2026年注冊化工工程師考試試題卷及答案1.某連續(xù)攪拌釜式反應(yīng)器（CSTR）中進(jìn)行一級不可逆液相反應(yīng)A→B，速率常數(shù)k=0.15min?1。進(jìn)料流量Q=25L·min?1，進(jìn)料濃度CA0=2.0mol·L?1。若要求出口轉(zhuǎn)化率xA=0.92，求反應(yīng)器有效體積V（L），并判斷若將反應(yīng)器改為兩個等體積CSTR串聯(lián)，總體積不變時能否達(dá)到相同轉(zhuǎn)化率。答案與解析單釜CSTR：τ=V/Q=CA0xA/(kCA0(1?xA))=xA/(k(1?xA))τ=0.92/(0.15×0.08)=76.67minV=Qτ=25×76.67=1917L兩等體積CSTR串聯(lián)，總體積仍為1917L，單釜體積V1=958.3L，τ1=38.33min。對第一釜：x1=kτ1/(1+kτ1)=0.15×38.33/(1+0.15×38.33)=0.852對第二釜：入口濃度CA1=CA0(1?x1)=0.296mol·L?1出口轉(zhuǎn)化率x2=1?CA1/(1+kτ1)=1?0.296/(1+0.15×38.33)=0.978總轉(zhuǎn)化率0.978>0.92，故可達(dá)到更高轉(zhuǎn)化率。若嚴(yán)格保持總轉(zhuǎn)化率0.92，則總體積可減小至約1360L。2.某精餾塔分離苯-甲苯混合物，相對揮發(fā)度α=2.46，進(jìn)料為飽和液體，xF=0.45（苯摩爾分?jǐn)?shù)），塔頂xD=0.98，塔底xB=0.03。采用全凝器，泡點回流，回流比R=1.8Rmin。求理論板數(shù)N（含再沸器），并指出最佳進(jìn)料位置。答案與解析平衡線：y=αx/(1+(α?1)x)q線：xF=0.45，q=1，q線垂直。Rmin：作圖得交點(x,y)=(0.45,0.669)Rmin=(xD?y)/(y?x*)=(0.98?0.669)/(0.669?0.45)=1.42R=1.8×1.42=2.56McCabe-Thiele圖解得：精餾段操作線截距xD/(R+1)=0.275從(xD,xD)開始階梯作圖，跨過q線后換至提餾段操作線，直至x≤xB。讀得理論板數(shù)N=11.3，取整12塊（含再沸器）。最佳進(jìn)料為從上往下第6塊理論板。3.某氣體吸收塔用純水逆流吸收氨，入口氣相y1=0.15，出口y2=0.005，入口液相x2=0，平衡關(guān)系y=1.2x，操作液氣比L/G=1.5(L/G)min。求傳質(zhì)單元數(shù)NOG，并計算若采用20%氨水作為吸收劑（x2=0.20）時，NOG的變化百分比。答案與解析(L/G)min=(y1?y2)/(y1/1.2?x2)=0.145/(0.125?0)=1.16L/G=1.5×1.16=1.74操作線：y=1.74(x?x2)+y2=1.74x+0.005出口液相x1=x2+(y1?y2)/(L/G)=0+0.145/1.74=0.0833NOG=∫_{y2}^{y1}dy/(y?y*)=∫_{0.005}^{0.15}dy/(y?1.2x)將x=(y?0.005)/1.4代入：y*=1.2(y?0.005)/1.74=0.690y?0.00345推動力y?y*=y?0.690y+0.00345=0.310y+0.00345NOG=∫_{0.005}^{0.15}dy/(0.310y+0.00345)=(1/0.310)ln[(0.310×0.15+0.00345)/(0.310×0.005+0.00345)]=3.226ln(0.04995/0.005)=3.226×2.303=7.43若x2=0.20，則(L/G)min=0.145/(0.125?0.20)無意義，說明操作線斜率必須大于平衡線斜率1.2，否則無法達(dá)到y(tǒng)2=0.005。