"MODULE TITLE:HEAT TRANSFER AND COMBUSTION TOPIC TITLE:COMBUSTION PROCESSES LESSON 2:HEATING VALUES AND THEIR USE IN COMBUSTION PROCESSES HTC - 4 - 2 © 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 combustion of fuels is mainly carried out to produce energy for use in chemical processes either directly or via the production of steam. Each fuel has an enthalpy change when it is burnt in air and values of this change are often given in standard reference books.We will calculate, using these values and a table of enthalpy values for combustion products at various temperatures, the temperature obtainable within the combustion flame (theoretical flame temperature) when the fuel is burnt. We will also look at furnace efficiency and illustrative temperature – enthalpy (T –H) diagrams. ________________________________________________________________________________________ YOUR AIMS ________________________________________________________________________________________ At the end of this lesson you should be able to: • explain the difference between gross calorific value (higher heating value) and net calorific value (lower heating value) • understand the term 'heat (enthalpy) of combustion' • use calorific values/heats of combustion to determine the heat released by single fuels and fuel mixtures • use temperature – enthalpy tables and/or graphs to obtain enthalpy values for the main products of combustion Teesside University Open Learning © Teesside University 2011 (Engineering)2 • use heat and mass balance to determine– flue gas composition– theoretical flame temperatures • produce illustrative temperature-enthalpy diagrams for combustion processes. ________________________________________________________________________________________ STUDY ADVICE ________________________________________________________________________________________ As with the last lesson, you should have already studied similar material when doing the Mass and Energy Balance module.If you have, then this lesson will act as a form of revision.It will not be covered in the detail that it was in Mass and Energy Balance and you may wish to refer to material in that unit if you do not fully understand this lesson. If you have not studied the Mass and Energy Balance, the work is explained in sufficient detail to answer the questions that will be asked in the Tutor Marked Assignment. The terms enthalpy and heat are used interchangeably in this lesson, the old term being heat and the new term being enthalpy.However, most text books still refer to heat and hence its use in this lesson. You will need a calculator and some graph paper when answering some of the questions. Teesside University Open Learning © Teesside University 2011 (Engineering)3 ________________________________________________________________________________________ CALORIFIC VALUES ________________________________________________________________________________________ The calorific value (CV) or heating value of a fuel is a measure of the heat 3 evolved when unit mass (1 kg in SI units) or volume (1 m in SI units) of a fuel is completely burned.Volume units are used for gases but, in these cases, the conditions of temperature and pressure must be defined and values are quoted generally at a temperature of 15°C (288 K)and a pressure of 1 bar (100 kPa). Note that the standard temperature used for calculating calorific values is 15°C (288 K) when normally temperatures of either 25°C or 0°C are used for other properties. Whenobtaining calorific values experimentally, the combustion products are cooled to 15ºC (288 K) and heat release includes the latent heat (enthalpy) of vaporisation of water (if present) as the water vapour produced will be condensed to liquid.The CV value obtained in this situation is known as the gross calorific value (higher heating value). In practice, products of combustion in most processes leave at higher temperatures than 15°C and any water vapour remains in the vapour state. Thus the latent heat of vaporisation is not recovered.The heat released is lower and we have a net calorific value (lower heating value).The net calorific value has to be determined from knowledge of gross calorific value, the fuel analysis and the equations for combustion.It does, however, more accurately represent the actual heat available from the combustion