化工基礎(chǔ)答案

1、化學(xué)工程基礎(chǔ)習(xí)題第二章.P69L 解：Pvac = P。- P 絕艮町 3.3xl(y3= 98.7x103 n P 絕=85.4x ICT , PaP=PP.=85.4x 10 3Pa-98.7xlQ 3Pa=-13.3xl()T Pa2. 解：de = 4x-2-打上L=d dQ =70ndx + nd 23. 解：對于穩(wěn)定流態(tài)的流體通過圓形管道，有九必?土- 2 2gW上吃di 2g64# L2U; H§2 _ d7U2p d2 2gw_LHfio 1 22x 4r=1616H"64" L 1 dxuip dx妒2g64#L2 “2%2 Pd22g%64&q

2、uot;LI("zb2d*" 2da 2g即產(chǎn)生的能量損失為原來的16倍。6.解：1)雷諾數(shù)Re = A4其中 p = lOOOAg -m -3, u = 1.0m-5 -13d = 25mm = 25xl0 m3jU = lep = 10 Ps -sND pud故 Re =1000x1.0x25x10-3 -.75IO=25000故為湍流。2)要使管中水層流，則 Re < 2000-1000x25xl0" 3m-w - _即 Re = - < 2000解得以 < 0.08m-5 -17. 解：取高位水槽液面為 1-1',A-A'

3、 截面為 2-2'截面，由伯努利方程2-工 P U _ , Z H f2 pg 2g - pg 2g f其中 q =10m,z 2 =2m;PI = P2; "I =O; H/二互2“c c Mo 16.15 M 則 10 = 2 + + 2x9.8 9.8解得1) A-A' 截面處流量 =%2) q y= Aup其中 A = =1x3.14x(100x10-3)2 4 4= 7.85xl0 -3m2q, = 7.85xl0Tx2.17x3600 = 61.32-X8. 解：對 1-1，截面和 2-2'截面，由伯努利方程得2 7p, Up2 U.Zy h1 -

4、 = A2 1 -Pg 2gpg 2g其中 Z =z 2,Pi=lm H2O = pgh Aj 02“2 = t 侃 1 ()2 x 0.5 = 2.0m .廣2 d ; 10.10 52_2A/? += = 0.1 9m2x9.8 2x9.815. W:選取貯槽液面為Id截面，高位槽液面為2-2'截面，由伯努利方程得,2 ,2勺 +業(yè) +土 + 乩=Z,+4+3+/Pg 2gpg 2g其中：Z = 2m, z。= 10m; u A = u, = 0P, = -lOOmmHg=-13.6xl0 15.468x2xAx(53xl0Jx980 x 9.8 x 0.1 = -13332.2p

5、 ap = 02c -13332.2 u s 19.6 , 1000、2 + + H =10 + (+ 4x)pg '9.8980p = He-qv-p102102=0.655kw17.解：取水池液面為IT'截面，高位截面為2-2'截面,由伯努利方程得ZEUP"其中：Z = 0, Z = 50m; Pi P2 = 0廣 8H, =50+ 竺=52.05'9.8P = LqvP = 52.05x36x1000 = 8 Q5kw102102x0.6x360019.解：取貯槽液面為IT,截面，蒸發(fā)器內(nèi)管路出口為2-2 '截面,由伯努利方程得乩=眼 8

6、 + 參 + 端=14.08 + 1.388 = 15.468Pg 2gPg 2g其中,Z =0,Z = 15m;=° ,p2 = 200x10 3 x 13.6x103 x 9.8 = -26656p aH / = 12%.8H =15+ 翊-圣人=24.97 '9.8 9.8x1200P = d 茫必爽 Q = 1.632kw102102x360020. W: 1）取貯水池液面為1-1,截面，出口管路壓力表所在液面為2-2，截面，由伯努利方程得其中，zj =0,Z2 =5.0m;巧=0, P2 = 2.5kgf .cm2.5x948 c . _ I=2.45x105 Pa

