EVERYDAY EXAMPLES OF ENGINEERING CONCEPTS

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1 EVEYDAY EXAMLES O ENGINEEING CONCETS T11: Combustion & reacting mixtures Copyrigt 014 Tis work is licensed under te Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. To view a copy o tis license, visit ttp://creativecommons.org/licenses/by-nc-nd/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, Caliornia, 94041, USA. Everyday Examples rom 1 o 8

2 Tis is an extract rom 'eal Lie Examples in Termodynamics: Lesson plans and solutions' edited by Eann A. atterson, irst publised in 010 (ISBN: ) and contains suggested exemplars witin lesson plans or Sopomore Termodynamics Courses. Tey were prepared as part o te NS-supported project (# ) entitled: Enancing Diversity in te Undergraduate Mecanical Engineering opulation troug Curriculum Cange". INTODUCTION (rom 'eal Lie Examples in Termodynamics: Lesson plans and solutions') Tese notes are designed to enance te teacing o a sopomore level course in termodynamics, increase te accessibility o te principles, and raise te appeal o te subject to students rom diverse backgrounds. Te notes ave been prepared as skeletal lesson plans using te principle o te 5Es: Engage, Explore, Explain, Elaborate and Evaluate. Te 5E outline is not original and was developed by te Biological Sciences Curriculum Study 1 in te 1980s rom work by Atkin & Karplus in 196. Today tis approac is considered to orm part o te constructivist learning teory 3. Tese notes are intended to be used by instructors and are written in a style tat addresses te instructor, owever tis is not intended to exclude students wo sould ind te notes and examples interesting, stimulating and opeully illuminating, particularly wen teir instructor is not utilizing tem. In te interest o brevity and clarity o presentation, standard derivations, common tables/carts, and deinitions are not included since tese are readily available in textbooks wic tese notes are not intended to replace but rater to supplement and enance. Similarly, it is anticipated tat tese lesson plans can be used to generate lectures/lessons tat supplement tose covering te undamentals o eac topic. Tis is te tird in a series o suc notes. Te oters are entitled eal Lie Examples in Mecanics o Solids (ISBN: ), eal Lie Examples in Dynamics (ISBN: ). Acknowledgements Many o tese examples ave arisen troug lively discussion in te consortium supported by te NS grant (# ) on Enancing Diversity in te Undergraduate Mecanical Engineering opulation troug Curriculum Cange and te input o tese colleagues is ceerully acknowledged as is te support o National Science oundation. Te comments on an early drat made by obert D. Handscombe o Handscombe Associates are grateully acknowledged. Eann A. atterson A.A. Griit Cair o Structural Materials and Mecanics Scool o Engineering, University o Liverpool, Liverpool, UK & oyal Society Wolson esearc Merit Award ecipient+ 1 Engleman, Laura (ed.), Te BSCS Story: A History o te Biological Sciences Curriculum Study. Colorado Springs: BSCS, 001. Atkin, J. M. and Karplus,. (196). Discovery or invention? Science Teacer 9(5): e.g. Trowbridge, L.W., and Bybee,.W., Becoming a secondary scool science teacer. Merrill ub. Co. Inc., Everyday Examples rom o 8

3 THEMODYNAMICS O CHEMICAL TANSOMATIONS 10. Topic: Combustion & reaction mixtures Engage: Take a cake wit a candle, a large glass jar and stop watc into class. Ligt te candle. Invite te students to make sealed bids on ow long it will take te candle to be extinguised wen it is covered wit te jar. Te prize or te nearest correct answer is te cake. A cup cake would be ine so tat you can cover te cake and candle wit te jar. Alternatively, i you are eeling generous, use a larger cake tat can support te weigt o te glass jar wen you put it over te candle; and award a slice to te nearest answers. Explore: Beore placing te jar over te candle, discuss te process tat is occurring as te candle burns. Te material o te candle is mainly solid parain, typically C 0 H 4, i.e. wit a generic cemical ormula o C n H n+ wic is te same as metane, CH 4 or octane, C 8 H 18 wic is te main component o gasoline; so not surprisingly it is very combustible. Combustion is a cemical reaction between a uel (parain) and an oxidant (oxygen in te air) wic results in a set o products (carbon dioxide and water) and eat C 0 H O eat 0CO + 1H O I suicient eat is generated, ten te products emit ligt generating te lame observed on te candle. Invite two students to join you in dowsing te candle; give one te stop-watc and te oter te glass jar. Explain: Wen te candle is burning in te room, tere is an excess o oxygen available or combustion. Tis is te state wen te jar is initially placed over te candle. Wen te candle goes out tere is no oxygen let to react wit te parain so combustion is arrested. Tere is a point somewere in between wen tere is exactly te rigt amount o oxygen; tis is known as te teoretical amount o air. It is usual practice to idealize air to be 1% oxygen and 79% nitrogen on a molar basis and ignoring te smaller percentages o oter gases present, suc as argon and carbon dioxide wic are less tan 1%. So or combustion in air te cemical reaction can be written as C 0 H 4 + a(o + [0.79/0.1] N ) eat bco + ch O + dn Everyday Examples rom 3 o 8