process. This lower value only applies for fuels that produce water as a product of combustion, i.e. those that contain hydrogen. Often industrial fuels are mixtures of different compounds and elements, and so calorific value data is normally determined on a complete fuel basis rather than on the individual constituents.However, many gases are obtainable pure, Teesside University Open Learning © Teesside University 2011 (Engineering)4 or nearly pure, and TABLE 1 gives a list of the gross calorific values of some pure gases at a temperature of 15°C and pressure of 1 bar. –3 Gas Gross CV MJ m Carbon monoxide 11.85 Hydrogen 11.92 Methane 37.07 Ethane 64.46 Propane 93.89 Butane 121.84 Ethylene (ethene) 58.12 Propylene (propene) 85.70 TABLE 1 3 You may notice that for gases there is an increase in calorific value per m with complexity of the molecule.Hydrocarbons burn to produce carbon dioxide, water and heat: the more carbon and hydrogen in the molecule the more heat there is to release.That is, since 1 mole of any gas occupies the same volume at the same conditions of temperature and pressure, a complex gaseous molecule will occupy the same volume as a simple molecule but contain more combustible material. For information only, approximate gross calorific values of some liquid and –1 solid fuels are listed in TABLE 2.Note that these are per kilogram (kg ) not 3 per m . Teesside University Open Learning © Teesside University 2011 (Engineering)5 –1Liquid fuels MJ kg (water free) Kerosene 46.5 Diesel 45.6 Light fuel oil 43.5 Medium fuel oil 43.1 Heavy fuel oil 42.6 –1 Solid fuels (dry, ash-free) MJ kg Lignite 26.7 High volatile coal 33.7 Low volatile coal 36.5 Athracite 36.3 Wood 19.8 TABLE 2 Calorific values of gaseous fuel mixtures may be estimated by multiplying the volume fraction of each component by its own CV, then summing the results. We shall calculate the gross CV of the following gas whose composition by volume is: hydrogen (H ) 50%,carbon monoxide (CO) 45%, methane (CH ) 5% 2 4 3 3 3 Using 1 m of gas as a basis, the fuel contains 0.5 m of H , 0.45 m of CO 2 3 and 0.05 m of CH .Using these volumes and the figures for CV from 4 TABLE 1, Teesside University Open Learning © Teesside University 2011 (Engineering)6 Gross CV=+ 50% of CV for H 455 % of CV for CO+ 5% of CV for CH 2 4 =× 05.. 1192+× 04. 5 118. 5 + 000.. 5 × 3707 ()()() =+ 59.. 6 533+18. 5 –3 = 13.14 MJ m To find the net CV of the same gas, we need to consider the individual reactions and find the water vapour produced by the fuel when it is burnt. Specific heat capacity and latent heat data will then enable us to calculate the heat released when the water vapour produced is cooled and condensed.Let's do this. For our example fuel, the three reaction equations areCO+? 0.5O CO 22 H+? 0.5O H O 22 2 CH+?22 O CO+ H O 42 2 2 From these equations we can see that 1 mole of H produces 1 mole of H O 2 2 and 1 mole of CH produces 2 moles of H O.(Note the carbon monoxide can 4 2 be ignored as it produces no water.)Since, for gases at the same temperature and pressure, volume % and mol % are interchangeable numerically, we can 3 3 3 say that 1 m of H will produce 1 m of water vapour and 1 m of CH will 2 4 3 3 produce 2 m of water vapour. For 1 m of this fuel, the volume of water 3 vapour produced is equal to the volume of H (0.5 m ) plus twice the volume 2 3 3 of CH (0.05 m ), i.e. 0.5+2× 0.05=0.6 m .This would be the volume 4 of water vapour produced at 15°C (288 K).We need to convert this volume first to kmoles and then to kg.We do this by using the ideal gas equation. Teesside University Open Learning © Teesside University 2011 (Engineering)7 pV n = RT wheren = number of moles p = pressure in Pa (1 bar = 100 000 Pa) 3 V = volume (m ) –– 11 R = univer rsal gas constant (8.314 J mol K ) T = temp perature in kelvin 100 000 × 0.6 n = == 25.1 mol 0.0251 kmol 8.314 × 288 Most data for specific heat capacities and specific latent heats are given per kilogram, so we now need to change kmol into kg (kg = kmoles × molecular mass), i.e. 0.0251× 18=0.452 kg of water are produced. –1 –1 The mean specific heat capacity of water from 100°C to 15°C = 4.20 kJ kg K and the latent heat of vaporisation of water at 100°C and 1 bar –1 –1 = 2258 kJ kg K (obtained from steam tables). Therefore the heat released during the condensation of water vapour and the cooling of the condensate to 15°C =× (mass latent heat)+× (mass specific heat c capacity × fall in temperature) =× 0.452 2258 ++× 0.. 