1、化工原理第三版陈敏恒下册习题解答1.解:查引忙 水的 P=4_24K:pa p=P-P3=lC1.34J4=97.0l5KPa 用十警=筮=歳riS笫xg込解:査25兀,COa-HaO系统 = 1,661x1樹加i设当地大气压为Iatm SP 1.033at,且不计S剂分压n巧=10+ 1.033= 11.0332 =1.0810(绝)= 0.2+1.033= 1.23S3fi = 1.21xl0Wf (绝)对稀濬痕其比质量分率代九.金1 0气址10今.X = 44x =47弘訂艇禺。187 Cyin7& = 44K = 1.73x10T辰Cq /炫禺018爪 解1邓忙 时,总二4.06x1朋
2、玖二406x1(/脸勺=1.38x10-x32x10 =427蛾4 解;筋0Cb系统,2邙C时,1 = 0.537xWA?Pa = 0.537xl0W(2廿竺=四沁2込LisioP101.3石厂 jt)二 1.09x10 (y - = 0.005764NC.耳二0尽0x1出血炉应二0,眈10哎灯匕-T=S= 兀-x= 0.47x1 qt _y-= 0.003685、解: 50x2x10 = 0.01(jr-jrJi = 0.025-0.01 = 0.015 (:z,-aX = (5-2)x10=3xWP 丫叫=叫=50 X = 2.5耳 2比=处龍=2.5x 2x10 7二 50x 10 抵土
3、二詈TgL(y -片b = 0.025-0.005 = 0.02(兀一xL=(10-2)x107 = Ec10喙牆山乐 解:查 2ZC 水,戸皿=2,3346肪= 3-2.3346=79 = 101.3-1 33 = 99 97Z?0Cr, = 0.2201020V 0=0,252幻旷/仏M負=2昌n鱼=旦(二)3母-阳) A 打 RT 巳 1 RW Pjr= 21375 = 0.59加霍 查 30叱,D = I s,p= 995.7Agy 饱和蒸汽压 备=?1一益忍屈/呂戸”1= 7吩0-31.82 = 723洒叨上馆P&1 P 760-728九= = 一预=744 演M/gIn 72EX貉
4、。2/Ayr = ,7= 001 +为M2 急叽-pG=蛊4筋1昨躊叽抨)=1 44x105 =1&7叱 =+ = + =+ = 0.542K护 3 &盘上严上用 64 6 16.6疋严=PKa = 54.9 亦(如幡)九严2 =竺=0轴g 疋严 5493 =磁小=-y,) L底浓度吸收儿=mi刖G / 、 =吧+(y丹)!_1Ji沁, 、- O-乃)y +F尸2 -狩2乃一科乃尸-1T :(1 一 p-);十(p-乃-吨2)ffiC?mi12、解.厂彩=呵I=0(人=淇旳Cj Cj21=丄乃 I-罕;.川鬲=丄由(1-丄)21+丄=_1M(1-丄)丄+丄=” 4 L1 /地才1-咅山肘_卩加
5、1-备(1-77)7?15 解:(1)旳=”(1-7) = 0 02x(1-0 95) = 0 001 = 1401-旳”住-jrJ = 13 K (0.02-0. OCHM 辔 - 0.00饲)=1/75 G=-|c;I-J) + J?-3 = (0.02-0.001)+0.0004 = 0.0113(2) &片=旳酬可= 0,02 1,2x00113 = 6,44X10= 0.001-1,2x0.0004 = 5.2x101严吧52x105铲豁r? n4G = = = 0.0门切规呛.w?M 29 = 0.052 = 0.2653X = H时 = 0.265x80.9= 2.146 解;(
6、1) TJ = 0 9=3何=1.3x = X.MmG_r!_ = =0.3551.17f? 1,171 冷(1-竺 丄+竺2 = n(l-0.S55)- + 0.855 = 5.76_竺 1 l-ry L 1-0.S55 1-0.5H 尺虫 Nr 二0.8x5,76= 4.1(刁 n= 0 99 = 1577 = 1 3x0.99/= 1.29wsG兰二亠丄=077L L29 岀 1 29疋二円怖,何为=0.兀140 = 11 2胡1 7Q啲 比校(1(2可知.当回收率提高,则吸收刑利用壘沖原来的匕竺= 1.1倍1. Vim由=1.3m!7可知,吸收刑利用量与回收率1成正比.17、解:厶=会
