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1绪论2.解:51001325.1Paatm1mNPa21mNJ3310mL∴2321001325.1mJmNmNatmL∴21001325.1JatmL以J·mol-1·K-1表示R的值R=0.08206×1.01325×102J﹒mol-1﹒K-1=8.315J﹒mol-1﹒K-1第一章流体流动1.表压=-真空度=-4.8×104Pa绝压=5.3×104Pa2.解:设右侧水面到B′点高为h3,根据流体静力学基本方程可知PB=PB′则ρ油gh2=ρ水gh3mmmkgmmmkgh4921000600820h3323水油h=h1+h3=892mm3.解:正U型管水银压差计由图可知PA=P1+(x+R1)ρ水gPB=P2+xρ水g∵P1-P2=2.472kPa∴PA-PB=2.472kPA+ρ水gR1又有PA=PCPC=PB+ρHggR1∴ρHggR1=2.472kPa+ρ水gR1∴mmmsmmkgR00.200200.081.9)100013600(2.472kPa231倒U型压差计设右侧管内水银高度为M∵指示流体为空气∴PC=PDP1=PC+ρ水g(R2+M)P2=PD+ρ水gM∴mmmsmmkgR0.2522520.081.910002.472kPa2324.(1)PB=-845.9Pa(表)2(2)R′=0.178m7.解:由公式AVsu可得Vs=uA=uπd2/4=0.8×π×(57-3.5×2)2×10-6=1.57×10-3m3/sWs=Vsρ=1.57×10-3×1840=2.89kg/ssmkguAWsG2/147218408.08.解:由题可知:1—1′截面:P1=1.013×105Pau=0以2—2′截面为基准水平面∴z1=3m2—2′截面:设管内流速为uz2=03—3′截面:u,z3=3m4—4′截面:u,z4=3+0.5=3.5m5—5′截面:u,z5=3m6—6′截面:u,z6=2m,P6=1.013×105Pa根据伯努利方程:We=0,∑hf=0有62611P2ugzPgz∵P1=P6∴u2/2=g(z1-z6)=9.8有222112PugzPgz9.8×3+1.013×105/1000=9.8+P2/1000∴P2=1.209×105Pa323112PugzPgz1.013×105/1000=9.8+P3/1000∴P3=0.915×105Pa424112PugzPgz9.8×3+1.013×105/1000=9.8×3.5+9.8+P4/1000∴P4=0.866×105Pa525112PugzPgz∴P5=0.915×105Pa9.(1)u=1.55m/sVh=10.95m3/h(2)Δz=2.86m3解:NeNNe=We﹒Ws取釜内液面为1—1′截面,高位槽内液面为2—2′截面根据伯努利方程:fhPugzWeuPgz22222111221—1′截面:z1=0,P1=-2.5×104(表压),u1=02—2′截面:z2=15m,P2=0(表压),AWsu2A=πd2/4=0.25×π×[(76-4×2)×10-3]2=3.63×10-3m2∴smhmu/46.1/3.524710501063.3102342173740106105046.11068Re43du>4000湍流又ε/d=0.3×10-3/68×10-3=4.41×10-3查图得λ=0.029kgJudlhf/7.22246.1068.050029.0222查表1—3得,ξ全开闸阀=0.17ξ半开截止阀=9.5ξ90°标准弯头=0.75ξ进=0.5ξ出=1∴hf′=(0.17+9.5+3×0.75+1.5)×1.462/2=14.2J/kg∴∑hf=22.7+14.2=36.9J/kg9.36246.1151050105.224gWeWe=208.87J/kgNe=208.87×2×104/3600=1.16kWN=1.16/0.7=1.66kW12.解:1—1′:高位槽液面2—2′:出口管内侧列伯努利方程fhPugzWeuPgz2222211122z2=0,z1=4m,P1=P2=0(表),u1=0,We=0∴∑hf+u22/2=4g∑hf=hf+hf′22udlhf查表1—3得,ξ半开截止阀=9.5ξ90°标准弯头=0.75hf′=∑ξ﹒u22/2=(9.5+0.75+0.5)×u22/2=10.75×u22/24∴gdu4)75.10201(222化简得(400λ+11.75)×u22/2=39.220℃时μ水=1.005λ=f(Re)=f(u2)需试差321075.49Reudu假设u0Reλ→u001.0d1.5766300.0391.661.6796000.0391.661.66825880.03881.66∴截止阀半开时该输水系统的u0=1.66m/sVs=uA=1.66×0.25π×0.052=0.00326m3/s∴Vs=11.73m3/h第二章流体输送设备1、解:分别取离心泵的进、出口截面为1-1′截面和2-2′截面,由伯努力方程得:21,2222211122fHguρgpZHguρgpZ21,212212122fHguuρgppZZH其中,12ZZ=0.4m;41109.1pPa(表压);52107.4pPa(表压);21uu;21,fH=0;20℃时水的密度3mkg2.998。