452 4 20× 85 ()() = 1182 kJ = 1.18 MJ Teesside University Open Learning © Teesside University 2011 (Engineering)8 We should, however, take into account that the calorific value would include the gaseous water heat content and we should deduct that from this value. The specific heat capacity of water vapour has a mean value of approximately –1 –1 1.88 kJ kg K (from steam tables) and hence the heat in water vapour =0.452× 1.88× 85=72 kJ. Thus total difference=1182–72=1110 kJ=1.11 MJ. –3 The net CV of the fuel is therefore = 13.14–1.11=12.03 MJ m . The value of 1.11 MJ for the difference in heat content between water vapour at 15ºC and water at 15ºC was for 0.452 kg of water.Thus for 1 kg water, the –1 difference would be 1.11÷ 0.452=2.46 MJ kg .[This value could have been estimateddirectly from steam tables by looking at the value ofthe difference between h and h for saturated steam at 15ºC, i.e. latent heat of g f water (h ) at 15ºC=2.465 MJ.]So for every kilogram of water formed in fg the combustion of a fuel, the net value of the CV will reduce by 2.46 MJ. Since 1 kilomole of water is 18 kg, the molar value of the heat given out will –1 –1 reduce by 2.46× 18=44 MJ kmol (= 44 kJ mol ). An alternative to using calorific values, for heat released from a fuel, is to use ° heat (enthalpy) of combustion data.This data is available in () ? H 298 textbooks for most single fuels (e.g. in Perry’s Chemical Engineer's Handbook there are 3 pages of values for many common hydrocarbon fuels). Some example heats of combustion are given in TABLE 3.The values given are negative values by convention as they are the heat lost by the fuel to its surroundings when it burns.These figures are quoted at a temperature of 25°C and 1 atmosphere pressure and are per mole of the fuel.As with calorific value, they have a lower and higher value depending on whether any water formed is in the liquid (higher) or gaseous (lower) state. They may also have different values based on the state of the fuel before combustion (see butane in the table which has a boiling point of – 0.5°C and is often stored as a liquid). Teesside University Open Learning © Teesside University 2011 (Engineering)9 –1 –1 Material Lower value (kJ mol ) Higher value (kJ mol ) Hydrogen (H ) –242 –286 2 (g) Methane (CH ) -803 -891 4 (g) Propane (C H ) -2046 -2222 3 8 (g) Liquid butane (C H ) -2639 -2859 4 10 (1) Gaseous butane (C H ) -2660 -2880 4 10 (g) Carbon (C ) -394 -394 (s) Carbon monoxide (CO ) -283 -283 (g) TABLE 3 From this table you may be able to see that for each mole of water product formed (or for each mole of H present in the material) the heat of reaction is 2 reduced by 44 kJ (i.e. H has a 44 kJdifference, CH has an 88 kJdifference, 2 4 C H has a 176 kJ difference, etc.).This is the value obtained earlier during 3 8 our CV calculations. Also, note that the carbon and carbon monoxide have no difference in higher and lower value as no water is formed in their combustion. We will now use some of this information and perform a heat balance to determine the maximum theoretical flame temperature (T ) reached when a F fuel is burnt.This occurswhen the temperature of products is such thatheat input to process = heat out of process We choose a datum temperature which could be any.In practice, it is often 25ºC as combustion data and enthalpy data sources are usually given at this datum (sometimes the datum in some textbooks is 15ºC, e.g. when dealing with calorific values). Teesside University Open Learning © Teesside University 2011 (Engineering)10 Heat input = sensible heat in fuel and air + heat r a eleased by combustion (relative to datum temp m erature) Heat out of process=× mass mean specific heat capacity × T ?() Ffor each product of co c mbustion = ?