7、o.or-'忽略出水管路水泵至壓力表之間的阻力損失，則：衡算系統(tǒng)的阻力損失主要為吸入管路的阻力損失h -0.2/ n f /9.836 =2.23600x :卜,uc 2.45 xio 5 2.2-2 220.2H =5.011e 1000x9.8 2x9.8 9.8=5.0 + 25 + 0.25 + 0.02 = 30.27p = Aj= 30.27x36x1000 =3 Qkw102102x3600 幻陣*琴=4.3kw3）取貯槽液面為IT，截面，水泵吸入管路上真空表處液面為 2-2，截面，Zi H 1 = G 1 F riPg 2g pg 2g其中： Z =0 工 2 =4.8

8、 秫 ;P1 = 0, P2 = ?忽略進水管路水泵中真空表至水泵之間的阻力損失 , 則：衡 算系統(tǒng)的阻力損失為吸入管路的阻力損失：H -0.2/ n f /9.8 0 2 2.2p, = (4.8 + + )x1000x9.8 = -49600pa- 2x9.8 9.8得真空表的讀數(shù)為 P = 49600Pa J- vac23. W： 1）取低位槽液面為 1-1'截面 高位槽液面為 2-2,截面 由伯努利方程得2 74+4+E+乩="業(yè) +2£+丑pg 2g - pg 2g f其中，Z = 0, z 2 = 20m; Pi = 0, P2 = 0H f = 5,

9、乩 =20 + 5 = 25 J ew e =25x9.8 = 245J/kg2）在管路 A、B 截面間列伯努利方程得：Ljr d 2gpg pg uC d (PHg Ph20)xR* Ph2O: 6=a2SX X( 6)VPH IO=J2gx - °x 0.504 = 2.03/ m?廣 V 6x0.027T025X2.03 X-X0.05 2x10003) P = =0.916kw1024) 根據(jù)靜力學(xué)基本方程式：PB+ PH1OSA+ H>= PHggR'+ Po=P B= PHs gR '+ Po 庭。8 (6 + H )Pa + PhW= Pb + P

10、h2oSa+ PhioSA-r) + p HggRPa = PB+ PHWSA-PHW8R+ pHgsRPA =PHggR'+Po PH2° g(6 + H)+PH2og6 PHZogR* PllggRN PA - Po = PHggR PH208A + (Png 戸刊0人8人=13.6x1.2 Ixl + (13.6 I)x0.04x9.8xl000=1.55x10。” = 2gPA PB = (Pttg - PH io)x g" + PHW x g x 6PA PB (%g - Pmo ) x7? + Pmo X 6PHIOSPHWplO5exl 解:A? t京

11、=5如角0.1 0.0於1150 30ex2 解:q'=4 人 2 人 3 1-07 0.14H20300 =451120= 1243W/m 20.901n+ & = 3.73 C °-m2? W 】£ ( R + R )°Ro =3.73 0.901 = 2.83 C? ?2 .W" 1(T,-7 ;)(T,-7 ;)lA vI!2 兀 LE-TQ 2”( 71-7 ;)f-+l"r2A-( 260-35)1 ,251 , 551 ,85l1 1 1 1 16 " 20 0.2 " 25 0.07 &quo

12、t; 552x3.14x225U130.223| 0.788 1.435 一0.0139+ 3.94 +6.214160.20.0071413=138.96W"10.168Ex4解220 + 180空氣的定性溫度T =200 °C22oo °nf空氣的物性參數(shù)為：p = 0.746Kg / m3CJ-034X10-X 2 6X 10-=()6823.931x10 2a = 0.0234(凡)。8(月)。,d=0.023x 3 §1x1。x(i,o9x 10 4)0'8 xO.68030.0254=53.8W/m 2-K40 + 20ex5解：水的

13、定性溫度 T = = 30 C230° ent水的物性參數(shù)為：p = 995.1 Kg / m1 1A = Q.6176W m K "=80.07x1() 5 4 ? sCp=4.174KJ ? KgL° C? = O.O23-(7?J 8(Pr)04d2JM = Im ? s 時，業(yè)=0.02 x lx 99=24868.6"80.07 x IO -53-5Cp" 4.174X10 X80.07X10Pr = =- =5.4120.61760-6176080,4a = 0.023x x (24868.6) x (5.41)0.002=4583

