Topics Adiabatic Combustion Adiabatic Flame Temperature Example Combustion Efficiency Second and First Law Efficiencies Contrasted Example Problem

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1 ME 410 Day 12 opis diabati Combustion diabati Flame emperature Example Combustion Eiieny Seond and First Law Eiienies Contrasted Example roblem 1. he Case o diabati Combustion Here is it assumed that there is no heat transer o signiiane during the ombustion proess. Hene adiabati. Constant Volume diabati Combustion 0 U U U U eatants roduts he temperature is alled the adiabati lame temperature or onstant pressure ombustion.

2 Constant ressure diabati Combustion 0 H H H H eatants roduts he temperature is alled the adiabati lame temperature or onstant pressure ombustion. Example with EES -- Just a modiiation o a previous Example "diabati Flame emperatures - ropane Combustion" "Equation o Combustion o ropane in ir C3H8 + 5 (O N2) 3 CO2 + 4 H2O + 5*3.773 N2" "Calulations will be based on mass o 1 kg C3H8" m_c3h8 1 m_o2 5*MOLMSS(O2)/MOLMSS(C3H8) m_co2 3*MOLMSS(CO2)/MOLMSS(C3H8) m_h2o4*molmss(h2o)/molmss(c3h8) m_n2 5*3.773*MOLMSS(N2)/MOLMSS(C3H8)

3 "Enthalpy o roduts and eatants" H_ m_c3h8*enhly(c3h8,1)+m_o2*enhly(o2,1) +m_n2*enhly(n2,1) H_ m_co2*enhly(co2,2)+m_h2o*enhly(h2o,2 )+m_n2*enhly(n2,2) DEL_H H_-H_ "his gets adiabati lame temp. or onstant pressure ombustion." DEL_H0 "Internal Energy o roduts and eatants" U_ m_c3h8*inenegy(c3h8,1)+m_o2*inenegy(o2, 1)+m_N2*INENEGY(N2,1) U_ m_co2*inenegy(co2,3)+m_h2o*inenegy(h2o, 3)+m_N2*INENEGY(N2,3) DEL_U U_-U_ "his gets adiabati lame temperature or onstant vol. ombustion" DEL_U

4 2. Combustion Eiieny Here s another eiieny to learn. he ombustion proess in IC engines is never peretly omplete. For example, very small amounts o unburned uel an hide out in revies. Or, more likely, the ombustion proess produes CO or soot, or any other produt o partial ombustion. Hene, not all the hemial energy in the uel is released by the ombustion proess. Let H be the enthalpy o the produts o ombustion and H the enthalpy o the reatants -- both evaluated at ambient temperature, and one atmosphere. Hene o H n i h ~, i and H n i h ~ ~ produts rea tan ts o where h, i represents the enthalpy o ormation and n i the number o moles. he atual hemial energy atually released is he total hemial energy that ould be released is m HV. hereore the ombustion eiieny is H H. mhv o, i H H.

5 3. Seond Law Eiieny eatants: H, S roduts: H, S W he piture shows what happens during one engine yle. First Law o hermo: W + mh mh in out 0 W mh mh out in H H H Seond Law o hermo (s, entropy per unit mass o & ) + ms ms + Sgen in out 0 I we say there is not any entropy generated - reversible ase - ND it must be true that. ms ms out in S S S S S So we an assume that this would orrespond to the ase where W is maximized.

6 Substitute into the irst law, S W, max H W,max H S B B is the steady low energy availability untion. It depends on the ambient onditions and the luid properties o the reatant and produt streams. Sine we have agreed that the luid streams are at ambient temperature, B ( H S ) ( H S ) in other words or this partiular situation B G where the Gibbs Free Energy is a property, or state variable. eall that we had H S H m HV. In the same way we let B G m a Where a might be alled the available energy o ombustion. Note it takes into aount the dierene in entropy in produts and reatants. he seond law eiieny is Work per yle Max ossible Work per W W a yle,max W m a

7 Contrast with the irst law eiieny Work per Chem Energy yle per yle W m LHV What we have alled the uel onversion eiieny. he seond law or availability eiieny is Not as easy to evaluate as the irst law eiieny. But this evaluation opens the door to availability analysis, whih is a ruitul way o looking at the dierent ators whih eet eiieny with an eye to improvement. For hydroarbon uels g and h are lose. Consequently the irst and seond law eiienies will be lose. We will ollow the text and ous on irst law eiienies. here is yet another eiieny whih omes into play at times. It takes into aount the inomplete ombustion proess. It is alled the thermal onversion eiieny. t W m LHV pparently, t.

8 Example roblem (3.6) he brake uel onversion eiieny is 0.3. he mehanial eiieny is 0.8. he ombustion eiieny is he heat losses to the oolant and oil are 60 kw. he uel hemial energy entering the engine per unit time!. m HV 190 kw What perentage o this energy beomes a) brake work or power b) rition work or power ) heat losses d) exhaust hemial energy e) exhaust sensible energy. Draw a diagram with the engine in a CV and these quantities indiated. Important: Let s assume that the rition work and the heat losses do not overlap. (ter all one is work, and the other heat) he sensible energy o the exhaust is what s let in the exhaust ater the exhaust hemial energy is removed.