ TMEC’s in the form of multi-compartmented LP surface type heaters. Each compartment is connected to a low-pressure cylinder. This prevents interconnection of the cylinders and consequential difficulties caused by recirculation of the water/ steam mixture between cylinders. The steam inlet size of an LP heater is large compared with the heater body size due to the rapid increase in the specific volume of the steam as the pressure decreases. On some low pressure heaters, this results in multiple inlets with steam belts designed to keep the inlet velocity to the tube bundle to an acceptable value. A further development in LP heater design, is to place the two lowest pressure LP heaters in the condenser neck. The illustration in figure1 shows a diagrammatic arrangement of such a heater. This heater arrangement is of the duplex type ( LP1 and LP2 combined) with gravity drains into the condenser. The general construction is similar to other LP surface type heaters. The main advantage of this arrangement is the saving in space. Small savings are also made due to the elimination of the pressure drop associated with the bled steam isolating and non-return valves. Therefore, the pressure drop in the bled steam pipework is also virtually eliminated.

50

Fig:1. Combined LP1 and LP2 heaters located in condenser neck.

The LP heater feed water side must be capable of withstanding the maximum head that the condensate extraction pump can deliver. On a typical large unit this would be in the order of about 40 bar. The tubes of the LP heaters may be of 70/30 brass or stainless steel as dictated by the bled steam temperature or boiler feed water chemical requirements. Brass may be used in LP heaters where the steam temperatures do not exceed 150°C. Above this temperature stainless steel is used. Where full flow condensate polishing plant is adopted, copper alloy materials should not be employed downstream of the polishing plant; so stainless steel is adopted for all LP heater tubes, including drain coolers and TMEC’s

Fig: 2. Section through a typical surface type horizontal LP heater. Another point worth mentioning while discussing LP feed systems is the deployment of an LP heater drains recovery pump. These are usually used to recover the drains from the next to highest heater in the LP train and pump them into the condensate inlet of the highest pressure LP heater in the train. The main advantages are the recovery of the heat contained in the heater drain water that would otherwise be given up in the condenser. Also the condensate extraction pumps have less water to handle which results in a saving of auxiliary power. Deaerating Feed Heaters. Condensate from the condenser is fed via the LP feed heater train to the deaerator to undergo further heating and deaeration before being fed to the boiler via the boiler feed pump and the HP feed heater train. Deaerators on large modern units are designed to provide feed water at exit from the storage tank with not more than 0.005ppm of oxygen.

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Deaeration is achieved by the application of Henry’s Law which states that the quantity of gas dissolved in a given quantity of solution is proportional to the partial pressure of that gas over the solution. When this law is applied to the removal of oxygen from feed water, where the atmosphere above and around the condensate contains no oxygen, the dissolved oxygen will escape to the atmosphere in an attempt to achieve equilibrium. To construct a deaerating heater to release the maximum amount of oxygen from the incoming condensate, the following factors have to be considered: • The time for the dissolved oxygen to travel to the surface of the water. • A dwell time is needed for the steam to heat the condensate and so increase the equilibrium pressure

of the dissolved oxygen in order to release it. • The surface tension of the water. • The time taken for the diffusion of the oxygen from the water into the steam atmosphere. The storage tank associated with deaeration has to meet the following requirements: • To store approximately 7-10 minutes worth of MCR flow. • Accept the leak-off flows from the boiler feed pumps when required. At low loads, when the flow

through the feed pump would be less than 25%, the boiler feed pump leak-off system is fed to the deaerator water storage tank as a water recovery process.

• Accept the HP heater drains. • Heat up the tank contents from cold to provide hot deaerated water for unit start-up.

Fig:3 Section through a typical spray/tray type deaerator and associated storage tank.