7、響=27沁宀氐加二)皿込l=L76xlCrHx278= 0一489如打啊2 卫 r 7R=695?w = 545* G 04此时,塔高不受限制.则可在塔顶达郢汽獗平衡,贝b1=沁1 =5453 2.5x10 =0.0136/X厂务= 2.5W一存呵込5心 低 斜丁(头=和(糸=(务 十3令迫-呢=扛二 +2匸丑=2C-r7j75 =2rj-rf 丹 巧:0 91=如-h得尸0.7 -(轧.=吨令=令=1代h首二1.4久血二1斗珈?7二1M 3汇0打阳二1洒说明操作线与平衡绽斜率相等,鬲推动力处处相等口V 一”一必-5g 1由=i尹土亠严 丑一用1丑 1-7 丁勺-號( 1-7-堪7 = 2 一
8、为1- U- /% = % = H = H兀-M加=1龙乂2 送3=2Sj50 i22.4x2901=2x0.386 = 0.772A L严q會弘需严爲+0加“19Z7 = 凭1)3.14x1Hg =且5% 7.193 =a117=0.659W3OG2)回收丙酮量玉= 117尿川嵌-h16珈辺丿W?卫=0.046说旳川/7?二了二 92* (0卫3 - Q.0026)二 4.36如o张二 2刃炯 f h2Ck 解土 m = = 3.592F 1013 = 码 =06丹喘= 0.592即撓作绽与平衡绽平行,此时, G 0.03叽=$L =叽=乃F =片H = = =Q.“ S 0.1叫忙-心得=
9、几=0上0221,解:fl)=”(1-77)= 0.01x11-0.9) = 0.001Ja-31 1 1 c “小 0.01-2x0.0002 . ,-= Inffl 0.667) X + 0.667 = 5.3S1-0.67 、 0.001-2x0.0002 当出上升时,由于H不变,Hog不变,也不&即532=二空2222哲 +0.6671-0.667 丿2-2x0000 芳幕 =0.0013g0i0吐丸刃0.01(2)当 K3=0.0002 Wi彳=#卜-旳)+ 龙空=1x(0.01-0.001) +(10002= 32 械100.00035 时,= )十亡2 =1江00100013)十
10、0 00035= 3 25況 10龙L 3丄=込T L;.碣十 比竺也空竺;协乃-呻 0X00322-1.0510x0.8当济率増加一倍时,號严二區具头呼因対塔高不变,则 H = g % = % %川陋二的 = 4.10 X2严=3勺:. 竺也空狙竺 0兀 一1.帖xlCT* xO.S解得启二0 000489第九章液体精馏h解,査安托因方程苹:log 鹫二&cm- 1即 1+220.8 甲狂 1绍母=&0和1羽亍启 + 21941711log = 6.031- = 2.3479C1J 10胪6 1OS4220.8= 222.80 屉帀弓= 94.00马= = 2 371 叫耳8fC =1 01
11、 1逸出=6回-苏芯严。84F=m.33kFa = 40,192) % = (ct + cs,)/2=(2.5% + 2J71)/2 = 2.434 Q5A 2484jv _p-ZH -r1+(24開-Ik求讦Xooss0.200.30Q.3P70.4S9计算y0 19303030 5150.6210.704实验y021:213700.5000.6130.710X05920.70C.030.9030.P301.00计算萨0.7S3aS530 9100.9590.9791.00冥验y0.7890 8530.9140.9570.9791.00最大请差0*1119拓0.2122、i才轻阴曽丸丽24。
12、-諾急=9.312 田I/ = 6.626“ S 9 312-5,626螟式正确,山呛3、解;门)设蛊度为S1.37(7飞黑囂65 = 17.6759 尿匾用=608232-晨務尺= 129153IS式正赢 = 31.370仔呼“4、解.因4 0忙时,=373J Ja E =117.1 匕jP 303.95.解! (1) In = Lln + cjln Iz略 a-l 旳 角由上空.丄兰亠如匕些畸 3-r 03 1-04% =100-6/7 = 31.3咖/一 脈 6S 77=1 +(商一吃)=0 4 +X (0/4-OR = 0.619% 31,33x03一 1 + -切1+(;3-1)x0.3 由(略-略勺+% 丁 =昭叼咏诃牙OOX牆弟=3&咲5解;2二&斥/24一小兀22孑F 心一斥 0 95-0.03r P 9(2)-= 0.667厂1+尺1 + 2-1= RD + FP = RD + F-护=(应 +1)D沪_ (R+0屮 _
copyright@ 2008-2022 冰豆网网站版权所有
经营许可证编号:鄂ICP备2022015515号-1