14、.5 W/ 己 K當“ =0.3時Re =蟲仝=> 7460.58 ,此時，2000<Re<i0000a = 0.023 x 38.85x1.965 x 7460.58 08 x(l -7460.58=1638W/ ?2 Cex7解：甲烷的定性溫度：T = 120 + 30 = 75 C 2o°C 條件下：p = 0.717Kg I 4 = 0.03W ?廣?,流體截面積 d = 4° 潤濕周邊-X0.19-37x-x0.019-n 0755；=1 / /ZO. /=4 乂 -44"0.19 + 37x7x0.019d dup 0.0255x10

15、x0.717 =1.03x10、20.03由于甲烷被冷卻，n =0.3? = 0.023 (7?J08(a)03 4=0.023 x -' ° x (17728.7)°8x (0.584)°3 0.0255=57.1W ?廠若甲烷在管內(nèi)流動：R “ = 0? °5X10X0.717 =1044175-1.03X10 5a = 64.2W'.或Ex8 AT ； =40-15 = 25 C ；AT, =130-33 = 97 C97-25cr / =53.1 c (逆流)I 97/25在按照折流校正P= 33-15 130-=0.1615 眼

16、心 0 = 5 33-15 7 ; =0.97x53.1 = 51.5ex9/> = Q mCp ( Tl-T2)=1.25x1.9x10 3x ( 80-30)=118750 八廣(2) M =30 20 = 10 ° ,AT 2 =80-40 = 40 C °7? = = + -a- + 人 + 7?.+7? 0 K ?0 %同列"，+ 0.21x10-3+0.176x10-3H 1.7x1030.85 xl0 3x 0.020.025-0.0245 xI 0.025"0.02=0.00251 己K = L = 398.91W ?廣 2 .K-

17、' R</>118750KMm-7?= = + + A + 7?+7?,”0SZS。K %。油 i An110.0025 CC1 3 ci* / 3=+ + +0.21x10 +0.176x10331.7 xlO 3 0.85 xlO 45K = 453.26A = 12.1 ImexlO(1) d0 = 16x 10-3m , dt - 13xl0-3m ,5 = 1.5x10 3 秫1.5 x IO -3 x 16 1)13x100016=1 1 1 40x14.590 J=1/(1.23lx 10 3 + 0.0414x 10 3 + 0.011)=81.5W ?廠

18、2 .R3(161.5x10 x161 )(13x100040x14.5180 J(16=11 (13x20001.5 X IO-3 X16 1)1- 140x14.590J=l/(0.616x IO -3 + 0.0414 x IO -3 + 0.011)2=85.78W - m - Kexll 解：(1)01 = 0"Cp (4 &) = GmCp (4 一萬)50 “ C=25q mC'pq , ” C p = 2qC p135-60 _92 5135，! W(P _ 50q “ C?頑人kN(2)0=q,"Cp(t-t2 ) = q"Cp7

19、0q"Cp =q"Cpk-t2135-3070g, “CAZ = = oy.ol , & =-| 135*2 也“ kM"30又流量及物性不變，幻=k270q"人 2 _ 4 州 _ 70x92.5 _ 6475X 500*/ 50x69.8 - 3490.7L A二=二=1.855 , L, =1.855/nexl2 解:Si QmdCpd (4 % )=193 以 x 2.88 x IO 'x5(0)0= 230x6xAt/JOUU=61.76x1()3 = 230 x 6 x A? mMm =44.75C90-r-32AT,=44.

20、7590 f"32t = 29.5C°a = 9i_ = 50q , G* kATm01 = q mdC pd (4 -心)=QmlSph (” 一, 2 )61.76x103 =s,x 4.2x( 29.5-18)細= 1.279x103 =4.6 奴 M吸收pl87ex?解：(DP; = 2026 P,L 7.821x10 3 / 100x10 3 vnn / tCa =/ ()= 2.300 mol.m*34/1000C； = 2300 =1 135x10 -3 mol mH = HP ； 2026E= P1000,=4.895 x IO ' R.=4.895

21、x10, a4.895 xlO 7= 483.221.013x105(4)總壓提高一倍，E、H 值均不變E 4.895 xlO 72.026 xlO 5= 242ex9 .解:y = *64= 0.02329_ 1 _29 64y_ y, = 0.02329_ 1-0.02329=0.02329第一解法：又紅=比土 = 165目 如谿m】n=L / 二Y九=1.65X|26.7'X、= ° °385 =0.00054又=Smol64q ,c? Xi = 8% c = 14814.8bn。/z = 266.66第二解法：設(shè)吸收率為r/貝ij,匕=(1 一)匕進氣量設(shè)為