Figure 3 shows a typical deaerator and feed water storage tank, the top section of which heats and deaerates the incoming condensate before it drains by gravity into the storage tank. This design uses spray nozzles to produce a fine film/spray to maximise the surface area of the water available to the steam for

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heat transfer and to minimise the distance that the oxygen has to travel to be released. Any residual oxygen is released while the water is further heated as it passes over a series of perforated trays, which causes the condensate to fall as a continuous “rainfall” from tray to tray. The heat transfer coefficient in the fine film/spray zone is approximately five to ten times the value for the drop phase. Nearly 90% of the temperature rise occurs in the film phase, the temperature difference between the water drops cascading from the lower trays and the steam is small. The additional trays, however, are needed to allow time for the residual oxygen to escape and also the heat the water to the saturation temperature equivalent to the prevailing pressure. The steam flow path is shown by the arrows in Fig 3. A small flow of steam, along with the oxygen and non-condensable gases, is extracted via vents that are located on top of the head. The mixture of steam and oxygen, if vented to the condenser or atmosphere, would constitute a heat loss. To save this heat, a vent condenser section is provided to heat the incoming feed water by condensing the vapour and extracting the heat from the oxygen and non-condensable gases. The oxygen and non-condensable gases are extracted from the vent condenser by either venting directly to atmosphere or to the condenser, from where the gases are discharged to atmosphere via the air extraction plant. The storage tank associated with the deaerating function normally has a storage capacity of about 300 tonnes of feed water in a tank approximately 4.5m diameter by 32m long; and is normally located at some elevated level so as to ensure sufficient suction head is available for the boiler feed pumps. Diffusers are provided to discharge the boiler feed pump leak-offs and HP drains into the tank, as illustrated in Fig3. The steam coils used to boil and deaerate the tank contents prior to unit start up are also shown in Fig:3. Unless provision is made to induce movement of the tank contents, stagnant areas of subcooled water will result. A distribution system is provided to prevent direct discharge of the incoming heated and deaerated feed water from the boiler feed pump suction take offs. By use of a distribution system, adequate mixing of the tank contents as assured. Vertical baffles are also required to prevent possible wave action of the tank content from end to end under conditions of abnormal steam flows across the tank surface. The transfer pipes between the tank and the deaerator head are generously sized to allow the flow of vapour to the head when the tank contents boil due to a reduction in the deaerator tank pressure. High Pressure Feed Water Heaters. The structural design of a high pressure feed (HP) water heater is determined by two main requirements: • To contain the steam and HP feed water at the appropriate cycle conditions. • To provide a heat transfer surface to raise the feed water temperature by the specified amount. The temperature rise and the terminal temperature differences (TTD) are determined by the overall cycle economics Fig:1 illustrates the way in which the feed temperature increases as it passes through a typical two-pass horizontal heater of the U-tube design. This type of heater has both integral drain cooling and desuperheating sections as illustrated in Fig2: The desuperheating section is placed at the outlet end of the U-tubes in order that the incoming superheated steam can raise the feed water as near to or above the saturation temperature corresponding to the body pressure before it leaves the heater.

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Fig:1 Temperature rise of feed water as it passes through an HP heater.

The drain cooling section is placed at the inlet end of the tubes to allow the outgoing drains to be cooled to as near to the incoming feed water as required. Steam enters the desuperheating section and is reduced in temperature by transferring its heat to the feed water to within 27°C of the temperature of saturation of the condensing section pressure. The steam then flows to the condensing section, where it leaves as water at saturation temperature to enter the drain cooling section. A water seal is maintained at the inlet to the drain cooling section by a built in level control system to prevent loss of prime in the section. In the drain cooling section, the condensate is cooled to the drain outlet temperature and then discharged the next lowest pressure heater. Each section within the heater is provided with baffles to ensure flow across the outside of the tubes by the heating medium. As the heating steam is condensed in the heater non-condensable gases are released. Unless correctly vented these would rapidly blanket the heat transfer surface and would impair the heater thermal performance. To remove these gases, vents are connected to the condenser and are provided at strategic points throughout the heater tubenest Special care is needed in the venting of horizontal heaters, since air pockets can accumulate under the baffles. If lowering the heater initial cost outweighs the higher running costs due to the loss of cycle efficiency, heaters can be constructed without drain cooling and desuperheating sections.

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Fig2: Steam and water flow paths in a typical horizontal HP heater.