實際L/G需>1.2，取L/G=1.74不變，則操作線y=1.74(x?0.20)+0.005，出口x1=0.20+0.145/1.74=0.283y*=1.2x=1.2(y?0.005)/1.74+0.24=0.690y+0.23655推動力y?y*=0.310y?0.23655在y=0.15處推動力為負(fù)，塔頂已超限，需提高L/G至2.5以上方可運(yùn)行。若強(qiáng)制L/G=2.5，則新NOG=4.12，比原工況降低44%。4.某套管換熱器用120°C飽和蒸汽冷凝加熱湍流流動的苯，苯入口25°C，出口75°C，質(zhì)量流量3.2kg·s?1，比熱容1.9kJ·kg?1·K?1。蒸汽側(cè)傳熱系數(shù)12kW·m?2·K?1，苯側(cè)1.8kW·m?2·K?1，管壁熱阻忽略。求所需傳熱面積A，并分析若苯流量增加30%而蒸汽溫度不變時，出口溫度能否達(dá)到70°C。答案與解析Q=mcpΔT=3.2×1.9×50=304kWΔTlm=(120?25)?(120?75)/ln[(120?25)/(120?75)]=95?45/ln(95/45)=66.2°C1/U=1/12+1/1.8=0.0833+0.5556=0.6389×10?3m2·K·W?1U=1.565kW·m?2·K?1A=Q/(UΔTlm)=304/(1.565×66.2)=2.93m2流量增加30%，m′=4.16kg·s?1，假設(shè)仍為湍流，hi∝u^0.8，故hi′=1.8×1.3^0.8=2.23kW·m?2·K?1新U′=1/(1/12+1/2.23)=1.88kW·m?2·K?1能量平衡：Q′=m′cp(Tout?25)=U′AΔTlm′ΔTlm′=[(120?25)?(120?Tout)]/ln[(120?25)/(120?Tout)]=(95?(120?Tout))/ln[95/(120?Tout)]聯(lián)立：4.16×1.9×(Tout?25)=1.88×2.93×ΔTlm′試差得Tout=68.4°C<70°C，故不能達(dá)到。5.某二級不可逆液相反應(yīng)A+B→C在35°C下進(jìn)行，速率方程?rA=kCACB，k=0.47L·mol?1·min?1。采用平推流反應(yīng)器（PFR），進(jìn)料體積流量18L·min?1，CA0=1.5mol·L?1，CB0=2.0mol·L?1。若要求CA降至0.20mol·L?1，求所需反應(yīng)器體積V，并討論若將一半B預(yù)先循環(huán)回反應(yīng)器入口（CB0不變）對體積的影響。答案與解析對二級反應(yīng)，設(shè)x=(CA0?CA)/CA0，則CA=CA0(1?x)，CB=CB0?CA0x?rA=kCA0(1?x)(CB0?CA0x)τ=V/Q=CA0∫_{0}^{x}dx/[kCA0(1?x)(CB0?CA0x)]=(1/k)∫_{0}^{x}dx/[(1?x)(CB0?CA0x)]令a=CB0/CA0=1.333，則τ=(1/(kCA0))∫_{0}^{x}dx/[(1?x)(a?x)]=(1/(kCA0(a?1)))ln[(a?x)/(a(1?x))]x=(1.5?0.2)/1.5=0.8667τ=(1/(0.47×1.5×0.333))ln[(1.333?0.8667)/(1.333×0.1333)]=4.25ln(0.4667/0.1777)=4.25×0.963=4.09minV=Qτ=18×4.09=73.6L若一半B循環(huán)，即實際新鮮B流量減半，但CB0仍保持2.0mol·L?1（因循環(huán)稀釋），反應(yīng)速率不變，故體積不變；然而若理解為新鮮B投料減半導(dǎo)致CB0降至1.0mol·L?1，則a=0.6667，重新計算得τ=11.4min，V=205L，體積增加178%。6.某廠采用單效蒸發(fā)器將10%NaOH水溶液濃縮至35%，進(jìn)料溫度25°C，流量5t·h?1，加熱蒸汽為150°C飽和蒸汽，冷凝水在飽和溫度下排出。蒸發(fā)室操作壓力20kPa（絕壓），對應(yīng)飽和溫度60.1°C，溶液沸點升高12°C。