22、a.kg/h/a.5%a -竺！ kn 叫/29x103 /h(竺)mi n=-M ??；啞=26.7 知 x；&X ；".5%(紅)-1.65( ) min = 1.65 x 26.7 x - 勺"Qn,B0" 29x10-3q.95%=14814-女口。卜 266。財An,CexlO. (1) Y= 0.02七=匕( 1 99%) = 0.0002 X 2 =0當液氣比為2.2時，q,、c _匕一匕-_0.02 0.0002 q,xxX|X| = 0.009匕=匕一匕 * =0.02 0.009 = 0.011匕=匕* =0.0002 K,N°

23、 GAy,-AK 2_ ,AK, - , 0.011 -mUEIn InAZ, 0.0002= =7 35 K,0.002695(2)當液氣比為1.25 nt,、，0.02x99%Xj =0.0158411.25Nnr =15.17 t/Cr匕一匕）+ X?=0.01584 + 0.00011 = 0.01595膈頃.03當液氣比為0.8,溶質(zhì)的最大回收率時溶液出口達到氣液平衡<7, c_r,-r 2_o.O2-r 2_？-丫2 = 0.004匕一匕 0.02-0.0016 g =- -=匕=80%0.02exii.K = 03 =0.030931 1-0.03匕=* (1 98%) =

24、 0.03093 X 0.02 = 0.0006X2 =01-67273你穴,= xx(l 3%) = " 22.4xlOT 303 q cmin = 12544 x 65.16 = 81.74mo/.sT qnc = 81.74x1.5 = 122.6mo/.sTmin =匕一匕=1.28 = 1.28x98% =1.2544MX?乂產(chǎn)竺(匕一匕fQn ,B=X (0.03093 - 0.0006) = 0.016122.6=0.03093-1.28x 0.016 = 0.01045AK, =n -K* =0.0006m AKIn匕捋蒙?° = 0.0034651 0.0

25、1045 In0.0006N =具° G 匕0.03093 0.0006 0.003465"皿=0.6503JS 60xA1.0H = H nr x Nnr = 8.8 x 0.6503 = 5.72m ULr12.0.05=0.05261-0.05些也=0.002637l-y2 1-0.0026361.2/58X| = (1000-61.2)/=0-也33X2 =0y* =2.ox匕=匕匕*=0.0526-2.0x0.02023 = 0.01214AK, =K,-K,* =0.002637K，i zn匕irr -InL0.002637AK,絲 4.00621.527。心=

26、8.060.00620.556 八心 273 Q, b =- Tx 0.95 x" B 22.4X1032981H “ _ °品 p 7x0212602=0.744KYa = 57.79mol .m 疽qn,c = K-*q ,B X 】一X20.0526 0.0026369=2.4698 0.02023qnC =2.4698 x 21.602=53.35mol.s -153.35x3600n=0.02023 X 53.35 X 3600=3885.37molm=58X3885.37=225.4kg(1)另解：q “ .二匕一匕q “ .Bq X2。億八。=2.46980.

27、02023Nogi1 In (1-地) B q,c q,cY-mX? ! mqnBY2mY, q nc1, zi 20.0526-0212 L 2.4698 0.002637 2.46982.4698-In (1-).8.028HogNog人一0.74748.028又H° G2'，6°2 _ 57.5 ?。/. ?3頂兀d0.7474 x 4若填料層增加3m, 則:N°GH'912.042m0.74741 】小 2、0.05260G 2 L 2.4698 匕，2.4698-2.4698匕'0.0011又液氣比一足,貝U :Y -Y '

28、;2.4698,X|' 0.02085Xn' X x'qnB 0.02085 x 53.35 x 36004004.45 ?。/m 58 ' 232.26kgAm 232.26-225.4 6.9A13解：100 ,x22.4匕320.07531000 - X22.432K, 0.0753x(1-0.98) 0.0015X 0.0196X2 00.278x1000 "八，! q?.B = 一元打一=12.41 ?。/$0"(匕-匕)=0 “(cX|-X2)12.41x(0.0753-0.0015)=<7, c (0.0196-0)qn