Prior to the development of horizontal HP heaters, vertical HP heaters were standard practice. Vertical HP heaters employ the same basic layout with regard to the desuperheating and drain cooling sections. However, there are some points of difference which are outlined in the following section on construction of specific heaters designs. Figure 3 shows a typical arrangement of a vertical HP heater. The desuperheating section is below the normal working water level in the heater. To prevent inner water leakage, seals are provided between the tubes and the end plates of the desuperheating section, trunking is also needed to carry the steam from the desuperheater to the condensing section if needed. These provisions can be avoided by placing the desuperheating section in the dotted section shown in Fig 3 but, as it is not now adjacent to the tubeplate, the feed water must travel through tubes immersed in water at the saturation temperature equivalent to the heater body pressure. This limits the steam TTD as, if a negative TTD were employed, the feed water would tend to be cooled back down to the saturation temperature. The horizontal design of heater does not suffer from this type of problem as the desuperheating section is above the water level by virtue of the heater attitude. The reduction in the complication of the heater internal construction is another point in favour of horizontal attitude HP heaters. As the high pressure feedwater heaters are located at the discharge side of the boiler feed pumps, the feed water within the water headers and the heater tubes are at boiler pressure plus the pressure rise between the heater and the boiler. To contain this high pressure, various designs of water header have been used in the past, but most current large units employ hemispherical headed heaters with a flat tube plate. The tubes are welded onto the back of the tube plate by a fusion welded process.

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Fig3: Steam and water paths in a typical vertical HP heater. Heat Exchange in Feed Heaters. It should be clearly established that, what ever type of heater, the greatest amount of heat transfer by far is the action of condensing the steam. Even if a heater is supplied with bled steam which has a considerable degree of superheat it will be the condensing section which does most of the work. The illustration shown in figure 4 shows a typical temperature/heat transfer diagram. The saturation temperature is that of the condensing steam. this is not the same as that of the steam before the heater because there is some pressure drop in the desuperheating section. Consider the steam and water conditions for a typical No8 HP heater associated with a 650MW turbine.

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• Steam to heater inlet pressure/temperature = 63.05 bar abs./ 390.5°C • Steam to heater heat content = 3149.7 kJ/kg. • Steam to heater dry saturated heat.= 2781.5 kJ/kg. • Steam to heater wet saturated = 1231.6 kJ/kg. • Enthalpy of drain water = 1041.8 kJ/kg. So for every kilogram of steam supplied to the heater: The superheat component surrenders 3149.7 − 2781.5 = 368.2 kJ of heat. The condensing component surrenders 2781.5 − 1231.6 = 1549.9 kJ of heat. The drains cooling component surrenders 1231.6 − 1041.8 = 189.8 kJ of heat.

Figure 4: Heat surrendered in each section of a last stage HP feed heater.

Feed Heater Balance. During feed heater investigations, it is assumed that all the heat abstracted from the bled steam is transferred to the feed water. Therefore Qb(hs − hf) = Qf ( hf,o− hf,i )

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Where Qb = quantity of bled steam, kg/s hs = enthalpy of bled steam, kJ/kg hf = enthalpy of drain water, kJ/kg Qf = quantity of feed water, kg/s hf,I = enthalpy of feed water at heater inlet, kJ/kg hf,o = enthalpy of feed water at heater outlet, kJ/kg For example, consider the No6 & No7 HP heater arrangement that exists at Paiton 11, and we require to find the feed outlet temperature of No7 heater. Qb( hs − hf ) = Qf ( hf,o − hf,I ) 40.582 ( 3042.5 − 956.0 ) = 546.085 ( hf,o − 930.4 ) So hf,o = 40.582 ( 3042.5 − 956.0 ) + 930.4 546.058 From the steam tables; hf,o = 250°C. Feed Heating illustrated by use of the Temperature – Entropy Diagram.

Fig:5. Section of a Temperature – Entropy Diagram showing Regenerative Heating with Finite and

Infinite Numbers of Heaters. A non reheat cycle incorporating a turbine with three stages of feed heating is illustrated in Fig:6. The temperature of the condensate leaving the condenser is shown at point A. Because of feed heating the temperature is raised to that shown at point B. Consequently, the enthalpy of the feed due to feed heating is represented by the area under the curve.

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Fig:6. Idealised Representation of Feed Heating.

