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第一章 燃烧的化学和物理基础 PowerPoint PPT Presentation


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第一章 燃烧的化学和物理基础. §1 燃烧反应热. 1.1 反应热、生成热和燃烧热. 此能量差值以热的形式向环境散发或从环境吸收,称 反应热 。显然,生成物所含能量少于反应物所含能量时,此差值为负值,表明有多余能量释放,称放热反应。相反,此差值为正值即要向系统加入能量,为吸热反应。在定温定容过程时,反应热等于系统内能变化,即Q=△U。在定温定压过程时,反应热等于系统焓的变化 ,即Q=△H。.

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第一章 燃烧的化学和物理基础

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1


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1.1

  • Q=UQ=H


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  • HfA1A2n1n2 Hf=HA-(n2H2+n1H1)

  • Hf0=HA.

  • 298K0.1013MpaH0c298(Mj/koml)


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1.2

  • 1.

  • HT0, HT0 =njHj-niHi (Mj/kmol)

    -njj

    -nii


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-HjjTK

-HiiTK

THT

:


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  • 2.

  • 1T12Tc.

  • njH0cj-niH0T1i=nj(H0Tc-H0298)j-ni(H0T1-H0298)+(njH0298j-niH0298i)=0


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2

  • 2.1

  • 1

  • ni=Ni/V (1/m3)

  • 2

  • Ci=Mi/V=Ni/NA/V (mol/m3)

  • (3)

  • i=Gi/V (kg/m3)


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  • 4i

  • Xi=Ni/Ni=ni/ni=Ci/Ci

  • 2.2

  • GG0


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  • H=U+PV

  • dH=dU+PdV+VdP

  • dU=dQ-dW=TdS-PdV

  • dH=TdS+VdP

  • GG=H-TS

  • dG=dH-TdS-SdT=VdP-SdT

  • n1,n2,n3..ni,TP

  • G=GTPn1,n2,n3..ni


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dn1=0,


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  • dG=-SdT+VdP+idni

  • dni=0,

  • dG=-SdT+VdP

  • dT=dP=0,dni0 dG=idni=0,dG=idni0,dG=idni0,

  • dG1P0P


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  • 2.3

  • CnHmOlNk(k,l)O2n+m/4-l/2.O2n+m/4-l/2/.HONH2OHCONOO2H2OCO2N2Ar12x1~x12x131


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  • =l/2+0

  • 0=(n+m/4-l/2)/

  • '=k/2+3.72740

  • ''=0.04440


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  • 1-27-1

  • C x6+x10=nx13(1-27-2)

  • H x1+2x4+x5+2x9=mx13 (1-27-3)

  • O x2+x5+x6+x7+2x8+x9+2x10=2x13

    (1-27-4)

  • N x3+x7+2x11=2x13(1-27-5)

  • Ar x12=x13 (1-27-6)

  • =1 (1-27-7)

  • x1~x1313


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N

k1~k10P()

1-27-81-27-14


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  • x1=K1x41/2/P1/2 (1-27-15)

  • x2=K2x81/2/P1/2(1-27-16)

  • x3=K3x111/2/P1/2 (1-27-17)

  • (1-27-18)

  • (1-27-19)

  • (1-27-20)

  • (1-27-21)

    1-27-2~1-27-71-27-15~1-27-211313x1~x13.


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OlikaraxiUihiPT

2.4

CO2CO2CO

  • CO+1/2O2CO2

  • aCO2a/2O2+aCO

  • a1CO2


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COM2P

  • M2=1-a+a+a/2=1+a/2

  • PCO2=(MCO2/M2)P\=(1-a)/(1+a/2)P

  • PCO=MCO/M2P=a/(1+a/2)P

  • PO2=MO2/M2P=a/2/(1+a/2) P

    CO+1/2O2CO2,K, KP=PCO2/(PCOPO21/2)


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  • TKPPCO2a,1-3


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3

  • 3.1


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  • (mol/m3s)