4 And te stoiciometric coeicients: a, b, c and d can be ound by applying te law o conservation o mass to carbon so b 0 ; to ydrogen so c 1; to oxygen so a b c and a 30.5 ; and inally to nitrogen so d 3.76a Tus C 0 H (O N ) eat 0CO + 1H O N So te number o moles o air required or complete combustion o parain is (= ) moles o air per mole o uel te teoretical amount o air. It can be turned into an air-uel mass ratio by multiplying it by te ratio o te molecular masses o air (8.97) and parain, (34 [=01+4]), i.e., 8.97 Air-uel mass ratio As te candle burns inside te jar, te oxygen is used up until tere is insuicient to allow complete combustion o te parain and so not all o te carbon is converted into carbon dioxide resulting in te black deposit on te inside o te glass, i.e., soot. Elaborate: An energy balance can be used to evaluate te eat generated during te reaction/combustion process. I it is assumed tat it is parain vapor evaporating rom te wick wic is reacting wit te surrounding air, ten tis allows a control volume to be considered in wic te reactants (parain vapor and air) are entering and te products o combustion (carbon dioxide, water and nitrogen) are leaving. Tus or te control volume applying te irst law o termodynamics to equate te cange in entalpy to te net eat transer and work done Q n W n n o ni o were n is te molar low rate o te uel, te subscripts i and o reer to te inlet and outlet gases respectively, and and to te products and reactants respectively. is te entalpy o ormation wic is te energy released or absorbed wen te compound is ormed rom its elements at te standard reerence state (5C, 1 atm) and is te cange in entalpy relative to te standard reerence state. So work, W is zero or te candle and given C 0 H O N eat 0CO + 1H O N te energy balance becomes i Q n ] [ CO ] -[ C0H 4 Everyday Examples rom 4 o 8

5 Te entalpy o ormation o parain (C 0 H 4 ) is -456,000kJ/kmol and treating it as an ideal gas suc tat C T and C. 13 kj/kg K. or carbon dioxide, oxygen, nitrogen and water p p te entalpy o ormation in can be ound in termodynamic tables 4. i.e. -393,50 kj/kgk; -41,80 kj/kgk; 0 C H Now or te reactants wic enter te combustion control volume at 400K: were CpT C0H4 kj/kgk C 0 H O N,684kJ/kmol values are obtained rom te ideal gas data or eac gas in te termodynamic tables 5. Similarly or te products given a temperature in te candle o 1400K: C CO H O N 6,911,81kJ/kmol Hence, rom te irst law o termodynamics, remembering W 0 Q ( 684) 6889 MJ/kmol K (C 0 H 4 ) n So i te candle (molecular mass = 34) burns at 1g/min ( kg/s) or kmols/s (= /34), ten te eat transer is 354W (= ). Evaluate: Invite students to attempt te ollowing examples: Example 11.1 Wat low rate o natural gas is required to raise two liters o water to boiling point in ive minutes in order to cook pasta? You may assume tat te gas ring is at a steady temperature o about 450K and te lame temperature is 1600K. Solution: Te irst law o termodynamics can be used to equate te energy to eat te pan and water to te eat supplied by te gas lame. Energy input required or water 4 e.g. ttp://v5.sdsu.edu/testome/test/solve/basics/tables/tablescomb/ormation.tml 5 e.g. ttp://v5.sdsu.edu/testome/test/solve/basics/tables/tablesig/ig.tml Everyday Examples rom 5 o 8

6 mc p T 4.18 A stainless steel pan large enoug to cook pasta in tis quantity o water would ave a mass o about 1.8kg, so energy input required or te pan mc p T And te eat transer rate, Q ignoring te eat losses rom te pan. Now, natural gas is mainly metane so we can approximate te combustion process to kw CH 4 + O N eat CO + H O N So applying an energy balance based on te irst law o termodynamics Q n [ 7.58 C kj kj ] -[ 7.58 CH4 ] Te entalpy o ormation o parain (C 0 H 4 ) is -74,850kJ/kmol (rom te termodynamic tables ) and treating it as an ideal gas suc tat c T and C. kj/kg K. Also -393,50 kj/kgk; -41,80 kj/kgk; 0 CO p p H Now or te reactants wic enter te combustion control volume at 450K CpT 7.58 CH kj/kgk CH 4 N kj/kmol Similarly or te products given a lame temperature o 1600K 7.58 C C applying te irst law o termodynamics to te lame in a control volume wit W 0 Q n combustion ( 31799) kJ/kmol K (C 0 H 4 ) Now applying te irst law o termodynamics to eating te water and pan Q required Q combustion Q n n Hence, Everyday Examples rom 6 o 8

7 so n Q Q n required And or metane, M 16, so m Mn kmol/s kg/s or 0.94 liters/s Example 11. Wen a car is cruising on te igway at 70mp, its uel consumption is 35mpg and te engine output is 60p, estimate te rate o eat transer rom te combustion process i te uel is assumed to be liquid octane and te temperature in te exaust is about 900K. Assume te ambient temperature is 5C. Solution: At 70mp te car will use gallons/our (= m 3 /r = m 3 /s) And, te density o octane is 703kg/m 3 so te mass low rate, m kg/s Octane is C 8 H 18 wit a molecular mass, M 13, so by deinition n m M Assuming complete combustion kmol/s C 8 H O + 47N eat 8CO + 9H O + 47N And applying te 1st law o termodynamics, te energy balance is Q n W n [ C ] -[ C8H18 ] Te entalpy o ormation o liquid octane is -49,950kJ/kmol and C.3 kj/kg K (rom te termodynamic tables ). Also -393,50 kj/kgk; -41,80 kj/kgk; 0 CO H O p H Now or te reactants wic enter te combustion control volume at 98K CpT C8H C H18 N 8 Similarly or te products given a lame temperature o 900K C kj/kmol kj/kgk Everyday Examples rom 7 o 8

8 C W 60 p W te energy balance becomes Q n W n 5 and Q n W kw So, wit Everyday Examples rom 8 o 8