求蒸汽消耗量S（kg·h?1）及蒸汽經(jīng)濟(jì)性，并計算若改為雙效并流蒸發(fā)，蒸汽經(jīng)濟(jì)性可提高多少倍。答案與解析總蒸發(fā)水量W=F(1?x0/x1)=5000(1?0.10/0.35)=3571kg·h?1溶液沸點=60.1+12=72.1°C取料液比熱容cp≈3.6kJ·kg?1·K?1熱量Q=Fcp(t?t0)+Wλλat60.1°C=2358kJ·kg?1Q=5000×3.6×(72.1?25)+3571×2358=847×103+8.42×10?=9.27×10?kJ·h?1蒸汽潛熱at150°C=2114kJ·kg?1S=Q/2114=4384kg·h?1蒸汽經(jīng)濟(jì)性=W/S=3571/4384=0.815雙效并流：設(shè)各效面積相等，溫度差損失相同，總有效ΔT=150?60.1?2×12=65.9°C，單效ΔT≈33°C，一效沸點≈117°C，二效≈72°C。近似估算蒸汽經(jīng)濟(jì)性≈1.6，提高倍數(shù)=1.6/0.815=1.96倍。7.某離心泵輸送20°C清水，特性曲線H=80?0.045Q2（Q單位m3·h?1，H單位m），管路系統(tǒng)曲線Hs=12+0.018Q2。現(xiàn)擬用變頻調(diào)速將流量調(diào)至80%原工作點流量，求新轉(zhuǎn)速n′與原轉(zhuǎn)速n之比，并計算年節(jié)電比例（假設(shè)功率∝QH/η，η在調(diào)節(jié)范圍內(nèi)不變）。答案與解析原工作點：80?0.045Q2=12+0.018Q2→0.063Q2=68→Q=32.9m3·h?1，H=71.2m新流量Q′=0.8×32.9=26.3m3·h?1新管路需求Hs′=12+0.018×26.32=12+12.5=24.5m泵相似律：H′/H=(n′/n)2，Q′/Q=n′/n故n′/n=Q′/Q=0.8驗證：H′=80×0.82?0.045×26.32=51.2?31.2=20.0m<24.5m，不滿足。需重新求泵曲線與新工作點交點：設(shè)n′/n=r，則新曲線H=80r2?0.045Q2與Hs=12+0.018Q2交于Q′：80r2?0.045Q′2=12+0.018Q′2且Q′=rQ=r×32.9代入：80r2?0.045×(32.9r)2=12+0.018×(32.9r)280r2?0.045×1082r2=12+0.018×1082r280r2?48.7r2=12+19.5r231.3r2=12+19.5r2→11.8r2=12→r=1.01表明原流量已接近最佳，降流量20%需r≈0.99，幾乎不降速。實際工程可取r=0.80，雖略偏離交點，但閥調(diào)節(jié)可彌補(bǔ)。功率比P′/P=(Q′H′)/(QH)=(0.8×24.5)/(1×71.2)=0.275年節(jié)電比例=1?0.275=72.5%。8.某催化裂化裝置再生器煙氣流量180000Nm3·h?1（干基），含CO8.5%（體積），擬用選擇性氧化催化劑將CO降至100mg·m?3（以CO計）。催化劑空速2500h?1，反應(yīng)器入口溫度280°C，出口420°C，反應(yīng)熱283kJ·mol?1CO。求催化劑床層體積Vc，并計算機(jī)組回收熱量相當(dāng)于多少噸3.82MPa、450°C過熱蒸汽（焓值3330kJ·kg?1，給水105°C焓440kJ·kg?1）。答案與解析干煙氣量180000Nm3·h?1=180000/22.4=8036kmol·h?1CO量=0.085×8036=683kmol·h?1=19.1t·h?1出口CO100mg·m3=0.1g·m3，煙氣量濕基近似不變，則出口CO摩爾流量=0.1×180000×1000/28/10?=0.643kmol·h?1轉(zhuǎn)化率=(683?0.643)/683=0.999催化劑體積Vc=氣體體積流量/空速=180000/2500=72m3反應(yīng)放熱Q=682.4kmol·h?1×283×103=1.93×10?kJ·h?1蒸汽產(chǎn)出D=Q/(3330?440)=1.93×10?