  • (1/m3s)

  • (1/s)

  • aA+bBeE+fF

  • A=-dCA/dt B=-dCB/dt

  • E=+dCE/dt P=+dCp/dt


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  • ABEF

  • ABEFABEF

  • 3.2


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  • 1-32

  • aA+bBeE+fF


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  • Aa(B)b

  • =-1/a[d(A)/dt]

  • = (1/e) [d(E)/dt ]=k(A)a(B)b

    kkABmol/m3sk


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3.3

  • HHHHHH

  • =k(H)3

  • H

  • H


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  • H

  • HNET0HNETH; HNET0,HNETHH

    3.4

  • I22Ii2=kiCi2,


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  • -dCA/dt=k2CACB

  • -dCA/dt=k2CA2(CA=CB)

  • =kCAaCBbCCc......

  • =a+b+c+......,


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4

  • 10C2~4 kt+10/kt=2~4Vant Hoff

  • lnk=-E/RT+lnA


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  • k=Ae(-E/RT)

  • logk=-E/2.303RT+B

    EABE

  • 4.2

  • EE


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  • 1-7

  • ACE

  • E=E1-E2=ED-EA-ED-EC=EC-EA

  • ECAH

  • 4.3

  • qZqZq


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  • A+BCBAuB rArBABrA+rB.

  • =rA+rB, BA2ABZ1B


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  • nAnBAB

  • AB

  • Z=ZAB=Z1B= (1-65)

  • u=8KbT/m*(1/2)

  • KbKb=1.38062210-17(uJ/K)

  • m*

  • m*=mAmB/(mA+mB)

  • mAmBAB(1-65)


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  • Z=ZAB=(nAnB/g)2(8KbT/m*)(1/2)

  • ABg=1;g=2rArB

  • A=2(8KbT/m*)(1/2)

  • Z=AnAnB

  • E


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  • -

  • nA'/nA=e(-EA/RT) (1-71) nB'/nB=e(-EB/RT) (1-72)

  • nA'nB'AB1-70nAnBnA'nB'

  • Zn=AnAnBe(-E/RT) (1-73)

  • E=EA+EB

  • =-dnA/dt= AnAnBe(-E/RT)


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  • -dnA/dt= knAnB (1-77)

  • 1-761-77

  • k=2(8KbT/m*)(1/2)e(-E/RT) 1-78

  • k/m3-1s-1

  • k=k0e(-E/RT) (1-79)

  • k0=2(8KbT/m*)(1/2) (1-80)

  • k0


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  • 1-691-80k0=A,1-791-60

  • (1-60

  • k=Z*T1/2e(-E/RT) (1-81)

  • Z*

  • K0=Z*T1/2


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5

  • 5.1

  • k0Ek0m*m*E


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  • 5.2

  • 1-801-791-80k0T1/2,1-79ke(-E/RT),


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  • 5.1

  • A+B+C+...... 1-83


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  • =-dCA/dt=k1CA

  • PXAA

  • =-dCA/dt=k1CA=k1XAP/RT

  • =-dCA/dt=k2CA2= k2XA2(P/RT)2

  • = -dCA/dt=kCA= kXA(P/RT)


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  • P


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6

  • 6.1


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  • 6.2

  • NO

  • d(N)/dt=0 d(O)/dt=0


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7

  • 7.1

  • 1-10ABABBACACA0CA0CAABBA


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  • A

  • (g/cm3sg/cm3/cm)

  • 1-98


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  • AJABADABABcm2/s1-98AA

  • 7.2

  • BBBTTw,TwT


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  • q

  • q=Q/A(Tw-T)/(J/cm3/sK/cm)


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  • QA

  • FourierJ/cm/k/s.


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  • 7.3

  • B


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  • Pas,u/y()


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  • 74

  • FDP

  • D


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  • Prandtl

  • DSchmidt)

  • Lewis)

  • Le=


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