"MODULE TITLE:HEAT TRANSFER AND COMBUSTION TOPIC TITLE:COMBUSTION PROCESSES LESSON 3:DEW POINT AND ITS EFFECT ON STACK TEMPERATURES HTC - 4 - 3 © Teesside University 2011Published by Teesside University Open Learning (Engineering) School of Science & Engineering Teesside University Tees Valley, UK TS1 3BA +44 (0)1642 342740 All rights reserved.No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the Copyright owner. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it ispublished and without a similar condition including this condition being imposed on the subsequent purchaser. 1 ________________________________________________________________________________________ INTRODUCTION ________________________________________________________________________________________ The products of a combustion reaction (flue gases) possess large amounts of energy (sensible heat) which can be recovered by using the gases to heat up other materials.One problem with this is the fact that these gases contain water vapour which, if the temperature falls below a certain value (dew point), can condense on surfaces and lead to poor heat transfer and possible corrosion.Another problem that can occur is that, if too much heat is removed, disposal of the waste gases into the atmosphere may not be effective as the hot gases fail to leave the chimney stack with sufficient 'lift' to take them away from the immediate area and into the wider air stream for dispersal.Instead they could fall back and lead to possible pollution close to the stack.In this lesson we will look at the concept and calculation of dew point and look at its implication for minimum stack temperatures. ________________________________________________________________________________________ YOUR AIMS ________________________________________________________________________________________ At the end of this lesson you should be able to: • state the factors that may affect stack temperatures • explain the meaning of dew point • calculate dew points for simple gas-water vapour mixes • state the effect of acid gases on the dew point of water. Teesside University Open Learning © Teesside University 2011 (Engineering)2 ________________________________________________________________________________________ STUDY ADVICE ________________________________________________________________________________________ This lesson will make use of 'steam tables' more correctly known as "Thermodynamic and Transport Properties of Fluids" by Rodgers and Mayhew.A version of these is currently on-line at:http://www.efunda.com/materials/water/steamtable_sat.cfm and these can be used to find 'steam' property values by entering either a known temperature or pressure of water vapour. You should have a hard copy of Steam Tables available (or have access to a computer with internet connection) in order to check the values given in this lesson and solve the Self-Assessment Questions. Teesside University Open Learning © Teesside University 2011 (Engineering)3 ________________________________________________________________________________________ FLUE GASES ________________________________________________________________________________________ In the last two lessons we looked at the combustion process and calculated amounts of air required for combustion, the composition of combustion products and theoretical flame temperatures. We are now going to look more closely at the combustion products often known as flue gases.These are formed at a high temperature (the same temperature as the flame) and are then used, generally, to heat up other materials, e.g. water to form steam. To make the most of the heat generated by the combustion process the flue gas temperature should be reduced to as low a value as possible.However, this can result in two problems. 1. As the temperature of the gases reduces, the water vapour (moisture) within the flue gases may condense.This is not necessarily a problem and can be a good thing – the latent heat of vaporisation within the water vapour is recovered and this will increase the overall efficiency of heat recovery.However, the liquid water formed may produce a film over the heat transfer surface and this would act as an extra resistance to heat transfer and could reduce the rate of heat transfer.Also the water formed may combine with some of the acid gases present in the flue gas (CO and 2 any SO /SO formed by the combustion of carbon or sulphur in the fuel) 2 3 and produce acids which will lead to possible corrosion of the heat transfer surfaces. The temperature at which the moisture condenses varies with the amount of moisture contained within the flue gases and the actual temperature and pressure of the gas/water mixture.The temperature where the water first condenses is known as the dew point temperature (or saturation temperature) of the gas/water mix.We will look in more detail as to how this is obtained later in this lesson. Teesside University Open Learning © Teesside University 2011 (Engineering)4 2. As the temperature of the flue gases reduces, the amount of lift created when the gases are discharged (via a chimney stack) to atmosphere is reduced.The gases leaving the end of the chimney, rather than been carried up and away from the stack and dispersing into the general environment (if the stack temperature is high enough), could fall and pollute the local