/2890=66.8t·h?19.某聚合物廠采用管式法生產(chǎn)LDPE，反應(yīng)壓力220MPa，入口乙烯溫度50°C，經(jīng)絕熱壓縮溫升至180°C，隨后進(jìn)入反應(yīng)段。反應(yīng)段內(nèi)徑25mm，長120m，單程轉(zhuǎn)化率18%，反應(yīng)熱95kJ·mol?1。假設(shè)流體物性近似純乙烯，cp=2.5kJ·kg?1·K?1，ρ=480kg·m?3，質(zhì)量流量8t·h?1。求沿管長溫度分布（忽略壓降），并判斷是否需設(shè)置冷卻段。答案與解析摩爾流量=8000/28=286kmol·h?1=79.4mol·s?1反應(yīng)速率沿管長變化，但絕熱溫升可估算：每轉(zhuǎn)化1mol乙烯升溫ΔT=95/(28×2.5)=1.36K總絕熱溫升=0.18×1.36×1000=244K出口溫度=180+244=424°C，超過聚乙烯分解溫度（≈350°C），必須設(shè)置冷卻段。采用分段計算：將120m分20段，每段6m，逐段求解物料與能量平衡，可得在約40m處溫度已達(dá)350°C，需在此前引入冷卻。10.某化工廠廢水含苯酚1200mg·L?1，流量60m3·h?1，擬用逆流萃取塔以苯為溶劑去除95%苯酚。平衡關(guān)系Y=0.85X（Y為質(zhì)量比g·g?1，X為g·g?1），溶劑含苯酚20mg·kg?1。要求溶劑用量為最小用量的1.4倍，求溶劑流量S（kg·h?1）及理論級數(shù)NTU。答案與解析水相：Xin=1200mg·L?1=1.2kg·t?1，Xout=0.06kg·t?1流量L=60t·h?1溶劑相：Yin=0.02kg·t?1最小溶劑比(S/L)min=(Xin?Xout)/(Yout*?Yin)Yout*=0.85×1.2=1.02kg·t?1(S/L)min=(1.2?0.06)/(1.02?0.02)=1.14S=1.4×1.14×60=95.8t·h?1實際Yout=Yin+(L/S)(Xin?Xout)=0.02+0.626×1.14=0.734kg·t?1NTU=∫dX/(X?X*)=∫_{0.06}^{1.2}dX/(X?Y/0.85)將Y=Yin+(L/S)(Xin?X)=0.02+0.626(1.2?X)X*=Y/0.85=0.0235+0.736(1.2?X)推動力X?X*=X?0.0235?0.736(1.2?X)=1.736X?0.9067NTU=∫_{0.06}^{1.2}dX/(1.736X?0.9067)=(1/1.736)ln[(1.736×1.2?0.9067)/(1.736×0.06?0.9067)]=0.576ln(1.176/?0.802)出現(xiàn)負(fù)值，說明計算方式需改為對數(shù)平均推動力：ΔX1=1.2?0.734/0.85=0.336，ΔX2=0.06?0.02/0.85=0.0365ΔXlm=(0.336?0.0365)/ln(0.336/0.0365)=0.139NTU=(1.2?0.06)/0.139=8.211.某固定床吸附器用活性炭回收空氣中的丙酮，入口濃度8000ppm，床層直徑1.2m，高0.9m，活性炭堆密度420kg·m?3，平衡吸附量q=0.052c^0.62（qg·g?1，cmg·m?3），操作溫度25°C，壓力101kPa，氣體流量1800m3·h?1。假設(shè)傳質(zhì)區(qū)長度LMTZ=0.15m，求穿透時間tb（出口濃度達(dá)5%入口），并計算若床層增高至1.5m時tb增加比例。答案與解析入口c0=8000ppm=8000×44/22.4×1000=15.7g·m?3=15700mg·m?3q0*=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052×15700^0.62=0.052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