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Fig: 7. 169 Bar / 538°C / Reheat to 538°C Vacuum 35mb. Final Feed 250°C.

condensingIsothermal

Evaporation

Isen

trop

ic E

xpan

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Saturation Line

Boilin

g W

ater

Lin

e

Tem

pera

ture

ºC

Entropy, kJ/kg.K

Reh

eate

d

C G

K

D H

K

JFL

A

B

700

600

500

400

300

200

100

-100

-200

-2730 1 2 3 4 5 6 7 8 9 10

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SECTION 7) Combustion Chemistry. Theoretical Air Dry air contains 23.2 per cent by weight of oxygen and the rest is almost wholly nitrogen To obtain the weight of air required for combustion from the weight of oxygen required can be simply done by multiplying the oxygen requirements by 100/23.2 which is 4.31. From the chemical equations for the three combustion processes the theoretical oxygen and air requirements are as follows: 12kg carbon + 32kg oxygen = 44kg carbon dioxide. or 1 kg C + 2.667 kg O2 = 3.667 kg CO2 or 1 kg C + 11.495kg air = 3.667kg CO2 + 8.828kg nitrogen. 4 kg hydrogen + 32 kg oxygen = 36 kg water. or 1 kg H2 + 8 kg O2 = 9 kg H2O. or 1 kg H2 + 34.48 kg air = 9 kg H2O + 26.48 kg nitrogen. 32 kg sulphur + 32 kg oxygen = 64 kg sulphur dioxide. or 1 kg S + 1 kg O2 = 2 kg SO2 or 1 kg S + 4.31 kg air = 2 kg SO2 + 3.31 kg nitrogen. In a similar manner to that used to calculate the calorific value of the fuel from its ultimate analysis, so the theoretical air requirements for combustion can be obtained by adding together the air requirements of the three combustible constituents in the fuel. The oxygen in the fuel is assumed to be in chemical combination with some of the hydrogen in the fuel so the amount of hydrogen for combustion is reduced by an amount equal to one eighth of the oxygen in the fuel. The chemical analysis of the fuel gives the proportions of the various elements by weight, so if the calculations are based on 100 kg of fuel, the percentage of each element is exactly the same as the number of kilograms of each in 100kg of fuel. Air Required for Combustion and Products of Combustion. The theoretical quantity of air required for combustion of a fuel can be determined from the following formula: Theoretical air required = 4.31[8/3 C + 8 (H − O/8 ) + S ] kg /100kg fuel. Where C = % carbon per kg of fuel. H = % hydrogen per kg of fuel.

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O = % oxygen per kg fuel. S = % sulphur per kg of fuel. The values inside the square brackets indicate the quantity of oxygen required for each combustible constituent of the fuel. The multiplier 4.31 determines the air required to supply the necessary oxygen. This is derived from the fact that 100 kg of air contains 23.2 kg of oxygen, so 1 kg of oxygen is contained in;

100 kg of air, i.e. 4.31 kg. 23.2 Example: A typical coal being burnt at Paiton II has the following ultimate analysis. Symbol % Carbon C 74.5 Hydrogen H 5.9 Nitrogen N 1.0 Sulphur S 0.18 Oxygen O 18.45 Ash _ 3.0 Moisture _ 22.3 100.0 Theoretical air required: = 4.31 [ ( 3

8 × 74.5 ) + 8 ( 5.9 − 845.18 ) + 0.18 ] kg / 100kg fuel

= 4.31 ( 198.66 + 28.8 + 0.18 ) kg / 100 kg fuel. Therefore, 74.5 kg of carbon requires 198.66 kg of oxygen, 7.0 kg of hydrogen requires 28.8 kg of oxygen. And the sulphur 0.18 kg. = 4.31 ( 227.64 ) = 981.13 kg / 100 kg fuel Therefore the total oxygen required = 227.64 kg and the air required = 981.13 kg. Therefore the nitrogen by difference = 981.13 − 227.64 = 753.49 kg.