environment close to the stack.The amount of pollution is not changed, but the effect of the discharge is increased as dilution of the gases is not carried out to as great an extent. The stack temperature should be such as to provide the necessary lift to disperse the gas effectively.Other factors are important here such as climate conditions (wind direction, air temperature, etc), geographical features (hills, valleys, closeness of towns, etc) and design of stack (height, diameter, pressure and flow rate within the stack, etc).The design of effluent disposal via stacks is a whole subject in its own right. We will not look at this problem any further in this module. Let's go back to the water condensing from the flue gases and look at this aspectin more detail. In the last lesson we looked at calculating the composition of the flue gases from the complete combustion of a fuel gas with the composition, by volume, of 90% CH (methane), 10% C H (propane), when using 30% excess air. 4 3 8 Using balanced chemical equations and the information given, we managed to determine the results in the table opposite. Teesside University Open Learning © Teesside University 2011 (Engineering)5 Feed O used Flue gases (moles) 2 (moles) Material moles CO produced H O produced N O 2 2 2 2 Methane CH 0.9 1.8 0.9 1.8 – – 4 Propane C H 0.1 0.5 0.3 0.4 3 8 Oxygen O 2.99 –2.99 0.69 2 Nitrogen N 11.24 0 11.24 2 Total –0.69 1.2 2.2 11.24 0.69 TABLE 1 From this we have the final flue gas composition as: 1.2 mol CO 2 2.2 mol H O2 11.24 mol N 2 0.69 mol O 2 The total number of moles=1.2+2.2+11.24+0.69=15.33. We can convert these compositions into mol % and hence mole fraction. Can you write the formulae which will enable us to do this in the space below? ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 (Engineering)6 mols A mols s A mol % l mole fraction== , mol % ×= 100 and mole fraction total mols total mols 100 Using these formulae we can determine the composition of the flue gases as mol % and mole fraction. mols CO 2 mol % CO=× 100 2 total mols 12 . = × 100 12.. + 22 +++ 11.. 24 0 69 12 . =× 100= 7.% 8 15.33 mol % and mole f fraction CO== 0.078 2 100 mols H O 2 mol % H O=× 100 2 total mols 22 . =× 100 0 = 14.% 4 15.33 mol % and mole fraction H O== 01 . 44 4 2 100 mols N 2 mol % N=× 100 2 total mols 11.24 =× 100 0 = 73.% 3 15.33 mol % and mole fraction N== 07 . 33 3 2 100 Teesside University Open Learning © Teesside University 2011 (Engineering)7 mols O 2 mol % O=× 100 2 total mols 06 . 9 =× 1004 0 = .%5 15.33 mol % and mole fraction O== 0.045 2 100 From earlier studies you may recall that the partial pressure of a gas can be determined from a knowledge of the mole fraction of a component and the total pressure of the gas system being looked at, i.e. partial pressure of gas = mole fraction of the e gas × total pressure within the sys stem Thus the partial pressure of water vapour (p ) within the flue gas can be w determined. In our flue gas analysis above, let's assume the total pressure of the flue gases is 1 bar (100 kPa).The partial pressure (p ) associated with the water vapour in this example is w therefore: p = mole fraction × total pressure () w water = 0.144 4 × 1bar = 0.144 bar Teesside University Open Learning © Teesside University 2011 (Engineering)8 From steam tables, Thermodynamic and Transport Properties of Fluids by Rodgers and Mayhew or currently available on-line at: http://www.efunda.com/materials/water/steamtable_sat.cfm we can find the temperature at which water will boil if this was the total pressure, known as the saturation temperature (T in hard copy steam tables s and T in the online version).If you use steam tables you will have to interpolate to find the result but if you use the web-based tables you need simply to input the pressure into the pressure box (making sure the value is in Find Sat Steam Properties bar!) and then click the box .The values will then appear below this box. The saturation temperature, in this case, is 53.1ºC (actually 53.13ºC). This boiling temperature is also the condensing temperature, i.e. the lowest temperature at which the water vapour at that partial pressure can exist.This is more commonly known as the dew point.Thus for the above flue gases, at a total pressure of 1 bar, the dew point temperature is 53.1ºC. If the temperature of the above gases falls below that figure, water will condense and be removed from the flue gases.As the amount of water in the gas falls, its mole fraction will fall and the partial pressure of the water vapour in the gases will also fall.As the pressure of water vapour falls the temperature at which it can exist also falls.For example, at 50ºC, from steam tables, the maximum partial pressure at which water vapour can exist is 0.1235 bar. So if the temperature of the original flue gases falls to 50ºC, the mole fraction of water vapour in the gases will to fall from 0.144 to 0.1235, provided the total pressure remains at 1 bar.The dew point of the remaining flue gases will now have fallen to 50ºC. Teesside University Open Learning © Teesside University 2011 (Engineering)9 If we now consider the original flue gases, but this time at a total pressure of 2 bar, can we determine what effect this will have on the dew point? The answer is yes.The partial pressure of water vapour in the flue gases is still given by p =(mole fraction) × total pressurebut this time the total w water pressure is 2 bar. p=× 0.144 2 bar= 0.288 bar Thus w From steam tables, the saturation temperature of water at that pressure is 68.2ºC (68.17ºC), so the dew point of the flue gases has been raised from 53.1 to 68.2ºC by a doubling of the total pressure. This will mean that, to avoid water condensing, a higher temperature will be required if the total pressure of the gases is increased whilst the composition remains the same. Try the following example for yourself. Determine the dew point of the flue gases that have the following analysis, if the total pressure is 1.1 bar. Flue gases (% by volume) CO H O N O 2 2 2 2 8 10 78 4 Remember that %volume for gases is the same as mol %. ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 (Engineering)10 mol % vol % 10 Mole fraction of H O== = = 01 . 2 100 100 100 0 p =() mole fraction × total pressure e w water =× 0.1 1.1 bar= 0.11 bar From steam tables, de ew point=° 47.7 C The first set of calculations were based on a fuel gas analysis where the only water present in the flue gas was that formed during combustion.In the in-text question we were given an actual flue gas analysis.Would this be any different?The answer is yes.The actual flue gas analysis would include any water present in the air entering the combustion process.In the first calculation we only worked out the water produced by combustion.In practice, the supply air will contain some water vapour.At an air temperature of 25ºC and a relative humidity of 50% (a typical summer’s day) the amount of water present in the –1 air is approximately 0.018 moles per mole of air (0.011 kg water [kg air] ), whilst in winter at 5ºC and 90% humidity this falls to approximately –1 0.008 moles water per mole of air (0.005 kg water [kg air] ). For our worked example: Feed O used Flue gases (moles) 2 (moles) Material moles CO produced H O produced N O 2 2 2 2 1.8 Methane CH 0.9 0.9 1.8 – – 4 0.5 Propane C H 0.1 0.3 0.4 3 8 –2.99 Oxygen O 2.99 0.69 2 0 Nitrogen N 11.24 11.24 2 –0.69 Total 1.2 2.2 11.24 0.69 Teesside University Open Learning © Teesside University 2011 (Engineering)11 The amount of air entering was approximately 14.2 moles (11.24 mols N 2 +2.99 mols O ) which will give an additional water content in the flue gases 2 (at the higher air supply temperature of 25ºC) of 14.2× 0.018 =approximately 0.26 moles.This would also mean that the total amount of flue gases would increase by this figure. The mole fraction of water in the flue gases would then bemols H O in flue gas 2 mol % H O = × 100 2 total mols of flue gas 22.. + 026 = × 100 15.. 33 + 0 26 24 . 6 = ×= 100 15.% 78 1555 . 9 mol % and mole fraction H O = = 0.158 2 100 1 p =() mole fraction × total pressure w water = 0.158 8×= 1bar 0.158 bar The dew point, in this case, would then be 55.1ºC (compared to the earlier 53.1ºC when the air was assumed to be completely dry). So far we have talked about the dew point for the whole of the gas.However, if the flue gas contacts a surface below the dew point, water will condense from the flue gas even if the main bulk of the flue gas is above the dew point. The water which is condensed at the surface will be removed from the flue gas. There will be a temperature gradient from the interface to the bulk of the gas. Whilst the flue gas in contact with the cold surface may be below the dew Teesside University Open Learning © Teesside University 2011 (Engineering)12 point, the bulk of the flue gases may still be above the dew point.Water will therefore only condense on the cold surface forming an insulating layer on the surface which in turn will reduce the heat transfer rate from the bulk gases. This is shown in FIGURE 1 below. Cold surface Temperature Bulk of flue gas Stagnant layer Water of flue gas layer FIG. 1 This liquid layer is undesirable as the lower temperature surface (below dew point) may be less effective at cooling the gases than a surface temperature slightly above the dew point which has no liquid layer present.(To prevent the liquid layer forming, the surface should be above the dew point.For effective heat transfer, the temperature difference between surface and gas should be at least 5°C and thus the gas should not be cooled to within at least 5°C of its dew point.) The liquid layer may also dissolve some carbon dioxide or other acid gases present in small concentrations within the flue gas and become an acidic layer which may cause corrosion. There is also a second effect of some acidic gases that we will deal with shortly. Teesside University Open Learning © Teesside University 2011 (Engineering)13 First, let's work through another example. The analysis of a flue gas leaving a combustion process was (by volume) CO 6%H O11.4%O 8.1%N 74.5% 2 2 2 2 Calculate the dew point of this gas if the pressure is 1 bar. Volume % is the same as mol % for (ideal) gases so the mol % of water (H O) = 11.4% 2 giving a mole fraction of 0.114. The partial pressure of water vapour: p = mole fraction × total pressure () w water = 0.114 4 × 1bar = 0.114 bar From steam tables, the saturation temperature of water at this pressure is 48.4ºC and hence the dew point temperature is 48.4ºC.In this case any water associated with the air has been accounted for in the analysis, so the dew point will be correct for the actual process conditions. Teesside University Open Learning © Teesside University 2011 (Engineering)"