34.500081.92.998109.1107.44.045HmkW7.13W1037.170.0360081.92.9987034.504egHQNN3、解:10,v0gfHhρgppH(1)20℃时:由附录2及附录7知,水的密度3mkg2.998,饱和蒸气压4v102335.0pPa。28.45.2381.92.998102335.01081.944gHm(2)85℃时:由附录2及附录7知,水的密度3mkg6.968,饱和蒸气压5v105788.0pPa。527.15.2381.96.968105788.01081.954gHm4、解:10,v0gfHhρgppH6.15.381.95301045.6106.655=-2.21m-1.5m故该泵不能正常工作。6、解:21,21221212e2fHguuρgppZZH81.91000109.600104=17m由于输送介质为水,结合13ehm100Q和17eHm,查附录21选IS100-80-125型泵,主要性能参数为:13hm100Q,20Hm,1minr2900n,78.0,0.7NkW故泵实际运行时的轴功率为0.7NkW,其中因阀门调节所多消耗的功率为kW05.1W104878.0360081.91000100)1720(gHQN第四章沉降与过滤2、解:tcSbLuV1Stcsm025.04036003600bLVu设颗粒沉降位于层流区,故18)(s2ctcgdum1075.181.9)06.13000(025.010218)(1855stccgud核算流型2023.010206.1025.01075.1e55tcctudR故假设成立,即颗粒沉降位于层流区。因此,可完全除去的最小颗粒直径为m1075.15。3、解:在操作温度下,气体量13Ssm54.127342727336002160V=5538.513hm设8μm颗粒沉降位于层流区,则613526s2ctcsm101.4104.31881.9)5.04000()108(18)(gdu由tcSNbLuV51101.41.48.154.13tcSbLuVN层(需50块隔板)082.0512.4NHhm核算流型2108.4104.35.0101.4108e4536tcctudR故假设成立,得降尘室内隔板间距和层数分别为0.082m和51层。注:此题原始数据有点问题。在该组数据下,气流在降尘室内已达到湍流。5、解:根据恒压过滤方程2e22KAVVV得6010)1.0()106.1(2)106.1(605)1.0()101(2)101(23e2323e23KVKV34em107V;127sm108K由下式可算得,15min收集到的总滤液量,即6015)1.0()108()107(22742VV33m1007.2V故再过5min所得的滤液量3333m1047.0106.11007.2V=0.47L6、解:(1)过滤面积22m86.4938281.0A滤框总容积32m623.038025.081滤饼充满滤框时滤液体积3m79.708.0623.0V过滤终了时23mm156.086.4979.7AVq恒压过滤方程Kqqqe2252105156.001.02156.0s5507(2)洗涤时间86.4910579.71.0)01.0156.0(8)(8dd5KAVqqVVe=416s(3)13DWhm2.1601541655008.079.736003600VQ(4)由于0s,故KppKKs2)(1。所以,由Kqqqe22可知27525505.0s第五章传热2.某燃烧炉的平壁由三种材料构成。最内层为耐火砖,厚度b1=200mm,导热系数1=1.07Wm-1oC-1;中间层为绝热砖,厚度b2=100mm,导热系数2=0.14Wm-1oC-1;最外层为普通钢板,厚度b3=6mm,导热系数3=45Wm-1oC-1。已知炉内壁表面温度t1=1000oC,钢板外表面温度t4=30oC,试计算:(1)通过燃烧炉平壁的热通量;(2)耐火砖与绝热砖以及绝热砖与普通钢板之间的界面温度。假设各层接触良好。解:(1))(2.107645006.014.01.007.12.0301000233221141mWbbbttSQq(2)321qqqq11211bttq,8.79807.12.02.10761000)(11112tbqtoC22322bttq,1.3014.01.02.10768.798)(22222tbqtoC3.在直径为894.5mm的蒸汽管道外包扎有两层绝热材料。已知管壁材料的导热系数为45Wm-1oC-1;内层绝热材料的厚度为40mm,导热系数为0.07Wm-1oC-1;外层绝热材料的厚度为20mm,导热系数为0.15Wm-1oC-1。现测得蒸汽管的内壁温度为180℃,最外层绝热材料的外表面温度为50℃。试计算每米管长的热损失及两绝热层间的界面温度。解:1801toC,504toC,mmr401,mmr5.442,mmr5.843,mmr5.10440845.01045.0ln15.010445.00845.0ln07.0104.00445
本文标题:制药化工原理课后习题答案
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