() enthalpies of each product of co ombustion (relative to datum temp perature) For heat out, one of the values in the equation is itself dependent on temperature (mean specific heat capacity) which, of course, is unknown.To solve the problem, we must first approximate to a solution, do the calculation, refine the estimate, and repeat until we obtain a balance.Fortunately, heat content (enthalpy) data [which is mean specific heat capacity over the range from 25ºC to the actual temperature multiplied by the actual temperature] is available – see Appendix – for the products of combustion at various temperatures, and we will make use of these values and a graph to obtain a reasonably accurate solution to the problem of finding the balance temperature. Hopefully this will become clearer as we do a calculation. Sometimes, the data is a table of mean molar specific heat capacities (C ) p mean for each product of combustion over the range of temperatures given.To calculate enthalpy in this case you multiply the C for the temperature by p mean the temperature.We have been kind by doing this for you!! Also, be careful with all data that the units are the correct ones for the calculation – sometimes enthalpies are given in kilocalories!!Teesside University Open Learning © Teesside University 2011 (Engineering)11 Before we use the data in a calculation of flame temperature, we will need to make some assumptions. • Above 1650°C, errors can arise due to dissociation of water vapour and carbon dioxide, giving lower flame temperatures, but we will not get involved with that difficulty here. • The process is adiabatic (no external heat is added to, or removed from, the process). • All the fuel is burnt (though in some calculations, combustion may not be completely to carbon dioxide and water and some carbon monoxide may be formed). • Fuel and air enter at datum conditions – usually 25ºC (though the calculation can be adjusted to any temperature). • Combustion products emerge at a single temperature. To calculate the flame temperature, we will first find the heat present in, and generated by, the fuel and air (heat in), then we will need to calculate the amounts and composition of the product gases.We can then find the enthalpy figures (heat out) for each product gas at 'selected' temperatures and thus the total heat out by estimation of a flame temperature.Finally, we need to refine and repeat our calculations until we find the 'actual' temperature where the heat in=heat out, i.e. they balance and this is the maximum theoretical flame temperature for the supply conditions of fuel and air. Let's do an example of a flame temperature calculation utilising this series of steps to help show this process. Teesside University Open Learning © Teesside University 2011 (Engineering)12 Example 1 A fuel gas has the composition 90% CH (methane), 10% C H (propane) by 4 3 8 volume.It is completely combusted with 30% excess air at a temperature of 25°C. The heats of combustion are °° –1 –1 ??HH CH== –, 803kJ mol C H– 2045 kJ mo ol 298 4 298 38 Calculate the theoretical flame temperature reached. ° The values for ? H are the lower values given in TABLE 3.Why? 298 ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ These are the lower values as we will have a flame temperature well above the condensing point of water and thus we will not get the benefit of the latent heat of vaporisation of water. Teesside University Open Learning © Teesside University 2011 (Engineering)13 Solution We will assume the datum temperature of 25°C which is very convenient as the temperature of fuel and air into the process is 25°C, the heat of combustion for each fuel is at 25°C and the enthalpy data given in the Appendix has a datum of 25°C!If the fuel and air were at a higher (or lower) temperature we would have to calculate their additional (reduced) heat content relative to 25°C using: Additional (reduced) heat content = mass× specific heat capacity ×temperature differenceThis would be, in effect, additional (reduced) heat supplied to the combustion process, i.e. additional (reduced) heat input. As the fuel and air in this question are at 25°C (our datum), the only heat input is from the heat of combustion of each fuel. Let's start with the assumption that we have 1 mole of fuel (we could choose any amount but by choosing 1 mole we make the calculations easier as most of the figures quoted in the question are per mole).As the composition of the fuel is given in volume %, which is the same as mole % for gaseous fuels, we can convert the volume into moles to give us a fuel composition of 0.9 mol CH and 0.1 mol C H . 4 3 8 Thus, heat released by combustion of 1 mole of the fuel °° =× 09..??HH CH+ 01× C H 29843 2988 =× 09.. 803+ 01× 20 045 = 927.k 2 J Teesside University Open Learning © Teesside University 2011 (Engineering)"