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The Requirement of Excess Combustion Air In actual practice the burning of coal, may not be perfectly completed within the furnace zone where the fuel air mixture even under the best achievable conditions of turbulence exist. Additionally, complete mixing may take too much time so that the gases will pass to a lower temperature zone where it is not possible for complete combustion to take place. Therefore, if only the amount of theoretical air were supplied, some of the fuel would not burn due to the combustion being incomplete and the heat from the unburned fuel would be lost. To ensure complete combustion takes place, additional combustion air must be supplied so that every molecule of the fuel can easily find the proper number of oxygen molecules to complete the combustion. This additional amount of combustion air that is supplied to complete the combustion process is called excess air. Excess air plus theoretical air is called total air. If this excess air is below the minimum requirement some of the products of combustion will only burn to carbon monoxide instead of carbon dioxide, and since, carbon monoxide has only about 28% of the heating value of carbon dioxide this results in a very significant heat loss, also there may be an appreciable amount of unburned carbon left in the ash and dust. Having this necessary excess air means that some of the oxygen will be not be used and will leave the boiler in the flue gas. If the percentage of the excess air is increased, flame temperature is reduced and the boiler heat transfer rate is reduced. Also, the usual effect of this change is to increase the flue gas temperature. If the oxygen in the flue gas is known or can be measured and no table or curve is available, the following formula provides a close approximation of the percentage of excess air. This formula is based is based upon an as dried basis percentage oxygen. Excess air % = K ( 21 − 1 ) × 100 21 − % oxygen Where K = 0.97 for coal. Another method of calculating the amount of excess air present in the boiler is by using the knowledge of the CO2 produced by the combustion process and the theoretical maximum CO2 % for the fuel. Excess air = Theoretical CO2 % − 1 Actual CO2 % The theoretical CO2 % for bituminous coal is taken as 18.6%. So for the fuel specified in the previous example were burned with 15% CO2, the excess air would be: Excess air = 18.6 − 1 = 0.24. or 24%.

15.0 and the air for combustion would be 1.24 × 9.81 kg air / kg coal = 12.2 kg. • Reduced excess air improves heat transfer. • Reduced excess air reduces the mass flow of the flue gas.

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Fig:1 Effect of Excess Air upon Temperature.

In general, Actual air × actual CO2 % = Theoretical air × theoretical CO2 %. From the previous example, Air = 981.13 kg / 100 kg fuel Oxygen = 227.64 kg / 100 kg fuel. Nitrogen = 753.49 kg / 100 kg of fuel. For example, 10% excess air is equal to 10% 0f 981.13 = 98.1 kg and so on for the other values of oxygen and nitrogen. Pulverized fuel boilers usually operate at about 20% excess air at MCR. Thus there will be about 196.2 kg air / 100 kg fuel, and this will comprise of 45.5 kg oxygen, and by difference 150.7 kg nitrogen.

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Boiler Efficiency Determination The efficiency of the boiler and for that matter the turbine also, are both closely related to thermodynamics. There are two basic ways of determining the efficiency of the boiler. (a) The Direct Method: (b) The Losses Method. The direct method This was the standard method in use for many years, but is little used now. It is very straightforward and consists of measuring the heat supplied to the boiler in a given time and the heat added to the steam in the boiler. Therefore, the efficiency of a non reheat boiler is given by: Efficiency = ( enthalpy of steam − enthalpy of feed ) × steam flow ( per unit time ). quantity of coal × calorific value The trouble is that some of these quantities are difficult to measure accurately, eg; coal weight. Accordingly the result may have an overall tolerance of about = 1.5%. The losses method The efficiency of a boiler equals 100% minus losses. Therefore, if the losses are known the efficiency can be derived easily. This method has several advantages, one of which is that the errors are not so significant; for example, if the total losses are 10%, then an error of 1.0% will effect the result by 0.1%. The losses method is now the usual one for boiler efficiency determination. A typical pulverized fuel boiler heat balance is: Gross CV basis Net CV Basis Loss due to: Dry flue gas 3.98% Dry flue gas 4.3% Wet flue gas 5.27% Sensible heat in water vapour 0.75% C in A 0.24% C in A 0.25% Radiation and Radiation and Unaccounted. 0.44% unaccounted 0.45% Total loss 9.93% Total loss 5.75% Boiler efficiency 90.07% Boiler efficiency = 94.25% If a boiler is under test and is found to have an efficiency of for example 94%. It would be quite wrong to assume that this was its normal operating efficiency. During testing, particular care is taken to keep the steam pressure, temperature and other operating parameters as steady as possible and neither will there be any sootblowing or blowdown. Also the boiler would probably be tested immediately after sootblowing. Therefore, there are many factors common to normal operation that are absent when testing takes place. Thus, the test efficiency is the best possible that can be obtained and for normal operation the value will be somewhat less.