29、c = 46.72 ?。/.$ '= 0.84 松(2)H=H G? N°GYi=0.073Y2=0.0015Xi =0.0196X2=0Y=1. 15X1=0.0225Y2*=1.5X2=0,AK = F"InF匕-匕*0.073-0.0025-0.0015 _ -,0.073-0.0025-='In0.00150.0753-0.0015 , =4.10.0180.5x1000絲一 =0.845/77 26.41 HogH = 4.1 x 0.845 = 3.46m14解:H OGN°GYl =0.025Y2=0.0045Xi =0.008X2=

30、0.Ay = d*)匕一匕0.025-0.012-0.0045 n In0.0045N = 口0G0.025-0.0045 y =2.56 0.008H nr = 3.91/ H°。2.56(2) 37/= 0.003 時 竺上匕匕一匕 0.008X| =0.0086匕=匕一匕=0.025-1.5x0.0086 = 0.0121AK2= 丫2-丫2= 0.003匕一匕i AKInAK=0.00650.0121 In0.003K-%OG K，0.0065H' = HOG-N OG=3.91x3.38 = 13.2mH = 13.2 10 = 3.2m第六章 精餡P244ex2.

31、(l)Qn,F = Qn,D + Qn,W<Qn,FXF = Qn,D Xd + Qn,W Xw lOO = 0"+4w100x 0.3 = qn D x 095 + qn W x 0.05qn D = 27.78 ? kmol - h xqnW =12.22-kmobh lqn L = R ? D = 27.78 x 3.5 = 97.23kmol ? h '$ = 1=97.23+ 100 = 197.23 饑 oUfiIqnV = q n,L+q n,D = 97.23 + 27.78 = 125.01Kkmol.h lqnyl = q nV = 125.01km

32、ol.h lQn,F = q ,D(3)<Qn,F XF =+ Qn,W Xw235 = qnD + q nW235 x 0.84 = 0.98/ 刀 + 0.002q 班Qn, = 201.4D kmol.h tD =33.67Q?n=, 1, 6 = 1Wl=+cQ,.L' = Qn.L +q“=.436.4 ? kmol.h q “ .V= Qn.LIn.D =402.8 qnv = q nV,=402.8 ? kmol.h1e (75.3 30)(04 X 2.68 + 0.6 x4.19) + 0.4 x 1055 + 0.6 x 232.(4) O =0.4x1055

33、 + 0.6x2320=1.094(5)精:r+=x +30 9=-x +=0.75x +0.225445 Xf y x3-13-1進：5 = 0.83= 1.03 = 1.2y = -4x + 0.25y = xf = 0.5y = 6x-2.51.5x y = I + 0.5x1.20.4yx0.20.2 xs = 0.42 ys = 0.521硫 Am = 4.30(8)y = 0.75x + 0.21y = -0.5x + 0.661.5if = 0.66 Xf = 0.44y = 0.48交點x = 0.36(10)y = 0.833x + 0.15 yr= x d = 0.898_

34、 3.0 駕 兒-2.0x;+1 0.898= 2.0x; +1x* = 0.746 1xoxi _ 0.898-尤 1 _0 6X0-X ; 0898-0.746 -'=0.806y = 0.833x0.806 + 0.15 = 0.821第八章化學(xué)反應(yīng)工程基本原理P340(1) -也= 2.58x10-6/dtE=pv = nRt一 =cv Rt言)=5(一"站J(cJ = k 2 Rt = k*2dtkc = kRt = 2.58 x 10 鄧 x 8.314 x 450 C P=9.65 x 10 -3m3 ? kmol ? hx krc2 = kmol - m-3

35、- hx C ANz kmol x o .，a，i?(=kmol - m - h I =83 .5% x (1 + 1 x 0.09 ) y r *,o(l 叫)以0 - V,40 (1 + 生辦)1-0.09=83.5% lx(l + lx0.09) m = m 3 ? kmol? hC QH 6 T C4 + H y a o = 1.0000yA =0.0900a 1+1-1,=1>A O - YA>A,O(I + AAyA)1-0.094NH3 + 5。2 = 4N0 + 6HQ + 0 主）4NH3 + 30 = 2% + 6HQ + Q （副）8NH3 + 8。2 = 4N0 + 2M + 12HQ + Q