化学工程习题答案

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1、43−o5700CP，，310×→aA2B+Cwk==116g，求x0.5时的，，，。cpycAA0AAAB解：211+−δδ===+=21yy11x+2xAA00AAAA15pA0310×Pa−3cm===37.085ol⋅mA0−−11RT8.314Jmol⋅⋅×+K(700273)K−3cxAA0(1−)37.085molm⋅×0.5−3cm===9.27olm⋅A1(++yxδ12×0.5)AAA0−−3−11p==cRT9.27molm⋅×⋅8.314JmolK⋅×+(700273)KAA4=×7.49910Payx(1−)(10.5)−y===AA00

2、.25A11++yxδ2×0.5AAA0cc−bx××cm===BA0a0A237.0850.518.54ol⋅m−3B11+yδx+×20.5AA0A144−ABC→+0.5kgA→0.26kgB，剩0.03kgA未反应，求，，xYSABB解：0.50.03−x==0.94A0.50.51000××gg0.261000n==10.87moln==9.286molAB046/gmol28/gmoln9.286Y===B0.854Bn10.87A0Y0.854S==B=0.909Bx0.94A245−ABCD+→+o−−723−1TC=−100，r=2.910×cm

3、ol⋅m⋅sAnn:===4.972:1，，w2400kgt'30minBA00c−3ρφ=⋅==750kgm，x0.5，0.7，求活塞流反应器V。混A解：设反应初始A的摩尔流量为qnA0AB+→+CD反应前qq4.97200nA00nA反应后q(1−−x)q(4.972xqx)qxnA00AnAAnA0AnA0A24001000×gqx==0.2395mols/nA0A116/gmol××243600sqm==0.2395ol//0.5sm0.479ol/snA0qm=×=4.9720.4792.381ol/snB0q=×0.479mols/60/gmol+×2

4、.381mols/74/gmolwT0=204.934/gsqwT0204.934/gs−33qm===×0.273210/sV033ρ75010×gm/qnA00.479mols/33c===×1.73510molm/A0−33qm0.273210×/sV00.5xxAAdxdx1xtc==AAAc=⋅rA00∫∫A2200−−rkc(1x)kc1−xAA00AAA01==1967s−732.910×××1.73510−33Vqtt=+(')=×0.273210ms/×+(19671800)sRV03=1.029m3VV==/φ1.029/0.71=.47mR3

5、48−同45−，改用全混流反应器，求V。xcxAfA0Afτ=⋅c=A022()−−rkcx(1)AfA0Af0.5==3934s−7322.910××××−1.75310(10.5)−33Vq==××=τ0.27321039341.08mV0413−同45−，改用活塞流反应器，求V。0.5dxAτ==cs1967A0∫0()−rA−33Vq==××=τ0.27321019670.54mV0在相同操作条件下，所需反应器体积从小到大为PFR<

6、13kq==1.7mkmo⋅l⋅s，，120m/(hSTP)x=0.7VA0求考虑或不考虑体积变化所需反应体积。（）不考虑体积变化1pA0101325Pa3cm===19.43ol/mA0−−11RT8.314Jmol⋅⋅×K627.2K0.7xAdxxAdx1xτ=ccAA==⋅AAA00∫0∫022−rkc(1−−xk)cx1AA0AAA000.7==70.64sh=0.019631−−131.7mkmols⋅⋅×0.01943kmolm/×0.3TK=627.231−−627.231qm=⋅120h×=275.54m⋅hV0273.23Vq=⋅=τ275.5

7、40.0196×=5.42mV0（）考虑体积变化2212+−δδ==0.51+=yx10.5+xAAA0AA22xxAAdxdx1xA⎛⎞10.5+xτ=cAA==cd⎜⎟AxA0∫∫00A02∫0A−−rk⎡⎤cx(1−)c⎝⎠1xAA0AkAA0⎢⎥⎣⎦10.5+xA0.7⎡⎤+212(10.5)xA=×⎢⎥20.5××（）1+0.5ln(1−xx)0.5++AAkc⎣⎦1−xA0A00.71⎡⎤2.25x=−⎢⎥1.5ln(1xx)0.25++AAA1.70.01943×−⎣⎦1xA0==109.56s0.0304h3Vq=⋅=τ275.540.0304×=

8、8.40m